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\begin{document}
	\title{Arithmetic: A Programmatic Approach}
	\author{Murisi Tarusenga}
	\date{\today{} \currenttime}
	\maketitle
	\section*{ReadMe}
    \subsection*{Introduction}
      Procedural mathematics is a method of doing mathematics that stemmed from my dissatisfaction with classical logic and in particular, universal quantification. My intuitive understanding of universal quantification is as an infinite conjunction, i.e. $P_1\land P_2\land P_3\land \cdots$. My uneasiness with the proposition is not deep, it's simply that the proposition's end has always felt inaccessible, out of reach, rarefied. Perhaps my discomfort stems from the fact that I was programming for nearly a decade before learning university-level mathematics, and so I had implicitly/intuitively developed a different more syntactical and restrained understanding of program correctness over an unbounded domain. Procedural mathematics is my attempt at articulating this programmatic understanding.
    \subsection*{Background of Procedural Mathematics}
      Perhaps it would be easier to describe procedural mathematics by first contrasting it to its alternatives. One method of doing mathematics is through argumentation. In this method, the objects under consideration are first defined, then a statement called a theorem is made, then a deductive argument for why the theorem holds called a proof is provided. Here, proofs can use theorems that have been proven earlier and likewise for definitions. This method is perfectly valid and is underpinned by such concepts as universal quantification for expressing generality, existential quantification for expressing existence, the domains over which they apply, and the logical rules governing their interactions with other logical operators such as conjunction and disjunction. The fact that this so far has been the only known general method for doing mathematics has, I believe, led to a situation where it has been applied in contexts (especially in education and technology) where alternative approaches may have been suitable.
      
      Another method for doing mathematics, albeit incomplete, goes by various names including Transparent Proofs, Transparent Pseudo Proofs, and Gneric Proofs. This method is essentially a specialization of the argumentation method presented above except that instead of giving a deductive argument for why the theorem holds, a deductive argument for why a particular case of a theorem holds is presented in such a way that the main ideas for the proof of the general case are communicated. The benefit of this method is that universal and existential quantification, domains of discourse, and the apparatus of mathematical logic are temporarily backgrounded whilst the techniques specific to the given proof are given center stage. The drawback of this method is that the outcome is not actually a proof because the line between the main ideas of the general proof and the ideas incidental to the particular case is not made explicit. And as soon as this line is made clear, the Transparent Pseudo Proof reverts to being an ordinary proof. Hence in the end, this method only tends to be used in brief proof sketches.
      
      Yet another incomplete method of doing mathematics goes by the name of Proof Without Words. A proof without words is understood to be a theorem statement followed by a diagram or a picture that demonstrates it to be self-evident. It is convenient to extend this definition to include unexplained algorithms because their persuasive effect can be similar to that of diagrams and pictures. Similar to the Transparent Pseudo Proofs mentioned above, this method of doing mathematics excels in emphasizing the content of a proof over its form. Unfortunately Proof Without Words are not quite proofs because the link between the "picture" and the theorem to be proved is necessarily implicit. That something is missing in these kinds of proofs is evidenced by the tendency for their authors to caption diagrams and pictures, and write proofs of correctness for unexplained algorithms so as to "complete" them. What is sought then is a method for doing proof based mathematics that does not depend on the concepts of mathematical logic, nor consequently, argumentation, and yet can scale enough to communicate topics like number theory, real and complex analysis, and linear algebra.
		\subsection*{Description of Procedural Mathematics}
      Overall, this method replaces proving stated mathematical theorems with inventing correct procedures to realize stated objectives. To achieve this, the first part of this method calls for the elucidation of the mathematical semantic rules being followed; their adherence is the criterion for “correctness”. Rules whose adherence is known to guarantee that procedure implementations meet stated objectives in practice (like those of elementary algebra) are what would typically be chosen here. The use of semantic rules in this method is similar to its use in computer science, though the types of rules are markedly different. There, type systems and ownership systems are the most common semantic rules and they serve to prevent syntactically valid programs from compiling when objects of an incorrect type are found in certain contexts within the source code or when object access patterns in the source code are incorrect respectively. The key point of semantic rules is that they serve to rule out grammatically correct instructions before they are even “executed”.
      
      So far with our semantic rules we have enough to “prove” “theorems” true in a programmatic manner. But just like machine code is hard to read and write for programmers, our newfangled “proofs” would be hard to read and write for mathemticians. The next parts of this method are about using commonplace software engineering techniques to structure our “proofs” better and hence allow them to scale enough to realize complex objectives. The second part of this method is the declaration of terminology that will later be used. Declarations are more similar to their namesake in computer programming than they are to definitions in proof-based mathematics. Their purpose is to simplify/make concise instructions that involve manipulating complex structures, just like in computer programming. Nevertheless, the conceptual role of declarations is analogous to that of definitions in mathematics: they give single a name to a group of related ideas; and this in turn enables comprehension of more complex structures.
      
      The third part of this method is announcement of the mathematical objective to be achieved. Some reasonable objectives might be to show that an arithmetical equality holds, construct an object with certain properties, or construct another procedure with a potentially different objective. Again, objectives are more similar to the comments that one might see above a procedure in a computer program’s source code than they are to a theorem statement in proof-based mathematics. Their purpose is to enable readers to understand more complex proofs by sometimes getting them to see certain parts of proofs as black boxes that have the effect given by their objective. For this reason procedure objectives could be said to play a role analogous to theorem statements in proof-based mathematics. However there are some conceptual differences: unlike the theorem statement of a proof, a procedure objective is not a fact nor is it a logical consequence of a group of axioms; rather it is merely a description of the intent of the associated implementation.
      
      The last part of this method is the implementation of a semantically valid procedure for achieving the stated objective. This procedure might in turn use previous procedures or even the current procedure (in which case, the procedure is said to be recursive) to achieve sub-objectives. Semantically valid procedures are very similar to source code written in a statically typed programming language because both comprise an unambiguous set of instructions that can be carried out by a practitioner and both are constructed according to some agreed upon some set of semantic rules. Their dissimilarity is that semantically valid procedures would usually be written in a natural language grammar for a human audience whereas source code is usually written in an artificial grammar primarily for execution on a computer. Semantically valid procedures are also similar to deductive proofs in that both are constructed according to rules which serve to bestow correctness upon the proof; however they are crucially different in that one is instructional whilst the other is argumentative in nature.
    \subsection*{Comparing Procedural Mathematics}
      \begin{tabular}{|p{3.6cm}|p{3.6cm}|p{3.6cm}|p{3.6cm}|p{3.6cm}|p{3.6cm}|}
        \hline & \textbf{Classical Mathematics} & \textbf{Transparent / Wordless Proofs} & \textbf{Procedural Mathematics} \\
        \hline \textbf{Grammatical Mood} & Factual because proofs establish theorems, which are facts about abstract objects. For example, claims that a proposition holds on every member of a domain are valid. & Factual because the theorems being proved are identical to their equivalents in classical mathematics. & Intentional because procedures are labelled only with their intent. I.e. no claim is made that procedures achieve the objectives on every member of a domain. \\
        \hline \textbf{Generality} & Full. Universal quantifiers are used to express the generality of propositions over infinite domains. & None becuase a theorem is proven for only one well-chosen case. That being said, most readers should be able to generalize proof to other cases. & Full. Symbols are used to indicate values that are only known at "run-time". At "run-time" these symbols take on a single value, rather than ranging over a set. \\
        \hline \textbf{Transparency} & Optional because the logical and object language can be inseparably mixed. For example, a non-constructive arithmetical proof generally cannot be put in purely arithmetical terms. & More than in classical mathematics because theorems are proven only on explicitly selected objects. I.e. transparent proofs enforce more terms in the object language than classical proofs. & Full because procedural mathematics is about providing instructions to do mathematics rather than "doing" mathematics. Hence procedure execution yields artifacts purely in the object language. \\
        \hline \textbf{Correctness Criterion} & All inferences leading to theorem must originate from stated inference rules and must trace back to stated axioms. & Almost the same as classical mathematics, but the generalization steps are not made explicit. & Adherence of procedure implementations to well-chosen mathematical semantic rules with the objective in a "conclusive" position. \\
        \hline \textbf{Expressiveness} & Full. Sufficient for doing pure and applied mathematics and includes topics like transfinite set theory. & Subset of classical mathematics where existential statements have witnesses because proofs are done on explicitly chosen witnesses. & Subset of classical mathematics. Captures topics like trigonometry and calculus. Fails to express topics like transfinite set theory. \\
        \hline \textbf{Imports} & Universal and existential quantification, transfinite set theory, inference rules, axioms. & Inference rules and axioms. Universal and existential quantification are made implicit. & Data structures, procedures, lambdas, recursion, static analysis. \\
        \hline \textbf{Treatment of Contradictions} & The derivation of a contradiction from an assumption implies that the assumption is false. Justified by the law of noncontradiction. & Largely outside the scope of transparent / wordless proofs. Where they do occur, interpretation is the same as that of classical mathematics. & Sentences are shown to be "impossible" by providing a procedure (with an effectively empty domain) to transform it into a more obviously "impossible" sentence.
      \end{tabular}
      
      The above table shows the similarities and differences between the four methods of doing mathematics presented above. Note that they are not all trying to do exactly the same thing, as is evidenced by the variations in their grammatical moods. Also note that while procedural mathematics is strictly less expressive than classical mathematics, it may just be expressive enough to conduct most fields of applied mathemtics within.
    \subsection*{Methodology of Mathematical Experiment}
      Above I have described a general method for doing mathematics but did not provide evidence that it is workable. The rest of this book intends to prove that this approach is indeed generally usable by reformulating the elementary parts of number theory, hard analysis, calculus, and linear algebra using the tools of procedural mathematics. So, while formal mathematics usually takes the format of definition-theorem-proof, this project has the format of declaration-procedure objective-procedure implementation. So where there usually would have been a statement and proof of Euler's totient theorem, \procedurehr{1.50} is provided, and where there would have been a definition of Euler's totient function, \declarationhr{1.16} is provided. Perhaps not surprisingly, software programming tools and concepts like lambdas, procedures, recursion, and modularity have turned out to be instrumental in rendering intelligable what could have been an indecipharable network of instructions/operations.
	\tableofcontents
	\twocolumn\printnoidxglossary[title={Declarations},sort=def]\onecolumn
	\clearpage
	\part{Integer Arithmetic}
	  \chapter{Integer Arithmetic}
	    \begin{multicols*}{2}
		    \declaration{1.22}
			    The phrase "\gls{integer}" will be used as a shorthand for an ordered pair of natural numbers.
		    \declaration{1.23}
			    The phrase "the positive part of $a$" and the notation \gls{intPo}, where $a$ is an integer, will be used as a shorthand for the first entry of $a$.
		    \declaration{1.24}
			    The phrase "the negative part of $a$" and the notation \gls{intNe}, where $a$ is an integer, will be used as a shorthand for the second entry of $a$.
		    \declaration{1.25}
			    The phrase "\gls{inta=b}", where $a,b$ are integers, will be used as a shorthand for "$\Po(a)+\Ne(b)=\Ne(a)+\Po(b)$".
		    \procedure{1.65}
			    \objective
				    Choose an integer $a$. The objective of the following instructions is to show that $a=a$.
			    \implementation
				    \begin{enumerate}
					    \item \textbf{Show that $a=a$ using \declarationhr{1.25} given that $\Po(a)+\Ne(a)=\Ne(a)+\Po(a)$.}
				    \end{enumerate}
		    \procedure{1.66}
			    \objective
				    Choose two integers $a,b$ such that $a=b$. The objective of the following instructions is to show that $b=a$.
			    \implementation
				    \begin{enumerate}
					    \item \textbf{Using \declarationhr{1.25}, show that $b=a$}
					    \begin{enumerate}
					      \item given that $\Po(b)+\Ne(a)=\Ne(b)+\Po(a)$
					      \item given that $\Po(a)+\Ne(b)=\Ne(a)+\Po(b)$
					      \item given that $a=b$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{1.67}
			    \objective
				    Choose three integers $a,b,c$ such that $a=b$ and $b=c$. The objective of the following instructions is to show that $a=c$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Po(a)+\Ne(b)=\Ne(a)+\Po(b)$ using \declarationhr{1.25}.
					    \item Show that $\Po(b)+\Ne(c)=\Ne(b)+\Po(c)$ using \declarationhr{1.25}.
					    \item \textbf{Hence show that $a=c$}
					    \begin{enumerate}
					      \item given that $\Po(a)+\Ne(c)=\Ne(a)+\Po(c)$
					      \item given that $\Po(a)+\Ne(b)+\Po(b)+\Ne(c)=\Ne(a)+\Po(b)+\Ne(b)+\Po(c)$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{1.26}
			    The notation \gls{inta+b}, where $a,b$ are integers, will be used as a shorthand for the pair $\langle\Po(a)+\Po(b),\Ne(a)+\Ne(b)\rangle$.
		    \procedure{1.68}
			    \objective
				    Choose four integers $a,b,c,d$ such that $a=c$ and $b=d$. The objective of the following instructions is to show that $a+b=c+d$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Po(a)+\Ne(c)=\Ne(a)+\Po(c)$ using \declarationhr{1.25}.
					    \item Show that $\Po(b)+\Ne(d)=\Ne(b)+\Po(d)$ using \declarationhr{1.25}.
					    \item Hence using \declarationhr{1.26}, show that $a+b$
					    \begin{enumerate}
						    \item $=\langle\Po(a),\Ne(a)\rangle+\langle\Po(b),\Ne(b)\rangle$
						    \item $=\langle\Po(a)+\Po(b),\Ne(a)+\Ne(b)\rangle$
						    \item $=\langle\Po(a)+\Po(b)+\Ne(c)+\Ne(d),\Ne(a)+\Ne(b)+\Ne(c)+\Ne(d)\rangle$
						    \item $=\langle(\Po(a)+\Ne(c))+(\Po(b)+\Ne(d)),\Ne(a)+\Ne(b)+\Ne(c)+\Ne(d)\rangle$
						    \item $=\langle(\Ne(a)+\Po(c))+(\Ne(b)+\Po(d)),\Ne(a)+\Ne(b)+\Ne(c)+\Ne(d)\rangle$
						    \item $=\langle\Ne(a)+\Ne(b)+\Po(c)+\Po(d),\Ne(a)+\Ne(b)+\Ne(c)+\Ne(d)\rangle$
						    \item $=\langle\Po(c)+\Po(d),\Ne(c)+\Ne(d)\rangle$
						    \item $=\langle\Po(c),\Ne(c)\rangle+\langle\Po(d),\Ne(d)\rangle$
						    \item $=c+d$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{1.69}
			    \objective
				    Choose three integers $a,b,c$. The objective of the following instructions is to show that $(a+b)+c=a+(b+c)$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{1.26}, show that $(a+b)+c$
					    \begin{enumerate}
						    \item $=\langle\Po(a)+\Po(b),\Ne(a)+\Ne(b)\rangle+\langle\Po(c),\Ne(c)\rangle$
						    \item $=\langle(\Po(a)+\Po(b))+\Po(c),(\Ne(a)+\Ne(b))+\Ne(c)\rangle$
						    \item $=\langle\Po(a)+(\Po(b)+\Po(c)),\Ne(a)+(\Ne(b)+\Ne(c))\rangle$
						    \item $=\langle\Po(a),\Ne(a)\rangle+\langle\Po(b)+\Po(c),\Ne(b)+\Ne(c)\rangle$
						    \item $=a+(b+c)$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{1.70}
			    \objective
				    Choose two integers $a,b$. The objective of the following instructions is to show that $a+b=b+a$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{1.26}, show that $a+b$
					    \begin{enumerate}
						    \item $=\langle\Po(a)+\Po(b),\Ne(a)+\Ne(b)\rangle$
						    \item $=\langle\Po(b)+\Po(a),\Ne(b)+\Ne(a)\rangle$
						    \item $=b+a$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{1.27}
			    The notation \gls{inta}, where $a$ is a natural number, will contextually be used as a shorthand for the pair $\langle a,0\rangle$.
		    \procedure{1.71}
			    \objective
				    Choose an integer $a$. The objective of the following instructions is to show that $0+a=a$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{1.26}, show that $0+a$
					    \begin{enumerate}
						    \item $=\langle 0,0\rangle+\langle\Po(a),\Ne(a)\rangle$
						    \item $=\langle 0+\Po(a),0+\Ne(a)\rangle$
						    \item $=\langle\Po(a),\Ne(a)\rangle$
						    \item $=a$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{1.28}
			    The notation \gls{int-a}, where $a$ is an integer, will be used as a shorthand for the pair $\langle\Ne(a),\Po(a)\rangle$.
		    \procedure{1.72}
			    \objective
				    Choose two integers $a,b$ such that $a=b$. The objective of the following instructions is to show that $-a=-b$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Po(a)+\Ne(b)=\Ne(a)+\Po(b)$ using \declarationhr{1.25}.
					    \item Hence using \declarationhr{1.28}, show that $-a$
					    \begin{enumerate}
						    \item $=\langle\Ne(a),\Po(a)\rangle$
						    \item $=\langle\Ne(a)+\Po(b),\Po(a)+\Po(b)\rangle$
						    \item $=\langle\Po(a)+\Ne(b),\Po(a)+\Po(b)\rangle$
						    \item $=\langle\Ne(b),\Po(b)\rangle$
						    \item $=-b$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{1.73}
			    \objective
				    Choose an integer $a$. The objective of the following instructions is to show that $-a+a=0$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{1.26}, show that $-a+a$
					    \begin{enumerate}
						    \item $=(-a)+a$
						    \item $=\langle\Ne(a),\Po(a)\rangle+\langle\Po(a),\Ne(a)\rangle$
						    \item $=\langle\Ne(a)+\Po(a),\Po(a)+\Ne(a)\rangle$
						    \item $=\langle 0,0\rangle$
						    \item $=0$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{1.29}
			    The notation \gls{intab}, where $a,b$ are integers, will be used as a shorthand for the pair $\langle\Po(a)\Po(b)+\Ne(a)\Ne(b),\Po(a)\Ne(b)+\Ne(a)\Po(b)\rangle$.
		    \procedure{1.74}
			    \objective
				    Choose four integers $a,b,c,d$ such that $a=c$ and $b=d$. The objective of the following instructions is to show that $ab=cd$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Po(a)+\Ne(c)=\Ne(a)+\Po(c)$ using \declarationhr{1.25}.
					    \item Show that $\Po(b)+\Ne(d)=\Ne(b)+\Po(d)$ using \declarationhr{1.25}.
					    \item Hence using \declarationhr{1.29}, show that $ab$
					    \begin{enumerate}
						    \item $=\langle\Po(a)\Po(b)+\Ne(a)\Ne(b),\Po(a)\Ne(b)+\Ne(a)\Po(b)\rangle$
						    \item $=\langle\Po(a)\Po(b)+\Ne(a)\Ne(b)+\Po(a)\Ne(d)+\Ne(c)\Po(d)+\Po(c)\Ne(d),\Po(a)\Ne(b)+\Ne(a)\Po(b)+\Po(a)\Ne(d)+\Ne(c)\Po(d)+\Po(c)\Ne(d)\rangle$
						    \item $=\langle\Po(a)(\Po(b)+\Ne(d))+\Ne(a)\Ne(b)+\Ne(c)\Po(d)+\Po(c)\Ne(d),\Po(a)\Ne(b)+\Ne(a)\Po(b)+\Po(a)\Ne(d)+\Ne(c)\Po(d)+\Po(c)\Ne(d)\rangle$
						    \item $=\langle\Po(a)(\Ne(b)+\Po(d))+\Ne(a)\Ne(b)+\Ne(c)\Po(d)+\Po(c)\Ne(d),\Po(a)\Ne(b)+\Ne(a)\Po(b)+\Po(a)\Ne(d)+\Ne(c)\Po(d)+\Po(c)\Ne(d)\rangle$
						    \item $=\langle(\Po(a)+\Ne(c))\Po(d)+\Ne(a)\Ne(b)+\Po(c)\Ne(d),\Ne(a)\Po(b)+\Po(a)\Ne(d)+\Ne(c)\Po(d)+\Po(c)\Ne(d)\rangle$
						    \item $=\langle(\Ne(a)+\Po(c))\Po(d)+\Ne(a)\Ne(b)+\Po(c)\Ne(d),\Ne(a)\Po(b)+\Po(a)\Ne(d)+\Ne(c)\Po(d)+\Po(c)\Ne(d)\rangle$
						    \item $=\langle\Ne(a)(\Po(d)+\Ne(b))+\Po(c)\Po(d)+\Po(c)\Ne(d),\Ne(a)\Po(b)+\Po(a)\Ne(d)+\Ne(c)\Po(d)+\Po(c)\Ne(d)\rangle$
						    \item $=\langle\Ne(a)(\Po(b)+\Ne(d))+\Po(c)\Po(d)+\Po(c)\Ne(d),\Ne(a)\Po(b)+\Po(a)\Ne(d)+\Ne(c)\Po(d)+\Po(c)\Ne(d)\rangle$
						    \item $=\langle(\Ne(a)+\Po(c))\Ne(d)+\Po(c)\Po(d),\Po(a)\Ne(d)+\Ne(c)\Po(d)+\Po(c)\Ne(d)\rangle$
						    \item $=\langle(\Po(a)+\Ne(c))\Ne(d)+\Po(c)\Po(d),\Po(a)\Ne(d)+\Ne(c)\Po(d)+\Po(c)\Ne(d)\rangle$
						    \item $=\langle\Ne(c)\Ne(d)+\Po(c)\Po(d),\Ne(c)\Po(d)+\Po(c)\Ne(d)\rangle$
						    \item $=cd$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{1.75}
			    \objective
				    Choose three integers $a,b,c$. The objective of the following instructions is to show that $(ab)c=a(bc)$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{1.29}, show that $(ab)c$
					    \begin{enumerate}
						    \item $=\langle\Po(a)\Po(b)+\Ne(a)\Ne(b),\Po(a)\Ne(b)+\Ne(a)\Po(b)\rangle\langle\Po(c),\Ne(c)\rangle$
						    \item $=\langle(\Po(a)\Po(b)+\Ne(a)\Ne(b))\Po(c)+(\Po(a)\Ne(b)+\Ne(a)\Po(b))\Ne(c),(\Po(a)\Po(b)+\Ne(a)\Ne(b))\Ne(c)+(\Po(a)\Ne(b)+\Ne(a)\Po(b))\Po(c)\rangle$
						    \item $=\langle\Po(a)(\Po(b)\Po(c)+\Ne(b)\Ne(c))+\Ne(a)(\Po(b)\Ne(c)+\Ne(b)\Po(c)),\Po(a)(\Po(b)\Ne(c)+\Ne(b)\Po(c))+\Ne(a)(\Po(b)\Po(c)+\Ne(b)\Ne(c))\rangle$
						    \item $=\langle\Po(a),\Ne(a)\rangle\langle\Po(b)\Po(c)+\Ne(b)\Ne(c),\Po(b)\Ne(c)+\Ne(b)\Po(c)\rangle$
						    \item $=a(bc)$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{1.76}
			    \objective
				    Choose two integers $a,b$. The objective of the following instructions is to show that $ab=ba$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{1.29}, show that $ab$
					    \begin{enumerate}
						    \item $=\langle\Po(a)\Po(b)+\Ne(a)\Ne(b),\Po(a)\Ne(b)+\Ne(a)\Po(b)\rangle$
						    \item $=\langle\Po(b)\Po(a)+\Ne(b)\Ne(a),\Po(b)\Ne(a)+\Ne(b)\Po(a)\rangle$
						    \item $=ba$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{1.77}
			    \objective
				    Choose an integer $a$. The objective of the following instructions is to show that $1a=a$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{1.29}, show that $1a$
					    \begin{enumerate}
						    \item $=\langle 1,0\rangle\langle\Po(a),\Ne(a)\rangle$
						    \item $=\langle 1\Po(a)+0\Ne(a),1\Ne(a)+0\Po(a)\rangle$
						    \item $=\langle\Po(a),\Ne(a)\rangle$
						    \item $=a$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{1.78}
			    \objective
				    Choose three integers $a,b,c$. The objective of the following instructions is to show that $a(b+c)=ab+ac$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{1.26} and \declarationhr{1.29}, show that $a(b+c)$
					    \begin{enumerate}
						    \item $=\langle\Po(a),\Ne(a)\rangle\langle\Po(b)+\Po(c),\Ne(b)+\Ne(c)\rangle$
						    \item $=\langle\Po(a)(\Po(b)+\Po(c))+\Ne(a)(\Ne(b)+\Ne(c)),\Po(a)(\Ne(b)+\Ne(c))+\Ne(a)(\Po(b)+\Po(c))\rangle$
						    \item $=\langle(\Po(a)\Po(b)+\Ne(a)\Ne(b))+(\Po(a)\Po(c)+\Ne(a)\Ne(c)),(\Po(a)\Ne(b)+\Ne(a)\Po(b))+(\Po(a)\Ne(c)+\Ne(a)\Po(c))\rangle$
						    \item $=\langle\Po(a)\Po(b)+\Ne(a)\Ne(b),\Po(a)\Ne(b)+\Ne(a)\Po(b)\rangle+\langle\Po(a)\Po(c)+\Ne(a)\Ne(c),\Po(a)\Ne(c)+\Ne(a)\Po(c)\rangle$
						    \item $=ab+ac$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{1.91}
			    \objective
				    Choose an integer $a$. The objective of the following instructions is to show that $(-1)^{2a}=1$ and $(-1)^{2a+1}=-1$.
			    \implementation
				    \begin{enumerate}
              \item Show that $(-1)^2=(-1)(-1)+1+(-1)=(-1)((-1)+1)+1=(-1)0+1=1$.
              \item \textbf{Hence show that $(-1)^{2a}=((-1)^2)^a=1^a=1$.}
					    \item \textbf{Also show that $(-1)^{2a+1}=(-1)^{2a}(-1)=1(-1)=-1$.}
				    \end{enumerate}
		    \declaration{1.30}
			    The phrases "\gls{inta<b}" and "$b>a$", where $a,b$ are rational numbers, will be used as a shorthand for "$\Po(a)+\Ne(b)<\Ne(a)+\Po(b)$".
		    \procedure{1.79}
			    \objective
				    Choose four integers $a,b,c,d$ such that $a<b$, $a=c$ and $b=d$. The objective of the following instructions is to show that $c<d$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Po(a)+\Ne(c)=\Ne(a)+\Po(c)$ using \declarationhr{1.25}.
					    \item Show that $\Po(b)+\Ne(d)=\Ne(b)+\Po(d)$ using \declarationhr{1.25}.
					    \item Show that $\Po(a)+\Ne(b)<\Ne(a)+\Po(b)$ using \declarationhr{1.30}.
					    \item Hence show that $\Po(c)+\Ne(d)$
					    \begin{enumerate}
						    \item $=(\Ne(a)+\Po(c))+(\Po(b)+\Ne(d))-\Ne(a)-\Po(b)$
						    \item $=(\Po(a)+\Ne(c))+(\Ne(b)+\Po(d))-\Ne(a)-\Po(b)$
						    \item $=(\Po(a)+\Ne(b))+\Ne(c)+\Po(d)-\Ne(a)-\Po(b)$
						    \item $<(\Ne(a)+\Po(b))+\Ne(c)+\Po(d)-\Ne(a)-\Po(b)$
						    \item $=\Ne(c)+\Po(d)$.
					    \end{enumerate}
					    \item \textbf{Hence show that $c<d$ using \declarationhr{1.30}.}
				    \end{enumerate}
		    \procedure{1.80}
			    \objective
				    Choose three integers $a,b,c$ such that $a<b$. The objective of the following instructions is to show that $a+c<b+c$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Po(a)+\Ne(b)<\Ne(a)+\Po(b)$ using \declarationhr{1.30}.
					    \item Hence show that $\Po(a+c)+\Ne(b+c)$
					    \begin{enumerate}
						    \item $=\Po(a)+\Po(c)+\Ne(b)+\Ne(c)$
						    \item $=(\Po(a)+\Ne(b))+\Po(c)+\Ne(c)$
						    \item $=(\Ne(a)+\Po(b))+\Po(c)+\Ne(c)$
						    \item $=\Ne(a)+\Ne(c)+\Po(b)+\Po(c)$
						    \item $=\Ne(a+c)+\Po(b+c)$.
					    \end{enumerate}
					    \item \textbf{Hence show that $a+c<b+c$ using \declarationhr{1.30}.}
				    \end{enumerate}
		    \procedure{1.81}
			    \objective
				    Choose two integers $a,b$. The objective of the following instructions is to show that $a<b$, $a=b$ and $a>b$.
			    \implementation
				    \begin{enumerate}
					    \item Show that either
					    \begin{enumerate}
					      \item $\Po(a)+\Ne(b)<\Ne(a)+\Po(b)$
					      \item $\Po(a)+\Ne(b)=\Ne(a)+\Po(b)$
					      \item $\Po(a)+\Ne(b)>\Ne(a)+\Po(b)$
					    \end{enumerate}
					    \item Hence show that either
					    \begin{enumerate}
					      \item $a<b$ using \declarationhr{1.30} given that $\Po(a)+\Ne(b)<\Ne(a)+\Po(b)$.
					      \item $a=b$ using \declarationhr{1.25} given that $\Po(a)+\Ne(b)=\Ne(a)+\Po(b)$.
					      \item $a>b$ using \declarationhr{1.30} given that $\Po(a)+\Ne(b)>\Ne(a)+\Po(b)$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{1.85}
			    \objective
				    Choose two integers $a,b$ such that $0<a$ and $0<b$. The objective of the following instructions is to show that $0<a+b$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Ne(a)=\Po(0)+\Ne(a)<\Ne(0)+\Po(a)=\Po(a)$ using \declarationhr{1.30}.
					    \item Show that $\Ne(b)=\Po(0)+\Ne(b)<\Ne(0)+\Po(b)=\Po(b)$ using \declarationhr{1.30}.
                        \item Show that $\Po(0)+\Ne(a+b)=\Ne(a+b)=\Ne(a)+\Ne(b)<\Po(a)+\Po(b)=\Po(a+b)=\Ne(0)+\Po(a+b)$.
					    \item \textbf{Hence show that $0<a+b$ given that $\Po(0)+\Ne(a+b)<\Ne(0)+\Po(a+b)$.}
				    \end{enumerate}
		    \procedure{1.86}
			    \objective
				    Choose two integers $a,b$ such that $0<a$ and $0<b$. The objective of the following instructions is to show that $0<ab$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Ne(a)=\Po(0)+\Ne(a)<\Ne(0)+\Po(a)=\Po(a)$ using \declarationhr{1.30}.
					    \item Hence show that $0<\Po(a)-\Ne(a)$.
					    \item Show that $\Ne(b)=\Po(0)+\Ne(b)<\Ne(0)+\Po(b)=\Po(b)$ using \declarationhr{1.30}.
					    \item Hence show that $0<\Po(b)-\Ne(b)$.
					    \item \textbf{Hence show that $0<ab$}
					    \begin{enumerate}
					      \item given that $\Po(0)+\Ne(ab)=\Ne(a)\Po(b)+\Po(a)\Ne(b)<\Po(a)\Po(b)+\Ne(a)\Ne(b)=\Ne(0)+\Po(ab)$
					      \item given that $\Ne(a)(\Po(b)-\Ne(b))<\Po(a)(\Po(b)-\Ne(b))$
					      \item given that $0<(\Po(a)-\Ne(a))(\Po(b)-\Ne(b))$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{1.34}
			    The notation \gls{int||a||} will be used as a shorthand for the following expression:
			    \begin{enumerate}
				    \item $-a$ if $a<0$
				    \item $a$ if $a\ge 0$
			    \end{enumerate}
		    \procedure{1.87}
			    \objective
				    Choose two integers $a,b$. The objective of the following instructions is to show that $\lVert ab\rVert=\lVert a\rVert\lVert b\rVert$.
			    \implementation
				    \begin{enumerate}
					    \item If $a\ge 0$ and $b\ge 0$, then do the following:
					    \begin{enumerate}
						    \item \textbf{Show that $\lVert ab\rVert=ab=\lVert a\rVert\lVert b\rVert$ given that $ab\ge 0$.}
					    \end{enumerate}
					    \item Otherwise if $a<0$ and $b\ge 0$, then do the following:
					    \begin{enumerate}
						    \item \textbf{Show that $\lVert ab\rVert=-(ab)=(-a)b=\lVert a\rVert\lVert b\rVert$ given that $ab<0$.}
					    \end{enumerate}
					    \item Otherwise if $a\ge 0$ and $b<0$, then do the following:
					    \begin{enumerate}
						    \item \textbf{Show that $\lVert ab\rVert=-(ab)=a(-b)=\lVert a\rVert\lVert b\rVert$ given that $ab<0$.}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
						    \item \textbf{Show that $\lVert ab\rVert=ab=(-a)(-b)=\lVert a\rVert\lVert b\rVert$.}
						    \begin{enumerate}
						      \item given that $ab>0$
						      \item given that $a<0$ and $b<0$.
						    \end{enumerate}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{1.88}
			    \objective
				    Choose two integers $a,b$. The objective of the following instructions is to show that $\lVert a+b\rVert\le\lVert a\rVert+\lVert b\rVert$.
			    \implementation
				    \begin{enumerate}
					    \item If $a+b\ge 0$, then do the following:
					    \begin{enumerate}
						    \item \textbf{Show that $\lVert a+b\rVert=a+b\le\lVert a\rVert+\lVert b\rVert$}
						    \begin{enumerate}
						      \item given that $a\le\lVert a\rVert$
						      \item and $b\le\lVert b\rVert$.
						    \end{enumerate}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
						    \item \textbf{Show that $\lVert a+b\rVert=-(a+b)=(-a)+(-b)\le\lVert a\rVert+\lVert b\rVert$}
						    \begin{enumerate}
						      \item given that $-a\le\lVert a\rVert$
						      \item and $-b\le\lVert b\rVert$
						      \item and $a+b<0$.
						    \end{enumerate}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{1.89}
			    \objective
				    Choose two integers $a,b$. The objective of the following instructions is to show that $\lVert a\rVert-\lVert b\rVert\le\lVert a-b\rVert$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\lVert a\rVert=\lVert b+(a-b)\rVert\le\lVert b\rVert+\lVert a-b\rVert$ using \procedurehr{1.88}.
					    \item \textbf{Hence show that $\lVert a\rVert-\lVert b\rVert\le\lVert a-b\rVert$.}
				    \end{enumerate}
		    \declaration{1.03}
			    The notation \gls{intsgn(a)} will be used as a shorthand for the following expression:
			    \begin{enumerate}
				    \item $-1$ if $a<0$
				    \item $0$ if $a=0$
				    \item $1$ if $a>0$
			    \end{enumerate}
			  \declaration{sun0902201144}
			    The notation \gls{inthsf(a)} will be used as a shorthand for the following expression:
			    \begin{enumerate}
				    \item $0$ if $a<0$
				    \item $1$ if $a\ge 0$
			    \end{enumerate}
		    \procedure{1.90}
			    \objective
				    Choose an integer $a$. The objective of the following instructions is to show that $a=\sgn(a)\lVert a\rVert$.
			    \implementation
				    \begin{enumerate}
					    \item If $a>0$, then do the following:
					    \begin{enumerate}
						    \item \textbf{Show that $a=1a=\sgn(a)\lVert a\rVert$}
						    \begin{enumerate}
						      \item given that $\lVert a\rVert=a$
						      \item and $\sgn(a)=1$.
						    \end{enumerate}
					    \end{enumerate}
					    \item If $a=0$, then do the following:
					    \begin{enumerate}
						    \item \textbf{Show that $a=0=\sgn(a)0=\sgn(a)\lVert a\rVert$ given that $\lVert a\rVert=a=0$.}
					    \end{enumerate}
					    \item Otherwise if $a<0$, then do the following:
					    \begin{enumerate}
						    \item \textbf{Show that $a=(-1)(-a)=\sgn(a)\lVert a\rVert$}
						    \begin{enumerate}
						      \item given that $\lVert a\rVert=-a$
						      \item and $\sgn(a)=-1$.
						    \end{enumerate}
					    \end{enumerate}
				    \end{enumerate}
	    \end{multicols*}
	  \chapter{Modular Arithmetic}
	    \begin{multicols*}{2}
		    \procedure{1.00}
			    \objective
				    Choose an integer $a$ and a positive integer $b$. The objective of the following instructions is to construct integers $n$ and $m$ such that $a=nb+m$ and $0\le m<b$.
			    \implementation
				    \begin{enumerate}
					    \item Let $n=0$.
					    \item While $(n+1)b\le a$, do the following:
					    \begin{enumerate}
						    \item Let $n$ receive $n+1$.
						    \item Show that $nb\le a$.
					    \end{enumerate}
					    \item While $nb>a$, do the following:
					    \begin{enumerate}
						    \item Let $n$ receive $n-1$.
						    \item Show that $(n+1)b>a$.
					    \end{enumerate}
					    \item Hence show that $nb\le a$ and $(n+1)b>a$.
					    \item Let $m=a-nb$.
					    \item \textbf{Now show that $b>a-nb=m\ge 0$ and $a=bn+a-nb=nb+m$.}
					    \item \textbf{Yield $\langle n,m\rangle$.}
				    \end{enumerate}
		    \declaration{1.00}
			    The notation \gls{intadivb} will be used to refer to the first part of the pair yielded by executing \procedurehr{1.00} on $\langle a,b\rangle$.
		    \declaration{1.01}
			    The notation \gls{intamodb} will be used to refer to the second part of the pair yielded by executing \procedurehr{1.00} on $\langle a,b\rangle$.
		    \declaration{1.02}
			    The notation \gls{inta=bmodc} will be used as a shorthand for "$a\bmod c=b\bmod c$".
		    \procedure{1.01}
			    \objective
				    Choose four integers $a,b,c,d$ and a positive integer $e$ in such a way that $a\equiv c\pmod{e}$ and $b\equiv d\pmod{e}$. The objective of the following instructions is to show that $a+b\equiv c+d\pmod{e}$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $a+b$
					    \begin{enumerate}
						    \item $\equiv(a\div e)e+(a\bmod e)+(b\div e)e+(b\bmod e)$
						    \item $\equiv(a\bmod e)+(b\bmod e)$
						    \item $\equiv(c\bmod e)+(d\bmod e)$
						    \item $\equiv(c\div e)e+(c\bmod e)+(d\div e)e+(d\bmod e)$
						    \item \textbf{$\equiv c+d\pmod{e}$.}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{1.02}
			    \objective
				    Choose four integers $a,b,c,d$ and a positive integer $e$ in such a way that $a\equiv c\pmod{e}$ and $b\equiv d\pmod{e}$. The objective of the following instructions is to show that $ab\equiv cd\pmod{e}$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $ab$
					    \begin{enumerate}
						    \item $\equiv((a\div e)e+(a\bmod e))((b\div e)e+(b\bmod e))$
						    \item $\equiv(a\div e)(b\div e)e^2+(a\div e)(b\bmod e)e+(a\bmod e)(b\div e)e+(a\bmod e)(b\bmod e)$
						    \item $\equiv(a\bmod e)(b\bmod e)$
						    \item $\equiv(c\bmod e)(d\bmod e)$
						    \item $\equiv(c\div e)(d\div e)e^2+(c\div e)(d\bmod e)e+(c\bmod e)(d\div e)e+(c\bmod e)(d\bmod e)$
						    \item \textbf{$\equiv cd\pmod{e}$.}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{1.03}
			    \objective
				    Choose an integer $a$ and two positive integers $b,c$. The objective of the following instructions is to show that $(a\bmod bc)\bmod b=a\bmod b$.
			    \implementation
				    \begin{enumerate}
					    \item \textbf{Show that $(a\bmod bc)\bmod b=(a-(a\div bc)bc)\bmod b=a\bmod b$.}
				    \end{enumerate}
		    \procedure{1.04}
			    \objective
				    Choose a positive integer $a$ and four integers $b_1,b_0,c_1,c_0$ such that $0\le b_0<a$, $0\le c_0<a$, and $b_1a+b_0=c_1a+c_0$. The objective of the following instructions is to show that $b_1=c_1$ and $b_0=c_0$.
			    \implementation
				    \begin{enumerate}
					    \item \textbf{Show that $b_0=b_0\bmod a=(b_1a+b_0)\bmod a=(c_1a+c_0)\bmod a=c_0\bmod a=c_0$.}
					    \item \textbf{Therefore show that $b_1=c_1$ given that $b_1a=c_1a$.}
				    \end{enumerate}
		    \procedure{1.05}
			    \objective
				    Choose an integer $a$ and two positive integers $b,c$. The objective of the following instructions is to show that $ca\bmod cb=c(a\bmod b)$ and that $ca\div cb=a\div b$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $bc(a\div b)+c(a\bmod b)=c(b(a\div b)+a\bmod b)=ca=cb(ca\div cb)+ca\bmod cb$.
					    \item Show that $0\le a\bmod b<b$.
					    \item Show that $0\le c(a\bmod b)<cb$.
					    \item Show that $0\le ca\bmod cb<cb$.
					    \item \textbf{Hence show that $c(a\bmod b)=ca\bmod cb$ and $a\div b=ca\div cb$ using \procedurehr{1.04}.}
				    \end{enumerate}
		    \procedure{1.06}
			    \objective
				    Choose two integers $a,b$ and a positive integer $c$ such that $a\bmod c+b\bmod c<c$. The objective of the following instructions is to show that $a\div c+b\div c=(a+b)\div c$ and $a\bmod c+b\bmod c=(a+b)\bmod c$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $a=c(a\div c)+a\bmod c$.
					    \item Show that $b=c(b\div c)+b\bmod c$.
					    \item Therefore show that $a+b=c(a\div c+b\div c)+(a\bmod c+b\bmod c)$.
					    \item Show that $0\le a\bmod c+b\bmod c<c$.
					    \item Also show that $a+b=((a+b)\div c)c+(a+b)\bmod c$.
					    \item Show that $0\le(a+b)\bmod c<c$.
					    \item \textbf{Hence show that $a\div c+b\div c=(a+b)\div c$ and $a\bmod c+b\bmod c=(a+b)\bmod c$ using \procedurehr{1.04}.}
				    \end{enumerate}
		    \procedure{1.07}
			    \objective
				    Choose two integers $a,b$ and a positive integer $c$ such that $a\bmod c+b\bmod c\ge c$. The objective of the following instructions is to show that $1+a\div c+b\div c=(a+b)\div c$ and $a\bmod c+b\bmod c-c=(a+b)\bmod c$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $a=c(a\div c)+a\bmod c$.
					    \item Show that $b=c(b\div c)+b\bmod c$.
					    \item Therefore show that $a+b=c(a\div c+b\div c)+a\bmod c+b\bmod c=c(1+a\div c+b\div c)+(a\bmod c+b\bmod c-c)$.
					    \item Show that $c\le a\bmod c+b\bmod c<2c$.
					    \item Therefore show that $0\le a\bmod c+b\bmod c-c<c$.
					    \item Also show that $a+b=c((a+b)\div c)+(a+b)\bmod c$.
					    \item Show that $0\le (a+b)\bmod c<c$.
					    \item \textbf{Therefore show that $1+a\div c+b\div c=(a+b)\div c$ and $a\bmod c+b\bmod c-c=(a+b)\bmod c$ using \procedurehr{1.04}.}
				    \end{enumerate}
		    \procedure{1.08}
			    \objective
				    Choose an integer $a$ and two positive integers $b,c$. The objective of the following instructions is to show that $a\div bc=(a\div b)\div c$ and $a\bmod bc=((a\div b)\bmod c)b+a\bmod b$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $a=(((a\div b)\div c)c+(a\div b)\bmod c)b+a\bmod b=((a\div b)\div c)bc+((a\div b)\bmod c)b+a\bmod b$
					    \begin{enumerate}
					      \item given that $a=(a\div b)b+a\bmod b$
					      \item given that $a\div b=((a\div b)\div c)c+(a\div b)\bmod c$.
					    \end{enumerate}
					    \item Show that $0\le((a\div b)\bmod c)b\le cb-b$ given that $0\le(a\div b)\bmod c\le c-1$.
					    \item Therefore show that $0\le ((a\div b)\bmod c)b+a\bmod b<cb$ given that $0\le a\bmod b<b$.
					    \item Now show that $a=(a\div bc)bc+a\bmod bc$ and $0\le a\bmod bc<bc$.
					    \item \textbf{Therefore show that $(a\div b)\div c=a\div bc$ and $((a\div b)\bmod c)b+a\bmod b=a\bmod bc$ using \procedurehr{1.04}.}
				    \end{enumerate}
		    \procedure{1.09}
			    \objective
				    Choose an integer $a$ and a non-negative integer $b$. The objective of the following instructions is to consruct integers $c,d,e,f,g$ such that $a=cd$, $b=ce$, $fa+gb=c$, and if $b=0$, then $c=\lvert a\rvert$, otherwise $0<c\le b$.
			    \implementation
				    \begin{enumerate}
					    \item If $b=0$, then do the following:
					    \begin{enumerate}
						    \item \textbf{Show that $a=\sgn(a)\lvert a\rvert$.}
						    \item \textbf{Show that $b=0\lvert a\rvert$.}
						    \item \textbf{Show that $\lvert a\rvert=\sgn(a)a+0b$.}
						    \item \textbf{Yield $\langle\lvert a\rvert,\sgn(a),0,\sgn(a),0\rangle$.}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
						    \item Show that $0\le a\bmod b<b$.
						    \item Use \procedurehr{1.09} on $\langle b,a\bmod b\rangle$ to construct $\langle c,d,e,f,g\rangle$ and show that:
						    \begin{enumerate}
						      \item $b=cd$
						      \item $a\bmod b=ce$
						      \item $c=\lVert b\rVert$ if $a\bmod b=0$, otherwise $0<c\le a\bmod b$
						      \item $fb+g(a\mod b)=c$.
						    \end{enumerate}
						    \item \textbf{Hence show that $a=(a\div b)b+(a\bmod b)=c(d(a\div b)+e)$.}
						    \item \textbf{Also show that $(f-g(a\div b))b+ga=fb+g(a-(a\div b)b)=fb+g(a\bmod b)=c$.}
						    \item If $a\bmod b=0$, then do the following:
						    \begin{enumerate}
							    \item \textbf{Show that $0<b=c\le b$ given that $b\ge 0$, $b\ne 0$, and $c=\lVert b\rVert=b$.}
						    \end{enumerate}
						    \item Otherwise do the following:
						    \begin{enumerate}
							    \item \textbf{Show that $0<c\le a\bmod b<b$ given $0<c\le a\bmod b$.}
						    \end{enumerate}
						    \item \textbf{Therefore yield $\langle c,d(a\div b)+e,d,g,f-g(a\div b)\rangle$.}
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{1.04}
			    The notation \gls{int(a,b)} will be used to refer to the first part of the quintuple constructed by using \procedurehr{1.09} on the pair $\langle a,b\rangle$.
		    \procedure{1.10}
			    \objective
				    Choose an integer $a$ and a positive integer $b$. Let $1\le c\le b$ be the largest integer such that $a\bmod c=0$ and $b\bmod c=0$. The objective of the following instructions is to either show that $0\ne 0$ or $(a,b)=c$.
			    \implementation
				    \begin{enumerate}
					    \item Use \procedurehr{1.09} on $\langle a,b\rangle$ to construct $\langle d,e,f,g,h\rangle$ and show that:
					    \begin{enumerate}
					      \item $a=ed$
					      \item $b=fd$
					      \item $ga+hb=d$
					      \item $0<d\le b$.
					    \end{enumerate}
					    \item If $d>c$, then do the following:
					    \begin{enumerate}
						    \item Show that $a\bmod d\ne 0$ or $b\bmod d\ne 0$ given that $0<d\le b$ is larger than the largest integer such that $a\bmod c=0$ and $b\bmod c=0$.
						    \item If $a\bmod d\ne 0$, then do the following:
						    \begin{enumerate}
							    \item Show that $a\bmod d=0$ given that $a=ed$.
							    \item \textbf{Hence show that $0\ne 0$ given that $a\bmod d\ne 0$ and $a\bmod d=0$.}
							    \item \textbf{Abort procedure.}
						    \end{enumerate}
						    \item Otherwise if $b\bmod d\ne 0$, then do the following:
						    \begin{enumerate}
							    \item Show that $b\bmod d=0$ given that $b=fd$.
							    \item \textbf{Hence show that $0\ne 0$ given that $b\bmod d\ne 0$ and $b\bmod d=0$.}
							    \item \textbf{Abort procedure.}
						    \end{enumerate}
					    \end{enumerate}
					    \item Otherwise if $d<c$, then do the following:
					    \begin{enumerate}
						    \item Show that $0\equiv gc(a\div c)+hc(b\div c)=g(c(a\div c)+a\bmod c)+h(c(b\div c)+b\bmod c)=ga+hb=d\not\equiv 0\pmod{c}$ given that:
						    \begin{enumerate}
						      \item $ga+hb=d$
						      \item $a\bmod c=0$
						      \item $b\bmod c=0$.
						    \end{enumerate}
						    \item \textbf{Hence show that $0\ne 0$.}
						    \item \textbf{Abort procedure.}
					    \end{enumerate}
					    \item \textbf{Otherwise show that $(a,b)=d=c$.}
				    \end{enumerate}
		    \procedure{1.11}
			    \objective
				    Choose integers $a,c,d,j$ and a non-negative integer $b$. Use \procedurehr{1.09} on $\langle a,b\rangle$ to construct $\langle e,f,g,h,i\rangle$. The objective of the following instructions is to show that $ca+db=(c+gj)a+(d-fj)b$.
			    \implementation
				    \begin{enumerate}
					    \item \textbf{Show that $(c+gj)a+(d-fj)b=ca+db+gja-fjb=ca+db+gjef-fjeg=ca+db$.}
				    \end{enumerate}
		    \procedure{1.12}
			    \objective
				    Choose integers $a,c,d$ and a non-negative integer $b$ such that $ca+db=(a,b)$. Use \procedurehr{1.09} on $\langle a,b\rangle$ to construct $\langle e,f,g,h,i\rangle$. The objective of the following instructions is to construct a $j$ such that $c=h+gj$ and $d=i-fj$.
			    \implementation
				    \begin{enumerate}
				      \item Use \procedurehr{1.09} on $\langle a,b\rangle$ to show that:
				      \begin{enumerate}
				        \item $a=ef$
				        \item $b=eg$
				        \item $ha+ib=e$.
				      \end{enumerate}
					    \item Show that $cf+dg=1$
					    \begin{enumerate}
					      \item given that $cef+deg=ca+db=(a,b)=e$
					      \item given that $a=ef$ and $b=eg$.
					    \end{enumerate}
					    \item Show that $hf+ig=1$
					    \begin{enumerate}
					      \item given that $hef+ieg=ha+ib=e$
					      \item given that $a=ef$ and $b=eg$.
					    \end{enumerate}
					    \item Let $j=ci-hd$.
					    \item \textbf{Show that $c=h+cig-hdg=h+g(ci-hd)=h+gj$}
					    \begin{enumerate}
					      \item given that $c-cig=c(1-ig)=chf=h(1-dg)=h-hdg$
					      \item given that $cf=1-dg$.
					    \end{enumerate}
					    \item \textbf{Show that $d=i-icf+dhf=i-f(ic-dh)=i-fj$}
					    \begin{enumerate}
					      \item given that $d-dhf=d(1-hf)=dig=i(1-cf)=i-icf$
					      \item given that $dg=1-cf$.
					    \end{enumerate}
					    \item \textbf{Yield $\langle j\rangle$.}
				    \end{enumerate}
		    \procedure{1.13}
			    \objective
				    Choose an integer $a$ and a positive integer $b$ such that $0<(a,b)<b$. The objective of the following instructions is to show that $0\ne 0$ or $a\bmod b\ne 0$.
			    \implementation
				    \begin{enumerate}
					    \item If $a\bmod b=0$, then do the following:
					    \begin{enumerate}
						    \item Show that $af\equiv 0f\equiv 0\pmod{b}$ given that $a\bmod b=0$.
						    \item Use \procedurehr{1.09} on $\langle a,b\rangle$ to construct $\langle c,d,e,f,g\rangle$ and show that:
						    \begin{enumerate}
						      \item $fa+gb=c=(a,b)$
						      \item $0<c=(a,b)\le b$.
						    \end{enumerate}
						    \item Hence show that $fa\equiv(a,b)\not\equiv 0\pmod{b}$ given that $0<(a,b)<b$.
						    \item Hence show that $0\ne 0$ given that $0\equiv af\not\equiv 0\pmod{b}$.
						    \item \textbf{Abort procedure.}
					    \end{enumerate}
					    \item \textbf{Otherwise show that $a\bmod b\ne 0$.}
				    \end{enumerate}
		    \procedure{1.14}
			    \objective
				    Choose five integers $a,d,e,f,g$ and two non-negative integers $b,c$ such that $a=cd$, $b=ce$, and $fa+gb=c$. The objective of the following instructions is to show that $0<0$ or $(a,b)=c$.
			    \implementation
				    \begin{enumerate}
					    \item Use \procedurehr{1.09} on $\langle a,b\rangle$ to construct $\langle u,v,x,y,z\rangle$ and show that:
					    \begin{enumerate}
					      \item $u\ge 0$
					      \item $a=uv$
					      \item $b=xu$
					      \item $u=ya+zb$.
					    \end{enumerate}
					    \item Hence show that $c=fa+gb=(fv+gx)u$.
					    \item If $u=0$, then do the following:
					    \begin{enumerate}
						    \item \textbf{Show that $c=(fv+gx)u=0=u=(a,b)$.}
						    \item \textbf{Yield.}
					    \end{enumerate}
					    \item Show that $u=ya+zb=(yd+ze)c$ given that $u=ya+zb$, $a=cd$, and $b=ce$.
					    \item If $c=0$, then do the following:
					    \begin{enumerate}
						    \item \textbf{Show that $(a,b)=u=(yd+ze)c=0=c$.}
						    \item \textbf{Yield.}
					    \end{enumerate}
					    \item Show that $fv+gx=yd+ze=\pm 1$
					    \begin{enumerate}
					      \item given that $(fv+gx)(yd+ze)=1$
					      \item given that $c=(fv+gx)u=(fv+gx)(yd+ze)c$ and $c>0$.
					    \end{enumerate}
					    \item If $fv+gx=yd+ze=-1$, then do the following:
					    \begin{enumerate}
						    \item Show that $u=(yd+ze)c=(-1)c<0$ given that $u=(yd+ze)c$ and $c>0$.
						    \item \textbf{Hence show that $0\le u<0$ given that $u\ge 0$.}
						    \item \textbf{Abort procedure.}
					    \end{enumerate}
					    \item Otherwise, do the following:
					    \begin{enumerate}
						    \item Show that $fv+gx=yd+ze=1$.
						    \item \textbf{Hence show that $c=(fv+gx)u=(1)u=(a,b)$ given that $c=(fv+gx)u$.}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{1.15}
			    \objective
				    Choose an integer $a$ and a non-negative integer $b$. The objective of the following instructions is to show that $0<0$ or $(a,b)=(-a,b)$.
			    \implementation
				    \begin{enumerate}
					    \item Use \procedurehr{1.09} on $\langle a,b\rangle$ to construct $\langle c,d,e,f,g\rangle$ and show that:
					    \begin{enumerate}
					      \item $a=dc$
					      \item $b=ec$
					      \item $fa+gb=c$.
					    \end{enumerate}
					    \item Hence show that $-a=(-d)c$.
					    \item Also show that $(-f)(-a)+gb=c$.
					    \item \textbf{Use \procedurehr{1.14} on $\langle -a,b,c,-d,e,-f,g\rangle$ to show that $(-a,b)=c=(a,b)$.}
				    \end{enumerate}
		    \procedure{1.16}
			    \objective
				    Choose two non-negative integers $a,b$. The objective of the following instructions is to show that $0<0$ or $(a,b)=(b,a)$.
			    \implementation
				    \begin{enumerate}
					    \item Use \procedurehr{1.09} on $\langle a,b\rangle$ to construct $\langle c,d,e,f,g\rangle$ and show that:
					    \begin{enumerate}
					      \item $b=ec$
					      \item $a=dc$
					      \item $gb+fa=c$.
					    \end{enumerate}
					    \item \textbf{Use \procedurehr{1.14} on $\langle b,a,c,e,d,g,f\rangle$ to show that $(b,a)=c=(a,b)$.}
				    \end{enumerate}
		    \procedure{1.17}
			    \objective
				    Choose two integers $a,b$ and a positive integer $c$ such that $a\equiv b\pmod{c}$. The objective of the following instructions is to show that $0<0$ or $(a,c)=(b,c)$.
			    \implementation
				    \begin{enumerate}
					    \item Use \procedurehr{1.09} on $\langle a,c\rangle$ to construct $\langle d,e,f,g,h\rangle$ and show that:
					    \begin{enumerate}
					      \item $a=ed$
					      \item $c=fd$
					      \item $ga+hc=d$.
					    \end{enumerate}
					    \item Let $j=b\div c-a\div c$.
					    \item Hence show that $b=a+jc=ed+jfd=(e+jf)d$.
					    \item Also show that $gb+(h-gj)c=g(a+jc)+(h-gj)c=ga+hc=d$ given that $b=a+jc$.
					    \item \textbf{Use \procedurehr{1.14} on $\langle b,c,d,e+jf,f,g,h-gj\rangle$ to show that $(b,c)=d=(a,c)$.}
				    \end{enumerate}
		    \procedure{1.18}
			    \objective
				    Choose an integer $a$ and two non-negative integers $b,c$. The objective of the following instructions is to show that either $0<0$ or $(ca,cb)=c(a,b)$.
			    \implementation
				    \begin{enumerate}
					    \item Use \procedurehr{1.09} on $\langle a,b\rangle$ to construct $\langle d,e,f,g,h\rangle$ and show that:
					    \begin{enumerate}
					      \item $a=ed$
					      \item $b=df$
					      \item $ga+hb=d$.
					    \end{enumerate}
					    \item Hence show that $ca=e(cd)$, $cb=f(cd)$, and $g(ca)+h(cb)=cd$.
					    \item \textbf{Use \procedurehr{1.14} on $\langle ca,cb,cd,e,f,g,h\rangle$ to show that $(ca,cb)=cd=c(a,b)$.}
				    \end{enumerate}
		    \procedure{1.19}
			    \objective
				    Choose an integer $a$ and two non-negative integers $b,c$. The objective of the following instructions is to show that either $0<0$ or $(a,(b,c))=((a,b),c)$.
			    \implementation
				    \begin{enumerate}
					    \item Use \procedurehr{1.09} on $\langle a,b\rangle$ to construct $\langle d_0,e_0,f_0,g_0,h_0\rangle$ and show that:
					    \begin{enumerate}
					      \item $a=d_0e_0$
					      \item $b=d_0f_0$
					      \item $g_0a+h_0b=d_0$.
					    \end{enumerate}
					    \item Use \procedurehr{1.09} on $\langle b,c\rangle$ to construct $\langle d_1,e_1,f_1,g_1,h_1\rangle$ and show that:
					    \begin{enumerate}
					      \item $b=d_1e_1$
					      \item $c=d_1f_1$
					      \item $g_1b+h_1c=d_1$.
					    \end{enumerate}
					    \item Use \procedurehr{1.09} on $\langle (a,b),c\rangle$ to construct $\langle d_2,e_2,f_2,g_2,h_2\rangle$ and show that:
					    \begin{enumerate}
					      \item $(a,b)=d_2e_2$
					      \item $c=d_2f_2$
					      \item $g_2(a,b)+h_2c=d_2$.
					    \end{enumerate}
					    \item Show that $a=d_0e_0=e_0(a,b)=e_0d_2e_2=e_0e_2((a,b),c)$.
					    \item Also show that $(b,c)$
					    \begin{enumerate}
						    \item $=g_1b+h_1c$
						    \item $=g_1d_0f_0+h_1d_2f_2$
						    \item $=g_1f_0(a,b)+h_1f_2((a,b),c)$
						    \item $=g_1f_0d_2e_2+h_1f_2((a,b),c)$
						    \item $=g_1f_0e_2((a,b),c)+h_1f_2((a,b),c)$
						    \item $=(g_1f_0e_2+h_1f_2)((a,b),c)$.
					    \end{enumerate}
					    \item Also show that $((a,b),c)$
					    \begin{enumerate}
						    \item $=d_2$
						    \item $=g_2(a,b)+h_2c$
						    \item $=g_2d_0+h_2d_1f_1$
						    \item $=g_2(g_0a+h_0b)+h_2f_1(b,c)$
						    \item $=g_2g_0a+g_2h_0d_1e_1+h_2f_1(b,c)$
						    \item $=g_2g_0a+g_2h_0e_1(b,c)+h_2f_1(b,c)$
						    \item $=g_2g_0a+(g_2h_0e_1+h_2f_1)(b,c)$.
					    \end{enumerate}
					    \item \textbf{Use \procedurehr{1.14} on $\langle a,(b,c),((a,b),c),e_0e_2,g_1f_0e_2+h_1f_2,g_2g_0,g_2h_0e_1+h_2f_1\rangle$ to show that $((a,b),c)=(a,(b,c))$.}
				    \end{enumerate}
		    \declaration{1.05}
			    The notation \gls{int(a_0,a_1,...,a_n-1)} will be used to contextually refer to one of the following integers:
			    \begin{enumerate}
				    \item $((a_0),(a_1,a_2,\cdots,a_{n-1}))$
				    \item $((a_0,a_1),(a_2,a_3,\cdots,a_{n-1}))$
				    \item $\vdots$
				    \item $((a_0,a_1,\cdots,a_{n-2}),(a_{n-1}))$
			    \end{enumerate}
		    \procedure{1.20}
			    \objective
				    Choose two integers $a,b$ and a non-negative integer $c$ such that $(a,c)=1$ and $(b,c)=1$. The objective of the following instructions is to show that either $0<0$ or $(ab,c)=1$.
			    \implementation
				    \begin{enumerate}
					    \item Use \procedurehr{1.09} on $\langle a,c\rangle$ to construct $\langle d,e,f,g,h\rangle$ and show that $ga+hc=d=(a,c)=1$.
					    \item Use \procedurehr{1.09} on $\langle b,c\rangle$ to construct $\langle t,u,v,w,x\rangle$ and show that $wb+xc=t=(b,c)=1$.
					    \item Hence show that $(gw)(ab)+(gax+wbh+hxc)c=(ga+hc)(wb+xc)=1$.
					    \item \textbf{Use \procedurehr{1.14} on $\langle ab,c,1,ab,c,gw,gax+wbh+hxc\rangle$ to show that $(ab,c)=1$.}
				    \end{enumerate}
		    \procedure{1.21}
			    \objective
				    Choose an integer $a$ and two non-negative integers $b,c$ such that $(a,bc)=1$. The objective of the following instructions is to show that either $0<0$ or $(a,b)=1$.
			    \implementation
				    \begin{enumerate}
					    \item Use \procedurehr{1.09} on $\langle a,bc\rangle$ to construct $\langle d,e,f,g,h\rangle$ and show that $ga+(hc)b=ga+h(bc)=d=(a,bc)=1$.
					    \item \textbf{Now use \procedurehr{1.14} on $\langle a,b,1,a,b,g,hc\rangle$ to show that $(a,b)=1$.}
				    \end{enumerate}
		    \declaration{1.06}
			    The phrase "\gls{intprime}" will be used to refer to integers $a$ such that $a>1$ and $a\bmod k\ne 0$ for $1<k<a$.
		    \procedure{1.22}
			    \objective
				    Choose an integer $a$ and a prime $b$ such that $a\bmod b\ne 0$. The objective of the following instructions is to show that either $0\ne 0$ or $(a,b)=1$.
			    \implementation
				    \begin{enumerate}
					    \item Use \procedurehr{1.09} on $\langle a,b\rangle$ to construct $\langle c,d,e,f,g\rangle$ and show that:
					    \begin{enumerate}
					      \item $a=cd$
					      \item $b=ce$
					      \item $0<c\le b$.
					    \end{enumerate}
					    \item If $c=b$, then do the following:
					    \begin{enumerate}
						    \item Show that $a\bmod b=0$ given that $a=cd=bd$.
						    \item \textbf{Hence show that $0\ne 0$ given that $a\bmod b\ne 0$.}
						    \item \textbf{Abort procedure.}
					    \end{enumerate}
					    \item Otherwise if $1<c<b$, then do the following:
					    \begin{enumerate}
						    \item Show that $b\bmod c=0$ given that $b=ce$.
						    \item \textbf{Hence show that $0\ne 0$ given that $b$ is prime.}
						    \item \textbf{Abort procedure.}
					    \end{enumerate}
					    \item Otherwise, do the following:
					    \begin{enumerate}
						    \item \textbf{Show that $(a,b)=c=1$.}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{1.23}
			    \objective
				    Choose two integers $a,b$ and a prime $c$ such that $a\bmod c\ne 0$ and $b\bmod c\ne 0$. The objective of the following instructions is to show that either $0\ne 0$ or $ab\bmod c\ne 0$.
			    \implementation
				    \begin{enumerate}
					    \item Use \procedurehr{1.22} on $\langle a,c\rangle$ to show that $(a,c)=1$.
					    \item Use \procedurehr{1.22} on $\langle b,c\rangle$ to show that $(b,c)=1$.
					    \item Use \procedurehr{1.20} on $\langle a,b,c\rangle$ to show that $0<(ab,c)=1<c$.
					    \item \textbf{Use \procedurehr{1.13} on $\langle ab,c\rangle$ to show that $ab\bmod c\ne 0$.}
				    \end{enumerate}
		    \declaration{1.07}
			    The notation \gls{int|a|} will be used to refer to the number of items in the list $a$.
		    \declaration{1.10}
			    The notation \gls{inta^b} will be used to refer to the list formed by concatenating $a$ and $b$.
		    \declaration{1.31}
			    The notation \gls{intf(R)}, where $R$ is a list and $f[r]$ is a function of $r$, will contextually be used as a shorthand for the list $\langle f(R_0),f(R_1),\cdots,f(R_{\lvert R\rvert-1})\rangle$.
		    \declaration{1.09}
			    The notation \gls{inta_*}, where $a$ is a list, will be used as a shorthand for $1$ if $a$ is empty, otherwise it will be a shorthand for the product of the entries of $a$.
		    \declaration{1.08}
			    The notation \gls{intpif(r)}, where $R$ is a list and $f[r]$ is a function of $r$, will be used as a shorthand for $f(R)_*$.
		    \procedure{1.24}
			    \objective
				    Choose a positive integer $a$. The objective of the following instructions is to construct a list of prime numbers $b$ such that $a=b_*$.
			    \implementation
				    \begin{enumerate}
					    \item If $a=1$, then do the following:
					    \begin{enumerate}
						    \item Show that $a=1=\langle\rangle_*$.
						    \item Hence yield $\langle\rangle$.
					    \end{enumerate}
					    \item Otherwsie, do the following:
					    \begin{enumerate}
						    \item Show that $a>1$.
						    \item If there is a $c\in[2:a]$ such that $a\bmod c=0$, then do the following:
						    \begin{enumerate}
							    \item Show that $a=(a\div c)c$.
							    \item Hence show that $1<a\div c<a$.
							    \item Use \procedurehr{1.24} on $\langle a\div c\rangle$ to construct $\langle d\rangle$ and show that:
							    \begin{enumerate}
							      \item every element of $d$ is prime.
							      \item $a\div c=d_*$.
							    \end{enumerate}
							    \item Hence show that $d$ is non-empty given that $1<a\div c=d_*$.
							    \item Use \procedurehr{1.24} on $\langle c\rangle$ to construct $\langle e\rangle$ and show that:
							    \begin{enumerate}
							      \item every element of $e$ is prime.
							      \item $c=e_*$.
							    \end{enumerate}
							    \item Hence show that $e$ is non-empty given that $1<c=e_*$.
							    \item \textbf{Hence show that $d^\frown e$ is a non-empty list of prime numbers such that $a=(a\div c)c=d_*e_*=(d^\frown e)_*$.}
							    \item \textbf{Yield $\langle d^\frown e\rangle$.}
						    \end{enumerate}
						    \item Otherwise do the following:
						    \begin{enumerate}
							    \item \textbf{Show that $a$ is prime.}
							    \item \textbf{Yield $\langle a\rangle$.}
						    \end{enumerate}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{1.25}
			    \objective
				    Choose a prime $a$ and a list of primes $b$ such that $b_*\equiv 0\pmod{a}$. The objective of the following instructions is to either show that $0=1$ or to construct a $k$ such that $a=b_k$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $a>1$ given that $a$ is prime.
					    \item If $\lvert b\rvert=0$, then do the following:
					    \begin{enumerate}
						    \item Show that $1=b_*\equiv 0\pmod{a}$.
						    \item \textbf{Hence show that $0=1$ given that $a>1$.}
						    \item \textbf{Abort procedure.}
					    \end{enumerate}
					    \item Otherwise if $0\not\in b\bmod a$, then do the following:
					    \begin{enumerate}
						    \item Show that $b_*\not\equiv 0 \pmod{a}$ using \procedurehr{1.23}.
						    \item \textbf{Hence show that $0\ne 0$ given that $b_*\equiv 0 \pmod{a}$.}
						    \item \textbf{Abort procedure.}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
						    \item Let $k$ be such that $b_k\bmod a=0$.
						    \item Show that $b_k=(b_k\div a)a$.
						    \item Hence show that $b_k\div a\ge 1$.
						    \item If $b_k\div a>1$, then do the following:
						    \begin{enumerate}
							    \item Show that $1<a<b_k$ given that:
							    \begin{enumerate}
							      \item $a>1$
							      \item $b_k\div a>1$
							      \item $b_k=(b_k\div a)a$.
							    \end{enumerate}
							    \item Hence show that $b_k\bmod a\ne 0$ given that $b_k$ is prime and $1<a<b_k$.
							    \item \textbf{Hence show that $0\ne b_k\bmod a=0$ given that $b_k\bmod a=0$.}
							    \item \textbf{Abort procedure.}
						    \end{enumerate}
						    \item Otherwise do the following:
						    \begin{enumerate}
							    \item \textbf{Show that $b_k=a$ given that $b_k\div a=1$.}
							    \item \textbf{Yield $\langle k\rangle$.}
						    \end{enumerate}
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{1.11}
			    The notation \gls{int[a:b]} will be used as a shorthand for the list:
			    \begin{enumerate}
				    \item $\langle a,a+1,\cdots,b-1\rangle$, if $b>a$
				    \item $\langle\rangle$, if $b=a$
				    \item $\langle a-1,a-2,\cdots,b\rangle$, if $b<a$
			    \end{enumerate}
		    \procedure{1.26}
			    \objective
				    Choose two lists of primes $a,b$ such that $a_*=b_*$. The objective of the following instructions is to show that either $1>1$ or $a$ is included in $b$.
			    \implementation
				    \begin{enumerate}
					    \item If $\lvert a\rvert=0$, then do the following:
					    \begin{enumerate}
						    \item \textbf{Show that $a$ is included in $b$.}
					    \end{enumerate}
					    \item Otherwise, do the following:
					    \begin{enumerate}
						    \item Show that $\lvert a\rvert>0$.
						    \item Show that $b_*\equiv a_*\equiv 0 \pmod{a_0}$.
						    \item Use \procedurehr{1.25} on $\langle a_0, b\rangle$ to construct $\langle k\rangle$ and show that $b_k=a_0$.
						    \item Now show that $(a_{[1:\lvert a\rvert]})_*=(b_{[0:k]^\frown[k+1:\lvert b\rvert]})_*$.
						    \item Now use \procedurehr{1.26} on $\langle a_{[1:\lvert a\rvert]},b_{[0:k]^\frown[k+1:\lvert b\rvert]}\rangle$ to show that $a_{[1:\lvert a\rvert]}$ is included in $b_{[0:k]^\frown[k+1:\lvert b\rvert]}\rangle$.
						    \item \textbf{Hence show that $a$ is included in $b$.}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{1.27}
			    \objective
				    Choose two lists of primes $a,b$ such that $a_*=b_*$. The objective of the following instructions is to show that either $1>1$ or $a$ is a rearrangement of $b$.
			    \implementation
				    \begin{enumerate}
					    \item Use \procedurehr{1.26} on $\langle a,b\rangle$ to show that $a$ is included in $b$.
					    \item Use \procedurehr{1.26} on $\langle b,a\rangle$ to show that $b$ is included in $a$.
					    \item \textbf{Hence show that $a$ is a rearrangement of $b$.}
				    \end{enumerate}
		    \procedure{1.28}
			    \objective
				    Choose a positive integer $a$. The objective of the following instructions is to either show that $0=1$ or to construct a prime $b$ such that $b>a$ and $[a+1:b]$ does not contain a prime.
			    \implementation
				    \begin{enumerate}
					    \item Show that $a!+1>1$.
					    \item Use \procedurehr{1.24} on $\langle a!+1\rangle$ to construct $\langle d\rangle$ and show that:
					    \begin{enumerate}
					      \item $a!+1=d_*$
					      \item every element of $d$ is prime.
					    \end{enumerate}
					    \item Hence show that $\lvert d\rvert>0$ given that $a!+1>1$.
					    \item Hence show that $(a!+1)\bmod d_0=0$.
					    \item If $d_0\in[2:a+1]$, then do the following:
					    \begin{enumerate}
					      \item Show that $a!+1\equiv 1 \pmod{d_0}$
					      \begin{enumerate}
					        \item given that $a!\pmod{d_0}\equiv 0$
					        \item given that $d_0\in[2:a+1]$.
					      \end{enumerate}
						    \item Show that $0\equiv a!+1\pmod{d_0}$
						    \begin{enumerate}
						      \item given that $(a!+1)\bmod d_0=0$
						      \item given that $a!+1=d_*$.
						    \end{enumerate}
						    \item \textbf{Hence show that $0=1$.}
						    \item \textbf{Abort procedure.}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
						    \item \textbf{Show that $d_0$ is prime given that every element of $d$ is prime.}
						    \item \textbf{Hence show that $d_0>a$ given that $d_0>1$ and $d_0\not\in[2:a+1]$.}
						    \item \textbf{Let $b$ be the least prime in $[a+1:d_0+1]$.}
						    \item \textbf{Yield $\langle b\rangle$.}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{1.29}
			    \objective
				    Choose a positive integer $a$. The objective of the following instructions is to construct a positive integer $b$ such that $[b+1:b+a]$ does not contain a prime.
			    \implementation
				    \begin{enumerate}
					    \item Let $b=a!+1$.
					    \item For $i\in[1:a]$, do the following:
					    \begin{enumerate}
						    \item Show that $b+i=a!+1+i=i!(i+1)(i+2)\cdots(a)+1+i=(1+i)(i!(i+2)(i+3)\cdots(a) +1)$.
						    \item Therefore show that $b+i\equiv 0 \pmod{i+1}$.
						    \item Also show that $b+i=a!+1+i>a!\ge a\ge i+1>1$.
						    \item \textbf{Hence show that $b+i$ is not prime.}
					    \end{enumerate}
					    \item \textbf{Yield $\langle b\rangle$.}
				    \end{enumerate}
		    \procedure{1.30}
			    \objective
				    Choose two lists of primes $a,b$ in such a way that their intersection is empty. The objective of the following instructions is to show that $0=1$ or $(a_*,b_*)=1$.
			    \implementation
				    \begin{enumerate}
					    \item Use \procedurehr{1.09} on $\langle a_*,b_*\rangle$ to construct $\langle c,d,e,f,g\rangle$ and show that:
					    \begin{enumerate}
					      \item $0<c\le b_*$
					      \item $a_*=cd$
					      \item $b_*=ce$.
					    \end{enumerate}
					    \item If $c>1$, then do the following:
					    \begin{enumerate}
						    \item Use \procedurehr{1.24} on $\langle c\rangle$ to construct $\langle h\rangle$ and show that $c=h_*$.
						    \item Hence show that $\lvert h\rvert>0$ given that $h_*=c>1$.
						    \item Now show that $a_*=dc=dh_*=dh_0(h_{[1:\vert h\rvert]})_*\equiv 0\pmod{h_0}$.
						    \item Use \procedurehr{1.25} on $\langle h_0,a\rangle$ to construct $\langle k\rangle$ and show that $h_0=a_k$.
						    \item Now show that $b_*=ec=eh_*=eh_0(h_{[1:\vert h\rvert]})_*\equiv 0\pmod{h_0}$.
						    \item Use \procedurehr{1.25} on $\langle h_0,b\rangle$ to construct $\langle m\rangle$ and show that $h_0=b_m$.
						    \item \textbf{Hence show that $a$ and $b$ intersect given that $a_k=h_0=b_m$.}
						    \item \textbf{Abort procedure.}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
						    \item \textbf{Show that $(a_*,b_*)=c=1$ given that $0<c\le b_*$ and $c\le 1$.}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{1.31}
			    \objective
				    Choose two lists of primes $a,b$. Let $c$ be the common sublist with multiplicity of $a$ and $b$. The objective of the following instructions is to show that either $0<0$ or $(a_*,b_*)=c_*$.
			    \implementation
				    \begin{enumerate}
					    \item Let $d$ be the result of removing with multiplicity elements of $c$ from $a$.
					    \item Show that $a_*=c_*d_*$.
					    \item Let $e$ be the result of removing with multiplicity elements of $c$ from $b$.
					    \item Show that $b_*=c_*e_*$.
					    \item Show that $d$ and $e$ share no common elements.
					    \item \textbf{Therefore show that $(a_*,b_*)=(c_*d_*,c_*e_*)=c_*(d_*,e_*)=c_*$ using \procedurehr{1.18} and \procedurehr{1.30}.}
				    \end{enumerate}
		    \procedure{1.32}
			    \objective
				    Choose an integer $a$ and a positive integer $b$. The objective of the following instructions is to construct integers $c,f,e$ such that $c=af$, $c=be$, $c(a,b)=ab$, and $\lvert a\rvert\le\lvert c\rvert\le\lvert a\rvert b$.
			    \implementation
				    \begin{enumerate}
					    \item Use \procedurehr{1.09} on $\langle a,b\rangle$ to construct $\langle d,e,f,g,h\rangle$ and show that:
					    \begin{enumerate}
					      \item $a=de$
					      \item $b=df$
					      \item $d>0$.
					    \end{enumerate}
					    \item \textbf{Let $c=af$.}
					    \item \textbf{Show that $c=af=def=be$.}
					    \item \textbf{Show that $c(a,b)=cd=afd=ab$.}
					    \item Show that $1\le f\le b$
					    \begin{enumerate}
					      \item given that $0<b=df$
					      \item and $d>0$.
					    \end{enumerate}
					    \item Therefore show that $\lvert a\rvert\le\lvert a\rvert f\le\lvert a\rvert b$.
					    \item \textbf{Therefore show that $\lvert a\rvert\le\lvert c\rvert\le\lvert a\rvert b$.}
					    \item \textbf{Yield the tuple $\langle c,f,e\rangle$.}
				    \end{enumerate}
		    \declaration{1.12}
			    The notation \gls{int[a,b]} will be used to refer to the first part of the triple yielded by executing \procedurehr{1.32} on $\langle a,b\rangle$.
		    \procedure{1.33}
			    \objective
				    Choose two positive integers $a,b$. The objective of the following instructions is to show that either $0<0$ or $[a,b]=[b,a]$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $(a,b)>0$.
					    \item Show that $[a,b](a,b)=ab=ba=[b,a](b,a)=[b,a](a,b)$ using \procedurehr{1.16}.
					    \item \textbf{Therefore show that $[a,b]=[b,a]$.}
				    \end{enumerate}
		    \procedure{1.34}
			    \objective
				    Choose an integer $a$ and two positive integers $b,c$. The objective of the following instructions is to show that either $0<0$ or $[ca,cb]=c[a,b]$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $(ca,cb)>0$.
					    \item Show that $[ca,cb](ca,cb)=cacb=c^2ab=c^2[a,b](a,b)=c[a,b](ca,cb)$ using \procedurehr{1.18}.
					    \item \textbf{Therefore show that $[ca,cb]=c[a,b]$.}
				    \end{enumerate}
		    \procedure{1.35}
			    \objective
				    Choose an integer $a$ and two positive integers $b,c$. The objective of the following instructions is to show that either $0<0$ or $[[a,b],c]=[a,[b,c]]$.
			    \implementation
				    \begin{enumerate}
					    \item Using \procedurehr{1.19}, show that $(a,b)(ab,(ac,bc))(b,c)[[a,b],c]$
					    \begin{enumerate}
						    \item $=(ab,(ac,bc))(b,c)[(a,b)[a,b],(a,b)c]$
						    \item $=(ab,(ac,bc))(b,c)[ab,(ac,bc)]$
						    \item $=ab(ac,bc)(b,c)$
						    \item $=abc(a,b)(b,c)$
						    \item $=bc(a,b)(ab,ac)$
						    \item $=(a,b)((ab,ac),bc)[(ab,ac),bc]$
						    \item $=(a,b)(ab,(ac,bc))[(ab,ac),bc]$
						    \item $=(a,b)(ab,(ac,bc))[a(b,c),[b,c](b,c)]$
						    \item $=(a,b)(ab,(ac,bc))(b,c)[a,[b,c]]$.
					    \end{enumerate}
					    \item Show that $(a,b)(ab,(ac,bc))(b,c)>0$.
					    \item \textbf{Therefore show that $[[a,b],c]=[a,[b,c]]$.}
				    \end{enumerate}
		    \declaration{1.13}
			    The notation \gls{int[a_0,a_1,...,a_n-1]} will be used to contextually refer to one of the following integers:
			    \begin{enumerate}
				    \item $[[a_0],[a_1,a_2,\cdots,a_{n-1}]]$
				    \item $[[a_0,a_1],[a_2,a_3,\cdots,a_{n-1}]]$
				    \item $\vdots$
				    \item $[[a_0,a_1,\cdots,a_{n-2}],[a_{n-1}]]$
			    \end{enumerate}
		    \procedure{1.36}
			    \objective
				    Choose three positive integers $a,b,c$. The objective of the following instructions is to show that either $0<0$ or $([a,b],c)=[(a,c),(b,c)]$.
			    \implementation
				    \begin{enumerate}
					    \item Using \procedurehr{1.32}, \procedurehr{1.18}, \procedurehr{1.19}, \procedurehr{1.16}, and \procedurehr{1.10}, show that $(a,b)((a,c),(b,c))([a,b],c)$
					    \begin{enumerate}
						    \item $=((a,c),(b,c))((a,b)[a,b],(a,b)c)$
						    \item $=((a,c),(b,c))(ab,(ac,bc))$
						    \item $=(a^2b,a^2c,c^2a,c^2b,b^2a,bac,b^2c)$
						    \item $=(a,b)(ab,ac,bc,c^2)$
						    \item $=(a,b)(a,c)(b,c)$
						    \item $=(a,b)((a,c),(b,c))[(a,c),(b,c)]$.
					    \end{enumerate}
					    \item Show that $(a,b)((a,c),(b,c))>0$.
					    \item \textbf{Therefore show that $([a,b],c)=[(a,c),(b,c)]$.}
				    \end{enumerate}
		    \procedure{1.37}
			    \objective
				    Choose three positive integers $a,b,c$. The objective of the following instructions is to show that either $0<0$ or $[(a,b),c]=([a,c],[b,c])$.
			    \implementation
				    \begin{enumerate}
					    \item Using \procedurehr{1.32}, \procedurehr{1.18}, \procedurehr{1.19}, \procedurehr{1.16}, and \procedurehr{1.10}, show that $((a,b),c)(a,c)(b,c)[(a,b),c]$
					    \begin{enumerate}
						    \item $=(a,c)(b,c)(a,b)c$
						    \item $=(ab,ac,cb,c^2)(a,b)c$
						    \item $=(a^2b,a^2c,ac^2,ab^2,abc,cb^2,bc^2)c$
						    \item $=(a,b,c)(ab,ac,bc)c$
						    \item $=((a,b),c)(ac(b,c),bc(a,c))$
						    \item $=((a,b),c)(a,c)(b,c)([a,c],[b,c])$.
					    \end{enumerate}
					    \item Show that $((a,b),c)(a,c)(b,c)>0$.
					    \item \textbf{Therefore show that $[(a,b),c]=([a,c],[b,c])$.}
				    \end{enumerate}
	    \end{multicols*}
	  \chapter{Congruence Equations}
	    \begin{multicols*}{2}
		    \declaration{1.14}
			    The notation \gls{intchibdac}, where $a,c$ are two integers and $b,d$ are two positive integers such that $a\equiv c\pmod{(b,d)}$,  will be used to refer to the result yielded by executing the following instructions:
			    \begin{enumerate}
				    \item Use \procedurehr{1.09} on $\langle b,d\rangle$ to construct $\langle f,g,h,i,j\rangle$.
				    \item \textbf{Yield the tuple $\langle (a+((c-a)\div(b,d))ib)\bmod[b,d]\rangle$.}
			    \end{enumerate}
		    \procedure{1.39}
			    \objective
				    Choose three integers $x,a,c$ and two positive integers $b,d$ such that $x\equiv a\pmod{b}$ and $x\equiv c\pmod{d}$. The objective of the following instructions is to show that $0\ne 0$ if $a\not\equiv c\pmod{(b,d)}$, otherwise $x\equiv\chi_{b,d}(a,c)\pmod{[b,d]}$.
			    \implementation
				    \begin{enumerate}
					    \item Use \procedurehr{1.09} on $\langle b,d\rangle$ to construct $\langle e,f,g,h,i\rangle$ and show that:
					    \begin{enumerate}
					      \item $b=ef$
					      \item $d=eg$
					      \item $hb+id=e$.
					    \end{enumerate}
					    \item Let $j=x\div b-a\div b$.
					    \item Show that $x=a+jb$ given that $x\equiv a\pmod{b}$.
					    \item Let $k=x\div d-c\div d$.
					    \item Show that $x=c+kd$ given that $x\equiv c\pmod{d}$.
					    \item Therefore show that $c-a=jb-kd$.
					    \item If $a\not\equiv c\pmod{(b,d)}$, then do the following:
					    \begin{enumerate}
						    \item Show that $0\not\equiv c-a=jb-kd=jef-keg\equiv 0\pmod{e}$.
						    \item \textbf{Therefore show that $0\ne 0$.}
						    \item \textbf{Abort procedure.}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
						    \item Let $l=(c-a)\div(b,d)$.
						    \item Show that $l(b,d)=le=c-a=jb-kd=jef-keg$ given that $c-a\equiv 0\pmod{(b,d)}$.
						    \item Hence show that $l\equiv jf\pmod{g}$ given that $l=jf-kg$.
						    \item Hence show that $fh\equiv 1\pmod{g}$
						    \begin{enumerate}
						      \item given that $fh+gi=1$
						      \item given that $efh+egi=bh+di=e$
						      \item given that $b=ef$, $d=eg$, and $hb+id=e$.
						    \end{enumerate}
						    \item Hence show that $lh\equiv jfh\equiv j\pmod{g}$
						    \begin{enumerate}
						      \item given that $l\equiv jf\pmod{g}$
						      \item and $fh\equiv 1\pmod{g}$.
						    \end{enumerate}
						    \item Hence show that $lhb\equiv jb\pmod{bg=[b,d]}$ using \procedurehr{1.05}.
						    \item \textbf{Hence show that $x=a+jb\equiv a+lhb\equiv\chi_{b,d}(a,c)\pmod{[b,d]}$.}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{1.40}
			    \objective
				    Choose two integers $a,c$ and two positive integers $b,d$ in such a way that $a\equiv c\pmod{(b,d)}$. The objective of the following instructions is to show that either $0<0$ or $\chi_{b,d}(a,c)=\chi_{d,b}(c,a)$.
			    \implementation
				    \begin{enumerate}
					    \item Use \procedurehr{1.09} on $\langle b,d\rangle$ to construct $\langle f,g,h,i,j\rangle$ and show that $ib+jd=f=(b,d)$.
					    \item Use \procedurehr{1.09} on $\langle d,b\rangle$ to construct $\langle k,l,m,n,p\rangle$ and show that $pb+nd=k=(d,b)=(b,d)$.
					    \item Use \procedurehr{1.12} on $\langle b,p,n,d\rangle$ to construct $\langle q\rangle$ and show that $n=j-qg$.
					    \item Now using \procedurehr{1.33}, show that $\chi_{b,d}(a,c)$
					    \begin{enumerate}
						    \item $=(a+((c-a)\div(b,d))ib)\bmod[b,d]$
						    \item $=(a+((c-a)\div(b,d))(f-jd))\bmod[b,d]$
						    \item $=(a+((c-a)\div(b,d))f+((a-c)\div(b,d))jd)\bmod[b,d]$
						    \item $=(a+(c-a)+((a-c)\div(b,d))jd)\bmod[b,d]$
						    \item $=(c+((a-c)\div(d,b))(n+qg)d)\bmod[b,d]$
						    \item $=(c+((a-c)\div(d,b))dn+((a-c)\div(d,b))q[b,d])\bmod[b,d]$
						    \item $=(c+((a-c)\div(d,b))dn)\bmod[b,d]$
						    \item $=(c+((a-c)\div(d,b))dn)\bmod[d,b]$
						    \item $=\chi_{d,b}(c,a)$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{1.41}
			    \objective
				    Choose three integers $x,a,c$ and two positive integers $b,d$ such that $a\equiv c\pmod{(b,d)}$ and $x\equiv\chi_{b,d}(a,c)\pmod{[b,d]}$. The objective of the following instructions is to show that $x\equiv a\pmod{b}$.
			    \implementation
				    \begin{enumerate}
					    \item Use \procedurehr{1.09} on $\langle b,d\rangle$ to construct $\langle e,f,g,h,i\rangle$.
					    \item Show that $[b,d]=bg$.
					    \item Hence show that $(x\bmod(bg))\bmod b=(\chi_{b,d}(a,c)\bmod(bg))\bmod b$
					    \begin{enumerate}
					      \item given that $x\bmod(bg)=\chi_{b,d}(a,c)\bmod(bg)$
					      \item given that $x\bmod[b,d]=\chi_{b,d}(a,c)\bmod[b,d]$.
					    \end{enumerate}
					    \item \textbf{Therefore using \procedurehr{1.03}, show that $x\bmod b=\chi_{b,d}(a,c)\bmod b=(a+((c-a)\div(b,d))hb)\bmod b=a\bmod b$.}
				    \end{enumerate}
		    \procedure{1.42}
			    \objective
				    Choose three integers $x,a,c$ and two positive integers $b,d$ such that $a\equiv c\pmod{(b,d)}$ and $x\equiv\chi_{b,d}(a,c)\pmod{[b,d]}$. The objective of the following instructions is to either show that $0<0$ or to show that $x\equiv a\pmod{b}$ and $x\equiv c\pmod{d}$.
			    \implementation
				    \begin{enumerate}
					    \item \textbf{Use \procedurehr{1.41} on $\langle x,a,c,b,d\rangle$ to show that $x\equiv a\pmod{b}$.}
					    \item Show that $x\equiv\chi_{b,d}(a,c)\equiv\chi_{d,b}(c,a)\pmod{[d,b]}$ using \procedurehr{1.40}.
					    \item \textbf{Use \procedurehr{1.41} on $\langle x,c,a,d,b\rangle$ to show that $x\equiv c\pmod{d}$.}
				    \end{enumerate}
		    \procedure{1.43}
			    \objective
				    Choose two integers $a,c$ and three positive integers $b,d,e$ such that $a\equiv c\pmod{(b,d)}$. The objective of the following instructions is to show that $\chi_{b,d}(ea,ec)=e\chi_{b,d}(a,c)$.
			    \implementation
				    \begin{enumerate}
					    \item Use \procedurehr{1.42} on $\langle\chi_{b,d}(a,c),a,c,b,d\rangle$ to show that:
					    \begin{enumerate}
					      \item $\chi_{b,d}(a,c)\equiv a\pmod{b}$
					      \item $\chi_{b,d}(a,c)\equiv c\pmod{d}$.
					    \end{enumerate}
					    \item Hence show that $e\chi_{b,d}(a,c)\equiv ea\pmod{b}$ using \procedurehr{1.05}.
					    \item Also show that $e\chi_{b,d}(a,c)\equiv ec\pmod{d}$ using \procedurehr{1.05}.
					    \item Also show that $ea\equiv ec\pmod{(b,d)}$ using using \procedurehr{1.02} given that $a\equiv c\pmod{(b,d)}$.
					    \item \textbf{Hence show that $e\chi_{b,d}(a,c)\equiv\chi_{b,d}(ea,ec)\pmod{[b,d]}$ using \procedurehr{1.39}.}
				    \end{enumerate}
		    \procedure{1.44}
			    \objective
				    Choose two integers $a,c$ and three positive integers $b,d,e$ such that $a\equiv c\pmod{(eb,ed)}$. The objective of the following instructions is to show that $\chi_{eb,ed}(a,c)\bmod[b,d]=\chi_{b,d}(a,c)$.
			    \implementation
				    \begin{enumerate}
				      \item Use \procedurehr{1.42} on $\langle\chi_{eb,ed}(a,c),a,c,eb,ed\rangle$ to show that:
				      \begin{enumerate}
				        \item $\chi_{eb,ed}(a,c)\equiv a\pmod{eb}$
				        \item $\chi_{eb,ed}(a,c)\equiv c\pmod{ed}$.
				      \end{enumerate}
					    \item Show that $\chi_{eb,ed}(a,c)\equiv a\pmod{b}$ using \procedurehr{1.03}.
					    \item Show that $\chi_{eb,ed}(a,c)\equiv c\pmod{d}$ using \procedurehr{1.03}.
					    \item Show that $a\equiv c\pmod{(b,d)}$ using \procedurehr{1.03} given that $a\equiv c\pmod{e(b,d)}$.
					    \item Hence show that $\chi_{eb,ed}(a,c)\equiv\chi_{b,d}(a,c)\pmod{[b,d]}$ using \procedurehr{1.39}.
					    \item \textbf{Hence show that $\chi_{eb,ed}(a,c)\bmod[b,d]=\chi_{b,d}(a,c)$.}
				    \end{enumerate}
		    \procedure{1.46}
			    \objective
				    Choose three integers $a,c,e$ and three positive integers $b,d,f$ such that $a\equiv c\pmod{(b,d)}$ and $\chi_{b,d}(a,c)\equiv e\pmod{([b,d],f)}$. The objective of the following instructions is to show that $0\ne 0$ if $c\not\equiv e\pmod{(d,f)}$ or $a\not\equiv\chi_{d,f}(c,e)\pmod{(b,[d,f])}$, otherwise $\chi_{[b,d],f}(\chi_{b,d}(a,c),e)=\chi_{b,[d,f]}(a,\chi_{d,f}(c,e))$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\chi_{[b,d],f}(\chi_{b,d}(a,c),e)\equiv e\pmod{f}$ using \procedurehr{1.42}.
					    \item Show that $\chi_{[b,d],f}(\chi_{b,d}(a,c),e)\equiv \chi_{b,d}(a,c)\pmod{[b,d]=gb=hd}$ using \procedurehr{1.42}.
					    \item Show that $\chi_{[b,d],f}(\chi_{b,d}(a,c),e)\equiv \chi_{b,d}(a,c)\equiv a\pmod{b}$ using \procedurehr{1.03} and \procedurehr{1.42}.
					    \item Show that $\chi_{[b,d],f}(\chi_{b,d}(a,c),e)\equiv \chi_{b,d}(a,c)\equiv c\pmod{d}$ using \procedurehr{1.03} and \procedurehr{1.42}.
					    \item Use \procedurehr{1.39} on $\langle\chi_{[b,d],f}(\chi_{b,d}(a,c),e),c,e,d,f\rangle$ to show that $0\ne 0$ if $c\not\equiv e\pmod{(d,f)}$, otherwise $\chi_{[b,d],f}(\chi_{b,d}(a,c),e)\equiv\chi_{d,f}(c,e)\pmod{[d,f]}$.
					    \item Use \procedurehr{1.39} on $\langle\chi_{[b,d],f}(\chi_{b,d}(a,c),e),a,\chi_{d,f}(c,e),b,[d,f]\rangle$ to show that $0\ne 0$ if $a\not\equiv\chi_{d,f}(c,e)\pmod{(b,[d,f])}$, otherwise $\chi_{[b,d],f}(\chi_{b,d}(a,c),e)\equiv\chi_{b,[d,f]}(a,\chi_{d,f}(c,e))\pmod{[b,[d,f]]=[[b,d],f]}$.
					    \item \textbf{Hence show that $\chi_{[b,d],f}(\chi_{b,d}(a,c),e)=\chi_{b,[d,f]}(a,\chi_{d,f}(c,e))$.}
				    \end{enumerate}
		    \declaration{1.15}
			    The notation \gls{intchib_0,b_1,...,b_n-1a_0a_1,...,a_n-1} will be used to contextually refer to one of the following integers:
			    \begin{enumerate}
				    \item $\chi_{b_0,[b_1,b_2,\cdots,b_{n-1}]}(a_0,\allowbreak\chi_{b_1,b_2,\cdots,b_{n-1}}(a_1,a_2,\cdots,a_{n-1}))$
				    \item $\chi_{[b_0,b_1],[b_2,b_3,\cdots,b_{n-1}]}(\chi_{b_0,b_1}(a_0,a_1),\allowbreak\chi_{b_2,b_3,\cdots,b_{n-1}}(a_2,a_3,\cdots,a_{n-1})$
				    \item $\vdots$
				    \item $\chi_{[b_0,b_1,\cdots,b_{n-2}],b_{n-1}}(\chi_{b_0,b_1,\cdots,b_{n-2}}(a_0,a_1,\cdots,a_{n-2}),\allowbreak a_{n-1})$
			    \end{enumerate}
		    \declaration{1.16}
			    The notation \gls{intphi(n)} will be used as a shorthand for the sublist of $[0:n]$ where each entry $x$ is such that $(x,n)=1$.
		    \procedure{1.47}
			    \objective
				    Choose an integer $a$ and a positive integer $b$ such that $(a,b)=1$. The objective of the following instructions is to either show that $0<0$ or to show that each element of $a\phi(b)\bmod b$ is in $\phi(b)$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $(a,b)=1$.
					    \item For $i$ in $[0:\lvert\phi(b)\rvert]$, do the following:
					    \begin{enumerate}
						    \item Show that $(\phi(b)_i,b)=1$ using \declarationhr{1.16}.
						    \item Use \procedurehr{1.20} on $\langle a,\phi(b)_i,b\rangle$ to show that $(a\phi(b)_i,b)=1$.
						    \item Use \procedurehr{1.17} on $\langle a\phi(b)_i\bmod b,a\phi(b)_i,b\rangle$ to show that $(a\phi(b)_i\bmod b,b)=(a\phi(b)_i,b)=1$.
						    \item Hence show that $a\phi(b)_i\bmod b$ is contained in the list $\phi(b)$ given that $0\le a\phi(b)_i\bmod b<b$.
					    \end{enumerate}
					    \item \textbf{Hence show that each element of $a\phi(b)\bmod b$ is in $\phi(b)$.}
				    \end{enumerate}
		    \procedure{1.48}
			    \objective
				    Choose an integer $a$ and a positive integer $b$ such that $(a,b)=1$. The objective of the following instructions is to either show that $0\ne 0$ or to show that each element of $a\phi(b)\bmod b$ is distinct.
			    \implementation
				    \begin{enumerate}
					    \item Use \procedurehr{1.09} on $\langle a,b\rangle$ to construct $\langle r,t,u,v,w\rangle$ and show that $va+wb=r=(a,b)=1$.
					    \item Hence show that $va\equiv 1\pmod{b}$.
					    \item Now for $i$ in $[0:\lvert\phi(b)\rvert]$, do the following:
					    \begin{enumerate}
						    \item For $j$ in $[i+1:\lvert\phi(b)\rvert]$, do the following:
						    \begin{enumerate}
							    \item If $a\phi(b)_i\equiv a\phi(b)_j\pmod{b}$, then do the following:
							    \begin{enumerate}
								    \item Show that $\phi(b)_i\equiv va\phi(b)_i\equiv va\phi(b)_j\equiv\phi(b)_j\pmod{b}$.
								    \item Hence show that $\phi(b)_i=\phi(b)_j$.
								    \item Show that $\phi(b)_i\ne\phi(b)_j$ using \declarationhr{1.16} given that $i\ne j$.
								    \item \textbf{Hence show that $\phi(b)_i\ne\phi(b)_i$ given that $\phi(b)_i=\phi(b)_j$ and $\phi(b)_i\ne\phi(b)_j$.}
								    \item \textbf{Abort procedure.}
							    \end{enumerate}
							    \item Otherwise, do the following:
							    \begin{enumerate}
								    \item Show that $a\phi(b)_i\not\equiv a\phi(b)_j\pmod{b}$.
							    \end{enumerate}
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Therefore show that $a\phi(b)\bmod b$ is composed of distinct elements.}
				    \end{enumerate}
		    \procedure{1.49}
			    \objective
				    Choose an integer $a$ and a positive integer $b$ such that $(a,b)=1$. The objective of the following instructions is to either show that $0<0$ or to show that $a\phi(b)\bmod b$ is a rearrangement of $\phi(b)$.
			    \implementation
				    \begin{enumerate}
					    \item Use \procedurehr{1.47} on $\langle a,b\rangle$ to show that each element of $a\phi(b)\bmod b$ is in $\phi(b)$.
					    \item Show that $\lvert a\phi(b)\bmod b\rvert=\lvert \phi(b)\rvert$.
					    \item Use \procedurehr{1.48} on $\langle a,b\rangle$ to show that $a\phi(b)\bmod b$ is composed of distinct elements.
					    \item \textbf{Hence show that $a\phi(b)\bmod b$ is a rearrangement of $\phi(b)$.}
				    \end{enumerate}
		    \procedure{1.50}
			    \objective
				    Choose an integer $a$ and a positive integer $b$ such that $(a,b)=1$. The objective of the following instructions is to show that either $0<0$ or $a^{\lvert\phi(b)\rvert}\equiv 1\pmod{b}$.
			    \implementation
				    \begin{enumerate}
					    \item For $i$ in $[0:\lvert\phi(b)\rvert]$, do the following:
					    \begin{enumerate}
						    \item Use \procedurehr{1.09} on $\langle\phi(b)_i,b\rangle$ to construct $\langle r_i,t_i,u_i,v_i,w_i\rangle$ and show that $v_i\phi(b)_i+w_ib=r_i=(\phi(b)_i,b)$.
						    \item Show that $v_i\phi(b)_i+w_ib=(\phi(b)_i,b)=1$ using \declarationhr{1.16}.
						    \item Hence show that $v_i\phi(b)_i\equiv 1\pmod{b}$.
					    \end{enumerate}
					    \item Hence using \procedurehr{1.49}, show that $\prod_i^{[0:{\lvert\phi(b)\rvert}]}\phi(b)_i$
					    \begin{enumerate}
					      \item $\equiv\prod_i^{[0:{\lvert\phi(b)\rvert}]}a\phi(b)_i$
					      \item $\equiv a^{\lvert\phi(b)\rvert}\prod_i^{[0:{\lvert\phi(b)\rvert}]}\phi(b)_i\pmod{b}$.
					    \end{enumerate}
					    \item \textbf{Hence show that $1$}
					    \begin{enumerate}
					      \item $\equiv\prod_i^{[0:{\lvert\phi(b)\rvert}]}(v_i\phi(b)_i)$
					      \item $=\prod_i^{[0:{\lvert\phi(b)\rvert}]}v_i\prod_i^{[0:{\lvert\phi(b)\rvert}]}\phi(b)_i$
					      \item $\equiv a^{\lvert\phi(b)\rvert}\prod_i^{[0:{\lvert\phi(b)\rvert}]}\phi(b)_i\prod_i^{[0:{\lvert\phi(b)\rvert}]}v_i$
					      \item $\equiv a^{\lvert\phi(b)\rvert}\pmod{b}$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{1.17}
			    The notation \gls{intatimesb} as a shorthand for the $\lvert a\rvert\times\lvert b\rvert$ matrix such that for $i$ in $[0:\lvert a\rvert]$, for $j$ in $[0:\lvert b\rvert]$, $(a\times b)_{i,j}=\langle a_i,b_j\rangle$.
		    \procedure{1.52}
			    \objective
				    Choose two positive integers $a,b$ such that $(a,b)=1$. The objective of the following instructions is to show that each entry of $\chi_{a,b}([0:a]\times[0:b])$ is in $[0:ab]$.
			    \implementation
				    \begin{enumerate}
					    \item Let $h=\chi_{a,b}([0:a]\times[0:b])$.
					    \item Show that $0\le h_{i,j}<[a,b]=[a,b](a,b)=ab$ for $i$ in $[0:a]$, for $j$ in $[0:b]$.
					    \item \textbf{Hence show that each entry of $h$ is in $[0:ab]$.}
				    \end{enumerate}
		    \procedure{1.53}
			    \objective
				    Choose two positive integers $a,b$ such that $(a,b)=1$. The objective of the following instructions is to either show that $0<0$ or to show that each entry of $\chi_{a,b}([0:a]\times[0:b])$ is distinct.
			    \implementation
				    \begin{enumerate}
					    \item Let $h=\chi_{a,b}([0:a]\times[0:b])$.
					    \item For each distinct unordered pair of index pairs $\langle i,j\rangle$ and $\langle k,l\rangle$ of $h$, do the following:
					    \begin{enumerate}
						    \item If $h_{i,j}=h_{k,l}$, then do the following:
						    \begin{enumerate}
							    \item Show that $\chi_{a,b}(i,j)=\chi_{a,b}([0:a]_i,[0:b]_j)=h_{i,j}=h_{k,l}=\chi_{a,b}([0:a]_k,[0:b]_l)=\chi_{a,b}(k,l)$.
							    \item Show that $i\equiv\chi_{a,b}(i,j)=\chi_{a,b}(k,l)\equiv k\pmod{a}$ using \procedurehr{1.42} given that $\chi_{a,b}(i,j)=\chi_{a,b}(k,l)$.
							    \item Hence show that $i=k$.
							    \item Show that $j\equiv\chi_{a,b}(i,j)=\chi_{a,b}(k,l)\equiv l\pmod{b}$ using \procedurehr{1.42} given that $\chi_{a,b}(i,j)=\chi_{a,b}(k,l)$.
							    \item Hence show that $j=l$.
							    \item Hence show that $\langle i,j\rangle=\langle k,l\rangle$.
							    \item \textbf{Hence show that $\langle i,j\rangle\ne\langle i,j\rangle$ given that $\langle i,j\rangle$ and $\langle k,l\rangle$ are distinct.}
							    \item \textbf{Abort procedure.}
						    \end{enumerate}
						    \item Otherwise do the following:
						    \begin{enumerate}
							    \item Show that $h_{i,j}\ne h_{k,l}$.
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Hence show that each entry of $h$ is distinct.}
				    \end{enumerate}
		    \procedure{1.54}
			    \objective
				    Choose two positive integers $a,b$ such that $(a,b)=1$. The objective of the following instructions is to show that either $0<0$ or $\chi_{a,b}([0:a]\times[0:b])$ is a rearrangement $[0:ab]$.
			    \implementation
				    \begin{enumerate}
					    \item Let $h=\chi_{a,b}([0:a]\times[0:b])$.
					    \item Use \procedurehr{1.52} on $\langle a,b\rangle$ to show that each element of $h$ is in $[0:ab]$.
					    \item Also show that $h$ has the same number of entries as $[0:ab]$.
					    \item Use \procedurehr{1.53} on $\langle a,b\rangle$ to show that $h$ is composed of distinct elements.
					    \item \textbf{Hence show that $h$ is a rearrangement of $[0:ab]$.}
				    \end{enumerate}
		    \procedure{1.55}
			    \objective
				    Choose two positive integers $a,b$ such that $(a,b)=1$. The objective of the following instructions is to either show that $0<0$ or to show that each entry of $\chi_{a,b}(\phi(a)\times\phi(b))$ is in $\phi(ab)$.
			    \implementation
				    \begin{enumerate}
					    \item Let $h=\chi_{a,b}(\phi(a)\times\phi(b))$.
					    \item Now, for each index pair $\langle i,j\rangle$ of $h$, do the following:
					    \begin{enumerate}
						    \item Show that $0\le h_{i,j}<[a,b]=[a,b](a,b)=ab$.
						    \item Show that $h_{i,j}=\chi_{a,b}(\phi(a)_i,\phi(b)_j)\equiv\phi(a)_i\pmod{a}$.
						    \item Hence use \procedurehr{1.17} on $\langle h_{i,j},\phi(a)_i,a\rangle$ to show that $(a,h_{i,j})=(h_{i,j},a)=(\phi(a)_i,a)=1$.
						    \item Also show that $h_{i,j}=\chi_{a,b}(\phi(a)_i,\phi(b)_j)\equiv\phi(b)_j\pmod{b}$.
						    \item Hence use \procedurehr{1.17} on $\langle h_{i,j},\phi(b)_j,b\rangle$ to show that $(b,h_{i,j})=(h_{i,j},b)=(\phi(b)_j,b)=1$.
						    \item Hence show that $(h_{i,j},ab)=(ab,h_{i,j})=1$.
						    \item Hence show that $h_{i,j}$ is in $\phi(ab)$.
					    \end{enumerate}
					    \item \textbf{Hence show that each entry of $\chi_{a,b}(\phi(a)\times\phi(b))$ is in $\phi(ab)$.}
				    \end{enumerate}
		    \procedure{1.56}
			    \objective
				    Choose two positive integers $a,b$ such that $(a,b)=1$. The objective of the following instructions is to either show that $0<0$ or to show that each entry of $\phi(ab)$ is in $\chi_{a,b}(\phi(a)\times\phi(b))$.
			    \implementation
				    \begin{enumerate}
					    \item For $i$ in $[0:\lvert\phi(ab)\rvert]$, do the following:
					    \begin{enumerate}
						    \item Show that $(\phi(ab)_i,ab)=1$.
						    \item Show that $\phi(ab)_i\equiv\phi(ab)_i\bmod a\pmod{a}$.
						    \item Hence show that $(\phi(ab)_i\bmod a,a)=(\phi(ab)_i,a)=1$ using \procedurehr{1.17}.
						    \item Hence show that $\phi(ab)_i\bmod a$ is amongst $\phi(a)$ given that $0\le \phi(ab)_i\bmod a<a$.
						    \item Show that $\phi(ab)_i\equiv\phi(ab)_i\bmod b\pmod{b}$.
						    \item Hence show that $(\phi(ab)_i\bmod b,b)=(\phi(ab)_i,b)=1$ using \procedurehr{1.17}.
						    \item Hence show that $\phi(ab)_i\bmod b$ is amongst $\phi(b)$ given that $0\le \phi(ab)_i\bmod b<b$.
						    \item Hence show that $\langle \phi(ab)_i\bmod a,\phi(ab)_i\bmod b\rangle$ is amongst $\phi(a)\times\phi(b)$.
						    \item Show that $\phi(ab)_i\equiv\chi_{a,b}(\phi(ab)_i\bmod a,\phi(ab)_i\bmod b)\pmod{[a,b]=[a,b](a,b)=ab}$ using \procedurehr{1.39} given that $\phi(ab)_i\equiv\phi(ab)_i\bmod a\pmod{a}$ and $\phi(ab)_i\equiv\phi(ab)_i\bmod b\pmod{b}$.
						    \item Hence show that $\phi(ab)_i=\chi_{a,b}(\phi(ab)_i\bmod a,\phi(ab)_i\bmod b)$.
						    \item Hence show that $\phi(ab)_i$ is amongst $\chi_{a,b}(\phi(a)\times\phi(b))$ given that $\langle \phi(ab)_i\bmod a,\phi(ab)_i\bmod b\rangle$ is amongst $\phi(a)\times\phi(b)$ and $\phi(ab)_i=\chi_{a,b}(\phi(ab)_i\bmod a,\phi(ab)_i\bmod b)$.
					    \end{enumerate}
					    \item \textbf{Hence show that each entry of $\phi(ab)$ is in $\chi_{a,b}(\phi(a)\times\phi(b))$.}
				    \end{enumerate}
		    \procedure{1.57}
			    \objective
				    Choose two positive integers $a,b$ such that $(a,b)=1$. The objective of the following instructions is to either show that $0<0$ or to show that $\phi(ab)$ is a rearrangement of $\chi_{a,b}(\phi(a)\times\phi(b))$ and that $\lvert\phi(ab)\rvert=\lvert\phi(a)\rvert\lvert\phi(b)\rvert$.
			    \implementation
				    \begin{enumerate}
					    \item Use \procedurehr{1.54} on $\langle a,b\rangle$ to show that $\chi_{a,b}([0:a]\times[0:b])$ is a rearrangement of $[0:ab]$.
					    \item Show that $\chi_{a,b}(\phi(a)\times\phi(b))$ is a submatrix of $\chi_{a,b}([0:a]\times[0:b])$.
					    \item Hence show that the entries of $\chi_{a,b}(\phi(a)\times\phi(b))$ are distinct.
					    \item Use \procedurehr{1.55} on $\langle a,b\rangle$ to show that the entries of $\chi_{a,b}(\phi(a)\times\phi(b))$ are in $\phi(ab)$.
					    \item Show that the entries of $\phi(ab)$ are distinct.
					    \item Use \procedurehr{1.56} on $\langle a,b\rangle$ to show that the entries of $\phi(ab)$ are in $\chi_{a,b}(\phi(a)\times\phi(b))$.
					    \item \textbf{Hence show that $\phi(ab)$ is a rearrangement of $\chi_{a,b}(\phi(a)\times\phi(b))$.}
					    \item \textbf{Hence show that $\lvert\phi(ab)\rvert=\lvert\chi_{a,b}(\phi(a)\times\phi(b))\rvert=\lvert\phi(a)\times\phi(b)\rvert=\lvert\phi(a)\rvert\lvert\phi(b)\rvert$.}
				    \end{enumerate}
		    \declaration{1.18}
			    The notation \gls{int[P]}, where $P$ is a condition, will be used as a shorthand for $1$ if $P$, otherwise it will stand for $0$.
		    \declaration{1.32}
			    The notation \gls{inta_+}, where $a$ is a list, will be used as a shorthand for $0$ if $a$ is empty, otherwise it will be a shorthand for the sum of the entries of $a$.
		    \declaration{1.19}
			    The notation \gls{intsigmaf(r)}, where $R$ is a list and $f[r]$ is a function of $r$, will be used as a shorthand for $f(R)_+$.
		    \procedure{1.58}
			    \objective
				    Choose a positive integer $a$ and a prime $b$. The objective of the following instructions is to show that either $0<0$ or $\lvert\phi(b^a)\rvert=b^a-b^{a-1}$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\sum_r^{[0:{b^a}]}[(r,b^a)=1]\le\sum_r^{[0:{b^a}]}[(r,b)=1]$ using \procedurehr{1.21}.
					    \item Show that $\sum_r^{[0:{b^a}]}[(r,b)=1]\le\sum_r^{[0:{b^a}]}[(r,b^a)=1]$ using \procedurehr{1.20}.
					    \item Hence show that $\sum_r^{[0:{b^a}]}[(r,b^a)=1]=\sum_r^{[0:{b^a}]}[(r,b)=1]$.
					    \item Show that $\sum_r^{[0:{b^a}]}[(r,b)=1]\le\sum_r^{[0:{b^a}]}[r\bmod b\ne 0]$ using \procedurehr{1.13}.
					    \item Show that $\sum_r^{[0:{b^a}]}[r\bmod b\ne 0]\le\sum_r^{[0:{b^a}]}[(r,b)=1]$ using \procedurehr{1.22}.
					    \item Hence show that $\sum_r^{[0:{b^a}]}[(r,b)=1]=\sum_r^{[0:{b^a}]}[r\bmod b\ne 0]$.
					    \item \textbf{Hence show that $\lvert\phi(b^a)\rvert=\sum_r^{[0:{b^a}]}[(r,b^a)=1]=\sum_r^{[0:{b^a}]}[(r,b)=1]=\sum_r^{[0:{b^a}]}[r\bmod b\ne 0]=\sum_r^{[0:{b^a}]}(1-[r\bmod b=0])=b^a-b^{a-1}$.}
				    \end{enumerate}
		    \procedure{1.59}
			    \objective
				    Choose a list of primes $a$. Let $b$ be the list of distinct primes in $a$. Let $c$ be a list such that $c_i$ is the multiplicity of $b_i$ in $a$ for $i=1$ to $i=\lvert b\rvert$. The objective of the following instructions is to show that either $0<0$ or $\lvert\phi(a_*)\rvert=\prod_i^{[0:{\lvert b\rvert}]}({b_i}^{c_i}-{b_i}^{c_i-1})$.
			    \implementation
				    \begin{enumerate}
					    \item If $a=\langle\rangle$, then do the following:
					    \begin{enumerate}
						    \item Show that $\lvert b\rvert=\lvert a\rvert=0$.
						    \item \textbf{Hence show that $\phi(a_*)=\phi(1)=1=\prod_i^{[0:{\lvert b\rvert}]}({b_i}^{c_i}-{b_i}^{c_i-1})$.}
					    \end{enumerate}
					    \item Otherwise, do the following:
					    \begin{enumerate}
						    \item Show that $a_*=\prod_i^{[0:{\lvert b\rvert}]} {b_i}^{c_i}$.
						    \item Show that $\lvert a\rvert>0$.
						    \item Hence show that $\lvert c\rvert=\lvert b\rvert>0$.
						    \item Hence show that $({b_0}^{c_0},\prod_i^{[1:{\lvert b\rvert}]}{b_i}^{c_i})=1$ using \procedurehr{1.30}.
						    \item Let $d$ be the list $a$ with all instances of $a_0$ removed.
						    \item Verify that $\lvert d\rvert<\lvert a\rvert$.
						    \item Now use \procedurehr{1.59} on $\langle d\rangle$ to show that $\phi(d_*)=\phi(\prod_i^{[1:{\lvert b\rvert}]}{b_i}^{c_i})=\prod_i^{[1:{\lvert b\rvert}]}({b_i}^{c_i}-{b_i}^{c_i-1})$.
						    \item \textbf{Hence show that $\lvert\phi(a_*)\rvert=\lvert\phi(\prod_i^{[0:{\lvert b\rvert}]} {b_i}^{c_i})\rvert=\lvert\phi({b_0}^{c_0}\prod_i^{[1:{\lvert b\rvert}]}{b_i}^{c_i})\rvert=\lvert\phi({b_0}^{c_0})\rvert\lvert\phi(\prod_i^{[1:{\lvert b\rvert}]}{b_i}^{c_i})\rvert=({b_0}^{c_0}-{b_0}^{c_0-1})\lvert\phi(\prod_i^{[1:{\lvert b\rvert}]}{b_i}^{c_i})\rvert=({b_0}^{c_0}-{b_0}^{c_0-1})\prod_i^{[1:{\lvert b\rvert}]}({b_i}^{c_i}-{b_i}^{c_i-1})=\prod_i^{[0:{\lvert b\rvert}]}({b_i}^{c_i}-{b_i}^{c_i-1})$ using \procedurehr{1.57} and \procedurehr{1.58} given that $({b_0}^{c_0},\prod_i^{[1:{\lvert b\rvert}]}{b_i}^{c_i})=1$ and $\phi(\prod_i^{[1:{\lvert b\rvert}]}{b_i}^{c_i})=\prod_i^{[1:{\lvert b\rvert}]}({b_i}^{c_i}-{b_i}^{c_i-1})$.}
					    \end{enumerate}
				    \end{enumerate}
	    \end{multicols*}
	  \chapter{Permutations and Combinations}
	    \begin{multicols*}{2}
		    \declaration{1.20}
			    The notation \gls{inta^ulb} will be used as a shorthand for $\prod_i^{[0:b]}(a-i)$.
		    \declaration{1.33}
			    The notation \gls{inta^olb} will be used as a shorthand for $\prod_i^{[0:b]}(a+i)$.
		    \procedure{1.60}
			    \objective
				    Choose a list of distinct elements $a$ and a non-negative integer $b$ such that $b\le\lvert a\rvert$. Let $c$ be a list of length-$b$ permutations of $a$. The objective of the following instructions is to show that $\lvert c\rvert=\lvert a\rvert^{\ul{b}}$.
			    \implementation
				    \begin{enumerate}
					    \item If $\lvert b\rvert>0$, then do the following:
					    \begin{enumerate}
						    \item For each entry $d$ in $a$, do the following:
						    \begin{enumerate}
							    \item Let $e$ be the list formed by removing $d$ from $a$.
							    \item Show that the entries of $e$ are distinct given that the entries of $a$ are distinct.
							    \item Show that $\lvert e\rvert=\lvert a\rvert-1$.
							    \item Now use \procedurehr{1.60} on $\langle e,b-1\rangle$ to show that the number of length-$b-1$ permutations of $e$ is $\lvert e\rvert^{\ul{b-1}}$.
							    \item Hence show that the number of length-$b$ permutations of $a$ beginning with $d$ is $\lvert e\rvert^{\ul{b-1}}=(\lvert a\rvert-1)^{\ul{b-1}}$.
						    \end{enumerate}
						    \item Hence show that the number of length-$b$ permutations of $a$ beginning with any entry of $a$ is $\lvert a\rvert(\lvert a\rvert-1)^{\ul{b-1}}=\lvert a\rvert^{\ul{b}}$.
						    \item Hence show that the number of length-$b$ permutations of $a$ are $\lvert a\rvert^{\ul{b}}$.
						    \item \textbf{Hence show that $\lvert c\rvert=\lvert a\rvert^{\ul{b}}$.}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
						    \item Show that $b=0$.
						    \item Show that the number of length-$0$ permutations of $a$ is $1$.
						    \item \textbf{Therefore show that $\lvert c\rvert=1=\lvert a\rvert^{\ul{0}}=\lvert a\rvert^{\ul{b}}$.}
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{1.21}
			    The notation \gls{intnchooser} will be used as a shorthand for $n^{\ul{r}}\div(r!)$.
		    \procedure{1.61}
			    \objective
				    Choose a list of distinct elements $n$ and a non-negative integer $r$ such that $r\le\lvert n\rvert$. Let $b$ be the largest list of length-$r$ sublists of $n$ such that no two of them are permutations of each other. The objective of the following instructions is to either show that $b$ contains at least two permutations of the same list, construct a list larger than $b$ that is also a list of length-$r$ sublists of $n$ such that no two of them are permutations of each other, or to show that $\lvert b\rvert=\binom{\lvert n\rvert}{r}$ and that $\lvert n\rvert^{\ul{r}}\bmod r!=0$.
			    \implementation
				    \begin{enumerate}
					    \item Let $a$ and $f$ be a list of all the permutations of $n$.
					    \item Show that $\lvert a\rvert=\lvert n\rvert^{\ul{\lvert n\rvert}}$ using \procedurehr{1.60}.
					    \item For each list $c$ in $b$, do the following:
					    \begin{enumerate}
						    \item Show that the number of permutations of $c$ is $r!$ using \procedurehr{1.60}.
						    \item Let $d$ be the list obtained by removing the elements of $c$ from $n$.
						    \item Show that the number of permutations of $d$ is $(n-r)!$ using \procedurehr{1.60}.
						    \item Let $e$ be the list of permutations of $n$ beginning with a permutation of $c$.
						    \item Show that $\lvert e\rvert=r!(\lvert n\rvert-r)!$ given that there are $r!$ possible choices for the first part of $e$ and $(\lvert n\rvert-r)!$ possible choices for the second part of $e$.
						    \item If $e$ is not a sublist of $a$, then do the following:
						    \begin{enumerate}
							    \item Let $g$ be a list in $e$ that is not in $a$.
							    \item Show that $e$ is a sublist of $f$.
							    \item Therefore show that $g$ was in $a$ but then was removed.
							    \item Therefore show that the variable $c$ was formerly equal to a permutation of the current $c$.
							    \item \textbf{Therefore show that $b$ contains at least two permutations of $c$.}
							    \item \textbf{Abort procedure.}
						    \end{enumerate}
						    \item Otherwise, do the following:
						    \begin{enumerate}
							    \item Show that $e$ is a sublist of $a$.
							    \item Remove the lists in $e$ from $a$.
						    \end{enumerate}
					    \end{enumerate}
					    \item If $a\ne\langle\rangle$, then do the following:
					    \begin{enumerate}
						    \item Let $g$ be a list in $a$.
						    \item Let $h$ be the sublist of $g$ corresponding to its first $r$ elements.
						    \item Therefore show that the permutations of $n$ beginning with a permutation of $h$ were never removed from $a$.
						    \item Therefore show that the variable $c$ was never equal to a permutation of $h$.
						    \item Therefore show that no permutation of $h$ is in $b$.
						    \item \textbf{Therefore show that $b^{\frown}\langle h\rangle$ is larger than $b$ and is also a list of length-$r$ sublists of $n$ such that no two of them are permutations of each other.}
						    \item \textbf{Abort procedure.}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
						    \item Show that $\lvert n\rvert!\bmod(r!(\lvert n\rvert-r)!)=0$.
						    \item \textbf{Therefore show that $n^{\ul{r}}\bmod r!$}
						    \begin{enumerate}
						      \item $=(\lvert n\rvert!\div(\lvert n\rvert-r)!)\bmod r!$
						      \item $=((\lvert n\rvert!\bmod(r!(\lvert n\rvert-r)!)r!(\lvert n\rvert-r)!)\div(\lvert n\rvert-r)!)\bmod r!$
						      \item $=((\lvert n\rvert!\div(r!(\lvert n\rvert-r)!))r!)\bmod r!$
						      \item $=0$.
						    \end{enumerate}
						    \item Also show that (3) iterated $\lvert n\rvert!\div(r!(\lvert n\rvert-r)!)$ times.
						    \item \textbf{Therefore using \procedurehr{1.08}, show that $\lvert b\rvert$}
						    \begin{enumerate}
						      \item $=\lvert n\rvert!\div(r!(\lvert n\rvert-r)!)$
						      \item $=(\lvert n\rvert!\div(\lvert n\rvert-r)!)\div(r!)$
						      \item $=n^{\ul{r}}\div(r!)$
						      \item $=\binom{n}{r}$.
						    \end{enumerate}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{1.62}
			    \objective
				    Choose two positive integers $a,b$. The objective of the following instructions is to show that $\binom{a}{b}=\binom{a-1}{b-1}+\binom{a-1}{b}$.
			    \implementation
				    \begin{enumerate}
					    \item Using \procedurehr{1.05} and \procedurehr{1.06}, show that $\binom{a-1}{b-1}+\binom{a-1}{b}$
					    \begin{enumerate}
						    \item $=(a-1)^{\ul{b-1}}\div(b-1)!+(a-1)^{\ul{b}}\div b!$
						    \item $=((a-1)^{\ul{b-1}}b)\div b!+(a-1)^{\ul{b}}\div b!$
						    \item $=((a-1)^{\ul{b-1}}b+(a-1)^{\ul{b}})\div b!$
						    \item $=((a-1)^{\ul{b-1}}b+(a-1)^{\ul{b-1}}(a-b))\div b!$
						    \item $=((a-1)^{\ul{b-1}}a)\div b!$
						    \item $=a^{\ul{b}}\div b!$
						    \item $=\binom{a}{b}$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{1.63}
			    \objective
				    Choose an integer $x$ and a non-negative integer $a$. The objective of the following instructions is to show that the $(1+x)^a=\sum_r^{[0:{a+1}]}\binom{a}{r}x^r$.
			    \implementation
				    \begin{enumerate}
					    \item If $a=0$, then do the following:
					    \begin{enumerate}
						    \item \textbf{Show that $(1+x)^a=(1+x)^0=1=\sum_r^{[0:1]}\binom{0}{r}x^r=\sum_r^{[0:{a+1}]}\binom{a}{r}x^r$.}
					    \end{enumerate}
					    \item Otherwise, do the following:
					    \begin{enumerate}
						    \item Show that $a>0$.
						    \item Therefore show that $a-1\ge 0$.
						    \item Use \procedurehr{1.63} on $\langle x,a-1\rangle$ to show that $(1+x)^{a-1}=\sum_r^{[0:{a}]}\binom{a-1}{r}x^r$.
						    \item Therefore using \procedurehr{1.62}, show that $(1+x)^a$
						    \begin{enumerate}
							    \item $=(1+x)(1+x)^{a-1}$
							    \item $=(1+x)\sum_r^{[0:{a}]}\binom{a-1}{r}x^r$
							    \item $=\sum_r^{[0:{a}]}\binom{a-1}{r}x^r+\sum_r^{[0:{a}]}\binom{a-1}{r}x^{r+1}$
							    \item $=\sum_r^{[0:{a+1}]}\binom{a-1}{r}x^r+\sum_r^{[1:{a+1}]}\binom{a-1}{r-1}x^{r}$
							    \item $=1+\sum_r^{[1:{a+1}]}(\binom{a-1}{r}+\binom{a-1}{r-1})x^{r}$
							    \item $=1+\sum_r^{[1:{a+1}]}\binom{a}{r}x^{r}$
							    \item $=\sum_r^{[0:{a+1}]}\binom{a}{r}x^{r}$.
						    \end{enumerate}
					    \end{enumerate}
				    \end{enumerate}
	    \end{multicols*}
	  \clearpage
	\part{Rational Arithmetic}
	  \chapter{Rational Arithmetic}
	    \begin{multicols*}{2}
		    \declaration{2.12}
			    The phrase "\gls{rational}" will be used as a shorthand for an ordered pair comprising an integer followed by a non-zero natural number.
		    \declaration{2.13}
			    The phrase "the numerator of $a$" and the notation \gls{ratNu}, where $a$ is a rational number, will be used as a shorthand for the first entry of $a$.
		    \declaration{2.14}
			    The phrase "the denominator of $a$" and the notation \gls{ratDe}, where $a$ is a rational number, will be used as a shorthand for the second entry of $a$.
		    \declaration{2.15}
			    The phrase "\gls{rata=b}", where $a,b$ are rational numbers, will be used as a shorthand for "$\Nu(a)\De(b)=\De(a)\Nu(b)$".
		    \procedure{2.27}
			    \objective
				    Choose a rational number $a$. The objective of the following instructions is to show that $a=a$.
			    \implementation
				    \begin{enumerate}
					    \item \textbf{Show that $a=a$ using \declarationhr{2.15} given that $\Nu(a)\De(a)=\De(a)\Nu(a)$.}
				    \end{enumerate}
		    \procedure{2.28}
			    \objective
				    Choose two rational numbers $a,b$ such that $a=b$. The objective of the following instructions is to show that $b=a$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Nu(a)\De(b)=\De(a)\Nu(b)$ using \declarationhr{2.15} given that $a=b$.
					    \item \textbf{Hence show that $b=a$ using \declarationhr{2.15} given that $\Nu(b)\De(a)=\De(b)\Nu(a)$.}
				    \end{enumerate}
		    \procedure{2.29}
			    \objective
				    Choose three rational numbers $a,b,c$ such that $a=b$ and $b=c$. The objective of the following instructions is to show that $a=c$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Nu(a)\De(b)=\De(a)\Nu(b)$ using \declarationhr{2.15} given that $a=b$.
					    \item Show that $\Nu(b)\De(c)=\De(b)\Nu(c)$ using \declarationhr{2.15} given that $b=c$.
					    \item If $\Nu(b)\ne 0$, then do the following:
					    \begin{enumerate}
						    \item Show that $\Nu(a)\De(b)\Nu(b)\De(c)=\De(a)\Nu(b)\De(b)\Nu(c)$.
						    \item Hence show that $\Nu(a)\De(c)=\De(a)\Nu(c)$.
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
                            \item Show that $\Nu(b)=0$.
						    \item Show that $\De(b)\ne 0$ using \declarationhr{2.12}.
						    \item Show that $\Nu(a)\De(b)=\De(a)\Nu(b)=0\De(a)=0$ given that $a=b$.
						    \item Hence show that $\Nu(a)=0$.
						    \item Show that $0=0\De(c)=\Nu(b)\De(c)=\De(b)\Nu(c)$.
						    \item Hence show that $\Nu(c)=0$.
						    \item Hence show that $\Nu(a)\De(c)=0\De(c)=\De(a)0=\De(a)\Nu(c)$.
					    \end{enumerate}
					    \item \textbf{Hence show that $a=c$.}
				    \end{enumerate}
		    \declaration{2.16}
			    The notation \gls{rata+b}, where $a,b$ are rational numbers, will be used as a shorthand for the pair $\langle\Nu(a)\De(b)+\De(a)\Nu(b),\De(a)\De(b)\rangle$.
		    \procedure{2.30}
			    \objective
				    Choose two rational numbers $a,b,c,d$ such that $a=c$ and $b=d$. The objective of the following instructions is to show that $a+b=c+d$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Nu(a)\De(c)=\De(a)\Nu(c)$ using \declarationhr{2.15} given that $a=c$.
					    \item Show that $\Nu(b)\De(d)=\De(b)\Nu(d)$ using \declarationhr{2.15} given that $b=d$.
					    \item Hence using \declarationhr{2.16}, show that $a+b$
					    \begin{enumerate}
						    \item $=\langle\Nu(a),\De(a)\rangle+\langle\Nu(b),\De(b)\rangle$
						    \item $=\langle\Nu(a)\De(b)+\De(a)\Nu(b),\De(a)\De(b)\rangle$
						    \item $=\langle\De(c)\De(d)(\Nu(a)\De(b)+\De(a)\Nu(b)),\De(c)\De(d)(\De(a)\De(b))\rangle$
						    \item $=\langle\Nu(a)\De(c)\De(b)\De(d)+\De(a)\De(c)\Nu(b)\De(d),\De(c)\De(d)\De(a)\De(b)\rangle$
						    \item $=\langle\De(a)\Nu(c)\De(b)\De(d)+\De(a)\De(c)\De(b)\Nu(d),\De(c)\De(d)\De(a)\De(b)\rangle$
						    \item $=\langle\De(a)\De(b)(\Nu(c)\De(d)+\De(c)\Nu(d)),\De(a)\De(b)(\De(c)\De(d))\rangle$
						    \item $=\langle\Nu(c)\De(d)+\De(c)\Nu(d),\De(c)\De(d)\rangle$
						    \item $=\langle\Nu(c),\De(c)\rangle+\langle\Nu(d),\De(d)\rangle$
						    \item $=c+d$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.31}
			    \objective
				    Choose three rational numbers $a,b,c$. The objective of the following instructions is to show that $(a+b)+c=a+(b+c)$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{2.16}, show that $(a+b)+c$
					    \begin{enumerate}
						    \item $=\langle\Nu(a)\De(b)+\De(a)\Nu(b),\De(a)\De(b)\rangle+\langle\Nu(c),\De(c)\rangle$
						    \item $=\langle(\Nu(a)\De(b)+\De(a)\Nu(b))\De(c)+(\De(a)\De(b))\Nu(c),(\De(a)\De(b))\De(c)\rangle$
						    \item $=\langle\Nu(a)(\De(b)\De(c))+\De(a)(\Nu(b)\De(c)+\De(b)\Nu(c)),\De(a)(\De(b)\De(c))\rangle$
						    \item $=\langle\Nu(a),\De(a)\rangle+\langle\Nu(b)\De(c)+\De(b)\Nu(c),\De(b)\De(c)\rangle$
						    \item $=a+(b+c)$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.32}
			    \objective
				    Choose two rational numbers $a,b$. The objective of the following instructions is to show that $a+b=b+a$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{2.16}, show that $a+b$
					    \begin{enumerate}
						    \item $=\langle\Nu(a)\De(b)+\De(a)\Nu(b),\De(a)\De(b)\rangle$
						    \item $=\langle\Nu(b)\De(a)+\De(b)\Nu(a),\De(b)\Nu(a)\rangle$
						    \item $=b+a$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{2.17}
			    The notation \gls{rata}, where $a$ is an integer, will contextually be used as a shorthand for the pair $\langle a,1\rangle$.
		    \procedure{2.33}
			    \objective
				    Choose a rational number $a$. The objective of the following instructions is to show that $0+a=a$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{2.16} and \declarationhr{2.17}, show that $0+a$
					    \begin{enumerate}
						    \item $=\langle 0,1\rangle+\langle\Nu(a),\De(a)\rangle$
						    \item $=\langle 0\De(a)+1\Nu(a),1\De(a)\rangle$
						    \item $=\langle\Nu(a),\De(a)\rangle$
						    \item $=a$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{2.18}
			    The notation \gls{rat-a}, where $a$ is a rational number, will be used as a shorthand for the pair $\langle-\Nu(a),\De(a)\rangle$.
		    \procedure{2.34}
			    \objective
				    Choose two rational numbers $a,b$ such that $a=b$. The objective of the following instructions is to show that $-a=-b$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Nu(a)\De(b)=\De(a)\Nu(b)$ using \declarationhr{2.15} given that $a=b$.
					    \item Hence using \declarationhr{2.18}, show that $-a$
					    \begin{enumerate}
						    \item $=\langle-\Nu(a),\De(a)\rangle$
						    \item $=\langle-\Nu(a)\De(b),\De(a)\De(b)\rangle$
						    \item $=\langle-\De(a)\Nu(b),\De(a)\De(b)\rangle$
						    \item $=\langle-\Nu(b),\De(b)\rangle$
						    \item $=-b$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.35}
			    \objective
				    Choose a rational number $a$. The objective of the following instructions is to show that $-a+a=0$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{2.16} and \declarationhr{2.18}, show that $-a+a$
					    \begin{enumerate}
						    \item $=(-a)+a$
						    \item $=\langle-\Nu(a),\De(a)\rangle+\langle\Nu(a),\De(a)\rangle$
						    \item $=\langle-\Nu(a)\De(a)+\De(a)\Nu(a),\De(a)^2\rangle$
						    \item $=\langle 0,\De(a)^2\rangle$
						    \item $=\langle 0,1\rangle$
						    \item $=0$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{2.19}
			    The notation \gls{ratab}, where $a,b$ are rational numbers, will be used as a shorthand for the pair $\langle\Nu(a)\Nu(b),\De(a)\De(b)\rangle$.
		    \procedure{2.36}
			    \objective
				    Choose two rational numbers $a,b,c,d$ such that $a=c$ and $b=d$. The objective of the following instructions is to show that $ab=cd$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Nu(a)\De(c)=\De(a)\Nu(c)$ using \declarationhr{2.15} given that $a=c$.
					    \item Show that $\Nu(b)\De(d)=\De(b)\Nu(d)$ using \declarationhr{2.15} given that $b=d$.
					    \item Hence using \declarationhr{2.19}, show that $ab$
					    \begin{enumerate}
						    \item $=\langle\Nu(a),\De(a)\rangle\langle\Nu(b),\De(b)\rangle$
						    \item $=\langle\Nu(a)\Nu(b),\De(a)\De(b)\rangle$
						    \item $=\langle(\De(c)\De(d))\Nu(a)\Nu(b),(\De(c)\De(d))\De(a)\De(b)\rangle$
						    \item $=\langle(\Nu(a)\De(c))(\Nu(b)\De(d)),\De(c)\De(d)\De(a)\De(b)\rangle$
						    \item $=\langle(\De(a)\Nu(c))(\De(b)\Nu(d)),\De(c)\De(d)\De(a)\De(b)\rangle$
						    \item $=\langle(\De(a)\De(b))\Nu(c)\Nu(d),(\De(a)\De(b))\De(c)\De(d)\rangle$
						    \item $=\langle\Nu(c)\Nu(d),\De(c)\De(d)\rangle$
						    \item $=\langle\Nu(c),\De(c)\rangle\langle\Nu(d),\De(d)\rangle$
						    \item $=cd$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.37}
			    \objective
				    Choose three rational numbers $a,b,c$. The objective of the following instructions is to show that $(ab)c=a(bc)$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{2.19}, show that $(ab)c$
					    \begin{enumerate}
						    \item $=\langle\Nu(a)\Nu(b),\De(a)\De(b)\rangle\langle\Nu(c),\De(c)\rangle$
						    \item $=\langle\Nu(a)\Nu(b)\Nu(c),\De(a)\De(b)\De(c)\rangle$
						    \item $=\langle\Nu(a),\De(a)\rangle\langle\Nu(b)\Nu(c),\De(b)\De(c)\rangle$
						    \item $=a(bc)$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.38}
			    \objective
				    Choose two rational numbers $a,b$. The objective of the following instructions is to show that $ab=ba$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{2.19}, show that $ab$
					    \begin{enumerate}
						    \item $=\langle\Nu(a)\Nu(b),\De(a)\De(b)\rangle$
						    \item $=\langle\Nu(b)\Nu(a),\De(b)\De(a)\rangle$
						    \item $=ba$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.39}
			    \objective
				    Choose a rational number $a$. The objective of the following instructions is to show that $1a=a$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{2.19}, show that $1a$
					    \begin{enumerate}
						    \item $=\langle 1,1\rangle\langle\Nu(a),\De(a)\rangle$
						    \item $=\langle 1\Nu(a),1\De(a)\rangle$
						    \item $=\langle\Nu(a),\De(a)\rangle$
						    \item $=a$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{2.20}
			    The notation \gls{rat1/a}, where $a$ is a rational number, will be used as a shorthand for the pair $\langle\De(a),\Nu(a)\rangle$ if $\Nu(a)>0$ and $\langle-\De(a),-\Nu(a)\rangle$ if $\Nu(a)<0$.
		    \procedure{2.40}
			    \objective
				    Choose two rational numbers $a,b$ such that $a=b$ and $a\ne 0$. The objective of the following instructions is to show that $\frac{1}{a}=\frac{1}{b}$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Nu(a)=\Nu(a)\De(0)\ne\De(a)\Nu(0)=0$ using \declarationhr{2.15} and \declarationhr{2.17} given that $a\ne 0$.
                        \item Show that $\Nu(a)\De(b)=\De(a)\Nu(b)$ using \declarationhr{2.15} given that $a=b$.
					    \item Hence show that $\De(a)\Nu(b)=\Nu(a)\De(b)\ne 0$ using \declarationhr{2.12} given that $\Nu(a)\ne 0$.
					    \item Hence show that $\Nu(b)\ne 0$.
					    \item If $\Nu(a)\Nu(b)>0$, then do the following:
					    \begin{enumerate}
						    \item Using \declarationhr{2.20}, show that $\frac{1}{a}$
						    \begin{enumerate}
							    \item $=\langle\De(a)\Nu(b),\Nu(a)\Nu(b)\rangle$
							    \item $=\langle\Nu(a)\De(b),\Nu(a)\Nu(b)\rangle$
							    \item $=\frac{1}{b}$.
						    \end{enumerate}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
						    \item Show that $\Nu(a)\Nu(b)<0$.
						    \item Hence using \declarationhr{2.20}, show that $\frac{1}{a}$
						    \begin{enumerate}
							    \item $=\langle-\De(a)\Nu(b),-\Nu(a)\Nu(b)\rangle$
							    \item $=\langle-\Nu(a)\De(b),-\Nu(a)\Nu(b)\rangle$
							    \item $=\frac{1}{b}$.
						    \end{enumerate}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.41}
			    \objective
				    Choose a rational number $a$ such that $a\ne 0$. The objective of the following instructions is to show that $\frac{1}{a}a=1$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Nu(a)=\Nu(a)\De(0)\ne\De(a)\Nu(0)=0$ using \declarationhr{2.15} and \declarationhr{2.17}, given that $a\ne 0$.
					    \item If $\Nu(a)>0$, then do the following:
					    \begin{enumerate}
						    \item Using \declarationhr{2.20}, show that $\frac{1}{a}a$
						    \begin{enumerate}
							    \item $=\langle\De(a),\Nu(a)\rangle\langle\Nu(a),\De(a)\rangle$
							    \item $=\langle\De(a)\Nu(a),\Nu(a)\De(a)\rangle$
							    \item $=\langle 1,1\rangle$
							    \item $=1$.
						    \end{enumerate}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
						    \item Show that $\Nu(a)<0$.
						    \item Hence using \declarationhr{2.20}, show that $\frac{1}{a}a$
						    \begin{enumerate}
							    \item $=\langle-\De(a),-\Nu(a)\rangle\langle\Nu(a),\De(a)\rangle$
							    \item $=\langle-\De(a)\Nu(a),-\Nu(a)\De(a)\rangle$
							    \item $=\langle 1,1\rangle$
							    \item $=1$.
						    \end{enumerate}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.42}
			    \objective
				    Choose three rational numbers $a,b,c$. The objective of the following instructions is to show that $a(b+c)=ab+ac$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{2.16} and \declarationhr{2.19}, show that $a(b+c)$
					    \begin{enumerate}
						    \item $=\langle\Nu(a),\De(a)\rangle\langle\Nu(b)\De(c)+\De(b)\Nu(c),\De(b)\De(c)\rangle$
						    \item $=\langle\Nu(a)(\Nu(b)\De(c)+\De(b)\Nu(c)),\De(a)(\De(b)\De(c))\rangle$
						    \item $=\langle\Nu(a)\Nu(b)\De(c)+\Nu(a)\De(b)\Nu(c),\De(a)\De(b)\De(c)\rangle$
						    \item $=\langle\De(a)(\Nu(a)\Nu(b)\De(c)+\Nu(a)\De(b)\Nu(c)),\De(a)(\De(a)\De(b)\De(c))\rangle$
						    \item $=\langle(\Nu(a)\Nu(b))(\De(a)\De(c))+(\De(a)\De(b))(\Nu(a)\Nu(c)),(\De(a)\De(b))(\De(a)\De(c))\rangle$
						    \item $=\langle\Nu(a)\Nu(b),\De(a)\De(b)\rangle+\langle\Nu(a)\Nu(c),\De(a)\De(c)\rangle$
						    \item $=ab+ac$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.09}
			    \objective
				    Choose an integer $a$. The objective of the following instructions is to show that $(-1)^{2a}=1$ and $(-1)^{2a+1}=-1$.
			    \implementation
				    Implementation is analogous to that of \procedurehr{1.91}.
		    \declaration{2.22}
			    The phrases "\gls{rata<b}" and "$b>a$", where $a,b$ are rational numbers, will be used as a shorthand for "$\Nu(a)\De(b)<\De(a)\Nu(b)$".
		    \procedure{2.43}
			    \objective
				    Choose four rational numbers $a,b,c,d$ such that $a<b$, $a=c$ and $b=d$. The objective of the following instructions is to show that $c<d$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Nu(a)\De(c)=\De(a)\Nu(c)$ using \declarationhr{2.15} given that $a=c$.
					    \item Show that $\Nu(b)\De(d)=\De(b)\Nu(d)$ using \declarationhr{2.15} given that $b=d$.
					    \item Show that $\Nu(a)\De(b)<\De(a)\Nu(b)$ using \declarationhr{2.22} given that $a<b$.
					    \item Hence show that $\Nu(c)\De(d)\De(a)\De(b)$
					    \begin{enumerate}
						    \item $=\Nu(a)\De(c)\De(d)\De(b)$
						    \item $<\De(a)\Nu(b)\De(c)\De(d)$
						    \item $=\De(b)\Nu(d)\De(a)\De(c)$.
					    \end{enumerate}
					    \item Hence show that $\Nu(c)\De(d)<\De(c)\Nu(d)$.
					    \item \textbf{Hence show that $c<d$ using \declarationhr{2.22}.}
				    \end{enumerate}
		    \procedure{2.44}
			    \objective
				    Choose three rational numbers $a,b,c$ such that $a<b$. The objective of the following instructions is to show that $a+c<b+c$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Nu(a)\De(b)<\De(a)\Nu(b)$ using \declarationhr{2.22} given that $a<b$.
					    \item Show that $0<\De(c)$ using \declarationhr{2.12}.
					    \item Hence show that $\Nu(a+c)\De(b+c)$
					    \begin{enumerate}
						    \item $=(\Nu(a)\De(c)+\De(a)\Nu(c))\De(b)\De(c)$
						    \item $=\Nu(a)\De(c)\De(b)\De(c)+\De(a)\Nu(c)\De(b)\De(c)$
						    \item $<\De(a)\De(c)\Nu(b)\De(c)+\De(a)\Nu(c)\De(b)\De(c)$
						    \item $=(\Nu(b)\De(c)+\Nu(c)\De(b))\De(a)\De(c)$
						    \item $=\Nu(b+c)\De(a+c)$.
					    \end{enumerate}
					    \item \textbf{Hence show that $a+c<b+c$.}
				    \end{enumerate}
		    \procedure{2.45}
			    \objective
				    Choose two rational numbers $a,b$. The objective of the following instructions is to show that either $a<b$, $a=b$ and $b<a$.
			    \implementation
				    \begin{enumerate}
					    \item Using \procedurehr{1.81}, show that either
					    \begin{enumerate}
					      \item $\Nu(a)\De(b)<\De(a)\Nu(b)$
					      \item $\Nu(a)\De(b)=\De(a)\Nu(b)$
					      \item $\Nu(a)\De(b)>\De(a)\Nu(b)$
					    \end{enumerate}
					    \item Hence show that either
					    \begin{enumerate}
					      \item $a<b$ using \declarationhr{2.22} given that $\Nu(a)\De(b)<\De(a)\Nu(b)$.
					      \item $a=b$ using \declarationhr{2.15} given that $\Nu(a)\De(b)=\De(a)\Nu(b)$.
					      \item $b>a$ using \declarationhr{2.22} given that $\Nu(b)\De(a)>\De(b)\Nu(a)$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.49}
			    \objective
				    Choose two rational numbers $a,b$ such that $0<a$ and $0<b$. The objective of the following instructions is to show that $0<a+b$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $0=\Nu(0)\De(a)<\De(0)\Nu(a)=\Nu(a)$ using \declarationhr{2.22} given that $0<a$.
					    \item Show that $0<\De(a)$ using \declarationhr{2.12}.
					    \item Show that $0=\Nu(0)\De(b)<\De(0)\Nu(b)=\Nu(b)$ using \declarationhr{2.22} given that $0<b$.
					    \item Show that $0<\De(b)$ using \declarationhr{2.12}.
					    \item Hence show that $\Nu(0)\De(a+b)=0<\Nu(a)\De(b)+\De(a)\Nu(b)=\De(0)\Nu(a+b)$.
					    \item \textbf{Hence show that $0<a+b$ using \declarationhr{2.22} given that $\Nu(0)\De(a+b)<\De(0)\Nu(a+b)$.}
				    \end{enumerate}
		    \procedure{2.50}
			    \objective
				    Choose two rational numbers $a,b$ such that $0<a$ and $0<b$. The objective of the following instructions is to show that $0<ab$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $0=\Nu(0)\De(a)<\De(0)\Nu(a)=\Nu(a)$ using \declarationhr{2.22} given that $0<a$.
					    \item Show that $0=\Nu(0)\De(b)<\De(0)\Nu(b)=\Nu(b)$ using \declarationhr{2.22} given that $0<b$.
					    \item Hence show that $\Nu(0)\De(ab)=0<\Nu(a)\Nu(b)=\De(0)\Nu(ab)$.
					    \item \textbf{Hence show that $0<ab$ using \declarationhr{2.22} given that $\Nu(0)\De(ab)<\De(0)\Nu(ab)$.}
				    \end{enumerate}
		    \procedure{2.81}
			    \objective
				    Choose two rational numbers $a,b$. The objective of the following instructions is to show that $\lVert ab\rVert=\lVert a\rVert\lVert b\rVert$.
			    \implementation
				    Implementation is analogous to that of \procedurehr{1.87}.
		    \procedure{2.82}
			    \objective
				    Choose two rational numbers $a,b$. The objective of the following instructions is to show that $\lVert a+b\rVert\le\lVert a\rVert+\lVert b\rVert$.
			    \implementation
				    Implementation is analogous to that of \procedurehr{1.88}.
		    \procedure{2.83}
			    \objective
				    Choose two rational numbers $a,b$. The objective of the following instructions is to show that $\lVert a\rVert-\lVert b\rVert\le\lVert a-b\rVert$.
			    \implementation
				    Implementation is analogous to that of \procedurehr{1.89}.
		    \procedure{2.84}
			    \objective
				    Choose a rational number $a$. The objective of the following instructions is to show that $a=\sgn(a)\lVert a\rVert$.
			    \implementation
				    Implementation is analogous to that of \procedurehr{1.90}.
            \procedure{thu3001201131}
                \objective
                    Choose two rational numbers $x,y$ such that $xy\le 0$. The objective of the following instructions is to show that $\lVert x\rVert\le\lVert y-x\rVert$ and $\lVert y\rVert\le\lVert y-x\rVert$.
                \implementation
                    \begin{enumerate}
                        \item Show that $-\frac{1}{2}(y-x)^2+\frac{1}{2}y^2+\frac{1}{2}x^2=xy\le 0$.
                        \item Hence show that $\frac{1}{2}(y^2+x^2)\le\frac{1}{2}(y-x)^2$.
                        \item \textbf{Hence show that $\lVert y\rVert\le\lVert y-x\rVert$ given that $y^2\le y^2+x^2\le(y-x)^2$.}
                        \item \textbf{Also show that $\lVert x\rVert\le\lVert y-x\rVert$ given that $x^2\le y^2+x^2\le(y-x)^2$.}
                    \end{enumerate}
		    \declaration{2.02}
			    The notation \gls{ratfloor}, where $a$ is a rational number, will be used as a shorthand for $\Nu(a)\div\De(a)$.
		    \declaration{2.03}
			    The notation \gls{ratceil}, where $a$ is a rational number, will be used as a shorthand for $(\Nu(a)\div\De(a))+1$.
		    \procedure{2.04}
			    \objective
				    Choose a rational number $r\ne 1$ and an integer $n\ge 0$. The objective of the following instructions is to show that $\sum_t^{[0:n]} r^t=\frac{1-r^n}{1-r}$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $r\sum_t^{[0:n]} r^t=\sum_t^{[0:n]} r^{t+1}=\sum_t^{[1:{n+1}]} r^t$.
					    \item Therefore show that $(1-r)\sum_t^{[0:n]} r^t=\sum_t^{[0:n]} r^t-\sum_t^{[1:{n+1}]} r^t=1-r^n$.
					    \item \textbf{Therefore show that $\sum_t^{[0:n]} r^t=\frac{1-r^n}{1-r}$.}
				    \end{enumerate}
		    \procedure{2.05}
			    \objective
				    Choose a rational $0<r<1$ and an integer $n\ge 0$. The objective of the following instructions is to show that $\sum_t^{[0:n]} r^t<\frac{1}{1-r}$.
			    \implementation
				    \begin{enumerate}
					    \item \textbf{Show that $\sum_t^{[0:n]} r^t=\frac{1-r^n}{1-r}<\frac{1}{1-r}$ using \procedurehr{2.04}.}
				    \end{enumerate}
		    \procedure{2.06}
			    \objective
				    Choose a non-negative integer $a$ and a rational number $x$. The objective of the following instructions is to show that $(1+x)^a=\sum_r^{[0:{a+1}]}\binom{a}{r}x^r$.
			    \implementation
				    Instructions are analogous to those of \procedurehr{1.63}.
		    \procedure{2.07}
			    \objective
				    Choose an integer $r\ge 0$ and a rational number $x\ge -1$. The objective of the following instructions is to show that $(1+x)^r\ge 1+rx$.
			    \implementation
				    \begin{enumerate}
					    \item If $-1\le x<0$, then do the following:
					    \begin{enumerate}
						    \item Using \procedurehr{2.04}, show that $(1+x)^r$
						    \begin{enumerate}
							    \item $=1+(1+x)^r-1$
							    \item $=1+x\frac{(1+x)^r-1}{(1+x)-1}$
							    \item $=1+x\sum_k^{[0:r]} (1+x)^k$
							    \item $\ge 1+x\sum_k^{[0:r]} 1$
							    \item $=1+rx$.
						    \end{enumerate}
					    \end{enumerate}
					    \item Otherwise, do the following:
					    \begin{enumerate}
						    \item Show that $x\ge 0$.
						    \item Now using \procedurehr{2.06}, show that $(1+x)^r$
						    \begin{enumerate}
							    \item $=\sum_k^{[0:{r+1}]}\binom{r}{k}x^k$
							    \item $\ge \binom{r}{0}x^0+\binom{r}{1}x^1$
							    \item $=1+rx$
						    \end{enumerate}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{wed2407191348}
	        \objective
	          Choose a non-negative integer $r$ and a rational number $x>-1$ such that $(r-1)x<1$. The objective of the following instructions is to show that $(1+x)^r\le\frac{1+x}{1-(r-1)x}$.
	        \implementation
	          \begin{enumerate}
	            \item Show that $1-\frac{x}{1+x}=\frac{1}{1+x}>0$.
	            \item Hence show that $(1-\frac{x}{1+x})^r\ge 1-\frac{rx}{1+x}$ using \procedurehr{2.07}.
	            \item Hence show that $(1-\frac{x}{1+x})^r\ge 1-\frac{rx}{1+x}>0$
                \begin{enumerate}
                    \item given that $0<\frac{1+x-rx}{1+x}=1-\frac{rx}{1+x}$
                    \item given that $0<1+x-rx$
                    \item given that $(r-1)x<1$.
                \end{enumerate}
	            \item Hence show that $(1+x)^r$
	            \begin{enumerate}
	              \item $=(\frac{1}{1+x})^{-r}$
	              \item $=(1-\frac{x}{1+x})^{-r}$
	              \item $\le(1-\frac{rx}{1+x})^{-1}$
	              \item $=\frac{1+x}{1-(r-1)x}$.
	            \end{enumerate}
	          \end{enumerate}
			\end{multicols*}
	  \chapter{Perplex Arithmetic}
			\begin{multicols*}{2}
			  \declaration{wed0502201651}
			    The phrase "\gls{perplex}" will be used as a shorthand for a pair of rational numbers.
			  \declaration{sun0902201114}
			    The phrase "the real part of $a$" and the notation \gls{perplexRe}, where $a$ is a perplex number, will be used as a shorthand for the first entry of $a$.
		    \declaration{sun0902201115}
			    The phrase "the imaginary part of $a$" and the notation \gls{perplexIm}, where $a$ is a perplex number, will be used as a shorthand for the second entry of $a$.
		    \declaration{sat0802201051}
			    The phrase "\gls{perplexa=b}", where $a,b$ are perplex numbers, will be used as a shorthand for "$\Re(a)=\Re(b)$ and $\Im(a)=\Im(b)$".
		    \procedure{sun0902201116}
			    \objective
				    Choose a perplex number $a$. The objective of the following instructions is to show that $a=a$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Re(a)=\Re(a)$.
					    \item Show that $\Im(a)=\Im(a)$.
					    \item \textbf{Hence show that $a=a$.}
				    \end{enumerate}
		    \procedure{sun0902201117}
			    \objective
				    Choose two perplex numbers $a,b$ such that $a=b$. The objective of the following instructions is to show that $b=a$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Re(b)=\Re(a)$ given that $\Re(a)=\Re(b)$.
					    \item Show that $\Im(b)=\Im(a)$ given that $\Im(a)=\Im(b)$.
					    \item \textbf{Hence show that $b=a$.}
				    \end{enumerate}
		    \procedure{sun0902201118}
			    \objective
				    Choose three perplex numbers $a,b,c$ such that $a=b$ and $b=c$. The objective of the following instructions is to show that $a=c$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Re(a)=\Re(c)$
					    \begin{enumerate}
					      \item given that $\Re(a)=\Re(b)$
					      \item and $\Re(b)=\Re(c)$.
					    \end{enumerate}
					    \item Show that $\Im(a)=\Im(c)$
					    \begin{enumerate}
					      \item given that $\Im(a)=\Im(b)$
					      \item and $\Im(b)=\Im(c)$.
					    \end{enumerate}
					    \item \textbf{Hence verify that $a=c$.}
				    \end{enumerate}
		    \declaration{sun0902201119}
			    The notation \gls{perplexa+b}, where $a,b$ are perplex numbers, will be used as a shorthand for the perplex number $\langle\Re(a)+\Re(b),\Im(a)+\Im(b)\rangle$.
		    \procedure{sun0902201120}
			    \objective
				    Choose two perplex numbers $a,b,c,d$ such that $a=c$ and $b=d$. The objective of the following instructions is to show that $a+b=c+d$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{sat0802201051}, show that
					    \begin{enumerate}
					      \item $\Re(a)=\Re(c)$
					      \item $\Im(a)=\Im(c)$
					      \item $\Re(b)=\Re(d)$
					      \item $\Im(b)=\Im(d)$.
					    \end{enumerate}
					    \item Hence show that $a+b$
					    \begin{enumerate}
						    \item $=\langle\Re(a),\Im(a)\rangle+\langle\Re(b),\Im(b)\rangle$
						    \item $=\langle\Re(a)+\Re(b),\Im(a)+\Im(b)\rangle$
						    \item $=\langle\Re(c)+\Re(d),\Im(c)+\Im(d)\rangle$
						    \item $=\langle\Re(c),\Im(c)\rangle+\langle\Re(d),\Im(d)\rangle$
						    \item $=c+d$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{sun0902201121}
			    \objective
				    Choose three perplex numbers $a,b,c$. The objective of the following instructions is to show that $(a+b)+c=a+(b+c)$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $(a+b)+c$
					    \begin{enumerate}
						    \item $=\langle\Re(a)+\Re(b),\Im(a)+\Im(b)\rangle+\langle\Re(c),\Im(c)\rangle$
						    \item $=\langle(\Re(a)+\Re(b))+\Re(c),(\Im(a)+\Im(b))+\Im(c)\rangle$
						    \item $=\langle\Re(a)+(\Re(b)+\Re(c)),\Im(a)+(\Im(b)+\Im(c))\rangle$
						    \item $=\langle\Re(a),\Im(a)\rangle+\langle\Re(b)+\Re(c),\Im(b)+\Im(c)\rangle$
						    \item $=a+(b+c)$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{sun0902201122}
			    \objective
				    Choose two perplex numbers $a,b$. The objective of the following instructions is to show that $a+b=b+a$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $a+b$
					    \begin{enumerate}
						    \item $=\langle\Re(a)+\Re(b),\Im(a)+\Im(b)\rangle$
						    \item $=\langle\Re(b)+\Re(a),\Im(b)+\Im(a)\rangle$
						    \item $=b+a$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{sun0902201123}
			    The notation \gls{perplexa}, where $a$ is a rational number, will contextually be used as a shorthand for the perplex number $\langle a,0\rangle$.
		    \procedure{sun0902201124}
			    \objective
				    Choose a perplex number $a$. The objective of the following instructions is to show that $0+a=a$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $0+a$
					    \begin{enumerate}
						    \item $=\langle 0,0\rangle+\langle\Re(a),\Im(a)\rangle$
						    \item $=\langle 0+\Re(a),0+\Im(a)\rangle$
						    \item $=\langle\Re(a),\Im(a)\rangle$
						    \item $=a$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{sun0902201125}
			    The notation \gls{perplex-a}, where $a$ is a perplex number, will be used as a shorthand for the pair $\langle-\Re(a),-\Im(a)\rangle$.
		    \procedure{sun0902201126}
			    \objective
				    Choose a perplex number $a$. The objective of the following instructions is to show that $-a+a=0$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $-a+a$
					    \begin{enumerate}
						    \item $=(-a)+a$
						    \item $=\langle-\Re(a),-\Im(a)\rangle+\langle\Re(a),\Im(a)\rangle$
						    \item $=\langle-\Re(a)+\Re(a),-\Im(a)+\Im(a)\rangle$
						    \item $=\langle 0,0\rangle$
						    \item $=0$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{sun0902201127}
			    The notation \gls{perplexab}, where $a,b$ are perplex numbers, will be used as a shorthand for the perplex number $\langle\Re(a)\Re(b)+\Im(a)\Im(b),\Re(a)\Im(b)+\Im(a)\Re(b)\rangle$.
		    \procedure{sun0902201128}
			    \objective
				    Choose four perplex numbers $a,b,c,d$ such that $a=c$ and $b=d$. The objective of the following instructions is to show that $ab=cd$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{sat0802201051}, show that
					    \begin{enumerate}
					      \item $\Re(a)=\Re(c)$
					      \item $\Im(a)=\Im(c)$
					      \item $\Re(b)=\Re(d)$
					      \item $\Im(b)=\Im(d)$.
					    \end{enumerate}
					    \item Hence show that $ab$
					    \begin{enumerate}
						    \item $=\langle\Re(a),\Im(a)\rangle\langle\Re(b),\Im(b)\rangle$
						    \item $=\langle\Re(a)\Re(b)+\Im(a)\Im(b),\Re(a)\Im(b)+\Im(a)\Re(b)\rangle$
						    \item $=\langle\Re(c)\Re(d)+\Im(c)\Im(d),\Re(c)\Im(d)+\Im(c)\Re(d)\rangle$
						    \item $=\langle\Re(c),\Im(c)\rangle\langle\Re(d),\Im(d)\rangle$
						    \item $=cd$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{sun0902201129}
			    \objective
				    Choose three perplex numbers $a,b,c$. The objective of the following instructions is to show that $(ab)c=a(bc)$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $(ab)c$
					    \begin{enumerate}
						    \item $=\langle\Re(a)\Re(b)+\Im(a)\Im(b),\Re(a)\Im(b)+\Im(a)\Re(b)\rangle\langle\Re(c),\Im(c)\rangle$
						    \item $=\langle(\Re(a)\Re(b)+\Im(a)\Im(b))\Re(c)+(\Re(a)\Im(b)+\Im(a)\Re(b))\Im(c),(\Re(a)\Re(b)+\Im(a)\Im(b))\Im(c)+(\Re(a)\Im(b)+\Im(a)\Re(b))\Re(c)\rangle$
						    \item $=\langle\Re(a)(\Re(b)\Re(c)+\Im(b)\Im(c))+\Im(a)(\Re(b)\Im(c)+\Im(b)\Re(c)),\Re(a)(\Re(b)\Im(c)+\Im(b)\Re(c))+\Im(a)(\Re(b)\Re(c)+\Im(b)\Im(c))\rangle$
						    \item $=\langle\Re(a),\Im(a)\rangle\langle\Re(b)\Re(c)+\Im(b)\Im(c),\Re(b)\Im(c)+\Im(b)\Re(c)\rangle$
						    \item $=a(bc)$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{sun0902201130}
			    \objective
				    Choose two perplex numbers $a,b$. The objective of the following instructions is to show that $ab=ba$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $ab$
					    \begin{enumerate}
						    \item $=\langle\Re(a)\Re(b)+\Im(a)\Im(b),\Re(a)\Im(b)+\Im(a)\Re(b)\rangle$
						    \item $=\langle\Re(b)\Re(a)+\Im(b)\Im(a),\Re(b)\Im(a)+\Im(b)\Re(a)\rangle$
						    \item $=ba$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{sun0902201131}
			    \objective
				    Choose a perplex number $a$. The objective of the following instructions is to show that $1a=a$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $1a$
					    \begin{enumerate}
						    \item $=\langle 1,0\rangle\langle\Re(a),\Im(a)\rangle$
						    \item $=\langle 1\Re(a)+0\Im(a),1\Im(a)+0\Re(a)\rangle$
						    \item $=\langle\Re(a),\Im(a)\rangle$
						    \item $=a$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{sat0802201553}
			    \objective
				    Choose three perplex numbers $a,b,c$. The objective of the following instructions is to show that $a(b+c)=ab+ac$.
			    \implementation
				    \begin{enumerate}
					    \item $a(b+c)$
					    \begin{enumerate}
						    \item $=\langle\Re(a),\Im(a)\rangle\langle\Re(b)+\Re(c),\Im(b)+\Im(c)\rangle$
						    \item $=\langle\Re(a)(\Re(b)+\Re(c))+\Im(a)(\Im(b)+\Im(c)),\Re(a)(\Im(b)+\Im(c))+\Im(a)(\Re(b)+\Re(c))\rangle$
						    \item $=\langle(\Re(a)\Re(b)+\Im(a)\Im(b))+(\Re(a)\Re(c)+\Im(a)\Im(c)),(\Re(a)\Im(b)+\Im(a)\Re(b))+(\Re(a)\Im(c)+\Im(a)\Re(c))\rangle$
						    \item $=\langle\Re(a)\Re(b)+\Im(a)\Im(b),\Re(a)\Im(b)+\Im(a)\Re(b)\rangle+\langle\Re(a)\Re(c)+\Im(a)\Im(c),\Re(a)\Im(c)+\Im(a)\Re(c)\rangle$
						    \item $=ab+ac$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{sun0902201132}
			    The notation \gls{perplexconj}, where $a$ is a perplex number, will be used as a shorthand for the perplex number $\langle\Re(a),-\Im(a)\rangle$.
		    \procedure{sun0902201133}
			    \objective
				    Choose two perplex numbers $a,b$. The objective of the following instructions is to show that $\conj{a+b}=\conj{a}+\conj{b}$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\conj{a+b}$
					    \begin{enumerate}
						    \item $=\langle\Re(a+b),-\Im(a+b)\rangle$
						    \item $=\langle\Re(a)+\Re(b),-\Im(a)-\Im(b)\rangle$
						    \item $=\conj{a}+\conj{b}$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{sun0902201134}
			    \objective
				    Choose two perplex numbers $a,b$. The objective of the following instructions is to show that $\conj{ab}=\conj{a}\conj{b}$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\conj{ab}$
					    \begin{enumerate}
						    \item $=\langle\Re(ab),-\Im(ab)\rangle$
						    \item $=\langle\Re(a)\Re(b)+\Im(a)\Im(b)),-\Re(a)\Im(b)-\Im(a)\Re(b)\rangle$
						    \item $=\langle\Re(a),-\Im(a)\rangle\langle\Re(b),-\Im(b)\rangle$
						    \item $=\conj{a}\conj{b}$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{sun0902201140}
			    The notation \gls{perplexabs^2}, where $a$ is a perplex number, will be used as a shorthand for $\Re(a)^2-\Im(a)^2$.
		    \procedure{sun0902201141}
			    \objective
				    Choose a perplex number $a$. The objective of the following instructions is to show that $a\conj{a}=\lVert a\rVert^2$.
			    \implementation
				    \begin{enumerate}
					    \item \textbf{Show that $a\conj{a}=\lVert a\rVert^2$.}
				    \end{enumerate}
		    \declaration{wed0502201719}
			    The notation $a<b$, where $a,b$ are perplex numbers, will be used as a shorthand for $0<\Re(b-a)$ and $\lVert b-a\rVert^2\ge 0$.
			  \procedure{sat0802200648}
			    \objective
				    Choose four perplex numbers $a,b,c,d$ such that $a<b$, $a=c$ and $b=d$. The objective of the following instructions is to show that $c<d$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Re(a)=\Re(c)$ and $\Im(a)=\Im(c)$ using \declarationhr{sat0802201051} given that $a=c$.
					    \item Show that $\Re(b)=\Re(d)$ and $\Im(b)=\Im(d)$ using \declarationhr{sat0802201051} given that $b=d$.
					    \item Show that $0<\Re(b-a)$ and $\lVert b-a\rVert^2\ge 0$ using \declarationhr{wed0502201719} given that $a<b$.
					    \item Hence show that $\Re(d-c)=\Re(b-a)>0$.
					    \item Also show that $\lVert d-c\rVert^2=\lVert b-a\rVert^2\ge 0$.
					    \item \textbf{Hence show that $c<d$ using \declarationhr{wed0502201719}.}
				    \end{enumerate}
		    \procedure{sun0902201135}
			    \objective
				    Choose three perplex numbers $a,b,c$ such that $a<b$. The objective of the following instructions is to show that $a+c<b+c$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $0<\Re(b-a)$ and $\lVert b-a\rVert^2\ge 0$ using \declarationhr{wed0502201719} given that $a<b$.
					    \item Hence show that $\Re((b+c)-(a+c))=\Re(b-a)>0$.
					    \item Also show that $\lVert(b+c)-(a+c)\rVert^2=\lVert b-a\rVert^2\ge 0$.
					    \item \textbf{Hence show that $a+c<b+c$ using \declarationhr{wed0502201719}.}
				    \end{enumerate}
		    \procedure{sun0902201138}
			    \objective
				    Choose two perplex numbers $a,b$ such that $0<a$ and $0<b$. The objective of the following instructions is to show that $0<a+b$.
			    \implementation
				    \begin{enumerate}
				      \item Show that $0<\Re(a)$ and $\lVert a\rVert^2\ge 0$ using \declarationhr{wed0502201719} given that $0<a$.
				      \item Show that $0<\Re(b)$ and $\lVert b\rVert^2\ge 0$ using \declarationhr{wed0502201719} given that $0<b$.
				      \item Hence show that $\Re(a+b)=\Re(a)+\Re(b)>0$ given that $0<\Re(a)$ and $0<\Re(b)$.
				      \item Also show that $\lVert a+b\rVert^2=\Re(a+b)^2-\Im(a+b)^2=\Re(a)^2-\Im(a)^2+\Re(b)^2-\Im(b)^2+2(\Re(a)\Re(b)-\Im(a)\Im(b))=\lVert a\rVert^2+\lVert b\rVert^2+(\Re(a)-\Im(a))(\Re(b)+\Im(b))+(\Re(a)+\Im(a))(\Re(b)-\Im(b))\ge 0$.
					    \item \textbf{Hence show that $0<a+b$ using \declarationhr{wed0502201719} given that $\Re(a+b)>0$ and $\lVert a+b\rVert^2\ge 0$.}
				    \end{enumerate}
		    \procedure{sun0902201139}
			    \objective
				    Choose two perplex numbers $a,b$ such that $0<a$ and $0<b$. The objective of the following instructions is to show that $0\le ab$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $0<\Re(a)$ and $\lVert a\rVert^2\ge 0$ using \declarationhr{wed0502201719} given that $0<a$.
				      \item Show that $0<\Re(b)$ and $\lVert b\rVert^2\ge 0$ using \declarationhr{wed0502201719} given that $0<b$.
				      \item Hence show that $\Re(ab)=\Re(a)\Re(b)+\Im(a)\Im(b)\ge\Re(a)\Re(b)-\Re(a)\Re(b)=0$.
				      \item Also show that $\lVert ab\rVert^2=\lVert a\rVert^2\lVert b\rVert^2\ge 0$.
					    \item If $\Re(ab)=0$, then do the following:
					    \begin{enumerate}
					      \item \textbf{Show that $ab=0$ using \declarationhr{sat0802201051} given that $\Im(ab)=0$ given that $\Im(ab)^2\le\Re(ab)^2=0$ given that $\lVert ab\rVert^2\ge 0$.}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
					      \item \textbf{Show that $ab>0$ using \declarationhr{wed0502201719} given that $\Re(ab)>0$ and $\lVert ab\rVert^2\ge 0$.}
					    \end{enumerate}
				    \end{enumerate}
			  \declaration{wed0502201655}
			    The notations $a\subset b$ and $b\supset a$, where $a,b$ are perplex numbers, will be used as shorthands for $\lVert b-a\rVert^2\le 0$ and $\Im(b-a)>0$.
			  \procedure{sun1602201324}
			    \objective
				    Choose four perplex numbers $a,b,c,d$ such that $a\subset b$, $a=c$ and $b=d$. The objective of the following instructions is to show that $c\subset d$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Re(a)=\Re(c)$ and $\Im(a)=\Im(c)$ using \declarationhr{sat0802201051} given that $a=c$.
					    \item Show that $\Re(b)=\Re(d)$ and $\Im(b)=\Im(d)$ using \declarationhr{sat0802201051} given that $b=d$.
					    \item Show that $0<\Im(b-a)$ and $\lVert b-a\rVert^2\le 0$ using \declarationhr{wed0502201655} given that $a\subset b$.
					    \item Hence show that $\Im(d-c)=\Im(b-a)>0$.
					    \item Also show that $\lVert d-c\rVert^2=\lVert b-a\rVert^2\le 0$.
					    \item \textbf{Hence show that $c\subset d$ using \declarationhr{wed0502201655}.}
				    \end{enumerate}
		    \procedure{sun1602201330}
			    \objective
				    Choose three perplex numbers $a,b,c$ such that $a\subset b$. The objective of the following instructions is to show that $a+c\subset b+c$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $0<\Im(b-a)$ and $\lVert b-a\rVert^2\le 0$ using \declarationhr{wed0502201655} given that $a\subset b$.
					    \item Hence show that $\Im((b+c)-(a+c))=\Im(b-a)>0$.
					    \item Also show that $\lVert(b+c)-(a+c)\rVert^2=\lVert b-a\rVert^2\le 0$.
					    \item \textbf{Hence show that $a+c\subset b+c$ using \declarationhr{wed0502201655}.}
				    \end{enumerate}
		    \procedure{sun1602201336}
			    \objective
				    Choose two perplex numbers $a,b$ such that $0\subset a$ and $0\subset b$. The objective of the following instructions is to show that $0\subset a+b$.
			    \implementation
				    \begin{enumerate}
				      \item Show that $0<\Im(a)$ and $\lVert a\rVert^2\le 0$ using \declarationhr{wed0502201655} given that $0\subset a$.
				      \item Show that $0<\Im(b)$ and $\lVert b\rVert^2\le 0$ using \declarationhr{wed0502201655} given that $0\subset b$.
				      \item Hence show that $\Im(a+b)=\Im(a)+\Im(b)>0$ given that $0<\Im(a)$ and $0<\Im(b)$.
				      \item Also show that $\lVert a+b\rVert^2=\Re(a+b)^2-\Im(a+b)^2=\Re(a)^2-\Im(a)^2+\Re(b)^2-\Im(b)^2+2(\Re(a)\Re(b)-\Im(a)\Im(b))=\lVert a\rVert^2+\lVert b\rVert^2+(\Re(a)-\Im(a))(\Re(b)+\Im(b))+(\Re(a)+\Im(a))(\Re(b)-\Im(b))\le 0$.
					    \item \textbf{Hence show that $0\subset a+b$ using \declarationhr{wed0502201655} given that $\Im(a+b)>0$ and $\lVert a+b\rVert^2\le 0$.}
				    \end{enumerate}
		    \procedure{sun1602201526}
			    \objective
				    Choose two perplex numbers $a,b$ such that $0\subset a$ and $0\subset b$. The objective of the following instructions is to show that $0\le ab$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $0<\Im(a)$ and $\lVert a\rVert^2\le 0$ using \declarationhr{wed0502201655} given that $0\subset a$.
				      \item Show that $0<\Im(b)$ and $\lVert b\rVert^2\le 0$ using \declarationhr{wed0502201655} given that $0\subset b$.
				      \item Hence show that $\Re(ab)=\Re(a)\Re(b)+\Im(a)\Im(b)\ge-\Im(a)\Im(b)+\Im(a)\Im(b)=0$.
				      \item Also show that $\lVert ab\rVert^2=\lVert a\rVert^2\lVert b\rVert^2\ge 0$.
					    \item If $\Re(ab)=0$, then do the following:
					    \begin{enumerate}
					      \item \textbf{Show that $ab=0$ using \declarationhr{sat0802201051} given that $\Im(ab)=0$ given that $\Im(ab)^2\le\Re(ab)^2=0$ given that $\lVert ab\rVert^2\ge 0$.}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
					      \item \textbf{Show that $ab>0$ using \declarationhr{wed0502201719} given that $\Re(ab)>0$ and $\lVert ab\rVert^2\ge 0$.}
					    \end{enumerate}
				    \end{enumerate}
			  \procedure{sun1602201531}
			    \objective
				    Choose two perplex numbers $a,b$ such that $0\subset a$ and $0<b$. The objective of the following instructions is to show that $0\subseteq ab$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $0<\Im(a)$ and $\lVert a\rVert^2\le 0$ using \declarationhr{wed0502201655} given that $0\subset a$.
				      \item Show that $0<\Re(b)$ and $\lVert b\rVert^2\ge 0$ using \declarationhr{wed0502201719} given that $0<b$.
				      \item Hence show that $\Im(ab)=\Re(a)\Im(b)+\Im(a)\Re(b)\ge-\Im(a)\Re(b)+\Im(a)\Im(b)=0$.
				      \item Also show that $\lVert ab\rVert^2=\lVert a\rVert^2\lVert b\rVert^2\le 0$.
					    \item If $\Im(ab)=0$, then do the following:
					    \begin{enumerate}
					      \item \textbf{Show that $ab=0$ using \declarationhr{sat0802201051} given that $\Re(ab)=0$ given that $\Re(ab)^2\le\Im(ab)^2=0$ given that $\lVert ab\rVert^2\le 0$.}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
					      \item \textbf{Show that $ab\supset 0$ using \declarationhr{wed0502201655} given that $\Im(ab)>0$ and $\lVert ab\rVert^2\le 0$.}
					    \end{enumerate}
				    \end{enumerate}
			  \procedure{sun0902201136}
			    \objective
				    Choose two perplex numbers $a,b$ such that $\lVert b-a\rVert^2\ne 0$. The objective of the following instructions is to show that either $a<b$, $a>b$, $a\subset b$, or $a\supset b$.
			    \implementation
				    \begin{enumerate}
				      \item Given that $\lVert b-a\rVert^2\ne 0$, show that either
				      \begin{enumerate}
				        \item $\lVert b-a\rVert^2>0$ and $\Re(b-a)>0$
				        \item $\lVert b-a\rVert^2>0$ and $\Re(b-a)<0$
				        \item $\lVert b-a\rVert^2<0$ and $\Im(b-a)>0$
				        \item $\lVert b-a\rVert^2<0$ and $\Im(b-a)<0$.
				      \end{enumerate}
					    \item Hence show that either:
					    \begin{enumerate}
					      \item $a<b$ using \declarationhr{wed0502201719} given that $\lVert b-a\rVert^2>0$ and $\Re(b-a)>0$
					      \item $a>b$ using \declarationhr{wed0502201719} given that $\lVert b-a\rVert^2>0$ and $\Re(b-a)<0$
					      \item $a\subset b$ using \declarationhr{wed0502201655} given that $\lVert b-a\rVert^2<0$ and $\Im(b-a)>0$
					      \item $a\supset b$ using \declarationhr{wed0502201655} given that $\lVert b-a\rVert^2<0$ and $\Im(b-a)<0$.
					    \end{enumerate}
				    \end{enumerate}
				\declaration{tue2502201203}
				  The phrases "proper perplex number" and "improper perplex number" will be used as shorthands for perplex numbers, $a$, such that $\Im(a)\ge 0$ and $\Im(a)\le 0$ respectively.
				\declaration{sun1602200918}
				  The phrases "$a$ and $b$ intersect", "$a$ and $b$ are disjoint", and "$a$ and $b$ adjoin", where $a$ and $b$ are perplex numbers, will be used to paraphrase $\lVert b-\conj{a}\rVert^2\le 0$, $\lVert b-\conj{a}\rVert^2\ge 0$, and $\lVert b-\conj{a}\rVert^2=0$ respectively.
		    \procedure{sat0802201500}
			    \objective
				    Choose a list of positive perplex numbers $a$. The objective of the following instructions is to show that $\lVert\sum_r^{[0:{\lvert a\rvert}]}a_r\rVert^2\ge\sum_r^{[0:{\lvert a\rvert}]}\lVert a_r\rVert^2$.
			    \implementation
				    \begin{enumerate}
				      \item Show that $\Re(\sum_r^{[0:\lvert a\rvert]}\sum_k^{[r+1:\lvert a\rvert]}a_r\conj{a_k})>0$ given that $\sum_r^{[0:\lvert a\rvert]}\sum_k^{[r+1:\lvert a\rvert]}a_r\conj{a_k}>0$.
					    \item Hence show that $\lVert\sum_r^{[0:{\lvert a\rvert}]}a_r\rVert^2$
					    \begin{enumerate}
						    \item $=\sum_r^{[0:{\lvert a\rvert}]}\sum_k^{[0:{\lvert a\rvert}]}a_r\conj{a_k}$
						    \item $=\sum_r^{[0:{\lvert a\rvert}]}\lVert a_r\rVert^2+\sum_r^{[0:\lvert a\rvert]}\sum_k^{[r+1:\lvert a\rvert]}(a_r\conj{a_k}+\conj{a_r\conj{a_k}})$
						    \item $=\sum_r^{[0:{\lvert a\rvert}]}\lVert a_r\rVert^2+2\sum_r^{[0:\lvert a\rvert]}\sum_k^{[r+1:\lvert a\rvert]}\Re(a_r\conj{a_k})$
						    \item $=\sum_r^{[0:{\lvert a\rvert}]}\lVert a_r\rVert^2+2\Re(\sum_r^{[0:\lvert a\rvert]}\sum_k^{[r+1:\lvert a\rvert]}a_r\conj{a_k})$
						    \item $\ge\sum_r^{[0:{\lvert a\rvert}]}\lVert a_r\rVert^2$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{sat0802201518}
			    \objective
				    Choose a non-empty list of positive perplex numbers $a$ such that $a_0>\sum_r^{[1:{\lvert a\rvert}]}a_r$. The objective of the following instructions is to show that $\lVert a_0\rVert^2-\sum_r^{[1:{\lvert a\rvert}]}\lVert a_r\rVert^2\ge\lVert a_0-\sum_r^{[1:{\lvert a\rvert}]}a_r\rVert^2$.
			    \implementation
				    \begin{enumerate}
					    \item Using \procedurehr{sat0802201500}, show that $\lVert a_0\rVert^2$
					    \begin{enumerate}
						    \item $=\lVert\sum_r^{[1:{\lvert a\rvert}]}a_r+(a_0-\sum_r^{[1:{\lvert a\rvert}]}a_r)\rVert^2$
						    \item $\ge\sum_r^{[1:{\lvert a\rvert}]}\lVert a_r\rVert^2+\lVert a_0-\sum_r^{[1:{\lvert a\rvert}]}a_r\rVert^2$
					    \end{enumerate}
					    \item \textbf{Therefore show that $\lVert a_0\rVert^2-\sum_r^{[1:{\lvert a\rvert}]}\lVert a_r\rVert^2\ge\lVert a_0-\sum_r^{[1:{\lvert a\rvert}]}a_r\rVert^2$.}
				    \end{enumerate}
		    \procedure{sat0802201359}
			    \objective
				    Choose a list of positive perplex numbers $a$ and a list of rational numbers $b$ such that $\lvert a\rvert=\lvert b\rvert$ and $\lVert a_i\rVert^2\ge {b_i}^2$ for each $i\in[0:\lvert a\rvert]$. The objective of the following instructions is to show that $\lVert\sum_r^{[0:{\lvert a\rvert}]}a_r\rVert^2\ge(\sum_r^{[0:{\lvert b\rvert}]}{b_r})^2$.
			    \implementation
				    \begin{enumerate}
					    \item If $\lvert a\rvert=0$, then do the following:
					    \begin{enumerate}
						    \item \textbf{Show that $\lVert\sum_i^{[0:\lvert a\rvert]} a_i\rVert^2=\lVert 0\rVert^2=(\sum_i^{[0:\lvert b\rvert]}b_i)^2$.}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
					      \item Show that $\lvert a\rvert>0$.
					      \item Show that $\lVert\sum_i^{[1:\lvert a\rvert]}a_i\rVert^2\ge(\sum_i^{[1:\lvert b\rvert]} b_i)^2$ using \procedurehr{sat0802201359} on $a_{[1:\lvert a\rvert]}$ and $b_{[1:\lvert b\rvert]}$.
					      \item Show that $\Re(\conj{a_0}\sum_i^{[1:\lvert a\rvert]}a_i)>0$
					      \begin{enumerate}
					        \item given that $\conj{a_0}\sum_i^{[1:\lvert a\rvert]}a_i>0$
					        \item given that $\sum_i^{[1:\lvert a\rvert]}a_i>0$
					        \item and $\conj{a_0}>0$.
					      \end{enumerate}
					      \item Show that $\Re(\conj{a_0}\sum_i^{[1:\lvert a\rvert]}a_i)^2$
					      \begin{enumerate}
					        \item $\ge\lVert\conj{a_0}\sum_i^{[1:\lvert a\rvert]}a_i\rVert^2$
					        \item $=\lVert\conj{a_0}\rVert^2\lVert\sum_i^{[1:\lvert a\rvert]}a_i\rVert^2$
					        \item $\ge{b_0}^2(\sum_i^{[1:\lvert a\rvert]}b_i)^2$.
					      \end{enumerate}
					      \item Hence show that $\lVert\sum_i^{[0:\lvert a\rvert]}a_i\rVert^2$
					      \begin{enumerate}
					        \item $=(a_0+\sum_i^{[1:\lvert a\rvert]}a_i)(\conj{a_0+\sum_i^{[1:\lvert a\rvert]}a_i})$
					        \item $=\lVert a_0\rVert^2+a_0\conj{\sum_i^{[1:\lvert a\rvert]}a_i}+\conj{a_0}\sum_i^{[1:\lvert a\rvert]}a_i+\lVert\sum_i^{[1:\lvert a\rvert]}a_i\rVert^2$
					        \item $\ge{b_0}^2+\conj{\conj{a_0}\sum_i^{[1:\lvert a\rvert]}a_i}+\conj{a_0}\sum_i^{[1:\lvert a\rvert]}a_i+(\sum_i^{[1:\lvert a\rvert]}b_i)^2$
					        \item $={b_0}^2+2\Re(\conj{a_0}\sum_i^{[1:\lvert a\rvert]}a_i)+(\sum_i^{[1:\lvert a\rvert]}b_i)^2$
					        \item $\ge{b_0}^2+2b_0\sum_i^{[1:\lvert a\rvert]} b_i+(\sum_i^{[1:\lvert a\rvert]}b_i)^2$
					        \item $=(b_0+\sum_i^{[1:\lvert a\rvert]}b_i)^2$
					        \item $=(\sum_i^{[0:\lvert a\rvert]}b_i)^2$.
					      \end{enumerate}
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{sun0902201142}
			    The notation \gls{perplex1/a}, where $a$ is a perplex number, will be used as a shorthand for the pair $\frac{1}{\lVert a\rVert^2}\conj{a}$.
		    \procedure{sun0902201143}
			    \objective
				    Choose a perplex number $a$ such that $\lVert a\rVert^2\ne 0$. The objective of the following instructions is to show that $\frac{1}{a}a=1$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\frac{1}{a}a$
					    \begin{enumerate}
						    \item $=(\frac{1}{\Vert a\rVert^2}\conj{a})a$
						    \item $=\frac{1}{\Vert a\rVert^2}(\conj{a}a)$
						    \item $=\frac{1}{\Vert a\rVert^2}\Vert a\rVert^2$
						    \item $=1$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{sat0802201702}
			    The notations \gls{perplexj} and \gls{perplexk} will be used as a shorthand for the perplex numbers $\langle 0,1\rangle$ and $\langle\frac{1}{2},\frac{1}{2}\rangle$ respectively.
		    \procedure{thu0602201510}
			    \objective
				    The objective of the following instructions is to show that $\ju^{2}=1$, $\ku^2=k$, $\kcu^2=\kcu$, $k\kcu=0$, $k+\kcu=1$, and $\ku-\kcu=\ju$.
			    \implementation
				    \begin{enumerate}
					    \item \textbf{Show that $\ju^2=1$.}
					    \item \textbf{Show that $\ku^2=k$.}
					    \item \textbf{Show that $\kcu^2=\kcu$.}
					    \item \textbf{Show that $\ku\kcu=0$.}
					    \item \textbf{Show that $\ku+\kcu=1$.}
					    \item \textbf{Show that $\ku-\kcu=\ju$.}
				    \end{enumerate}
				\procedure{fri1402201203}
				  \objective
				    Choose a perplex number $a$. The objective of the following instructions is to show that $a=\Re(a)+\Im(a)\ju$.
				  \implementation
				    \begin{enumerate}
				      \item Show that $a$
				      \begin{enumerate}
				        \item $=\langle\Re(a),\Im(a)\rangle$
				        \item $=\langle\Re(a),0\rangle+\langle 0,\Im(a)\rangle$
				        \item $=\Re(a)+\Im(a)\ju$.
				      \end{enumerate}
				    \end{enumerate}
				\procedure{fri1402201147}
				  \objective
				    Choose a perplex number $a$. The objective of the following instructions is to show that $a=(\Re(a)+\Im(a))\ku+(\Re(a)-\Im(a))\kcu$.
				  \implementation
				    \begin{enumerate}
				      \item Using \procedurehr{thu0602201510}, show that $a$
				      \begin{enumerate}
				        \item $=\Re(a)+\Im(a)\ju$
				        \item $=\Re(a)(\ku+\kcu)+\Im(a)(\ku-\kcu)$
				        \item $=(\Re(a)+\Im(a))\ku+(\Re(a)-\Im(a))\kcu$.
				      \end{enumerate}
				    \end{enumerate}
				\procedure{sat0802201559}
			    \objective
				    Choose rational numbers $a,b,c,d$. The objective of the following instructions is to show that $(a\ku+b\kcu)+(c\ku+d\kcu)=(a+c)\ku+(b+d)\kcu$.
			    \implementation
				    \begin{enumerate}
				      \item Show that $(a\ku+b\kcu)+(c\ku+d\kcu)=(a+c)\ku+(b+d)\kcu$.
				    \end{enumerate}
				\procedure{sun0902201144}
			    \objective
				    Choose rational numbers $a,b,c,d$. The objective of the following instructions is to show that $(a\ku+b\kcu)(c\ku+d\kcu)=ac\ku+bd\kcu$.
			    \implementation
				    \begin{enumerate}
				      \item Show that $(a\ku+b\kcu)(c\ku+d\kcu)$
				      \begin{enumerate}
				        \item $=a\ku(c\ku+d\kcu)+b\kcu(c\ku+d\kcu)$
				        \item $=a\ku c\ku+a\ku d\kcu+b\kcu c\ku+b\kcu d\kcu$
				        \item $=ac\ku+0ad+0bc+bd\kcu$
				        \item $=ac\ku+bd\kcu$.
				      \end{enumerate}
				    \end{enumerate}
	    \end{multicols*}
	  \chapter{Polynomial Arithmetic}
	    \begin{multicols*}{2}
	      \declaration{2.08}
			    The notation \gls{ratminc}, where $c$ is a list, will be used as a shorthand for $\infty$ if $c$ is empty, otherwise it will stand for the minimum entry of $c$. The related notation \gls{ratmincr}, where $R$ is a list and $c[r]$ is a function of $r$, will be used as a shorthand for $\min(c(R))$.
		    \declaration{2.11}
			    The notation \gls{ratmaxc}, where $c$ is a list, will be used as a shorthand for $-\infty$ if $c$ is empty, otherwise it will stand for the maximum entry of $c$. The related notation \gls{ratmaxcr}, where $R$ is a list and $c[r]$ is a function of $r$, will be used as a shorthand for $\max(c(R))$.
		    \declaration{2.25}
			    The phrase "\gls{ratpolynomial}" will be used as a shorthand for a list of rational numbers.
		    \declaration{2.26}
			    The notation \gls{ratpolya_i}, where $a$ is a polynomial and $i$ is a natural number such that $i\ge\lvert a\rvert$, will be used as a shorthand for $0$.
		    \declaration{2.27}
			    The phrase "\gls{ratpolya=b}", where $a,b$ are polynomials, will be used as a shorthand for "$a_i=b_i$ for each $i\in[0:\max(\lvert a\rvert,\lvert b\rvert)]$".
		    \declaration{2.28}
			    The notation \gls{ratpolyLambda} will be used as a shorthand for $\sum_r^{[0:\lvert a\rvert]}a_rb^r$.
		    \procedure{2.51}
			    \objective
				    Choose two polynomials $a,b$ and a rational number $c$ such that $a=b$. The objective of the following instructions is to show that $\Lambda(a,c)=\Lambda(b,c)$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{2.27} and \declarationhr{2.28}, show that $\Lambda(a,c)$
					    \begin{enumerate}
						    \item $=\sum_r^{[0:\lvert a\rvert]}a_rc^r$
						    \item $=\sum_r^{[0:\max(\lvert a\rvert,\lvert b\rvert)]}a_rc^r$
						    \item $=\sum_r^{[0:\max(\lvert a\rvert,\lvert b\rvert)]}b_rc^r$
						    \item $=\sum_r^{[0:\lvert b\rvert]}b_rc^r$
						    \item $=\Lambda(b,c)$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.52}
			    \objective
				    Choose a natural number $c$ and two polynomials $a,b$ such that $a=b$. The objective of the following instructions is to show that $a_c=b_c$.
			    \implementation
				    \begin{enumerate}
					    \item If $c<\max(\lvert a\rvert,\lvert b\rvert)$, then do the following:
					    \begin{enumerate}
						    \item \textbf{Show that $a_c=b_c$.}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
						    \item \textbf{Show that $a_c=0=b_c$ given that $c\ge\max(\lvert a\rvert,\lvert b\rvert)$.}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.53}
			    \objective
				    Choose a polynomial $a$. The objective of the following instructions is to show that $a=a$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $a_i=a_i$ for each $i\in[0:\max(\lvert a\rvert,\lvert a\rvert)]$.
					    \item \textbf{Hence show that $a=a$ using \declarationhr{2.27}.}
				    \end{enumerate}
		    \procedure{2.54}
			    \objective
				    Choose two polynomials $a,b$ such that $a=b$. The objective of the following instructions is to show that $b=a$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $a_i=b_i$ for each $i\in[0:\max(\lvert a\rvert,\lvert b\rvert)]$ using \declarationhr{2.27}.
					    \item Hence show that $b_i=a_i$ for each $i\in[0:\max(\lvert b\rvert,\lvert a\rvert)]$.
					    \item \textbf{Hence show that $b=a$ using \declarationhr{2.27}.}
				    \end{enumerate}
		    \procedure{2.55}
			    \objective
				    Choose three polynomials $a,b,c$ such that $a=b$ and $b=c$. The objective of the following instructions is to show that $a=c$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $a_i=b_i$ for each $i\in[0:\max(\lvert a\rvert,\lvert b\rvert,\lvert c\rvert)]$ using \declarationhr{2.27}.
					    \item Show that $b_i=c_i$ for each $i\in[0:\max(\lvert a\rvert,\lvert b\rvert,\lvert c\rvert)]$ using \declarationhr{2.27}.
					    \item Hence show that $a_i=c_i$ for each $i\in[0:\max(\lvert a\rvert,\lvert b\rvert,\lvert c\rvert)]$.
					    \item \textbf{Hence verify that $a=c$ using \declarationhr{2.27}.}
				    \end{enumerate}
		    \declaration{2.37}
			    The notation \gls{ratlistcomp}, where $f[j]$ is a function of $j$ and $R$ is a list, will be used as a shorthand for $\langle f(R)\rangle$.
		    \declaration{2.29}
			    The notation \gls{ratpolya+b}, where $a,b$ are polynomials, will be used as a shorthand for the list $\langle a_i+b_i\for i\in[0:\max(\lvert a\rvert,\lvert b\rvert)]\rangle$.
		    \procedure{2.56}
			    \objective
				    Choose two polynomials $a,b$ and a rational number $c$. The objective of the following instructions is to show that $\Lambda(a+b,c)=\Lambda(a,c)+\Lambda(b,c)$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{2.28} and \declarationhr{2.29}, show that $\Lambda(a+b,c)$
					    \begin{enumerate}
						    \item $=\Lambda(\langle a_r+b_r\for r\in[0:\max(\lvert a\rvert,\lvert b\rvert)]\rangle,c)$
						    \item $=\sum_r^{[0:\max(\lvert a\rvert,\lvert b\rvert)]}(a_r+b_r)c^r$
						    \item $=\sum_r^{[0:\max(\lvert a\rvert,\lvert b\rvert)]}a_rc^r+\sum_r^{[0:\max(\lvert a\rvert,\lvert b\rvert)]}b_rc^r$
						    \item $=\sum_r^{[0:\lvert a\rvert]}a_rc^r+\sum_r^{[0:\lvert b\rvert]}b_rc^r$
						    \item $=\Lambda(a,c)+\Lambda(b,c)$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.57}
			    \objective
				    Choose a natural number $c$ and two polynomials $a,b$. The objective of the following instructions is to show that $(a+b)_c=a_c+b_c$.
			    \implementation
				    \begin{enumerate}
					    \item If $c<\max(\lvert a\rvert,\lvert b\rvert)$, then do the following:
					    \begin{enumerate}
						    \item \textbf{Show that $(a+b)_c=a_c+b_c$ using \declarationhr{2.29}.}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
						    \item Show that $c\ge\max(\lvert a\rvert,\lvert b\rvert)$.
						    \item Hence show that $a_c=0$, $b_c=0$, and $(a+b)_c=0$ using \declarationhr{2.26}.
						    \item \textbf{Hence show that $(a+b)_c=a_c+b_c$.}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.58}
			    \objective
				    Choose four polynomials $a,b,c,d$ such that $a=c$ and $b=d$. The objective of the following instructions is to show that $a+b=c+d$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $a_i=c_i$ for each $i\in[0:\max(\lvert a\rvert,\lvert b\rvert,\lvert c\rvert,\lvert d\rvert)]$ using \declarationhr{2.27} given that $a=c$.
					    \item Verify that $b_i=d_i$ for each $i\in[0:\max(\lvert a\rvert,\lvert b\rvert,\lvert c\rvert,\lvert d\rvert)]$ using \declarationhr{2.27} given that $b=d$.
					    \item Hence using \declarationhr{2.29}, show that $a+b$
					    \begin{enumerate}
						    \item $=\langle a_i+b_i\for i\in[0:\max(\lvert a\rvert,\lvert b\rvert,\lvert c\rvert,\lvert d\rvert)]\rangle$
						    \item $=\langle c_i+d_i\for i\in[0:\max(\lvert a\rvert,\lvert b\rvert,\lvert c\rvert,\lvert d\rvert)]\rangle$
						    \item $=c+d$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.59}
			    \objective
				    Choose three polynomials $a,b,c$. The objective of the following instructions is to show that $(a+b)+c=a+(b+c)$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{2.29}, show that $(a+b)+c$
					    \begin{enumerate}
						    \item $\langle(a+b)_i+c_i\for i\in[0:\max(\lvert a+b\rvert,\lvert c\rvert)]\rangle$
						    \item $\langle(a_i+b_i)+c_i\for i\in[0:\max(\lvert a\rvert,\lvert b\rvert,\lvert c\rvert)]\rangle$
						    \item $\langle a_i+(b_i+c_i)\for i\in[0:\max(\lvert a\rvert,\lvert b+c\rvert)]\rangle$
						    \item $\langle a_i+(b+c)_i\for i\in[0:\max(\lvert a\rvert,\lvert b+c\rvert)]\rangle$
						    \item $=a+(b+c)$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.60}
			    \objective
				    Choose two polynomials $a,b$. The objective of the following instructions is to show that $a+b=b+a$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{2.29}, show that $a+b$
					    \begin{enumerate}
						    \item $=\langle a_i+b_i\for i\in[0:\max(\lvert a\rvert,\lvert b\rvert)]\rangle$
						    \item $=\langle b_i+a_i\for i\in[0:\max(\lvert b\rvert,\lvert a\rvert)]\rangle$
						    \item $=b+a$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{2.30}
			    The notation \gls{ratpolya}, where $a$ is a rational number, will contextually be used as a shorthand for the list $\langle a\rangle$.
		    \procedure{2.61}
			    \objective
				    Choose a polynomial $a$. The objective of the following instructions is to show that $0+a=a$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{2.29} and \declarationhr{2.30}, show that $0+a$
					    \begin{enumerate}
						    \item $=\langle 0_i+a_i\for i\in[0:\lvert a\rvert]\rangle$
						    \item $=\langle 0+a_i\for i\in[0:\lvert a\rvert]\rangle$
						    \item $=a$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{2.31}
			    The notation \gls{ratpoly-a}, where $a$ is a polynomial, will be used as a shorthand for the list $\langle-a_i\for i\in[0:\lvert a\rvert]\rangle$.
		    \procedure{2.00}
			    \objective
				    Choose a polynomial $a$ and a rational number $b$. The objective of the following instructions is to show that $\Lambda(-a,b)=-\Lambda(a,b)$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{2.28} and \declarationhr{2.31}, show that $\Lambda(-a,b)$
					    \begin{enumerate}
						    \item $=\Lambda(\langle-a_i\for i\in[0:\lvert a\rvert]\rangle,b)$
						    \item $=\sum_j^{[0:\lvert a\rvert]}(-a_j)b^j$
						    \item $=-\sum_j^{[0:\lvert a\rvert]}a_jb^j$
						    \item $=-\Lambda(a,b)$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.62}
			    \objective
				    Choose two polynomials $a,b$ such that $a=b$. The objective of the following instructions is to show that $-a=-b$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $a_i=b_i$ for $i\in[0:\max(\lvert a\rvert,\lvert b\rvert)]$ using \declarationhr{2.27} given that $a=b$.
					    \item Hence using \declarationhr{2.31}, show that $-a$
					    \begin{enumerate}
						    \item $=\langle-a_i\for i\in[0:\max(\lvert a\rvert,\lvert b\rvert)]\rangle$
						    \item $=\langle-b_i\for i\in[0:\max(\lvert a\rvert,\lvert b\rvert)]\rangle$
						    \item $=-b$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.63}
			    \objective
				    Choose a polynomial $a$. The objective of the following instructions is to show that $-a+a=0$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{2.29} and \declarationhr{2.31}, show that $-a+a$
					    \begin{enumerate}
						    \item $=(-a)+a$
						    \item $=\langle-a_i\for i\in[0:\lvert a\rvert]\rangle+\langle a_i\for i\in[0:\lvert a\rvert]\rangle$
						    \item $=\langle-a_i+a_i\for i\in[0:\lvert a\rvert]\rangle$
						    \item $=\langle 0\for i\in[0:\lvert a\rvert]\rangle$
						    \item $=0$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{2.32}
			    The notation \gls{ratpolyab}, where $a,b$ are integers, will be used as a shorthand for the list $\langle\sum_r^{[0:i+1]}a_rb_{i-r}\for i\in[0:\lvert a\rvert+\lvert b\rvert-1]\rangle$.
		    \procedure{2.64}
			    \objective
				    Choose two polynomials $a,b$ and a rational number $c$. The objective of the following instructions is to show that $\Lambda(ab,c)=\Lambda(a,c)\Lambda(b,c)$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{2.28} and \declarationhr{2.32}, show that $\Lambda(ab,c)$
					    \begin{enumerate}
						    \item $=\Lambda(\langle\sum_r^{[0:j+1]}a_rb_{j-r}\for j\in[0:\lvert a\rvert+\lvert b\rvert-1]\rangle,c)$
						    \item $=\sum_j^{[0:\lvert a\rvert+\lvert b\rvert-1]}(\sum_r^{[0:j+1]}a_rb_{j-r})c^j$
						    \item $=\sum_j^{[0:\lvert a\rvert+\lvert b\rvert-1]}\sum_r^{[0:j+1]}a_rc^rb_{j-r}c^{j-r}$
						    \item $=\sum_r^{[0:\lvert a\rvert+\lvert b\rvert-1]}\sum_j^{[r:\lvert a\rvert+\lvert b\rvert-1]}a_rc^rb_{j-r}c^{j-r}$
						    \item $=\sum_r^{[0:\lvert a\rvert+\lvert b\rvert-1]}a_rc^r\sum_j^{[r:\lvert a\rvert+\lvert b\rvert-1]}b_{j-r}c^{j-r}$
						    \item $=\sum_r^{[0:\lvert a\rvert+\lvert b\rvert-1]}a_rc^r\sum_j^{[0:\lvert a\rvert+\lvert b\rvert-1-r]}b_{j}c^{j}$
						    \item $=\sum_r^{[0:\lvert a\rvert]}a_rc^r\sum_j^{[0:\lvert a\rvert+\lvert b\rvert-1-r]}b_{j}c^{j}$
						    \item $=\sum_r^{[0:\lvert a\rvert]}a_rc^r\sum_j^{[0:\lvert b\rvert]}b_{j}c^{j}$
						
						    \item $=(\sum_j^{[0:\lvert a\rvert]}a_jc^j)(\sum_j^{[0:\lvert b\rvert]}b_jc^j)$
						    \item $=\Lambda(a,c)\Lambda(b,c)$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.65}
			    \objective
				    Choose a natural number $c$ and two polynomials $a,b$. The objective of the following instructions is to show that $(ab)_c=\sum_r^{[0:c+1]}a_rb_{c-r}$.
			    \implementation
				    \begin{enumerate}
					    \item If $c<\lvert a\rvert+\lvert b\rvert-1$, then do the following:
					    \begin{enumerate}
						    \item \textbf{Show that $(ab)_c=\sum_r^{[0:c+1]}a_rb_{c-r}$ using \declarationhr{2.32}.}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
						    \item Show that $c\ge\lvert a\rvert+\lvert b\rvert-1$.
						    \item Hence using \declarationhr{2.26}, show that $(ab)_c$
						    \begin{enumerate}
							    \item $=0$
							    \item $=\sum_r^{[0:\lvert a\rvert]}0a_r+\sum_r^{[\lvert a\rvert:c+1]}0b_{c-r}$
							    \item $=\sum_r^{[0:\lvert a\rvert]}a_rb_{c-r}+\sum_r^{[\lvert a\rvert:c+1]}a_rb_{c-r}$
							    \item $=\sum_r^{[0:c+1]}a_rb_{c-r}$.
						    \end{enumerate}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.66}
			    \objective
				    Choose four polynomials $a,b,c,d$ such that $a=c$ and $b=d$. The objective of the following instructions is to show that $ab=cd$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $a_i=c_i$ for $i\in[0:\max(\lvert a\rvert,\lvert c\rvert)+\max(\lvert b\rvert,\lvert d\rvert)-1]$ using \procedurehr{2.52} given that $a=c$.
					    \item Show that $b_i=d_i$ for $i\in[0:\max(\lvert a\rvert,\lvert c\rvert)+\max(\lvert b\rvert,\lvert d\rvert)-1]$ using \procedurehr{2.52} given that $b=d$.
					    \item Hence using \declarationhr{2.32}, show that $ab$
					    \begin{enumerate}
						    \item $=\langle\sum_r^{[0:i+1]}a_rb_{i-r}\for i\in[0:\max(\lvert a\rvert,\lvert c\rvert)+\max(\lvert b\rvert,\lvert d\rvert)-1]\rangle$
						    \item $=\langle\sum_r^{[0:i+1]}c_rd_{i-r}\for i\in[0:\max(\lvert a\rvert,\lvert c\rvert)+\max(\lvert b\rvert,\lvert d\rvert)-1]\rangle$
						    \item $=cd$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.67}
			    \objective
				    Choose three polynomials $a,b,c$. The objective of the following instructions is to show that $(ab)c=a(bc)$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{2.32}, show that $(ab)c$
					    \begin{enumerate}
						    \item $=\langle\sum_t^{[0:j+1]}(ab)_tc_{j-t}\for j\in[0:\lvert ab\rvert+\lvert c\rvert-1]\rangle$
						    \item $=\langle\sum_t^{[0:j+1]}\langle\sum_r^{[0:i+1]}a_rb_{i-r}\for i\in[0:\lvert a\rvert+\lvert b\rvert-1]\rangle_tc_{j-t}\for j\in[0:\lvert a\rvert+\lvert b\rvert+\lvert c\rvert-2]\rangle$
						    \item $=\langle\sum_t^{[0:j+1]}\sum_r^{[0:t+1]}a_rb_{t-r}c_{j-t}\for j\in[0:\lvert a\rvert+\lvert b\rvert+\lvert c\rvert-2]\rangle$
						    \item $=\langle\sum_r^{[0:j+1]}\sum_t^{[r:j+1]}a_rb_{t-r}c_{j-t}\for j\in[0:\lvert a\rvert+\lvert b\rvert+\lvert c\rvert-2]\rangle$
						    \item $=\langle\sum_r^{[0:j+1]}a_r\sum_t^{[r:j+1]}b_{t-r}c_{j-t}\for j\in[0:\lvert a\rvert+\lvert b\rvert+\lvert c\rvert-2]\rangle$
						    \item $=\langle\sum_r^{[0:j+1]}a_r\sum_t^{[0:j-r+1]}b_{t}c_{j-r-t}\for j\in[0:\lvert a\rvert+\lvert b\rvert+\lvert c\rvert-2]\rangle$
						    \item $=\langle\sum_r^{[0:j+1]}a_r\langle\sum_t^{[0:i+1]}b_{t}c_{i-t}\for i\in[0:\lvert b\rvert+\lvert c\rvert-1]\rangle_{j-r}\for j\in[0:\lvert a\rvert+\lvert b\rvert+\lvert c\rvert-2]\rangle$
						    \item $=\langle\sum_r^{[0:j+1]}a_r(bc)_{j-r}\for j\in[0:\lvert a\rvert+\lvert bc\rvert-1]\rangle$
						    \item $=a(bc)$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.68}
			    \objective
				    Choose two polynomials $a,b$. The objective of the following instructions is to show that $ab=ba$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{2.32}, show that $ab$
					    \begin{enumerate}
						    \item $=\langle\sum_r^{[0:i+1]}a_rb_{i-r}\for i\in[0:\lvert a\rvert+\lvert b\rvert-1]\rangle$
						    \item $=\langle\sum_r^{[0:i+1]}b_ra_{i-r}\for i\in[0:\lvert a\rvert+\lvert b\rvert-1]\rangle$
						    \item $=ba$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.69}
			    \objective
				    Choose a polynomial $a$. The objective of the following instructions is to show that $1a=a$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{2.30} and \declarationhr{2.32}, show that $1a$
					    \begin{enumerate}
						    \item $=\langle\sum_r^{[0:i+1]}1_ra_{i-r}\for i\in[0:\lvert 1\rvert+\lvert a\rvert-1]\rangle$
						    \item $=\langle 1_0a_{i-0}\for i\in[0:\lvert a\rvert]\rangle$
						    \item $=\langle a_{i}\for i\in[0:\lvert a\rvert]\rangle$
						    \item $=a$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.70}
			    \objective
				    Choose three polynomials $a,b,c$. The objective of the following instructions is to show that $a(b+c)=ab+ac$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{2.29} and \declarationhr{2.32}, show $a(b+c)$
					    \begin{enumerate}
						    \item $=\langle\sum_r^{[0:i+1]}a_r(b+c)_{i-r}\for i\in[0:\lvert a\rvert+\lvert b+c\rvert-1]\rangle$
						    \item $=\langle\sum_r^{[0:i+1]}a_r(b_{i-r}+c_{i-r})\for i\in[0:\lvert a\rvert+\lvert b+c\rvert-1]\rangle$
						    \item $=\langle\sum_r^{[0:i+1]}(a_rb_{i-r}+a_rc_{i-r})\for i\in[0:\lvert a\rvert+\lvert b+c\rvert-1]\rangle$
						    \item $=\langle\sum_r^{[0:i+1]}a_rb_{i-r}+\sum_r^{[0:i+1]}a_rc_{i-r}\for i\in[0:\lvert a\rvert+\lvert b+c\rvert-1]\rangle$
						    \item $=\langle\sum_r^{[0:i+1]}a_rb_{i-r}\for i\in[0:\lvert a\rvert+\lvert b\rvert-1]\rangle+\langle\sum_r^{[0:i+1]}a_rc_{i-r}\for i\in[0:\lvert a\rvert+\lvert c\rvert-1]\rangle$
						    \item $=ab+ac$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{2.33}
			    The notation \gls{ratpolylambda} will be used as a shorthand for the list $\langle 0,1\rangle$.
		    \procedure{2.71}
			    \objective
				    Choose a polynomial $a$. The objective of the following instructions is to show that $\lambda a=\langle 0\rangle^{\frown}a$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\lvert\lambda a\rvert=\lvert\lambda\rvert+\lvert a\rvert-1=\lvert a\rvert+1$ using \declarationhr{2.32}.
					    \item For $j\in[1:\lvert a\rvert+1]$, do the following:
					    \begin{enumerate}
						    \item Using \declarationhr{2.32}, show that $(\lambda a)_j$
						    \begin{enumerate}
							    \item $=\sum_r^{[0:j+1]}\lambda_ra_{j-r}$
							    \item $=\sum_r^{[0:j+1]}[r=1]a_{j-r}$
							    \item $=a_{j-1}$
						    \end{enumerate}
					    \end{enumerate}
					    \item Hence using \declarationhr{2.32}, show that $(\lambda a)_0=\sum_r^{[0:1]}\lambda_ra_{0-r}=\lambda_0a_{0}=0$.
					    \item \textbf{Hence show that $\lambda a=\langle 0\rangle^{\frown}a$.}
				    \end{enumerate}
		    \procedure{2.72}
			    \objective
				    Choose a natural number $n$. The objective of the following instructions is to show that $\lambda^n=\langle[j=n]\for j\in[0:n+1]\rangle$.
			    \implementation
				    \begin{enumerate}
					    \item If $n=0$, then do the following:
					    \begin{enumerate}
						    \item Show that $\lambda^n$
						    \begin{enumerate}
							    \item $=\lambda^0$
							    \item $=\langle 1\rangle$
							    \item $=\langle[j=0]\for j\in[0:1]\rangle$
							    \item $=\langle[j=n]\for j\in[0:n+1]\rangle$.
						    \end{enumerate}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
						    \item Use \procedurehr{2.25} on $\langle n-1\rangle$ to show that $\lambda^{n-1}=\langle[j=n-1]\for j\in[0:n]\rangle$.
						    \item Hence using \procedurehr{2.71}, show that $\lambda^n$
						    \begin{enumerate}
							    \item $=\lambda\lambda^{n-1}$
							    \item $=\lambda\langle[j=n-1]\for j\in[0:n]\rangle$
							    \item $=\langle 0\rangle^{\frown}\langle[j=n-1]\for j\in[0:n]\rangle$
							    \item $=\langle[j=n]\for j\in[0:n+1]\rangle$.
						    \end{enumerate}
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{2.34}
			    The notation \gls{ratpolydeg}, where $a$ is a polynomial such that $a\ne 0$, will be used as a shorthand for the largest natural number $j<\lvert a\rvert$ such that $a_j\ne 0$.
		    \procedure{2.73}
			    \objective
				    Choose two polynomials $a,b$ such that $a=b$ and $a\ne 0$. The objective of the following instructions is to show that $\deg(a)=\deg(b)$.
			    \implementation
				    \begin{enumerate}
					    \item For $j\in[\max(\lvert a\rvert,\lvert b\rvert):0]$, do the following:
					    \begin{enumerate}
						    \item If $a_j=0$, then do the following:
						    \begin{enumerate}
							    \item Show that $0=a_j=b_j$ using \declarationhr{2.27} given that $a=b$.
						    \end{enumerate}
						    \item Otherwise do the following:
						    \begin{enumerate}
							    \item Show that $0\ne a_j=b_j$ using \declarationhr{2.27} given that $a=b$.
							    \item Show that $j<\min(\lvert a\rvert,\lvert b\rvert)$.
							    \item \textbf{Hence show that $\deg(a)=j=\deg(b)$.}
                                \item \textbf{Yield.}
						    \end{enumerate}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.74}
			    \objective
				    Let $\deg(0)=-1$. Choose two polynomials $a,b$ such that $\deg(a)<\deg(b)$. The objective of the following instructions is to show that $\deg(a+b)=\deg(b)$.
			    \implementation
				    \begin{enumerate}
					    \item For $j\in[\max(\lvert a\rvert,\lvert b\rvert):\deg(b)+1]$, do the following:
					    \begin{enumerate}
						    \item Show that $j>\deg(b)>\deg(a)$.
						    \item Hence show that $a_j=b_j=0$ using \declarationhr{2.34}.
						    \item Hence show that $(a+b)_j=a_j+b_j=0$.
					    \end{enumerate}
					    \item Show that $(a+b)_{\deg(b)}=a_{\deg(b)}+b_{\deg(b)}=0+b_{\deg(b)}=b_{\deg(b)}\ne 0$ using \declarationhr{2.34} given that $\deg(b)>\deg(a)$.
					    \item \textbf{Hence show that $\deg(a+b)=\deg(b)$.}
				    \end{enumerate}
		    \procedure{2.75}
			    \objective
				    Let $\deg(0)=-1$. Choose two polynomials $a,b$. The objective of the following instructions is to show that $\deg(a+b)\le\max(\deg(a),\deg(b))$.
			    \implementation
				    \begin{enumerate}
					    \item For $j\in[\max(\lvert a\rvert,\lvert b\rvert):\max(\deg(a),\deg(b))+1]$, do the following:
					    \begin{enumerate}
						    \item Show that $a_j=b_j=0$ using \declarationhr{2.34} given that $j>\deg(a)$ and $j>\deg(b)$.
						    \item Hence show that $(a+b)_j=a_j+b_j=0$ using \declarationhr{2.29}.
					    \end{enumerate}
					    \item \textbf{Hence show that $\deg(a+b)\le\max(\deg(a),\deg(b))$ using \declarationhr{2.34}.}
				    \end{enumerate}
		    \procedure{2.76}
			    \objective
				    Let $\deg(0)=-1$. Choose a polynomial $a$. The objective of the following instructions is to show that $\deg(-a)=\deg(a)$.
			    \implementation
				    \begin{enumerate}
					    \item For $j\in[\lvert a\rvert:\deg(a)+1]$, do the following:
					    \begin{enumerate}
						    \item Show that $a_j=0$ using \declarationhr{2.34} given that $j>\deg(a)$.
						    \item Hence show that $(-a)_j=-(a_j)=-0=0$ using \declarationhr{2.31}.
					    \end{enumerate}
					    \item Show that $(-a)_{\deg(a)}=-(a_{\deg(a)})\ne 0$ given that $a_{\deg(a)}\ne 0$.
					    \item \textbf{Hence show that $\deg(-a)=\deg(a)$ using \declarationhr{2.34}.}
				    \end{enumerate}
		    \procedure{2.77}
			    \objective
				    Choose two polynomials $a,b$ such that $a\ne 0$ and $b\ne 0$. The objective of the following instructions is to show that $(ab)_{\deg(a)+\deg(b)}=a_{\deg(a)}b_{\deg(b)}\ne 0$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $a_{\deg(a)}\ne 0$ given that $a\ne 0$.
					    \item Show that $b_{\deg(b)}\ne 0$ given that $b\ne 0$.
					    \item Hence using \declarationhr{2.32}, show that $(ab)_{\deg(a)+\deg(b)}$
					    \begin{enumerate}
						    \item $=\sum_r^{[0:\deg(a)+\deg(b)+1]}a_rb_{\deg(a)+\deg(b)-r}$
						    \item $=\sum_r^{[0:\deg(a)]}a_rb_{\deg(a)+\deg(b)-r}+a_{\deg(a)}b_{\deg(a)+\deg(b)-\deg(a)}+\sum_r^{[\deg(a)+1:\deg(a)+\deg(b)+1]}a_rb_{\deg(a)+\deg(b)-r}$
						    \item $=\sum_r^{[0:\deg(a)]}0a_r+a_{\deg(a)}b_{\deg(b)}+\sum_r^{[\deg(a)+1:\deg(a)+\deg(b)+1]}0b_{\deg(a)+\deg(b)-r}$
						    \item $=a_{\deg(a)}b_{\deg(b)}$
						    \item $\ne 0$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.78}
			    \objective
				    Choose two polynomials $a,b$ such that $a\ne 0$ and $b\ne 0$. The objective of the following instructions is to show that $\deg(ab)=\deg(a)+\deg(b)$.
			    \implementation
				    \begin{enumerate}
					    \item For $j\in[\deg(a)+\deg(b)+1:\lvert a\rvert+\lvert b\rvert-1]$, do the following:
					    \begin{enumerate}
						    \item Using \declarationhr{2.32}, show that $(ab)_j$
						    \begin{enumerate}
							    \item $=\sum_r^{[0:j+1]}a_rb_{j-r}$
							    \item $=\sum_r^{[0:\deg(a)+1]}a_rb_{j-r}+\sum_r^{[\deg(a)+1:j+1]}a_rb_{j-r}$
							    \item $=\sum_r^{[0:\deg(a)+1]}0a_r+\sum_r^{[\deg(a)+1:j+1]}0b_{j-r}$
							    \item $=0$.
						    \end{enumerate}
					    \end{enumerate}
					    \item Now show that $(ab)_{\deg(a)+\deg(b)}=a_{\deg(a)}b_{\deg(b)}\ne 0$ using \procedurehr{2.77}.
					    \item \textbf{Hence show that $\deg(ab)=\deg(a)+\deg(b)$ using \declarationhr{2.34}.}
				    \end{enumerate}
		    \declaration{2.00}
			    The phrase "\gls{ratpolymonic}" will be used to refer to polynomials $p$ such that $p\ne 0$ and $p_{\deg(p)}=1$.
		    \declaration{2.01}
			    The notation \gls{ratpolymon}, where $p$ is a polynomial such that $p\ne 0$, will be used as a shorthand for $\frac{p}{p_{\deg(p)}}$.
		    \procedure{2.25}
			    \objective
				    Choose two polynomials, $a,b$ such that $b\ne 0$. The objective of the following instructions is to construct two polynomials $u,w$ such that $a=ub+w$ and $\deg(w)<\deg(b)$.
			    \implementation
				    \begin{enumerate}
					    \item If $\deg(a)\ge\deg(b)$, then do the following:
					    \begin{enumerate}
						    \item Let $y=\frac{a_{\deg(a)}}{b_{\deg(b)}}\lambda^{\deg(a)-\deg(b)}$
						    \item Let $e=a-yb$.
						    \item Show that $\deg(e)<\deg(a)$.
						    \item Use \procedurehr{2.25} on $\langle e,b\rangle$ to construct $\langle c,d\rangle$ and show that:
                            \begin{enumerate}
                                \item $cb+d=e$.
                                \item \textbf{$\deg(d)<\deg(b)$.}
                            \end{enumerate}
						    \item Hence show that $cb+d=a-yb$ given that $cb+d=e$ and $e=a-yb$.
						    \item \textbf{Hence show that $(y+c)b+d=a$.}
						    \item \textbf{Now yield the tuple $\langle y+c, d\rangle$.}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
						    \item \textbf{Show that $0b+a=a$ and $\deg(a)<\deg(b)$.}
						    \item \textbf{Yield the tuple $\langle 0,a\rangle$.}
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{2.35}
			    The notation \gls{ratpolydiv}, where $a,b$ are polynomials, will be used to refer to the first part of the pair yielded by executing \procedurehr{2.25} on $\langle a,b\rangle$.
		    \declaration{2.36}
			    The notation \gls{ratpolymod}, where $a,b$ are polynomials, will be used to refer to the second part of the pair yielded by executing \procedurehr{2.25} on $\langle a,b\rangle$.
		    \procedure{2.79}
			    \objective
				    Choose a polynomial $a$ and a rational number $b$. The objective of the following instructions is to show that $a\bmod(\lambda-b)=\Lambda(a,b)$.
			    \implementation
				    \begin{enumerate}
					    \item Let $d=\lambda-b$.
					    \item Show that $d\ne 0$.
					    \item Let $c=a\div d$.
					    \item Using \procedurehr{2.25}, show that:
              \begin{enumerate}
                \item $a=cd+(a\bmod d)$
                \item $\deg(a\bmod d)<\deg(d)=1$.
              \end{enumerate}
					    \item Hence show that $\deg(a\bmod d)=0$.
					    \item Now using \procedurehr{2.56} and \procedurehr{2.64}, show that $\Lambda(a,b)$
					    \begin{enumerate}
						    \item $=\Lambda(cd+(a\bmod d),b)$
						    \item $=\Lambda(cd,b)+\Lambda(a\bmod d,b)$
						    \item $=\Lambda(c,b)\Lambda(d,b)+\Lambda(a\bmod d,b)$
						    \item $=\Lambda(c,b)(-b+b)+\Lambda(a\bmod d,b)$
						    \item $=0\Lambda(c,b)+\Lambda(a\bmod d,b)$
						    \item $=\Lambda(a\bmod d,b)$
						    \item $=a\bmod d$
                \item $=a\bmod(\lambda-b)$.
					    \end{enumerate}
				    \end{enumerate}
				\procedure{fri1402201125}
				  \objective
				    Choose a polynomial $a$ and a perplex number $b$. The objective of the following instructions is to show that $\Re(\Lambda(a,b))=\frac{1}{2}(\Lambda(a,\Re(b)+\Im(b))+\Lambda(a,\Re(b)-\Im(b)))$.
				  \implementation
				    \begin{enumerate}
				      \item Show that $\Re(\Lambda(a,b))$
				      \begin{enumerate}
				        \item $=\Re(\sum_r^{[0:\lvert a\rvert]}a_rb^r)$
				        \item $=\Re(\sum_r^{[0:\lvert a\rvert]}a_r((\Re(b)+\Im(b))\ku+(\Re(b)-\Im(b))\kcu)^r)$
				        \item $=\sum_r^{[0:\lvert a\rvert]}a_r\Re((\Re(b)+\Im(b))^r\ku+(\Re(b)-\Im(b))^r\kcu)$
				        \item $=\frac{1}{2}\sum_r^{[0:\lvert a\rvert]}a_r((\Re(b)+\Im(b))^r+(\Re(b)-\Im(b))^r)$
				        \item $=\frac{1}{2}(\sum_r^{[0:\lvert a\rvert]}a_r(\Re(b)+\Im(b))^r+\sum_r^{[0:\lvert a\rvert]}a_r(\Re(b)-\Im(b))^r)$
				        \item $=\frac{1}{2}(\Lambda(a,\Re(b)+\Im(b))+\Lambda(a,\Re(b)-\Im(b)))$.
				      \end{enumerate}
				    \end{enumerate}
				\procedure{fri1402201210}
				  \objective
				    Choose a polynomial $a$ and a perplex number $b$. The objective of the following instructions is to show that $\Im(\Lambda(a,b))=\frac{1}{2}(\Lambda(a,\Re(b)+\Im(b))-\Lambda(a,\Re(b)-\Im(b)))$.
				  \implementation
				    The implementation is analogous to that of \procedurehr{fri1402201125}.
				\procedure{fri1402201213}
				  \objective
				    Choose a polynomial $a$ and a perplex number $b$. The objective of the following instructions is to show that $\lVert\Lambda(a,b)\rVert^2=\Lambda(a,\Re(b)-\Im(b))\Lambda(a,\Re(b)+\Im(b))$.
				  \implementation
				    \begin{enumerate}
				      \item Using \declarationhr{sun0902201140}, \procedurehr{fri1402201125}, and \procedurehr{fri1402201210}, show that $\lVert\Lambda(a,b)\rVert^2$
				      \begin{enumerate}
				        \item $=\Re(\Lambda(a,b))^2-\Im(\Lambda(a,b))^2$
				        \item $=(\frac{1}{2}(\Lambda(a,\Re(b)+\Im(b))+\Lambda(a,\Re(b)-\Im(b))))^2-(\frac{1}{2}(\Lambda(a,\Re(b)+\Im(b))-\Lambda(a,\Re(b)-\Im(b))))^2$
				        \item $=\Lambda(a,\Re(b)-\Im(b))\Lambda(a,\Re(b)+\Im(b))$.
				      \end{enumerate}
				    \end{enumerate}
				\procedure{mon1702200807}
				  \objective
				    Choose a polynomial $a$ and a perplex number $b$. The objective of the following instructions is to show that $\conj{\Lambda(a,b)}=\Lambda(a,\conj{b})$.
				  \implementation
				    \begin{enumerate}
				      \item Show that $\conj{\Lambda(a,b)}$
				      \begin{enumerate}
				        \item $=\conj{\sum_r^{[0:\lvert a\rvert]}a_rb^r}$
				        \item $=\sum_r^{[0:\lvert a\rvert]}\conj{a_rb^r}$
				        \item $=\sum_r^{[0:\lvert a\rvert]}a_r\conj{b}^r$
				        \item $=\Lambda(a,\conj{b})$.
				      \end{enumerate}
				    \end{enumerate}
				\procedure{mon1702200743}
				  \objective
				    Choose a polynomial $a$ and two adjoint perplex numbers $b,c$. The objective of the following instructions is to show that $\Lambda(a,b)$ and $\Lambda(a,c)$ are adjoint.
				  \implementation
				    \begin{enumerate}
				      \item Show that $\lVert c-\conj{b}\rVert^2=0$ given that $b$ and $c$ are adjoint.
				      \item Using \procedurehr{mon1702200807}, show that $\lVert\Lambda(a,c)-\conj{\Lambda(a,b)}\rVert^2$
				      \begin{enumerate}
				        \item $=\lVert\Lambda(a,c)-\Lambda(a,\conj{b})\rVert^2$
				        \item $=\lVert\sum_r^{[0:\lvert a\rvert]}a_rc^r-\sum_r^{[0:\lvert a\rvert]}a_r\conj{b}^r\rVert^2$
				        \item $=\lVert\sum_r^{[0:\lvert a\rvert]}a_r(c^r-\conj{b}^r)\rVert^2$
				        \item $=\lVert\sum_r^{[0:\lvert a\rvert]}a_r(c-\conj{b})\sum_t^{[0:r]}c^t\conj{b}^{r-1-t}\rVert^2$
				        \item $=\lVert c-\conj{b}\rVert^2\lVert\sum_r^{[0:\lvert a\rvert]}a_r\sum_t^{[0:r]}c^t\conj{b}^{r-1-t}\rVert^2$
				        \item $=0\lVert\sum_r^{[0:\lvert a\rvert]}a_r\sum_t^{[0:r]}c^t\conj{b}^{r-1-t}\rVert^2$
				        \item $=0$.
				      \end{enumerate}
				      \item \textbf{Hence show that $\Lambda(a,b)$ and $\Lambda(a,c)$ are adjoint.}
				    \end{enumerate}
	    \end{multicols*}
	  \chapter{Polynomial Sign Changes}
	    \begin{multicols*}{2}
		    \procedure{2.80}
			    \objective
				    Choose a polynomial $p\ne 0$ and rational numbers $a_0<a_1<\cdots<a_{\deg(p)-2}<a_{\deg(p)-1}$ in such a way that $\Lambda(p,a_i)=0$ for $i\in[0:\deg(p)]$. The objective of the following instructions is to show that $p=p_{\deg(p)}\prod_j^{[0:\deg(p)]}(\lambda-a_j)$.
			    \implementation
				    \begin{enumerate}
					    \item Let $n=\deg(p)$.
					    \item If $n=0$, then do the following:
					    \begin{enumerate}
					      \item \textbf{Show that $p=p_0=p_{\deg(p)}\prod_j^{[0:{n}]}(\lambda-a_j)$.}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
					      \item Show that $p\bmod(\lambda-a_{n-1})=\Lambda(p,a_{n-1})=0$ using \procedurehr{2.79} given that $\Lambda(p,a_{n-1})=0$.
					      \item Let $q=p\div(\lambda-a_{n-1})$.
					      \item Hence show that $p=(\lambda-a_{n-1})q+p\bmod(\lambda-a_{n-1})=(\lambda-a_{n-1})q$.
					      \item For $i\in[0:n-1]$, do the following:
					      \begin{enumerate}
					        \item Show that $0$
					        \begin{enumerate}
					          \item $=\Lambda(p,a_i)$
					          \item $=\Lambda((\lambda-a_{n-1})q,a_i)$
					          \item $=\Lambda(\lambda-a_{n-1},a_i)\Lambda(q,a_i)$
					          \item $=(a_i-a_{n-1})\Lambda(q,a_i)$.
					        \end{enumerate}
					        \item Hence show that $\Lambda(q,a_i)=0$ given that $a_i-a_{n-1}\ne 0$.
					      \end{enumerate}
					      \item Hence use \procedurehr{2.80} on $\langle q,a_{[0:n-1]}\rangle$ to show that $q=q_{\deg(q)}\prod_j^{[0:{n-1}]}(\lambda-a_j)$.
					      \item Now show that $p_{\deg(p)}=(\lambda-a_{n-1})_{\deg(\lambda-a_{n-1})}q_{\deg{q}}=1q_{\deg{q}}=q_{\deg{q}}$ using \procedurehr{2.77} given that $p=(\lambda-a_{n-1})q$.
					      \item \textbf{Hence show that $p=(\lambda-a_{n-1})q=q_{\deg{q}}(\lambda-a_{n-1})\prod_j^{[0:{n-1}]}(\lambda-a_j)=p_{\deg{p}}\prod_j^{[0:{n}]}(\lambda-a_j)$.}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.16}
			    \objective
				    Choose a polynomial $p\ne 0$ and rational numbers $a_0<a_1<\cdots<a_{\deg(p)-1}<a_{\deg(p)}$ in such a way that $\Lambda(p,a_i)=0$ for $i\in[0:\deg(p)+1]$. The objective of the following instructions is to show that $0\ne 0$.
			    \implementation
				    \begin{enumerate}
					    \item Let $n=\deg(p)$.
					    \item Use \procedurehr{2.80} on $\langle p,a_{[0:n]}\rangle$ to show that $p=p_n\prod_j^{[0:{n}]}(\lambda-a_j)$.
					    \item Hence show that $\Lambda(p,a_{n})=\Lambda(q_0\prod_j^{[0:{n}]}(\lambda-a_j),a_{n})=\Lambda(q_0,a_{n})\prod_j^{[0:{n}]}\Lambda(\lambda-a_j,a_{n})=q_0\prod_j^{[0:{n}]}(a_{n}-a_j)\ne 0$.
					    \item Hence show that $0=\Lambda(p,a_{n})\ne 0$ given that $\Lambda(p,a_{n})=0$.
					    \item \textbf{Abort procedure.}
				    \end{enumerate}
            \procedure{thu2001191149}
                \objective
                    Choose a polynomial $p$ and a rational number $X$. The objective of the following instructions is to construct a rational number $a$ and a procedure $q(y)$ to show that $\lVert\Lambda(p,y)\rVert\le a$ when a rational number $y$ such that $\lVert y\rVert\le X$ is chosen.
                \implementation
                    \begin{enumerate}
                        \item Let $a=\sum_r^{[0:\lvert p\rvert]}\lVert p_r\rVert X^r$.
                        \item Let $q(y)$ be the following procedure:
                        \begin{enumerate}
                            \item Given that $\lVert y\rVert\le X$, show that $\lVert\Lambda(p,y)\rVert$
                            \begin{enumerate}
                                \item $=\lVert\sum_r^{[0:\lvert p\rvert]}p_ry^r\rVert$
                                \item $\le\sum_r^{[0:\lvert p\rvert]}\lVert p_ry^r\rVert$
                                \item $=\sum_r^{[0:\lvert p\rvert]}\lVert p_r\rVert\lVert y\rVert^r$
                                \item $\le\sum_r^{[0:\lvert p\rvert]}\lVert p_r\rVert X^r$
                                \item $=a$.
                            \end{enumerate}
                        \end{enumerate}
                        \item \textbf{Yield the tuple $\langle a,q\rangle$.}
                    \end{enumerate}
		    \procedure{2.15}
			    \objective
				    Choose a polynomial $p$ and a rational number $X$. The objective of the following instructions is to construct a rational number $a$ and a procedure $q(z)$ to show that $\lVert\Im(\Lambda(p,z))\rVert\le a\lVert\Im(z)\rVert$ when a proper perplex number $z$ such that $z\subseteq X\ju$ are chosen.
			    \implementation
				    \begin{enumerate}
					    \item Let $a=\sum_{r}^{[1:\lvert p\rvert]}r\lVert p_r\rVert X^{r-1}$.
					    \item Let $q(y,z)$ be the following procedure:
					    \begin{enumerate}
						    \item Show that $\lVert\Im(\Lambda(p,z))\rVert$
						    \begin{enumerate}
							    \item $=\lVert\Im(\sum_{r}^{[0:\lvert p\rvert]} p_rz^r)\rVert$
							    \item $=\lVert\Im(\sum_{r}^{[0:\lvert p\rvert]} p_r((\Re(z)-\Im(z))\kcu+(\Re(z)+\Im(z))\ku)^r)\rVert$
							    \item $=\lVert\sum_{r}^{[0:\lvert p\rvert]} p_r\Im((\Re(z)-\Im(z))^r\kcu+(\Re(z)+\Im(z))^r\ku)\rVert$
							    \item $=\lVert\frac{1}{2}\sum_{r}^{[0:\lvert p\rvert]} p_r((\Re(z)+\Im(z))^r-(\Re(z)-\Im(z))^r)\rVert$
							    \item $=\lVert\sum_{r}^{[1:\lvert p\rvert]} p_r\Im(z)\sum_t^{[0:r]}(\Re(z)+\Im(z))^t(\Re(z)-\Im(z))^{r-1-t}\rVert$
							    \item $\le\sum_{r}^{[1:\lvert p\rvert]}\lVert p_r\rVert\lVert\Im(z)\rVert\lVert\sum_t^{[0:r]}(\Re(z)+\Im(z))^t(\Re(z)-\Im(z))^{r-1-t}\rVert$
							    \item $\le\sum_{r}^{[1:\lvert p\rvert]}\lVert p_r\rVert\lVert\Im(z)\rVert\sum_t^{[0:r]}\lVert(\Re(z)+\Im(z))^t\rVert\lVert(\Re(z)-\Im(z))^{r-1-t}\rVert$
							    \item $\le\sum_{r}^{[1:\lvert p\rvert]}\lVert p_r\rVert\lVert\Im(z)\rVert\sum_t^{[0:r]}X^tX^{r-1-t}$
							    \item $=\lVert\Im(z)\rVert\sum_{r}^{[1:\lvert p\rvert]}\lVert p_r\rVert\sum_t^{[0:r]}X^{r-1}$
							    \item $=\lVert\Im(z)\rVert\sum_{r}^{[1:\lvert p\rvert]}r\lVert p_r\rVert X^{r-1}$
							    \item $=a\lVert\Im(z)\rVert$
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,q\rangle$.}
				    \end{enumerate}
        \procedure{thu3001201111}
          \objective
            Choose a polynomial $p$ and a rational number $X$. The objective of the following instructions is to construct a rational number $a>0$ and a procedure $q(z)$ to show that $\lVert\Lambda(p,\Re(z)\pm\Im(z))\rVert\le a\lVert\Im(z)\rVert$ when a proper perplex number $z$ such that $z\subseteq X\ju$, and $\lVert\Lambda(p,z)\rVert^2\le 0$ are chosen.
          \implementation
            \begin{enumerate}
              \item Use \procedurehr{2.15} on $\langle p,X\rangle$ to construct $\langle a_1,q_1\rangle$.
              \item Let $a=2a_1$.
              \item Let $q(y,z)$ be the following procedure:
              \begin{enumerate}
                \item Show that $\lVert\Im(\Lambda(p,z))\rVert\le a_1\lVert\Im(z)\rVert$ using procedure $q_1$.
                \item Show that $\Lambda(p,\Re(z)+\Im(z))\Lambda(p,\Re(z)-\Im(z))=\lVert\Lambda(p,z)\rVert^2\le 0$ using \procedurehr{fri1402201213}.
                \item Hence using \procedurehr{thu3001201131} show that $\lVert\Lambda(p,\Re(z)\pm\Im(z))\rVert$
                \begin{enumerate}
                  \item $\le\lVert\Lambda(p,\Re(z)+\Im(z))\rVert+\lVert\Lambda(p,\Re(z)-\Im(z))\rVert$
                  \item $=\lVert\Lambda(p,\Re(z)+\Im(z))-\Lambda(p,\Re(z)-\Im(z))\rVert$
                  \item $=2\lVert\Im(\Lambda(p,z))\rVert$
                  \item $\le 2a_1\lVert\Im(z)\rVert$
                  \item $\le a\lVert\Im(z)\rVert$.
                \end{enumerate}
              \end{enumerate}
              \item \textbf{Yield the tuple $\langle a,q\rangle$.}
            \end{enumerate}
        \procedure{sat0102201050}
			    \objective
				    Choose a polynomial $f$, a proper perplex number $c$, and a rational number $B$ such that $\lVert\Lambda(f,c)\rVert^2\le 0$ and $B>0$. The objective of the following instructions is to construct a proper perplex number $e$ such that $e\subseteq c$, $\Im(e)<B$ and $\lVert\Lambda(f,e)\rVert^2\le 0$.
			    \implementation
				    \begin{enumerate}
					    \item If $\lVert\Im(c)\rVert<B$, then do the following:
					    \begin{enumerate}
					      \item \textbf{Yield the tuple $\langle c\rangle$.}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
						    \item Let $g=(\Re(c)-\Im(c))\kcu+\Re(c)\ku$ and show that $g\subseteq c$ and $\lVert\Im(g)\rVert=\frac{1}{2}\lVert\Im(c)\rVert$.
						    \item Let $m=\Re(c)\kcu+(\Re(c)+\Im(c))\ku$ and show that $m\subseteq c$ and $\lVert\Im(m)\rVert=\frac{1}{2}\lVert\Im(c)\rVert$.
						    \item Show that $\Re(c)-\Im(c)=\Re(g)-\Im(g)$, $\Re(g)+\Im(g)=\Re(m)-\Im(m)$, and $\Re(m)+\Im(m)=\Re(c)+\Im(c)$.
						    \item If $\lVert\Lambda(f,g)\rVert^2\le 0$, then do the following:
						    \begin{enumerate}
							    \item Use \procedurehr{sat0102201050} on $\langle f,g,B\rangle$ to construct $\langle e,h\rangle$ and show that:
							    \begin{enumerate}
							      \item $e\subseteq g\subseteq c$
							      \item $\lVert\Im(e)\rVert<B$
							      \item $\lVert\Lambda(f,e)\rVert^2\le 0$.
							    \end{enumerate}
						    \end{enumerate}
						    \item Otherwise do the following:
						    \begin{enumerate}
						      \item Given that $\lVert\Lambda(f,c)\rVert^2\le 0$ and $\lVert\Lambda(f,g)\rVert^2>0$, show that $\lVert\Lambda(f,m)\rVert^2$
						      \begin{enumerate}
						        \item $=\Lambda(f,\Re(m)-\Im(m))\Lambda(f,\Re(m)+\Im(m))$
						        \item $=\frac{\Lambda(f,\Re(m)-\Im(m))}{\Lambda(f,\Re(g)-\Im(g))}\Lambda(f,\Re(g)-\Im(g))\Lambda(f,\Re(m)+\Im(m))$
						        \item $=\frac{\Lambda(f,\Re(g)+\Im(g))}{\Lambda(f,\Re(g)-\Im(g))}\lVert\Lambda(f,c)\rVert^2$
						        \item $\le 0$.
						      \end{enumerate}
							    \item Hence use \procedurehr{sat0102201050} on $\langle f,m,B\rangle$ to construct $\langle e\rangle$ and show that:
							    \begin{enumerate}
							      \item $e\subseteq m\subseteq c$
							      \item $\lVert\Im(e)\rVert<B$
							      \item $\lVert\Lambda(f,e)\rVert^2\le 0$.
							    \end{enumerate}
						    \end{enumerate}
						    \item \textbf{Yield the tuple $\langle e\rangle$.}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.17}
			    \objective
				    Choose a polynomial $f$, a proper perplex number $a$, and a rational number $B$ such that $\lVert\Lambda(f,a)\rVert^2\le 0$ and $B>0$. The objective of the following instructions is to construct a proper perplex number $d$ such that $d\subseteq a$, $\lVert\Lambda(f,d)\rVert^2\le 0$, and $\lVert\Im(\Lambda(f,d))\rVert<B$.
			    \implementation
				    \begin{enumerate}
					    \item Use \procedurehr{2.15} on $\langle f,\lVert\Re(a)\rVert+\lVert\Im(b)\rVert\rangle$ to construct $\langle G,q\rangle$.
					    \item Use \procedurehr{sat0102201050} on $\langle f,a,\frac{B}{G}\rangle$ to construct $\langle c\rangle$ and show that:
					    \begin{enumerate}
					      \item $c\subseteq a$
					      \item $\lVert\Im(c)\rVert\le\frac{B}{G}$
					      \item $\lVert\Lambda(f,c)\rVert^2\le 0$.
					    \end{enumerate}
					    \item \textbf{Use procedure $q$ on $\langle c\rangle$ to show that $\lVert\Im(\Lambda(f,c))\rVert\le G\lVert\Im(c)\rVert\le G\frac{B}{G}=B$.}
					    \item \textbf{Yield the tuple $\langle c\rangle$.}
				    \end{enumerate}
				\declaration{tue2502201328}
	        The notation \gls{ratperplexmu}, where $x$ is a perplex number and $p$ is a perplex polynomial, will be used as a shorthand for $x$ if $\Im(\Lambda(p,x))>0$ and $\conj{x}$ if $\Im(\Lambda(p,x))<0$.
	      \procedure{tue2502201349}
	        \objective
	          Choose a perplex polynomial $p$ and an increasing list of pairwise disjoint proper perplex numbers $r$ such that $\lvert r\rvert=\deg(p)$ and $-\Lambda(p,r_i)\sim\pm\ju\sim\Lambda(p,\conj{r_i})$ for $0<i<\lvert r\rvert$. The objective of the following instructions is to construct a list of perplex numbers $t$ such that $\lvert t\rvert=\lvert r\rvert$ and $\mu_\lambda(t_i)\subseteq r_i$ for $0<i<\lvert r\rvert$, and a procedure $q(x)$ to show that $\Lambda(p,x)\supseteq p_{\lvert t\rvert}\prod_{m}^{[0:\lvert t\rvert]}(x-\pmconj{t_m})$ when a perplex number $x$ that is disjoint from $r$ is chosen.
	        \implementation
	          \begin{instructions}
	            \item If $\deg(p)=0$, then do the following:
	            \begin{instructions}
	              \item Let $t=\langle\rangle$.
	              \item Let $q(x)$ be the following procedure:
	              \begin{instructions}
	                \item Show that $\Lambda(p,x)$
	                \begin{instructions}
	                  \item $=p_0$
	                  \item $\supseteq p_{\lvert t\rvert}\prod_{m}^{[0:\lvert t\rvert]}(x-\pmconj{t_m})$.
	                \end{instructions}
	              \end{instructions}
	              \item \textbf{Yield the tuple $\langle t,q\rangle$.}
	            \end{instructions}
	            \item Otherwise do the following:
	            \begin{instructions}
	              \item Let $z=\max(\Lambda(p,\mu_p(r))\conj{\ku})+\min(\Lambda(p,\mu_p(r))\ku)$.
	              \item Let $y=\min(-(\Re(z)-\Im(z)),\Re(z)+\Im(z))$.
	              \item Use \procedurehr{2.17} on $\langle p,r_0,\frac{1}{2}y\rangle$ to construct $t_0$ and show that:
	              \begin{instructions}
	                \item $t_0$ is a proper perplex number
	                \item $t_0\subseteq r_0$
	                \item $\lVert\Lambda(p,t_0)\rVert^2\le 0$
	                \item $\lVert\Im(\Lambda(p,t_0))\rVert<\frac{1}{2}y$.
	              \end{instructions}
	              \item Hence show that $\Lambda(p,\mu_p(t_0))\subseteq y\ju\subseteq z$.
	              \item Let $d=\sum_m^{[0:\lvert r\rvert+1]}p_m\sum_n^{[0:m]}\lambda^n{t_0}^{m-1-n}$.
	              \item For $i$ in $[1:\lvert r\rvert]$, do the following:
	              \begin{instructions}
	                \item Show that $\Lambda(p,\mu_p(t_0))\subseteq z\subseteq\Lambda(p,\mu_p(r_i))$.
	                \item Hence show that $-(r_i-t_0)\Lambda(d,r_i)$
	                \begin{instructions}
	                  \item $=-\Lambda(\lambda-t_0,r_i)\Lambda(d,r_i)$
	                  \item $=-\Lambda((\lambda-t_0)d,r_i)$
	                  \item $=-(\Lambda(p,r_i)-\Lambda(p,t_0))$
	                  \item $\sim-\Lambda(p,r_i)$
	                  \item $\sim\pm\ju$
	                  \item $\sim\Lambda(p,\conj{r_i})$
	                  \item $\sim\Lambda(p,\conj{r_i})-\Lambda(p,t_0)$
	                  \item $=\Lambda((\lambda-t_0)d,\conj{r_i})$
	                  \item $=\Lambda(\lambda-t_0,\conj{r_i})\Lambda(d,\conj{r_i})$
	                  \item $=(\conj{r_i}-t_0)\Lambda(d,\conj{r_i})$
	                \end{instructions}
	                \item Hence show that $-\Lambda(d,r_i)\sim\pm\ju\sim\Lambda(d,\conj{r_i})$
	                \begin{instructions}
	                  \item given that $r_i-t_0>0$
	                  \item and $\conj{r_i}-t_0>0$.
	                \end{instructions}
	              \end{instructions}
	              \item Use \procedurehr{tue2502201349} on $\langle d,r_{[1:\lvert r\rvert]}\rangle$ to construct $\langle t_{[1:\lvert r\rvert]},u\rangle$.
	              \item Use \procedurehr{thu2702201419} on $\langle d,r_{[1:\lvert r\rvert]}\rangle$ to construct $\langle t_{[1:\lvert r\rvert]},w\rangle$.
	              \item Let $q(x)$ be the following procedure:
	              \begin{instructions}
	                \item If $x>\mu_p(t_0)$, then do the following:
	                \begin{instructions}
	                  \item Using procedure $u$, show that $\Lambda(d,x)$
	                  \begin{instructions}
	                    \item $\supseteq d_{\lvert t\rvert-1}\prod_{m}^{[1:\lvert t\rvert]}(x-\pmconj{t_m})$
	                    \item $=p_{\lvert t\rvert}\prod_{m}^{[1:\lvert t\rvert]}(x-\pmconj{t_m})$.
	                  \end{instructions}
	                  \item Given that $x-\mu_p(t_0)>0$, show that $\Lambda(p,x)$
	                  \begin{instructions}
	                    \item $=\Lambda((\lambda-\mu_p(t_0))d,x)+\Lambda(p,\mu_p(t_0))$
	                    \item $\supseteq\Lambda((\lambda-\mu_p(t_0))d,x)$
	                    \item $=(x-\mu_p(t_0))\Lambda(d,x)$
	                    \item $\supseteq(x-\mu_p(t_0))p_{\lvert t\rvert}\prod_{m}^{[1:\lvert t\rvert]}(x-\pmconj{t_m})$
	                    \item $=p_{\lvert t\rvert}\prod_{m}^{[0:\lvert t\rvert]}(x-\pmconj{t_m})$.
	                  \end{instructions}
	                \end{instructions}
	                \item Otherwise if $x<\mu_p(t_0)$, then do the following:
	                \begin{instructions}
	                  \item Using procedure $w$, show that $\Lambda(d,x)$
	                  \begin{instructions}
	                    \item $\subseteq d_{\lvert t\rvert-1}\prod_{m}^{[1:\lvert t\rvert]}(x-\pmconj{t_m})$
	                    \item $=p_{\lvert t\rvert}\prod_{m}^{[1:\lvert t\rvert]}(x-\pmconj{t_m})$.
	                  \end{instructions}
	                  \item Given that $x-\mu_p(t_0)<0$, show that $\Lambda(p,x)$
	                  \begin{instructions}
	                    \item $=\Lambda((\lambda-\mu_p(t_0))d,x)+\Lambda(p,\mu_p(t_0))$
	                    \item $\supseteq\Lambda((\lambda-\mu_p(t_0))d,x)$
	                    \item $=(x-\mu_p(t_0))\Lambda(d,x)$
	                    \item $\supseteq(x-\mu_p(t_0))p_{\lvert t\rvert}\prod_{m}^{[1:\lvert t\rvert]}(x-\pmconj{t_m})$
	                    \item $=p_{\lvert t\rvert}\prod_{m}^{[0:\lvert t\rvert]}(x-\pmconj{t_m})$.
	                  \end{instructions}
	                \end{instructions}
	              \end{instructions}
	              \item \textbf{Yield the tuple $\langle t,q\rangle$.}
	            \end{instructions}
	          \end{instructions}
	      \procedure{thu2702201419}
	        \objective
	          Choose a perplex polynomial $p$ and an increasing list of pairwise disjoint proper perplex numbers $r$ such that $\lvert r\rvert=\deg(p)$ and $-\Lambda(p,r_i)\sim\pm\ju\sim\Lambda(p,\conj{r_i})$ for $0<i<\lvert r\rvert$. The objective of the following instructions is to construct a list of perplex numbers $t$ such that $\lvert t\rvert=\lvert r\rvert$ and $\mu_\lambda(t_i)\subseteq r_i$ for $0<i<\lvert r\rvert$, and a procedure $q(x)$ to show that $\Lambda(p,x)\subseteq p_{\lvert t\rvert}\prod_{m}^{[0:\lvert t\rvert]}(x-\pmconj{t_m})$ when a perplex number $x$ that is disjoint from $r$ is chosen.
	        \implementation
	          Implementation is analogous to that of \procedurehr{tue2502201349}.
		    \procedure{2.18}
			    \objective
				    Choose a polynomial $f\ne 0$ and pairs of rational numbers $(a_{\deg(f)},b_{\deg(f)}),(a_{\deg(f)-1},b_{\deg(f)-1}),\cdots,(a_0,b_0)$ in such a way that:
				    \begin{enumerate}
					    \item $a_{\deg(f)}<b_{\deg(f)}\le a_{\deg(f)-1}<b_{\deg(f)-1}\le\cdots\le a_1<b_1\le a_0<b_0$.
					    \item $\sgn(\Lambda(f,a_i))=-\sgn(\Lambda(f,b_i))$ for $i\in[0:\deg(f)+1]$.
				    \end{enumerate}
				    The objective of the following instructions is to show that $1=-1$.
			    \implementation
				    \begin{enumerate}
					    \item If $\deg(f)>0$:
					    \begin{enumerate}
						    \item Let $B=\min_k^{[0:\deg(f)-1]}\min(\lvert\Lambda(f,a_k)\rvert,\lvert\Lambda(f,b_k)\rvert)$.
						    \item For $k\in[0:\deg(f)]$, verify that $\lvert\Lambda(f,a_k)\rvert\ge B$.
						    \item Execute \procedurehr{2.17} on the formal polynomial $f$, interval $(a_{\deg(f)}, b_{\deg(f)})$, and target of $B$. Let the tuple $\langle d\rangle$ receive the result.
						    \item Verify that $\lvert\Lambda(f,d)\rvert<B$.
						    \item Let $h=f\div(\lambda-d)$.
						    \item Execute \procedurehr{2.79} on $\langle f,d\rangle$.
						    \item Hence verify that $f=(\lambda-d)h+f\bmod(\lambda-d)=(\lambda-d)h+\Lambda(f,d)$.
						    \item Hence verify that $0\ne f-\Lambda(f,d)=(\lambda-d)h$.
						    \item Hence verify that $h\ne 0$.
						    \item Hence verify that $\deg(f)=\deg(f-\Lambda(f,d))=\deg((\lambda-d)h)=\deg(\lambda-d)+\deg(h)=1+\deg(h)$.
						    \item Hence verify that $\deg(h)=\deg(f)-1$.
						    \item For $k\in[0:\deg(h)+1]$, do the following:
						    \begin{enumerate}
							    \item If $\Lambda(f,a_k)\ge B$, in-order verify that:
							    \begin{enumerate}
								    \item $\Lambda(f,a_k)\ge B>\lvert\Lambda(f,d)\rvert\ge\Lambda(f,d)$.
								    \item $\Lambda(f,a_k)-\Lambda(f,d)>0$.
								    \item $(a_k-d)\Lambda(h,a_k)>0$.
								    \item \textbf{$\Lambda(h,a_k)>0$.}
								    \item $\Lambda(f,b_k)\le-B<-\lvert\Lambda(f,d)\rvert\le\Lambda(f,d)$.
								    \item $\Lambda(f,b_k)-\Lambda(f,d)<0$.
								    \item $(b_k-d)\Lambda(h,b_k)<0$.
								    \item \textbf{$\Lambda(h,b_k)<0$.}
							    \end{enumerate}
							    \item Otherwise, if $\Lambda(f,a_k)\le -B$, do the following:
							    \begin{enumerate}
								    \item \textbf{Using steps analogous to (ji), verify that $\Lambda(h,a_k)<0$.}
								    \item \textbf{Using steps analogous to (ji), verify that $\Lambda(h,b_k)>0$.}
							    \end{enumerate}
						    \end{enumerate}
						    \item Execute \procedurehr{2.18} on $h$ and $a_{\deg(h)}<b_{\deg(h)}\le a_{\deg(h)-1}<b_{\deg(h)-1}\le\cdots\le a_1<b_1\le a_0<b_0$.
					    \end{enumerate}
					    \item Otherwise, do the following:
					    \begin{enumerate}
						    \item Verify that $\deg(f)=0$.
						    \item Therefore verify that $f=f_0\ne 0$.
						    \item Therefore verify that $\sgn(f_0)=\sgn(\Lambda(f,a_0))=-\sgn(\Lambda(f,b_0))=-\sgn(f_0)$.
						    \item \textbf{Therefore verify that $1=-1$.}
						    \item \textbf{Abort procedure.}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.19}
			    \objective
				    Choose two lists of polynomials $s,q$ in such a way that:
				    \begin{enumerate}
					    \item $\lvert s\rvert>1$.
					    \item For $i$ in $[0:\lvert s\rvert]$, $\deg(s_i)=i$.
					    \item For $i$ in $[0:\lvert s\rvert]$, $\sgn((s_i)_i)=\sgn((s_m)_m)$.
					    \item For $i$ in $[1:\lvert s\rvert-1]$, $s_{i-1}+s_{i+1}=q_is_i$.
				    \end{enumerate}
				    The objective of the following instructions is to construct lists of polynomials $g,h$ such that $g_is_{i+1}+h_is_i=1$ for $i$ in $[0:\lvert s\rvert-1]$.
			    \implementation
				    \begin{enumerate}
					    \item Let $m=\lvert s\rvert-1$
					    \item Let $g=h=\langle\rangle$.
					    \item If $m>1$, do the following:
					    \begin{enumerate}
						    \item Verify that $q_{m-1}s_{m-1}-s_{m}=s_{m-2}$.
						    \item Execute \procedurehr{2.19} on $s_{[0:m]}$ and $q_{[1:m-1]}$ and let the tuple $\langle,,g,h\rangle$ receive.
						    \item Verify that $g_{m-2}s_{m-1}+h_{m-2}s_{m-2}=1$.
						    \item Let $g_{m-1}=-h_{m-2}$.
						    \item Let $h_{m-1}=g_{m-2}+h_{m-2}q_{m-1}$.
						    \item Therefore verify that $g_{m-1}s_{m}+h_{m-1}s_{m-1}$
						    \begin{enumerate}
							    \item $=g_{m-2}s_{m-1}+h_{m-2}(q_{m-1}s_{m-1}-s_{m})$
							    \item $=g_{m-2}s_{m-1}+h_{m-2}s_{m-2}$
							    \item $=1$.
						    \end{enumerate}
					    \end{enumerate}
					    \item Otherwise, if $m=1$ do the following:
					    \begin{enumerate}
						    \item Let $g_0=0$.
						    \item Let $h_0=\frac{1}{s_0}$.
						    \item \textbf{Therefore verify that $g_0s_1+h_0s_0=1$.}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle s,q,g,h\rangle$.}
				    \end{enumerate}
				\procedure{fri3101200641}
				  \objective
				    Choose polynomials $g,h,p,q$ and a rational number $X$ such that $gp+hq=1$. The objective of the following instructions is to construct a rational numbers $a$ and a procedure $r(y,z)$ to show that $\Lambda(p,y)\Lambda(p,z)>0$ when two rational numbers $y,z$ such that $\lVert y\rVert\le X$, $\lVert z\rVert\le X$, $\lVert y-z\rVert\le a$, and $\Lambda(q,y)\Lambda(q,z)\le 0$ are chosen.
				  \implementation
				    \begin{enumerate}
				      \item Use \procedurehr{thu3001201111} on $\langle p,X\rangle$ to construct $\langle a_1,r_1\rangle$.
				      \item Use \procedurehr{thu3001201111} on $\langle q,X\rangle$ to construct $\langle a_2,r_2\rangle$.
				      \item Use \procedurehr{thu2001191149} on $\langle g,X\rangle$ to construct $\langle a_3,r_3\rangle$.
				      \item Use \procedurehr{thu2001191149} on $\langle h,X\rangle$ to construct $\langle a_4,r_4\rangle$.
				      \item Let $a=\frac{1}{a_1a_3+a_2a_4+1}$.
				      \item Let $r(y,z)$ be the following procedure:
				      \begin{enumerate}
				        \item If $\Lambda(p,y)\Lambda(p,z)\le 0$, then do the following:
				        \begin{enumerate}
				          \item Show that $\lVert\Lambda(p,y)\rVert\le a_1\lVert z-y\rVert\le a_1a$ using procedure $r_1$.
				          \item Show that $\lVert\Lambda(q,y)\rVert\le a_2\lVert z-y\rVert\le a_2a$ using procedure $r_2$.
				          \item Show that $\lVert\Lambda(g,y)\rVert\le a_3$ using procedure $r_3$.
				          \item Show that $\lVert\Lambda(h,y)\rVert\le a_4$ using procedure $r_4$.
				          \item Given that $gp+hq=1$, show that $\Lambda(gp,y)+\Lambda(h,y)\Lambda(q,y)$
				          \begin{enumerate}
				            \item $=\Lambda(gp+hq,y)$
				            \item $=\Lambda(1,y)$
				            \item $=1$.
				          \end{enumerate}
				          \item Hence show that $\Lambda(gp,y)$
				          \begin{enumerate}
				            \item $=1-\Lambda(h,y)\Lambda(q,y)$
				            \item $\ge 1-a_4a_2a$
				            \item $=\frac{a_1a_3+1}{a_1a_3+a_2a_4+1}$
				            \item $=(a_1a_3+1)a$
				            \item $>a_1a_3a$
				            \item $\ge\lVert\Lambda(p,y)\rVert\lVert\Lambda(g,y)\rVert$
				            \item $\ge\Lambda(p,y)\Lambda(g,y)$
				            \item $=\Lambda(pg,y)$.
				          \end{enumerate}
				          \item Hence show that $0>0$.
				          \item \textbf{Abort procedure.}
				        \end{enumerate}
				        \item Otherwise do the following:
				        \begin{enumerate}
				          \item Show that $\Lambda(p,y)\Lambda(p,z)>0$.
				        \end{enumerate}
				      \end{enumerate}
				      \item \textbf{Yield the tuple $\langle a,r\rangle$.}
				    \end{enumerate}
				\procedure{fri3101200730}
				  \objective
				    Choose polynomials $g,h,j,p,q,r$ and a rational number $X$ such that $hq+jr=1$ and $p+r=gq$. The objective of the following instructions is to construct a rational number $a$ and a procedure $t(y,z)$ to show that $\Lambda(p,y)\Lambda(r,y)<0$ and $\Lambda(j,y)\ne 0$ when two rational numbers $y,z$ such that $\lVert y\rVert\le X$, $\lVert z\rVert\le X$, $\lVert y-z\rVert\le a$, and $\Lambda(q,y)\Lambda(q,z)\le 0$ are chosen.
				  \implementation
				    \begin{enumerate}
				      \item Use \procedurehr{thu2001191149} on $\langle h,X\rangle$ to construct $\langle a_1,t_1\rangle$.
				      \item Use \procedurehr{thu2001191149} on $\langle g,X\rangle$ to construct $\langle a_2,t_2\rangle$.
				      \item Use \procedurehr{thu2001191149} on $\langle j,X\rangle$ to construct $\langle a_3,t_3\rangle$.
				      \item Use \procedurehr{thu3001201111} on $\langle q,X\rangle$ to construct $\langle a_4,t_4\rangle$.
				      \item Let $a=\frac{1}{(a_1+a_2a_3)a_4+1}$.
				      \item Let $t(y,z)$ be the following procedure:
				      \begin{enumerate}
				        \item Show that $\lVert\Lambda(h,y)\rVert\le a_1$ using procedure $t_1$.
				        \item Show that $\lVert\Lambda(g,y)\rVert\le a_2$ using procedure $t_2$.
				        \item Show that $\lVert\Lambda(j,y)\rVert\le a_3$ using procedure $t_3$.
				        \item Show that $\lVert\Lambda(q,y)\rVert\le a_4\lVert z-y\rVert\le a_4a$ using procedure $t_4$.
				        \item Show that $jr=1-hq$ given that $hq+jr=1$.
				        \item Hence show that $\lVert\Lambda(j,y)\rVert\lVert\Lambda(r,y)\rVert$
				        \begin{enumerate}
				          \item $=\lVert\Lambda(jr,y)\rVert$
				          \item $=\lVert\Lambda(1-hq,y)\rVert$
				          \item $=\lVert\Lambda(1,y)\rVert-\lVert\Lambda(h,y)\Lambda(q,y)\rVert$
				          \item $\ge 1-a_1a_4\lVert y-z\rVert$
				          \item $=1-a_1a_4a$
				          \item $=\frac{a_2a_3a_4+1}{(a_1+a_2a_3)a_4+1}$
				          \item $=(a_2a_3a_4+1)a$
				          \item $>a_2a_3a_4a$
				          \item $\ge\lVert\Lambda(q,y)\rVert\lVert\Lambda(g,y)\rVert\lVert\Lambda(j,y)\rVert$
				          \item $\ge\lVert\Lambda(qg,y)\rVert\lVert\Lambda(j,y)\rVert$.
				        \end{enumerate}
				        \item Hence show that $\lVert\Lambda(r,y)\rVert>\lVert\Lambda(qg,y)\rVert\ge 0$
				        \begin{enumerate}
				          \item given that $\Lambda(j,y)\ne 0$
				          \item given that $\lVert\Lambda(j,y)\rVert\lVert\Lambda(r,y)\rVert>\lVert\Lambda(qg,y)\rVert\lVert\Lambda(j,y)\rVert$.
				        \end{enumerate}
				        \item Show that $p=gq-r$ given that $p+r=gq$.
				        \item If $\Lambda(r,y)>0$, then do the following:
				        \begin{enumerate}
				          \item Show that $\Lambda(p,y)$
				          \begin{enumerate}
				            \item $=\Lambda(gq-r,y)$
				            \item $=\Lambda(gq,y)-\Lambda(r,y)$
				            \item $\le\lVert\Lambda(gq,y)\rVert-\lVert\Lambda(r,y)\rVert$
				            \item $<0$.
				          \end{enumerate}
				          \item \textbf{Hence show that $\Lambda(p,y)\Lambda(r,y)<0$.}
				        \end{enumerate}
				        \item Otherwise do the following:
				        \begin{enumerate}
				          \item Given that $\Lambda(r,y)<0$, show that $\Lambda(p,y)$
				          \begin{enumerate}
				            \item $=\Lambda(gq-r,y)$
				            \item $=\Lambda(gq,y)-\Lambda(r,y)$
				            \item $\ge-\lVert\Lambda(gq,y)\rVert+\lVert\Lambda(r,y)\rVert$
				            \item $>0$.
				          \end{enumerate}
				          \item \textbf{Hence show that $\Lambda(p,y)\Lambda(r,y)<0$.}
				        \end{enumerate}
				      \end{enumerate}
				      \item \textbf{Yield the tuple $\langle a,t\rangle$.}
				    \end{enumerate}
				\procedure{fri3101200807}
				  \objective
				    Choose polynomials $g,h,j,p,q,r$ and a rational number $X$ such that $hq+jr=1$ and $p+r=gq$. The objective of the following instructions is to construct a rational number $a$ and a procedure $t(y,z)$ to show that $\Lambda(p,y)\Lambda(r,y)<0$,$\Lambda(p,z)\Lambda(r,z)<0$,$\Lambda(r,y)\Lambda(r,z)>0$, and $\Lambda(p,y)\Lambda(p,z)>0$ when two rational numbers $y,z$ such that $\lVert y\rVert\le X$, $\lVert z\rVert\le X$, $\lVert y-z\rVert\le a$, and $\Lambda(q,y)\Lambda(q,z)\le 0$ are chosen.
				  \implementation
				    \begin{enumerate}
				      \item Use \procedurehr{fri3101200730} on $\langle g,h,j,p,q,r,X\rangle$ to construct $\langle a_1,t_1\rangle$.
				      \item Use \procedurehr{fri3101200641} on $\langle j,h,r,q,X\rangle$ to construct $\langle a_2,t_2\rangle$.
				      \item Show that $(j+jg)q+(-j)p=1$ given that $hq+jr=1$ and $r=gq-p$.
				      \item Use \procedurehr{fri3101200641} on $\langle -j,h+jg,p,q,X\rangle$ to construct $\langle a_3,t_3\rangle$.
				      \item Let $a=\min(a_1,a_2,a_3)$.
				      \item Let $t(y,z)$ be the following procedure:
				      \begin{enumerate}
				        \item \textbf{Show that $\Lambda(p,y)\Lambda(r,y)<0$ using procedure $t_1$.}
				        \item \textbf{Show that $\Lambda(r,y)\Lambda(r,z)>0$ using procedure $t_2$.}
				        \item \textbf{Show that $\Lambda(p,y)\Lambda(p,z)>0$ using procedure $t_3$.}
				        \item \textbf{Hence show that $\Lambda(p,z)\Lambda(r,z)=\frac{\Lambda(p,z)}{\Lambda(p,y)}\cdot\frac{\Lambda(r,z)}{\Lambda(r,y)}\Lambda(p,y)\Lambda(r,y)<0$.}
				      \end{enumerate}
				      \item \textbf{Yield the tuple $\langle a,t\rangle$.}
				    \end{enumerate}
				\declaration{2.10}
			    The notation \gls{ratpolyJ}, where $s$ is a list of polynomials and $x$ is a rational number, will be used as a shorthand for the number of changes observed when the list $\hsf(\Lambda(s,x))$ is iterated through in order.
				\procedure{fri3101200839}
				  \objective
				    Choose polynomials $g,h,j,p,q,r$ and a rational number $X$ such that $hq+jr=1$ and $p+r=gq$. The objective of the following instructions is to construct a rational number $a$ and a procedure $t(y,z)$ to show that $\jsc_{\langle p,q,r\rangle}(y)=\jsc_{\langle p,q,r\rangle}(z)=1$ when two rational numbers $y,z$ such that $\lVert y\rVert\le X$, $\lVert z\rVert\le X$, $\lVert y-z\rVert\le a$, and $\Lambda(q,y)\Lambda(q,z)\le 0$ are chosen.
				  \implementation
				    \begin{enumerate}
				      \item Use \procedurehr{fri3101200807} on $\langle g,h,j,p,q,r,X\rangle$ to construct $\langle a,t_1\rangle$.
				      \item Let $t(y,z)$ be the following procedure:
				      \begin{enumerate}
				        \item Use procedure $t_1$ to show that:
				        \begin{enumerate}
				          \item $\Lambda(p,y)\Lambda(r,y)<0$
				          \item $\Lambda(r,y)\Lambda(r,z)>0$
				          \item $\Lambda(p,y)\Lambda(p,z)>0$
				          \item $\Lambda(p,z)\Lambda(r,z)<0$.
				        \end{enumerate}
				        \item Now show that $\hsf(\Lambda(p,y))\le\hsf(\Lambda(q,y))\le\hsf(\Lambda(r,y))$ or $\hsf(\Lambda(r,y))\le\hsf(\Lambda(q,y))\le\hsf(\Lambda(p,y))$ given that $\Lambda(p,y)\Lambda(r,y)<0$.
				        \item Hence using \procedurehr{thu3001201131}, show that $\jsc_{\langle p,q,r\rangle}(y)$
				        \begin{enumerate}
				          \item $=\lVert\hsf(\Lambda(q,y))-\hsf(\Lambda(p,y))\rVert+\lVert\hsf(\Lambda(r,y))-\hsf(\Lambda(q,y))\rVert$
				          \item $=\lVert\hsf(\Lambda(r,y))-\hsf(\Lambda(p,y))\rVert$
				          \item $=1$.
				        \end{enumerate}
				        \item Also show that $\hsf(\Lambda(p,z))\le\hsf(\Lambda(q,z))\le\hsf(\Lambda(r,z))$ or $\hsf(\Lambda(r,z))\le\hsf(\Lambda(q,z))\le\hsf(\Lambda(p,z))$ given that $\Lambda(p,z)\Lambda(r,z)<0$.
				        \item Hence using \procedurehr{thu3001201131}, show that $\jsc_{\langle p,q,r\rangle}(z)$
				        \begin{enumerate}
				          \item $=\lVert\hsf(\Lambda(q,z))-\hsf(\Lambda(p,z))\rVert+\lVert\hsf(\Lambda(r,z))-\hsf(\Lambda(q,z))\rVert$
				          \item $=\lVert\hsf(\Lambda(r,z))-\hsf(\Lambda(p,z))\rVert$
				          \item $=1$.
				        \end{enumerate}
				        \item \textbf{Hence show that $\jsc_{\langle p,q,r\rangle}(y)=1=\jsc_{\langle p,q,r\rangle}(z)$.}
				      \end{enumerate}
				      \item \textbf{Yield the tuple $\langle a,t\rangle$.}
				    \end{enumerate}
				\procedure{fri3101201221}
				  \objective
				    Choose a list of polynomials $s$, a rational number $r$, and a natural number $k$ such that $k<\lvert s\rvert$. The objective of the following instructions is to show that $\jsc_s(r)=\jsc_{s_{[0:k+1]}}(r)+\jsc_{s_{[k:\lvert s\rvert]}}(r)$.
				  \implementation
				    \begin{enumerate}
				      \item Show that $\jsc_s(r)$
				      \begin{enumerate}
				        \item $=\sum_t^{[0:\lvert s\rvert-1]}\lVert\hsf(\Lambda(s_{t+1},r))-\hsf(\Lambda(s_t,r))\rVert$
				        \item $=\sum_t^{[0:k]}\lVert\hsf(\Lambda(s_{t+1}))-\hsf(\Lambda(s_t,r))\rVert$
				        \item $=\sum_t^{[k:\lvert s\rvert-1]}\lVert\hsf(\Lambda(s_{t+1},r))-\hsf(\Lambda(s_t,r))\rVert$
				        \item $=\jsc_{s_{[0:k+1]}}(r)+\jsc_{s_{[k:\lvert s\rvert]}}(r)$.
				      \end{enumerate}
				    \end{enumerate}
				\declaration{fri3101201236}
				  The phrase "\gls{ratpolySturm}" will be used as a shorthand for a non-empty list of polynomials $s$ such that:
				  \begin{enumerate}
				    \item For $i$ in $[0:\lvert s\rvert]$, $\deg(s_i)=i$.
				    \item For $i$ in $[0:\lvert s\rvert-1]$, $\sgn((s_i)_i)=\sgn_((s_{i+1})_{i+1})$
				    \item For $i$ in $[1:\lvert s\rvert-1]$, $s_{i-1}+s_{i+1}\bmod s_i=0$.
				  \end{enumerate}
				\procedure{fri3101201247}
				  \objective
				    Choose a Sturm chain $s$, and a natural number $k$ such that $0<k\le\lvert s\rvert$. The objective of the following instructions is to show that $s_{[0:k]}$ is also a Sturm chain.
				  \implementation
				    \begin{enumerate}
				      \item For $i$ in $[0:k]$, show that $\deg(s_i)=i$.
				      \item For $i$ in $[0:k-1]$, show that $\sgn((s_i)_i)=\sgn((s_{i+1})_{i+1})$.
				      \item For $i$ in $[1:k-1]$, show that $s_{i-1}+s_{i+1}\bmod s_i=0$.
				      \item \textbf{Hence show that $s_{[0:k]}$ is a Sturm chain.}
				    \end{enumerate}
		    \procedure{2.20}
			    \objective
				    Choose a Sturm chain $s$ and a rational number $X$. The objective of the following instructions is to construct a rational number $l$ and a procedure $u(c,d)$ to show that either $0<0$ or $\lvert\jsc_{s}(d)-\jsc_{s}(c)\rvert=\lVert\hsf(\Lambda(s_{\lvert s\rvert-1},c))-\hsf(\Lambda(s_{\lvert s\rvert-1},d))\rVert$, when rational numbers $c,d$ such that $\lvert c\rvert\le X$, $\lvert d\rvert\le X$, and $\lvert d-c\rvert\le l$ are chosen.
			    \implementation
				    \begin{enumerate}
				      \item If $\lvert s\rvert>2$, then do the following:
				      \begin{enumerate}
				        \item Use \procedurehr{2.20} on $\langle s_{[0:\lvert s\rvert-2]},X\rangle$ to construct $\langle l_1,u_1\rangle$.
				        \item Use \procedurehr{2.20} on $\langle s_{[0:\lvert s\rvert-1]},X\rangle$ to construct $\langle l_2,u_2\rangle$.
				        \item Use \procedurehr{2.19} on $\langle s_{[0:\lvert s\rvert-1]}\rangle$ to construct $\langle g,h\rangle$ and show that $\langle gs_{\lvert s\rvert-2}+hs_{\lvert s\rvert-3}=1$.
				        \item Use \procedurehr{fri3101200807} on $\langle(s_{\lvert s\rvert-1}+s_{\lvert s\rvert-3})\div s_{\lvert s\rvert-2},g,h,s_{\lvert s\rvert-1},s_{\lvert s\rvert-2},s_{\lvert s\rvert-3},X\rangle$ to construct $\langle a_4,u_4\rangle$.
				        \item Use \procedurehr{fri3101200839} on $\langle(s_{\lvert s\rvert-1}+s_{\lvert s\rvert-3})\div s_{\lvert s\rvert-2},g,h,s_{\lvert s\rvert-1},s_{\lvert s\rvert-2},s_{\lvert s\rvert-3},X\rangle$ to construct $\langle a_5,u_5\rangle$.
				        \item Let $l=\min(l_1,l_2,a_4,a_5)$.
				      \end{enumerate}
				      \item Otherwise do the following:
				      \begin{enumerate}
				        \item Let $l=1$.
				      \end{enumerate}
				      \item Let $u(c,d)$ be the following procedure:
				      \begin{enumerate}
				        \item If $\lvert s\rvert=1$, then do the following:
				        \begin{enumerate}
				          \item Show that $\lVert\jsc_s(d)-\jsc_s(c)\rVert$
				          \begin{enumerate}
				            \item $=\lVert\sum_r^{[0:\lvert s\rvert-1]}\lVert\hsf(\Lambda(s_{r+1},d))-\hsf(\Lambda(s_r,d))\rVert-\sum_r^{[0:\lvert s\rvert-1]}\lVert\hsf(\Lambda(s_{r+1},c))-\hsf(\Lambda(s_r,c))\rVert\rVert$
				            \item $=\lVert 0-0\rVert$
				            \item $=\lVert\hsf((s_{\lvert s\rvert-1})_0)-\hsf((s_{\lvert s\rvert-1})_0)\rVert$
				            \item $=\lVert\hsf(\Lambda(s_{\lvert s\rvert-1},c))-\hsf(\Lambda(s_{\lvert s\rvert-1},d))\rVert$.
				          \end{enumerate}
				        \end{enumerate}
				        \item Otherwise if $\lvert s\rvert=2$, then do the following:
				        \begin{enumerate}
				          \item Show that $\lVert\jsc_s(d)-\jsc_s(c)\rVert$
				          \begin{enumerate}
				            \item $=\lVert\sum_r^{[0:\lvert s\rvert-1]}\lVert\hsf(\Lambda(s_{r+1},d))-\hsf(\Lambda(s_r,d))\rVert-\sum_r^{[0:\lvert s\rvert-1]}\lVert\hsf(\Lambda(s_{r+1},c))-\hsf(\Lambda(s_r,c))\rVert\rVert$
				            \item $=\lVert\lVert\hsf(\Lambda(s_1,d))-\hsf(\Lambda(s_0,d))\rVert-\lVert\hsf(\Lambda(s_1,c))-\hsf(\Lambda(s_0,c))\rVert\rVert$
				            \item $=\lVert\lVert\hsf(\Lambda(s_1,d))-\hsf((s_0)_0)\rVert-\lVert\hsf(\Lambda(s_1,c))-\hsf((s_0)_0)\rVert\rVert$
				            \item $=\lVert\hsf(\Lambda(s_1,d))-\hsf(\Lambda(s_1,c))\rVert$.
				          \end{enumerate}
				        \end{enumerate}
				        \item Otherwise if $\hsf(\Lambda(s_{\lvert s\rvert-2},c))=\hsf(\Lambda(s_{\lvert s\rvert-2},d))$, then do the following:
				        \begin{enumerate}
				          \item Use procedure $u_2$ to show that $\lVert\jsc_{s_{[0:\lvert s\rvert-1]}}(d)-\jsc_{s_{[0:\lvert s\rvert-1]}}(c)$
				          \begin{enumerate}
				            \item $=\lVert\hsf(\Lambda(s_{\lvert s\rvert-2},c))-\hsf(\Lambda(s_{\lvert s\rvert-2},d))\rVert$
				            \item $=0$.
				          \end{enumerate}
				          \item Hence show that $\lVert\jsc_s(d)-\jsc_s(c)\rVert$
				          \begin{enumerate}
				            \item $=\lVert(\jsc_{s_{[0:\lvert s\rvert-1]}}(d)+\jsc_{s_{[\lvert s\rvert-2:\lvert s\rvert]}}(d))-(\jsc_{s_{[0:\lvert s\rvert-1]}}(c)+\jsc_{s_{[\lvert s\rvert-2:\lvert s\rvert]}}(c))\rVert$
				            \item $=\lVert\jsc_{s_{[\lvert s\rvert-2:\lvert s\rvert]}}(c)-\jsc_{s_{[\lvert s\rvert-2:\lvert s\rvert]}}(d)\rVert$
				            \item $=\lVert\lVert\hsf(\Lambda(s_{\lvert s\rvert-1},c))-\hsf(\Lambda(s_{\lvert s\rvert-2},c))\rVert-\lVert\hsf(\Lambda(s_{\lvert s\rvert-1},d))-\hsf(\Lambda(s_{\lvert s\rvert-2},d))\rVert\rVert$
				            \item $=\lVert\hsf(\Lambda(s_{\lvert s\rvert-1},c))-\hsf(\Lambda(s_{\lvert s\rvert-1},d))\rVert$.
				          \end{enumerate}
				        \end{enumerate}
				        \item Otherwise do the following:
				        \begin{enumerate}
				          \item Show that $\hsf(\Lambda(s_{\lvert s\rvert-2},c))\ne\hsf(\Lambda(s_{\lvert s\rvert-2},d))$.
				          \item Show that $\Lambda(s_{\lvert s\rvert-1},c)\Lambda(s_{\lvert s\rvert-1},d)>0$ and $\Lambda(s_{\lvert s\rvert-3},c)\Lambda(s_{\lvert s\rvert-3},d)>0$ using procedure $u_4$.
				          \item Hence show that $\hsf(\Lambda(s_{\lvert s\rvert-1},c))=\hsf(\Lambda(s_{\lvert s\rvert-1},d))$ and $\hsf(\Lambda(s_{\lvert s\rvert-3},c))=\hsf(\Lambda(s_{\lvert s\rvert-3},d))$.
				          \item Use procedure $u_1$ to show that $\lVert\jsc_{s_{[0:\lvert s\rvert-2]}}(d)-\jsc_{s_{[0:\lvert s\rvert-2]}}(c)\rVert=\lVert\hsf(\Lambda(s_{\lvert s\rvert-3},d))-\hsf(\Lambda(s_{\lvert s\rvert-3},c))\rVert=0$ given that $\hsf(\Lambda(s_{\lvert s\rvert-3},c))=\hsf(\Lambda(s_{\lvert s\rvert-3},d))$.
				          \item Use procedure $u_5$ to show that $\jsc_{s_{[\lvert s\rvert-3:\lvert s\rvert]}}(c)=\jsc_{s_{[\lvert s\rvert-3:\lvert s\rvert]}}(d)=1$ given that $\Lambda(s_{\lvert s\rvert-2},c)\Lambda(s_{\lvert s\rvert-2},d)<0$.
				          \item Hence given that $\hsf(\Lambda(s_{\lvert s\rvert-1},c))=\hsf(\Lambda(s_{\lvert s\rvert-1},d))$ show that $\lVert\jsc_s(d)-\jsc_s(c)\rVert$
				          \begin{enumerate}
				            \item $=\lVert(\jsc_{s_{[0:\lvert s\rvert-2]}}(d)+\jsc_{s_{[\lvert s\rvert-3:\lvert s\rvert]}}(d))-(\jsc_{s_{[0:\lvert s\rvert-2]}}(c)+\jsc_{s_{[\lvert s\rvert-3:\lvert s\rvert]}}(c))\rVert$
				            \item $=\lVert 0+(1-1)\rVert$
				            \item $=0$
				            \item $=\lVert\hsf(\Lambda(s_{\lvert s\rvert-1},d))-\hsf(\Lambda(s_{\lvert s\rvert-1},c))\rVert$.
				          \end{enumerate}
				        \end{enumerate}
				      \end{enumerate}
				      \item \textbf{Yield the tuple $\langle l,u\rangle$.}
				    \end{enumerate}
		    \procedure{2.21}
			    \objective
				    Choose a polynomial $p\ne 0$. Choose a rational number $k>1+\max_{i}^{[0:\deg(p)]}\lvert\frac{p_i}{p_{\deg(p)}}\rvert$. The objective of the following instructions is to show that $\sgn(\Lambda(p,k))=\sgn(p_{\deg(p)})$.
			    \implementation
				    \begin{enumerate}
					    \item Let $n=\deg(p)$.
					    \item In reverse order verify the following:
					    \begin{enumerate}
						    \item \textbf{$\sgn(\Lambda(p,k))=\sgn(p_{\deg(p)})$}
						    \item $\sgn(p_nk^n+p_{n-1}k^{n-1}+\cdots+p_0k^0)=\sgn(p_{n})$
						    \item $\sgn(k^n+\frac{p_{n-1}}{p_n}k^{n-1}+\cdots+\frac{p_0}{p_n}k^0)=1$
						    \item $k^n+\frac{p_{n-1}}{p_n}k^{n-1}+\cdots+\frac{p_0}{p_n}k^0>0$
						    \item $k^n>-(\frac{p_{n-1}}{p_n}k^{n-1}+\cdots+\frac{p_0}{p_n}k^0)$
						    \item $k^n>\lvert \frac{p_{n-1}}{p_n}k^{n-1}+\cdots+\frac{p_0}{p_n}k^0\rvert$
						    \item $k^n>\lvert\max_i^{[0:n]}\lvert\frac{p_i}{p_n}\rvert(k^{n-1}+\cdots+k^0)\rvert$
						    \item $k^n>\max_i^{[0:n]}\lvert \frac{p_i}{p_n}\rvert\frac{k^n-1}{k-1}$
						    \item $k^{n+1}-k^n>\max_i^{[0:n]}\lvert\frac{p_i}{p_n}\rvert(k^n-1)$
						    \item $k^{n+1}-(1+\max_i^{[0:n]}\lvert\frac{p_i}{p_n}\rvert)k^n+\max_i^{[0:n]}\lvert\frac{p_i}{p_n}\rvert>0$
						    \item $k>1+\max_i^{[0:n]}\lvert\frac{p_i}{p_n}\rvert$
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.22}
			    \objective
				    Choose a polynomial $p\ne 0$. Choose a rational number $k<-(1+\max_i^{[0:\deg(p)]}\lvert\frac{p_i}{p_{\deg(p)}}\rvert)$. The objective of the following instructions is to show that $\sgn(\Lambda(p,k))=(-1)^{\deg(p)}\sgn(p_{\deg(p)})$.
			    \implementation
				    \begin{enumerate}
					    \item Let $t=\deg(p)$.
					    \item Let $q=\langle(-1)^{t-i}p_i\for i\in[0:t+1]\rangle$.
					    \item Verify that $k<-(1+\max_i^{[1:t+1]}\lvert\frac{q_i}{q_{\deg(q)}}\rvert)$.
					    \item Therefore verify that $-k>1+\max_i^{[0:t]}\lvert\frac{q_i}{q_{\deg(q)}}\rvert$.
					    \item Execute \procedurehr{2.21} on $\langle q,-k\rangle$.
					    \item Hence verify that $(-1)^t\sgn(\Lambda(p,k))$
					    \begin{enumerate}
						    \item $=\sgn((-1)^t\Lambda(p,k))$
						    \item $=\sgn((-1)^t\sum_i^{[0:{t+1}]} p_ik^i)$
						    \item $=\sgn(\sum_i^{[0:{t+1}]} (-1)^i(-1)^{t-i}p_ik^i)$
						    \item $=\sgn(\sum_i^{[0:{t+1}]} q_i(-k)^i)$
						    \item $=\sgn(\Lambda(q,-k))$
						    \item $=\sgn(q_t)$
						    \item $=\sgn(p_t)$.
					    \end{enumerate}
					    \item \textbf{Therefore verify that $\sgn(\Lambda(p,k))=(-1)^t(-1)^t\sgn(\Lambda(p,k))=(-1)^t\sgn(p_t)$.}
				    \end{enumerate}
		    \procedure{2.23}
			    \objective
				    Choose a list of polynomials, $s$, and rational numbers $a,l,c$ such that $a<c$ and $l>0$. The objective of the following instructions is to either show that $0<0$ or to construct a list of rational numbers, $b$, such that $a=b_0<b_1<\cdots<b_{\lvert b\rvert-1}=c$, $b_{i}-b_{i-1}\le l$ for $i$ in $[1:\lvert b\rvert]$, and $0\not\in\Lambda(s,b_i)$ for $i$ in $[1:\lvert b\rvert-1]$.
			    \implementation
				    \begin{enumerate}
					    \item Let $e=\langle\langle\rangle,\langle\rangle,\cdots,\langle\rangle\rangle$.
					    \item Let $f=\sum_r^{[0:{\lvert s\rvert}]}\deg(s_r)$.
					    \item Let $b=\langle a\rangle$.
					    \item Let $d=b_1$.
					    \item While $d+l<c$, do the following:
					    \begin{enumerate}
						    \item Let $m=l$.
						    \item While $0\in\Lambda(s,d+m)$ and $\sum\lvert e\rvert\le f$, do the following:
						    \begin{enumerate}
							    \item Let $0\le i<\lvert s\rvert$ be an integer such that $\Lambda(s_i,d+m)=0$.
							    \item Append $d+m$ onto $e_i$.
							    \item Set $m=\frac{m}{2}$
						    \end{enumerate}
						    \item If $\sum\lvert e\rvert>f$, then do the following:
						    \begin{enumerate}
							    \item If $\lvert e_i\rvert\le\deg(s_i)$ for $0\le i<\lvert s\rvert$, then do the following:
							    \begin{enumerate}
								    \item Verify that $\sum\lvert e\rvert\le f$.
								    \item Therefore using (c), verify that $\sum\lvert e\rvert\le f<\sum\lvert e\rvert$.
								    \item \textbf{Abort procedure.}
							    \end{enumerate}
							    \item Otherwise, do the following:
							    \begin{enumerate}
								    \item Let $0\le i<\lvert s\rvert$ be an integer such that $\lvert e_i\rvert>\deg(s_i)$.
								    \item Execute \procedurehr{2.16} on $s_i$ and a sorted $e_i$.
								    \item \textbf{Abort procedure.}
							    \end{enumerate}
						    \end{enumerate}
						    \item Otherwise, do the following:
						    \begin{enumerate}
							    \item \textbf{Verify that $0\not\in\Lambda(s,d+m)$.}
							    \item Append $d+m$ onto $b$.
							    \item \textbf{Verify that $0<b_{\lvert b\rvert-1}-b_{\lvert b\rvert-2}=m\le l$.}
							    \item Set $d$ to $d+m$.
							    \item Using (5), verify that $d<c$.
						    \end{enumerate}
					    \end{enumerate}
					    \item Verify that $d<c$.
					    \item Verify that $d+l\ge c$.
					    \item \textbf{Therefore verify that $0<c-d\le l$.}
					    \item Append $c$ onto $b$.
					    \item \textbf{Yield $\langle b\rangle$.}
				    \end{enumerate}
		    \procedure{2.24}
			    \objective
				    Execute \procedurehr{2.19} and let $\langle s,q,g,h\rangle$ receive. Let $m=\lvert s\rvert-1$. The objective of the following instructions is to either show that $0<0$ or to construct two lists of rational numbers $c,d$ such that $c_0<d_0\le c_1<d_1\le\cdots\le c_{m-1}<d_{m-1}$ and $0\ne\sgn(\Lambda(s_m,c_i))=-\sgn(\Lambda(s_m,d_i))$ for $i$ in $[0:m]$.
			    \implementation
				    \begin{enumerate}
					    \item Let $U=1+\max_i^{[0:\lvert s\rvert]}\left(1+\max_j^{[1:i+1]}\lvert\frac{(s_i)_{i-j}}{(s_i)_i}\rvert\right)$
					    \item Using \procedurehr{2.21}, verify that $J(U)=0$.
					    \item Using \procedurehr{2.22}, verify that $J(-U)=m$.
					    \item Execute \procedurehr{2.20} on the tuple $\langle s,q,U\rangle$ and let $\langle l,u\rangle$ receive.
					    \item Execute \procedurehr{2.23} on $s$ with endpoints $-U,U$ and a step size of $l$ and let $\langle e\rangle$ receive the result.
					    \item Let $c=d=\langle\rangle$.
					    \item For $i=1$ to $i=\lvert e\rvert-1$:
					    \begin{enumerate}
						    \item Execute procedure $u$ on the tuple $\langle e_{i-1},e_{i}\rangle$.
						    \item If $\jsc_m(e_{i-1})\ne \jsc_m(e_{i})$, then do the following:
						    \begin{enumerate}
							    \item Append $e_{i-1}$ to $c$.
							    \item Append $e_{i}$ to $d$.
							    \item Verify that $0\ne\lvert \jsc_{s}(d_{\lvert d\rvert-1})-\jsc_{s}(c_{\lvert c\rvert-1})\rvert=[\sgn(\Lambda(s_{\lvert s\rvert-1},c_{\lvert c\rvert-1}))\ne\sgn(\Lambda(s_{\lvert s\rvert-1},d_{\lvert d\rvert-1}))]$.
							    \item Therefore verify that $\sgn(s_m(c_{\lvert c\rvert-1}))\ne\sgn(s_m(d_{\lvert d\rvert-1}))$.
							    \item Therefore verify that $\lvert \jsc_m(d_{\lvert d\rvert-1})-\jsc_m(c_{\lvert c\rvert-1})\rvert=1$.
							    \item Also verify that $0\not\in\Lambda(s,c_{\lvert c\rvert-1})$.
							    \item Hence verify that $\Lambda(s_m,c_{\lvert c\rvert-1})\ne 0$.
							    \item Also verify that $0\not\in\Lambda(s,d_{\lvert d\rvert-1})$.
							    \item Hence verify that $\Lambda(s_m,d_{\lvert d\rvert-1})\ne 0$.
							    \item \textbf{Therefore verify that $0\ne\sgn(s_m(c_{\lvert c\rvert-1}))=-\sgn(s_m(d_{\lvert d\rvert-1}))$.}
							    \item \textbf{Also verify that $d_{\lvert d\rvert-2}\le c_{\lvert c\rvert-1}<d_{\lvert d\rvert-1}$.}
						    \end{enumerate}
					    \end{enumerate}
					    \item If $\lvert c\rvert=\lvert d\rvert<m$, then do the following:
					    \begin{enumerate}
						    \item Verify that each change of $\jsc_m(x)$ over the course of (7) was by $1$.
						    \item Verify that $\jsc_m(x)$ changed less than $m$ times over the course of (12).
						    \item Therefore verify that $\lvert \jsc_m(U)-\jsc_m(-U)\rvert<m$.
						    \item Therefore using (2) and (3), verify that $m=\lvert \jsc_m(U)-\jsc_m(-U)\rvert<m$.
						    \item \textbf{Abort procedure.}
					    \end{enumerate}
					    \item Otherwise, do the following:
					    \begin{enumerate}
						    \item \textbf{Verify that $m\le\lvert c\rvert=\lvert d\rvert$.}
						    \item \textbf{Yield the tuple $\langle c,d\rangle$.}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{2.26}
			    \objective
				    Choose two lists of polynomials $s,q$ and a non-negative integer $k$ in such a way that, letting $m=\lvert s\rvert-1$,
				    \begin{enumerate}
					    \item $k<m$.
					    \item For $k\le i\le m$, $\deg(s_i)=i$.
					    \item For $k<i<m$, $s_{i-1}+s_{i+1}=q_is_i$.
				    \end{enumerate}
				    Let $\deg(0)=-1$. The objective of the following instructions is to construct polynomials $g,h$ such that $s_k=gs_{m-1}+hs_m$, $\deg(g)=m-1-k$, and $\deg(h)=m-2-k$.
			    \implementation
				    \begin{enumerate}
					    \item If $k<m-2$, do the following:
					    \begin{enumerate}
						    \item Verify that $s_k+s_{k+2}=q_{k+1}s_{k+1}$.
						    \item Therefore verify that $s_k=q_{k+1}s_{k+1}-s_{k+2}$.
						    \item Execute \procedurehr{2.26} on $s,q,k+1$ and let the tuple $\langle g_1,h_1\rangle$ receive.
						    \item Verify that $s_{k+1}=g_1s_{m-1}+h_1s_m$.
						    \item Verify that $\deg(g_1)=m-1-(k+1)=m-k-2$.
						    \item Verify that $\deg(h_1)=m-2-(k+1)=m-k-3$.
						    \item Execute \procedurehr{2.26} on $s,q,k+2$ and let the tuple $\langle g_2,h_2\rangle$ receive.
						    \item Verify that $s_{k+2}=g_2s_{m-1}+h_2s_m$.
						    \item Verify that $\deg(g_2)=m-1-(k+2)=m-k-3$.
						    \item Verify that $\deg(h_2)=m-2-(k+2)=m-k-4$.
						    \item Let $g=q_{k+1}g_1-g_2$.
						    \item \textbf{Verify that $\deg(g)=\max(1+(m-k-2),m-k-3)=m-1-k$.}
						    \item Let $h=q_{k+1}h_1-h_2$.
						    \item \textbf{Verify that $\deg(h)=\max(1+(m-k-3),m-k-4)=m-2-k$.}
						    \item \textbf{Verify that $s_k=q_{k+1}(g_1s_{m-1}+h_1s_m)-(g_2s_{m-1}+h_2s_m)=(q_{k+1}g_1-g_2)s_{m-1}+(q_{k+1}h_1-h_2)s_m=gs_{m-1}+hs_m$.}
					    \end{enumerate}
					    \item Otherwise, if $k=m-2$ do the following:
					    \begin{enumerate}
						    \item Verify that $s_{m-2}+s_m=q_{m-1}s_{m-1}$.
						    \item Let $g=q_{m-1}$.
						    \item \textbf{Verify that $\deg(g)=1=m-1-k$.}
						    \item Let $h=-1$.
						    \item \textbf{Verify that $\deg(h)=0=m-2-k$.}
						    \item \textbf{Therefore verify that $s_k=s_{m-2}=q_{m-1}s_{m-1}-s_m=gs_{m-1}+hs_m$.}
					    \end{enumerate}
					    \item Otherwise, if $k=m-1$ do the following:
					    \begin{enumerate}
						    \item Let $g=1$.
						    \item \textbf{Verify that $\deg(g)=0=m-1-k$.}
						    \item Let $h=0$.
						    \item \textbf{Verify that $\deg(h)=-1=m-2-k$.}
						    \item \textbf{Verify that $s_k=s_{m-1}=gs_{m-1}+hs_m$.}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle g,h\rangle$.}
				    \end{enumerate}
	    \end{multicols*}
	  \clearpage
	\part{Complex Arithmetic}
	  \chapter{Complex Arithmetic}
	    \begin{multicols*}{2}
		    \declaration{3.19}
			    The phrase "\gls{complex}" will be used as a shorthand for an ordered pair of rational numbers.
		    \declaration{3.20}
			    The phrase "the real part of $a$" and the notation \gls{complexRe}, where $a$ is a complex number, will be used as a shorthand for the first entry of $a$.
		    \declaration{3.21}
			    The phrase "the imaginary part of $a$" and the notation \gls{complexIm}, where $a$ is a complex number, will be used as a shorthand for the second entry of $a$.
		    \declaration{3.22}
			    The phrase "\gls{complexa=b}", where $a,b$ are complex numbers, will be used as a shorthand for "$\Re(a)=\Re(b)$ and $\Im(a)=\Im(b)$".
		    \procedure{3.68}
			    \objective
				    Choose a complex number $a$. The objective of the following instructions is to show that $a=a$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Re(a)=\Re(a)$.
					    \item Show that $\Im(a)=\Im(a)$.
					    \item \textbf{Hence show that $a=a$.}
				    \end{enumerate}
		    \procedure{3.69}
			    \objective
				    Choose two complex numbers $a,b$ such that $a=b$. The objective of the following instructions is to show that $b=a$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Re(b)=\Re(a)$ given that $\Re(a)=\Re(b)$.
					    \item Show that $\Im(b)=\Im(a)$ given that $\Im(a)=\Im(b)$.
					    \item \textbf{Hence show that $b=a$.}
				    \end{enumerate}
		    \procedure{3.70}
			    \objective
				    Choose three complex numbers $a,b,c$ such that $a=b$ and $b=c$. The objective of the following instructions is to show that $a=c$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Re(a)=\Re(c)$
					    \begin{enumerate}
					      \item given that $\Re(a)=\Re(b)$
					      \item and $\Re(b)=\Re(c)$.
					    \end{enumerate}
					    \item Show that $\Im(a)=\Im(c)$
					    \begin{enumerate}
					      \item given that $\Im(a)=\Im(b)$
					      \item and $\Im(b)=\Im(c)$.
					    \end{enumerate}
					    \item \textbf{Hence verify that $a=c$.}
				    \end{enumerate}
		    \declaration{3.23}
			    The notation \gls{complexa+b}, where $a,b$ are complex numbers, will be used as a shorthand for the pair $\langle\Re(a)+\Re(b),\Im(a)+\Im(b)\rangle$.
		    \procedure{3.71}
			    \objective
				    Choose two complex numbers $a,b,c,d$ such that $a=c$ and $b=d$. The objective of the following instructions is to show that $a+b=c+d$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{3.22}, show that
					    \begin{enumerate}
					      \item $\Re(a)=\Re(c)$
					      \item $\Im(a)=\Im(c)$
					      \item $\Re(b)=\Re(d)$
					      \item $\Im(b)=\Im(d)$.
					    \end{enumerate}
					    \item Hence show that $a+b$
					    \begin{enumerate}
						    \item $=\langle\Re(a),\Im(a)\rangle+\langle\Re(b),\Im(b)\rangle$
						    \item $=\langle\Re(a)+\Re(b),\Im(a)+\Im(b)\rangle$
						    \item $=\langle\Re(c)+\Re(d),\Im(c)+\Im(d)\rangle$
						    \item $=\langle\Re(c),\Im(c)\rangle+\langle\Re(d),\Im(d)\rangle$
						    \item $=c+d$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{3.72}
			    \objective
				    Choose three complex numbers $a,b,c$. The objective of the following instructions is to show that $(a+b)+c=a+(b+c)$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $(a+b)+c$
					    \begin{enumerate}
						    \item $=\langle\Re(a)+\Re(b),\Im(a)+\Im(b)\rangle+\langle\Re(c),\Im(c)\rangle$
						    \item $=\langle(\Re(a)+\Re(b))+\Re(c),(\Im(a)+\Im(b))+\Im(c)\rangle$
						    \item $=\langle\Re(a)+(\Re(b)+\Re(c)),\Im(a)+(\Im(b)+\Im(c))\rangle$
						    \item $=\langle\Re(a),\Im(a)\rangle+\langle\Re(b)+\Re(c),\Im(b)+\Im(c)\rangle$
						    \item $=a+(b+c)$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{3.73}
			    \objective
				    Choose two complex numbers $a,b$. The objective of the following instructions is to show that $a+b=b+a$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $a+b$
					    \begin{enumerate}
						    \item $=\langle\Re(a)+\Re(b),\Im(a)+\Im(b)\rangle$
						    \item $=\langle\Re(b)+\Re(a),\Im(b)+\Im(a)\rangle$
						    \item $=b+a$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{3.24}
			    The notation \gls{complexa}, where $a$ is a rational number, will contextually be used as a shorthand for the pair $\langle a,0\rangle$.
		    \procedure{3.74}
			    \objective
				    Choose a complex number $a$. The objective of the following instructions is to show that $0+a=a$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $0+a$
					    \begin{enumerate}
						    \item $=\langle 0,0\rangle+\langle\Re(a),\Im(a)\rangle$
						    \item $=\langle 0+\Re(a),0+\Im(a)\rangle$
						    \item $=\langle\Re(a),\Im(a)\rangle$
						    \item $=a$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{3.25}
			    The notation \gls{complex-a}, where $a$ is a complex number, will be used as a shorthand for the pair $\langle-\Re(a),-\Im(a)\rangle$.
		    \procedure{3.75}
			    \objective
				    Choose a complex number $a$. The objective of the following instructions is to show that $-a+a=0$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $-a+a$
					    \begin{enumerate}
						    \item $=(-a)+a$
						    \item $=\langle-\Re(a),-\Im(a)\rangle+\langle\Re(a),\Im(a)\rangle$
						    \item $=\langle-\Re(a)+\Re(a),-\Im(a)+\Im(a)\rangle$
						    \item $=\langle 0,0\rangle$
						    \item $=0$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{3.26}
			    The notation \gls{complexab}, where $a,b$ are complex numbers, will be used as a shorthand for the pair $\langle\Re(a)\Re(b)-\Im(a)\Im(b),\Re(a)\Im(b)+\Im(a)\Re(b)\rangle$.
		    \procedure{3.76}
			    \objective
				    Choose four complex numbers $a,b,c,d$ such that $a=c$ and $b=d$. The objective of the following instructions is to show that $ab=cd$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{3.22}, show that
					    \begin{enumerate}
					      \item $\Re(a)=\Re(c)$
					      \item $\Im(a)=\Im(c)$
					      \item $\Re(b)=\Re(d)$
					      \item $\Im(b)=\Im(d)$.
					    \end{enumerate}
					    \item Hence show that $ab$
					    \begin{enumerate}
						    \item $=\langle\Re(a),\Im(a)\rangle\langle\Re(b),\Im(b)\rangle$
						    \item $=\langle\Re(a)\Re(b)-\Im(a)\Im(b),\Re(a)\Im(b)+\Im(a)\Re(b)\rangle$
						    \item $=\langle\Re(c)\Re(d)-\Im(c)\Im(d),\Re(c)\Im(d)+\Im(c)\Re(d)\rangle$
						    \item $=\langle\Re(c),\Im(c)\rangle\langle\Re(d),\Im(d)\rangle$
						    \item $=cd$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{3.77}
			    \objective
				    Choose three complex numbers $a,b,c$. The objective of the following instructions is to show that $(ab)c=a(bc)$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $(ab)c$
					    \begin{enumerate}
						    \item $=\langle\Re(a)\Re(b)-\Im(a)\Im(b),\Re(a)\Im(b)+\Im(a)\Re(b)\rangle\langle\Re(c),\Im(c)\rangle$
						    \item $=\langle(\Re(a)\Re(b)-\Im(a)\Im(b))\Re(c)-(\Re(a)\Im(b)+\Im(a)\Re(b))\Im(c),(\Re(a)\Re(b)-\Im(a)\Im(b))\Im(c)+(\Re(a)\Im(b)+\Im(a)\Re(b))\Re(c)\rangle$
						    \item $=\langle\Re(a)(\Re(b)\Re(c)-\Im(b)\Im(c))-\Im(a)(\Re(b)\Im(c)+\Im(b)\Re(c)),\Re(a)(\Re(b)\Im(c)+\Im(b)\Re(c))+\Im(a)(\Re(b)\Re(c)-\Im(b)\Im(c))\rangle$
						    \item $=\langle\Re(a),\Im(a)\rangle\langle\Re(b)\Re(c)-\Im(b)\Im(c),\Re(b)\Im(c)+\Im(b)\Re(c)\rangle$
						    \item $=a(bc)$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{3.78}
			    \objective
				    Choose two complex numbers $a,b$. The objective of the following instructions is to show that $ab=ba$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $ab$
					    \begin{enumerate}
						    \item $=\langle\Re(a)\Re(b)-\Im(a)\Im(b),\Re(a)\Im(b)+\Im(a)\Re(b)\rangle$
						    \item $=\langle\Re(b)\Re(a)-\Im(b)\Im(a),\Re(b)\Im(a)+\Im(b)\Re(a)\rangle$
						    \item $=ba$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{3.79}
			    \objective
				    Choose a complex number $a$. The objective of the following instructions is to show that $1a=a$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $1a$
					    \begin{enumerate}
						    \item $=\langle 1,0\rangle\langle\Re(a),\Im(a)\rangle$
						    \item $=\langle 1\Re(a)-0\Im(a),1\Im(a)+0\Re(a)\rangle$
						    \item $=\langle\Re(a),\Im(a)\rangle$
						    \item $=a$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{3.82}
			    \objective
				    Choose three complex numbers $a,b,c$. The objective of the following instructions is to show that $a(b+c)=ab+ac$.
			    \implementation
				    \begin{enumerate}
					    \item $a(b+c)$
					    \begin{enumerate}
						    \item $=\langle\Re(a),\Im(a)\rangle\langle\Re(b)+\Re(c),\Im(b)+\Im(c)\rangle$
						    \item $=\langle\Re(a)(\Re(b)+\Re(c))-\Im(a)(\Im(b)+\Im(c)),\Re(a)(\Im(b)+\Im(c))+\Im(a)(\Re(b)+\Re(c))\rangle$
						    \item $=\langle(\Re(a)\Re(b)-\Im(a)\Im(b))+(\Re(a)\Re(c)-\Im(a)\Im(c)),(\Re(a)\Im(b)+\Im(a)\Re(b))+(\Re(a)\Im(c)+\Im(a)\Re(c))\rangle$
						    \item $=\langle\Re(a)\Re(b)-\Im(a)\Im(b),\Re(a)\Im(b)+\Im(a)\Re(b)\rangle+\langle\Re(a)\Re(c)-\Im(a)\Im(c),\Re(a)\Im(c)+\Im(a)\Re(c)\rangle$
						    \item $=ab+ac$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{3.02}
			    The notation \gls{complexconj}, where $a$ is a complex number, will be used as a shorthand for $\langle\Re(a),-\Im(a)\rangle$.
		    \procedure{3.00}
			    \objective
				    Choose two complex numbers $a,b$. The objective of the following instructions is to show that $\conj{a+b}=\conj{a}+\conj{b}$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\conj{a+b}$
					    \begin{enumerate}
						    \item $=\langle\Re(a+b),-\Im(a+b)\rangle$
						    \item $=\langle\Re(a)+\Re(b),-\Im(a)-\Im(b)\rangle$
						    \item $=\conj{a}+\conj{b}$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{3.01}
			    \objective
				    Choose two complex numbers $a,b$. The objective of the following instructions is to show that $\conj{ab}=\conj{a}\conj{b}$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\conj{ab}$
					    \begin{enumerate}
						    \item $=\langle\Re(ab),-\Im(ab)\rangle$
						    \item $=\langle\Re(a)\Re(b)-\Im(a)\Im(b)),-\Re(a)\Im(b)-\Im(a)\Re(b)\rangle$
						    \item $=\langle\Re(a),-\Im(a)\rangle\langle\Re(b),-\Im(b)\rangle$
						    \item $=\conj{a}\conj{b}$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{3.03}
			    The notation \gls{complexabs^2}, where $a$ is a complex number, will be used as a shorthand for $\Re(a)^2+\Im(a)^2$.
		    \procedure{3.02}
			    \objective
				    Choose a complex number $a$. The objective of the following instructions is to show that $a\conj{a}=\lVert a\rVert^2$.
			    \implementation
				    \begin{enumerate}
					    \item \textbf{Show that $a\conj{a}=\lVert a\rVert^2$.}
				    \end{enumerate}
		    \procedure{3.04}
			    \objective
				    Choose a list of complex numbers $a$. The objective of the following instructions is to show that $\lVert\sum_r^{[0:{\lvert a\rvert}]}a_r\rVert^2\le\lvert a\rvert\sum_r^{[0:{\lvert a\rvert}]}\lVert a_r\rVert^2$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\lVert\sum_r^{[0:{\lvert a\rvert}]}a_r\rVert^2$
					    \begin{enumerate}
						    \item $=\sum_r^{[0:{\lvert a\rvert}]}\sum_k^{[0:{\lvert a\rvert}]}a_r\conj{a_k}$
						    \item $=\sum_r^{[0:{\lvert a\rvert}]}\lVert a_r\rVert^2+2\sum_r^{[0:\lvert a\rvert]}\sum_k^{[r+1:\lvert a\rvert]}(\Re(a_r)\Re(a_k)+\Im(a_r)\Im(a_k))$
						    \item $=\sum_r^{[0:{\lvert a\rvert}]}\lVert a_r\rVert^2+\sum_r^{[0:\lvert a\rvert]}\sum_k^{[r+1:\lvert a\rvert]}(\Re(a_r)^2-(\Re(a_r)-\Re(a_k))^2+\Re(a_k)^2+\Im(a_r)^2-(\Im(a_r)-\Im(a_k))^2+\Im(a_k)^2)$
						    \item $\le\sum_r^{[0:{\lvert a\rvert}]}\lVert a_r\rVert^2+\sum_r^{[0:\lvert a\rvert]}\sum_k^{[r+1:\lvert a\rvert]}(\Re(a_r)^2+\Re(a_k)^2+\Im(a_r)^2+\Im(a_k)^2)$
						    \item $=\sum_r^{[0:{\lvert a\rvert}]}\lVert a_r\rVert^2+\sum_r^{[0:\lvert a\rvert]}\sum_k^{[r+1:\lvert a\rvert]}(\lVert a_r\rVert^2+\lVert a_k\rVert^2)$
						    \item $=\sum_r^{[0:{\lvert a\rvert}]}\lVert a_r\rVert^2+\frac{1}{2}\sum_r^{[0:\lvert a\rvert]}\sum_k^{[0:r]^{\frown}[r+1:\lvert a\rvert]}(\lVert a_r\rVert^2+\lVert a_k\rVert^2)$
						    \item $=\sum_r^{[0:{\lvert a\rvert}]}\lVert a_r\rVert^2+\frac{1}{2}(\sum_r^{[0:{\lvert a\rvert}]}(\lvert a\rvert-1)\lVert a_r\rVert^2+\sum_k^{[0:{\lvert a\rvert}]}(\lvert a\rvert-1)\lVert a_k\rVert^2)$
						    \item $=\sum_r^{[0:{\lvert a\rvert}]}\lVert a_r\rVert^2+\sum_r^{[0:{\lvert a\rvert}]}(\lvert a\rvert-1)\lVert a_r\rVert^2$
						    \item $=\lvert a\rvert\sum_r^{[0:{\lvert a\rvert}]}\lVert a_r\rVert^2$
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{3.05}
			    \objective
				    Choose a list of complex numbers $a$. The objective of the following instructions is to show that $\frac{\lVert a_0\rVert^2}{\lvert a\rvert}-\sum_r^{[1:{\lvert a\rvert}]}\lVert a_r\rVert^2\le\lVert a_0-\sum_r^{[1:{\lvert a\rvert}]}a_r\rVert^2$.
			    \implementation
				    \begin{enumerate}
					    \item Using \procedurehr{3.04}, show that $\lVert a_0\rVert^2$
					    \begin{enumerate}
						    \item $=\lVert\sum_r^{[1:{\lvert a\rvert}]}a_r+(a_0-\sum_r^{[1:{\lvert a\rvert}]}a_r)\rVert^2$
						    \item $\le \lvert a\rvert\sum_r^{[1:{\lvert a\rvert}]}\lVert a_r\rVert^2+\lvert a\rvert\lVert a_0-\sum_r^{[1:{\lvert a\rvert}]}a_r\rVert^2$
					    \end{enumerate}
					    \item \textbf{Therefore show that $\frac{\lVert a_0\rVert^2}{\lvert a\rvert}-\sum_r^{[1:{\lvert a\rvert}]}\lVert a_r\rVert^2\le\lVert a_0-\sum_r^{[1:{\lvert a\rvert}]}a_r\rVert^2$.}
				    \end{enumerate}
		    \procedure{3.04aa}
			    \objective
				    Choose a list of complex numbers $a$ and a list of rational numbers $b$ such that $\lvert a\rvert=\lvert b\rvert$ and $\lVert a_i\rVert^2\le {b_i}^2$ for each $i\in[0:\lvert a\rvert]$. The objective of the following instructions is to show that $\lVert\sum_r^{[0:{\lvert a\rvert}]}a_r\rVert^2\le(\sum_r^{[0:{\lvert b\rvert}]}{b_r})^2$.
			    \implementation
				    \begin{enumerate}
					    \item If $\lvert a\rvert=0$, then do the following:
					    \begin{enumerate}
						    \item \textbf{Show that $\lVert\sum_i^{[0:\lvert a\rvert]} a_i\rVert^2=\lVert 0\rVert^2=(\sum_i^{[0:\lvert b\rvert]}b_i)^2$.}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
					      \item Show that $\lvert a\rvert>0$.
					      \item Show that $\lVert\sum_i^{[1:\lvert a\rvert]}a_i\rVert^2\le(\sum_i^{[1:\lvert b\rvert]} b_i)^2$ using \procedurehr{3.04aa} on $a_{[1:\lvert a\rvert]}$ and $b_{[1:\lvert b\rvert]}$.
					      \item Show that $\Re(\conj{a_0}\sum_i^{[1:\lvert a\rvert]}a_i)^2$
					      \begin{enumerate}
					        \item $\le\lVert\conj{a_0}\sum_i^{[1:\lvert a\rvert]}a_i\rVert^2$
					        \item $=\lVert\conj{a_0}\rVert^2\lVert\sum_i^{[1:\lvert a\rvert]}a_i\rVert^2$
					        \item $\le{b_0}^2(\sum_i^{[1:\lvert a\rvert]}b_i)^2$.
					      \end{enumerate}
					      \item Hence show that $\lVert\sum_i^{[0:\lvert a\rvert]}a_i\rVert^2$
					      \begin{enumerate}
					        \item $=(a_0+\sum_i^{[1:\lvert a\rvert]}a_i)(\conj{a_0+\sum_i^{[1:\lvert a\rvert]}a_i})$
					        \item $=\lVert a_0\rVert^2+a_0\conj{\sum_i^{[1:\lvert a\rvert]}a_i}+\conj{a_0}\sum_i^{[1:\lvert a\rvert]}a_i+\lVert\sum_i^{[1:\lvert a\rvert]}a_i\rVert^2$
					        \item $\le{b_0}^2+\conj{\conj{a_0}\sum_i^{[1:\lvert a\rvert]}a_i}+\conj{a_0}\sum_i^{[1:\lvert a\rvert]}a_i+(\sum_i^{[1:\lvert a\rvert]}b_i)^2$
					        \item $={b_0}^2+2\Re(\conj{a_0}\sum_i^{[1:\lvert a\rvert]}a_i)+(\sum_i^{[1:\lvert a\rvert]}b_i)^2$
					        \item $\le{b_0}^2+2b_0\sum_i^{[1:\lvert a\rvert]} b_i+(\sum_i^{[1:\lvert a\rvert]}b_i)^2$
					        \item $=(b_0+\sum_i^{[1:\lvert a\rvert]}b_i)^2$
					        \item $=(\sum_i^{[0:\lvert a\rvert]}b_i)^2$.
					      \end{enumerate}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{sat1708191238}
		      \objective
		        Choose two complex numbers $a,d$ and two rational numbers $b,c$ such that $\lVert a\rVert^2\le b^2<c^2\le\lVert d\rVert^2$. The objective of the following instructions is to show that $\lVert d-a\rVert^2\ge(c-b)^2$.
		      \implementation
		        \begin{enumerate}
		          \item Show that $\Re(\frac{a}{d})^2$
		          \begin{enumerate}
		            \item $=\Re(\frac{a\conj{d}}{\lVert d\rVert^2})^2=\frac{\Re(a\conj{d})^2}{\lVert d\rVert^4}\le\frac{\lVert a\conj{d}\rVert^2}{\lVert d\rVert^4}$
		            \item $=\frac{\lVert a\rVert^2\lVert d\rVert^2}{\lVert d\rVert^4}=\frac{\lVert a\rVert^2}{\lVert d\rVert^2}\le\frac{b^2}{c^2}=(\frac{b}{c})^2$.
		          \end{enumerate}
		          \item Hence show that $\Re(\frac{a}{d})\le\frac{b}{c}<1$.
		          \item Hence show that $\lVert d-a\rVert^2$
		          \begin{enumerate}
		            \item $=\lVert\frac{d-a}{d}\rVert^2\lVert d\rVert^2$
		            \item $=(\Re(1-\frac{a}{d})^2+\Im(1-\frac{a}{d})^2)\lVert d\rVert^2$
		            \item $\ge\Re(1-\frac{a}{d})^2\lVert d\rVert^2$
		            \item $=(1-\Re(\frac{a}{d}))^2\lVert d\rVert^2$
		            \item $\ge(1-\frac{b}{c})^2c^2$
		            \item $=(c-b)^2$.
		          \end{enumerate}
		        \end{enumerate}
		    \declaration{3.27}
			    The notation \gls{complex1/a}, where $a$ is a complex number, will be used as a shorthand for the pair $\frac{1}{\lVert a\rVert^2}\conj{a}$.
		    \procedure{3.81}
			    \objective
				    Choose a complex number $a$ such that $a\ne 0$. The objective of the following instructions is to show that $\frac{1}{a}a=1$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Re(a)\ne\Re(0)=0$ or $\Im(a)\ne\Im(0)=0$ using \declarationhr{3.22}.
					    \item Hence show that $\lVert a\rVert^2=\Re(a)^2+\Im(a)^2>0$.
					    \item Hence show that $\frac{1}{a}a$
					    \begin{enumerate}
						    \item $=(\frac{1}{\Vert a\rVert^2}\conj{a})a$
						    \item $=\frac{1}{\Vert a\rVert^2}(\conj{a}a)$
						    \item $=\frac{1}{\Vert a\rVert^2}\Vert a\rVert^2$
						    \item $=1$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{3.28}
			    The notation \gls{complexi} will be used as a shorthand for $\langle 0,1\rangle$.
		    \procedure{3.03}
			    \objective
				    Choose an integer $a$. The objective of the following instructions is to show that $\iu^{4a}=1$, $\iu^{4a+1}=i$, $\iu^{4a+2}=-1$, and $\iu^{4a+3}=-i$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\iu^2=-1$.
					    \item Hence show that $\iu^4=(-1)^2=1$.
					    \item \textbf{Hence show that}
					    \begin{enumerate}
					      \item $\iu^{4a}=(\iu^4)^a=1^a=1$
					      \item $\iu^{4a+1}=\iu^{4a}\iu=1\iu=\iu$
					      \item $\iu^{4a+2}=\iu^{4a+1}\iu=\iu^2=-1$
					      \item $\iu^{4a+3}=\iu^{4a+2}\iu=(-1)\iu=-\iu$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{sun2107190636}
			    \objective
				    Choose a non-negative integer $a$ and a complex number $x$. The objective of the following instructions is to show that $(1+x)^a=\sum_r^{[0:{a+1}]}\binom{a}{r}x^r$.
			    \implementation
				    Instructions are analogous to those of \procedurehr{1.63}.
		    \declaration{mon1908191749}
		      The notation \gls{complexerr} will be used as a shorthand for $\lVert b-a\rVert^2\le\lVert c_1\rVert^2\le\lVert c_2\rVert^2\le\cdots\le\lVert c_n\rVert^2$.
		    \procedure{mon1908191916}
		      \objective
		        Choose four complex numbers $a,b,c,d$ in such a way that $a\equiv b\err{c}$ and $\lVert c\rVert^2\le\lVert d\rVert^2$. The objective of the following instructions is to show that $a\equiv b\err{c}\err{d}$.
		      \implementation
		        \begin{enumerate}
		          \item Show that $\lVert b-a\rVert\le\lVert c\rVert^2\le\lVert d\rVert^2$.
		          \item \textbf{Hence show that $a\equiv b\err{c}\err{d}$ using \declarationhr{mon1908191749}.}
		        \end{enumerate}
		    \procedure{mon1908191825}
		      \objective
		        Choose three complex numbers $a,b,c$ and two non-negative rational numbers $d,e$ in such a way that $a\equiv b\err{d}$ and $b\equiv c\err{e}$. The objective of the following instructions is to show that $a\equiv c\err{d+e}$.
		      \implementation
		        \begin{enumerate}
		          \item Show that $\lVert b-a\rVert^2\le\lVert d\rVert^2=d^2$.
		          \item Show that $\lVert c-b\rVert^2\le\lVert e\rVert^2=e^2$.
		          \item Hence show that $\lVert c-a\rVert^2=\lVert(c-b)+(b-a)\rVert^2\le(e+d)^2$ using \procedurehr{3.04aa}.
		          \item \textbf{Hence show that $a\equiv c\err{e+d}$ using \declarationhr{mon1908191749}.}
		        \end{enumerate}
		    \procedure{mon1908191839}
		      \objective
		        Choose four complex numbers $a,b,c,d$ and two non-negative rational numbers $e,f$ in such a way that $a\equiv b\err{e}$ and $c\equiv d\err{f}$. The objective of the following instructions is to show that $a+c\equiv b+d\err{e+f}$.
		      \implementation
		        \begin{enumerate}
		          \item Show that $\lVert b-a\rVert^2\le\lVert e\rVert^2=e^2$.
		          \item Show that $\lVert d-c\rVert^2\le\lVert f\rVert^2=f^2$.
		          \item Hence show that $\lVert(b+d)-(a+c)\rVert^2=\lVert(b-a)+(d-c)\rVert^2\le(e+f)^2$ using \procedurehr{3.04aa}.
		          \item \textbf{Hence show that $a+c\equiv b+d\err{e+f}$ using \declarationhr{mon1908191749}.}
		        \end{enumerate}
		    \procedure{mon1908191849}
		      \objective
		        Choose three complex numbers $a,b,c$ in such a way that $a\equiv b\err{c}$. The objective of the following instructions is to show that $-a\equiv -b\err{c}$.
		      \implementation
		        \begin{enumerate}
		          \item Show that $\lVert(-b)-(-a)\rVert^2=\lVert b-a\rVert\le\lVert c\rVert^2$.
		          \item \textbf{Hence show that $-a\equiv-b\err{c}$ using \declarationhr{mon1908191749}.}
		        \end{enumerate}
		    \procedure{mon1908191857}
		      \objective
		        Choose four complex numbers $a,b,c,d$ in such a way that $a\equiv b\err{c}$. The objective of the following instructions is to show that $ad\equiv bd\err{cd}$.
		      \implementation
		        \begin{enumerate}
		          \item Show that $\lVert b-a\rVert^2\le\lVert c\rVert^2$.
		          \item Hence show that $\lVert bd-ad\rVert^2\le\lVert cd\rVert^2$.
		          \item \textbf{Hence show that $ad\equiv bd\err{cd}$ using \declarationhr{mon1908191749}.}
		        \end{enumerate}
		    \procedure{mon1908191905}
		      \objective
		        Choose two complex numbers $a,b,c$ in such a way that $a\ne 0$, $b\ne 0$, and $a\equiv b\err{c}$. The objective of the following instructions is to show that $\frac{1}{a}\equiv\frac{1}{b}\err{\frac{c}{ab}}$.
		      \implementation
		        \begin{enumerate}
		          \item Show that $\lVert b-a\rVert\le\lVert c\rVert^2$.
		          \item Hence show that $\lVert\frac{1}{b}-\frac{1}{a}\rVert^2=\lVert\frac{a-b}{ba}\rVert^2\le\lVert\frac{c}{ba}\rVert^2$.
		          \item \textbf{Hence show that $\frac{1}{a}\equiv\frac{1}{b}\err{\frac{c}{ab}}$ using \declarationhr{mon1908191749}.}
		        \end{enumerate}
	    \end{multicols*}
	  \chapter{Exponential and Trigonometric Functions}
	    \begin{multicols}{2}
		    \declaration{3.05}
			    The notation \gls{complexexp}, where $a$ is a complex number, will be used as a shorthand for $(1+\frac{a}{n})^n$.
		    \procedure{3.08}
			    \objective
				    Choose a rational number $a$ and a positive integer $n$ such that $-n<a<1$. The objective of the following instructions is to show that $\exp_n(a)\le\frac{1}{1-a}$.
			    \implementation
				    \begin{enumerate}
					    \item Using \procedurehr{2.07}, show that $\exp_n(a)$
					    \begin{enumerate}
						    \item $=(\frac{n+a}{n})^n$
						    \item $=(\frac{n}{n+a})^{-n}$
						    \item $=\frac{1}{(1+\frac{-a}{n+a})^n}$
						    \item $\le\frac{1}{1+\frac{-an}{n+a}}$
						    \item $\le\frac{1}{1-a}$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{3.09}
			    \objective
				    Choose a rational number $a$ and a positive integer $n$ such that $a>-n$. The objective of the following instructions is to show that $\frac{\exp_{n+1}(a)}{\exp_n(a)}\ge 1$.
			    \implementation
				    \begin{enumerate}
					    \item Using \procedurehr{2.07}, show that $\frac{\exp_{n+1}(a)}{\exp_n(a)}$
					    \begin{enumerate}
						    \item $=\frac{(\frac{n+1+a}{n+1})^{n}}{(\frac{n+a}{n})^{n}}(1+\frac{a}{n+1})$
						    \item $=(\frac{(n+1+a)n}{(n+1)(n+a)})^n(1+\frac{a}{n+1})$
						    \item $=(\frac{n^2+n+na}{n^2+an+n+a})^n(1+\frac{a}{n+1})$
						    \item $=(1-\frac{a}{(n+1)(n+a)})^n(1+\frac{a}{n+1})$
						    \item $\ge(1-\frac{an}{(n+1)(n+a)})(1+\frac{a}{n+1})$
						    \item $=1+\frac{a(n+a)}{(n+1)(n+a)}-\frac{an}{(n+1)(n+a)}-\frac{a^2n}{(n+1)^2(n+a)}$
						    \item $=1+\frac{a^2}{(n+1)(n+a)}-\frac{a^2n}{(n+1)^2(n+a)}$
						    \item $=1+\frac{a^2}{(n+1)^2(n+a)}$
						    \item $\ge 1$
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{3.10}
			    \objective
				    Choose a rational number $X\ge 0$. The objective of the following instructions is to construct positive rational numbers $a,b$ such that $a>1$, and a procedure, $p(x,r,n)$, to show that $(1+\frac{x}{n})^r\le a^2$ when given a rational number $x$ and a non-negative integers $r,n$ such that $r\le n$, $n\ge b$ and $x^2\le X^2$ are chosen.
			    \implementation
				    \begin{enumerate}
					    \item Let $a=2^{\lceil X\rceil}$.
					    \item Let $b=X$.
					    \item Let $p(x,r,n)$ be the following procedure:
					    \begin{enumerate}
						    \item Show that $0\le 1+\frac{x}{n}\le 2$
						    \begin{enumerate}
						      \item given that $-1\le\frac{x}{n}\le 1$
						      \item given that $-X\le x\le X$
						      \item given that $x^2\le X^2$.
						    \end{enumerate}
						    \item Hence using \procedurehr{3.08} and \procedurehr{3.09}, show that $(1+\frac{x}{n})^r$
						    \begin{enumerate}
						      \item $\le(1+\frac{X}{n})^r$
							    \item $\le\exp_n(X)$
							    \item $\le(1+\frac{X}{2\lceil X\rceil n})^{2\lceil X\rceil n}$
							    \item $=((1+\frac{\frac{X}{2\lceil X\rceil}}{n})^n)^{2\lceil X\rceil}$
							    \item $=\exp_n(\frac{X}{2\lceil X\rceil})^{2\lceil X\rceil}$
							    \item $\le(\frac{1}{1-\frac{X}{2\lceil X\rceil}})^{2\lceil X\rceil}$
							    \item $\le 2^{2\lceil X\rceil}$
							    \item $=a^2$.
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,b,p\rangle$.}
				    \end{enumerate}
		    \procedure{3.11}
			    \objective
				    Choose a rational number $X\le 0$. The objective of the following instructions is to construct two rational numbers $a>0,b$, and a procedure $p(x,r,n)$ to show that $(1+\frac{x}{n})^r\ge a^2$ when a rational number $x$ and non-negative integers $r,n$ such that $X\le x\le 0$, $r\le n$, and $n>b$ are chosen.
			    \implementation
				    \begin{enumerate}
					    \item Use \procedurehr{3.10} on $\langle -2X\rangle$ to construct $\langle c,d,q\rangle$.
					    \item Let $a=c^{-1}$.
					    \item Let $b=\max(-2X,d)$.
					    \item Let $p(x,r,n)$ be the following procedure:
					    \begin{enumerate}
						    \item Show that $0\le -2x\le -2X$
						    \begin{enumerate}
						      \item given that $2X\le 2x\le 0$
						      \item given that $X\le x\le 0$.
						    \end{enumerate}
						    \item Show that $(1+\frac{-2x}{n})^r\le c^2$ using procedure $q$.
						    \item Show that $\frac{n}{2}\le n+x<n$
						    \begin{enumerate}
						      \item given that $-\frac{n}{2}\le x\le 0$
						      \item given that $n>b\ge -2X\ge -2x\ge 0$.
						    \end{enumerate}
						    \item Hence show that $(1+\frac{x}{n})^r$
						    \begin{enumerate}
							    \item $=(\frac{n+x}{n})^r$
							    \item $=(\frac{n}{n+x})^{-r}$
							    \item $=(1-\frac{x}{n+x})^{-r}$
							    \item $\ge(1-\frac{x}{\frac{1}{2}n})^{-r}$
							    \item $=(1-\frac{2x}{n})^{-r}$
							    \item $=((1+\frac{-2x}{n})^{r})^{-1}$
							    \item $\ge (c^2)^{-1}$
							    \item $=a^2$.
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,b,p\rangle$.}
				    \end{enumerate}
		    \procedure{3.12}
			    \objective
				    Choose a rational number $X\ge 0$. The objective of the following instructions is to construct two rational numbers $a>0,b$, and a procedure $p(x,r,n)$ to show that $(1+\frac{x}{n})^r\ge a^2$ when a rational number $x$ and non-negative integers $r,n$ such that $x^2\le X^2$, $r\le n$, and $n>b$ are chosen.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{3.11} on $\langle -X\rangle$ and let $\langle c,b,q\rangle$ receive.
					    \item Let $a=\min(1,c)$.
					    \item Let $p(x,r,n)$ be the following procedure:
					    \begin{enumerate}
						    \item If $x<0$, then do the following:
						    \begin{enumerate}
							    \item Show that $-X\le x\le 0$ given that $x^2\le X^2$.
							    \item \textbf{Hence show that $(1+\frac{x}{n})^r\ge c^2\ge a^2$ using procedure $q$.}
						    \end{enumerate}
						    \item Otherwise do the following:
						    \begin{enumerate}
							    \item Verify that $x\ge 0$.
							    \item \textbf{Show that $(1+\frac{x}{n})^r\ge 1+\frac{rx}{n}\ge 1\ge a^2$ using \procedurehr{2.07}.}
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,b,p\rangle$.}
				    \end{enumerate}
		    \procedure{3.13}
			    \objective
				    Choose a rational number $X\ge 0$. The objective of the following instructions is to construct positive rational numbers $a,b$ such that $a>1$, and a procedure, $p(x,r,n)$, to show that $\lVert(1+\frac{x}{n})^r\rVert^2\le a^2$ when a complex number $x$ and non-negative integers $r,n$ such that $\lVert x\rVert^2\le X^2$, $r\le n$, and $n>b$ are chosen.
			    \implementation
				    \begin{enumerate}
					    \item Let $c=2X+X^2$.
					    \item Execute \procedurehr{3.10} on $\langle c\rangle$ and let $\langle a,b,q\rangle$.
					    \item Let $p(x,r,n)$ be the following procedure:
					    \begin{enumerate}
						    \item Let $y=2\lvert\Re(x)\rvert+\lVert x\rVert^2$.
						    \item Show that $\lvert y\rvert=y\le 2X+X^2=c$
						    \begin{enumerate}
						      \item given that $\lvert\Re(x)\rvert\le X$
						      \item given that $\lvert\Re(x)\rvert^2\le\lVert x\rVert^2\le X^2$.
						    \end{enumerate}
						    \item Hence show that $(1+\frac{y}{n})^r\le a^2$ using procedure $q$.
						    \item Now using \procedurehr{3.02} show that $\lVert(1+\frac{x}{n})^r\rVert^2$
						    \begin{enumerate}
							    \item $=(1+\frac{x}{n})^r\conj{(1+\frac{x}{n})^r}$
							    \item $=(1+\frac{x}{n})^r(1+\frac{\conj{x}}{n})^r$
							    \item $=(1+\frac{2\Re(x)}{n}+\frac{\lVert x\rVert^2}{n^2})^r$
							    \item $\le(1+\frac{2\lvert\Re(x)\rvert}{n}+\frac{\lVert x\rVert^2}{n^2})^r$
							    \item $\le(1+\frac{2\lvert\Re(x)\rvert+\lVert x\rVert^2}{n})^r$
							    \item $=(1+\frac{y}{n})^r$
							    \item $\le a^2$.
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,b,p\rangle$.}
				    \end{enumerate}
		    \procedure{3.14}
			    \objective
				    Choose a rational number $X\ge 0$. The objective of the following instructions is to construct two rational numbers $a,b$ and a procedure, $p(x,r,n)$, to show that $\lVert(1+\frac{x}{n})^r\rVert^2\ge a^2$ when a rational number $x$ and non-negative integers $r,n$ such that $\lVert x\rVert^2\le X^2$, $r\le n$ and $n>b$ are chosen.
			    \implementation
				    \begin{enumerate}
					    \item Let $c=2X+X^2$.
					    \item Execute \procedurehr{3.12} on $\langle c\rangle$ and let $\langle a,d,q\rangle$ receive.
					    \item Let $b=\max(c,d)$.
					    \item Let $p(x,r,n)$ be the following procedure:
					    \begin{enumerate}
						    \item Let $y=2\lvert\Re(x)\rvert+\lVert x\rVert^2$.
						    \item Verify that $\lvert -y\rvert=y\le 2X+X^2=c$.
						    \item Hence show that $(1+\frac{-y}{n})^r\ge a^2$ using procedure $q$.
						    \item Also, show that $n>b\ge c\ge y$.
						    \item Hence show that $\lVert(1+\frac{x}{n})^r\rVert^2$
						    \begin{enumerate}
							    \item $=(1+\frac{x}{n})^r\conj{(1+\frac{x}{n})^r}$
							    \item $=(1+\frac{x}{n})^r(1+\frac{\conj{x}}{n})^r$
							    \item $=(1+\frac{2\Re(x)}{n}+\frac{\lVert x\rVert^2}{n^2})^r$
							    \item $\ge(1-\frac{2\lvert\Re(x)\rvert}{n}-\frac{\lVert x\rVert^2}{n^2})^r$
							    \item $\ge(1-\frac{2\lvert\Re(x)\rvert+\lVert x\rVert^2}{n})^r$
							    \item $=(1+\frac{-y}{n})^r$
							    \item $\ge a^2$.
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,b,p\rangle$.}
				    \end{enumerate}
		    \procedure{3.15}
			    \objective
				    Choose a rational number $X\ge 0$. The objective of the following instructions is to construct rational numbers $a,b$ such that $a>0$, and a procedure, $p$, to show that $\exp_n(x+y)\equiv\exp_n(x)\exp_n(y)\err{\frac{axy}{n}}\err{\frac{aX^2}{n}}$ when two complex numbers $x,y$ and a positive integer $n>b$ such that $\lVert x\rVert^2\le X^2$, $\lVert y\rVert^2\le X^2$ are chosen.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{3.13} on $\langle 2X\rangle$ and let $\langle c,b,q\rangle$ receive.
					    \item Let $a=c^3$.
					    \item Let $p(x,y,n)$ be the following procedure:
					    \begin{enumerate}
						    \item Using procedure $q$, show that $\exp_n(x+y)\equiv\exp_n(x)\exp_n(y)$
						    \begin{enumerate}
						      \item $\err{\exp_n(x)\exp_n(y)-\exp_n(x+y)}$
							    \item $\err{(1+\frac{x}{n})^n(1+\frac{y}{n})^n-(1+\frac{x+y}{n})^n}$
							    \item $\err{(1+\frac{x+y}{n}+\frac{xy}{n^2})^n-(1+\frac{x+y}{n})^n}$
							    \item $\err{\frac{xy}{n^2}\sum_r^{[0:n]}(1+\frac{x+y}{n}+\frac{xy}{n^2})^r(1+\frac{x+y}{n})^{n-1-r}}$
							    \item $\err{\frac{xy}{n^2}\sum_r^{[0:n]}(1+\frac{x}{n})^r(1+\frac{y}{n})^r(1+\frac{x+y}{n})^{n-1-r}}$
							    \item $\err{\frac{xy}{n^2}\sum_r^{[0:n]}c^3}$
							    \item $\err{\frac{axy}{n}}$
							    \item $\err{\frac{aX^2}{n}}$.
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,b,p\rangle$.}
				    \end{enumerate}
		    \procedure{thu2507191359}
		      \objective
		        Choose a rational number $X\ge 0$. The objective of the following instructions is to construct rational numbers $a,b$ such that $a>0$ and a procedure $p(x,y,n)$ to show that $\exp_n(x-y)\equiv\frac{\exp_n(x)}{\exp_n(y)}\err{\frac{a}{n}}$ when two complex numbers $x,y$ and a positive integer $n$ such that $\lVert x\rVert^2\le X$, $\lVert y\rVert^2\le X$, and $n>b$ are chosen.
		      \implementation
		        \begin{enumerate}
		          \item Execute \procedurehr{3.15} on $\langle X\rangle$ and let $\langle c,d,q\rangle$ receive.
		          \item Execute \procedurehr{3.14} on $\langle X\rangle$ and let $\langle e,f,r\rangle$ receive.
		          \item Execute \procedurehr{3.13} on $\langle X\rangle$ and let $\langle g,h,t\rangle$ receive.
		          \item Let $b=\max(d,f,h)$.
		          \item Let $a=c(1+\frac{g}{e})X^2$.
		          \item Let $p(x,y,n)$ be the following procedure:
		          \begin{enumerate}
		            \item Using procedures $q,r,t$, show that $\exp_n(x-y)$
		            \begin{enumerate}
		              \item $\equiv\exp_n(x)\exp_n(-y)\err{\frac{cX^2}{n}}$
		              \item $=\exp_n(x)\frac{\exp_n(y)\exp_n(-y)}{\exp_n(y)}$
		              \item $\equiv\exp_n(x)\frac{\exp_n(0)}{\exp_n(y)}\err{\frac{g}{e}\cdot\frac{cX^2}{n}}$
		              \item $=\frac{\exp_n(x)}{\exp_n(y)}$.
		            \end{enumerate}
		            \item \textbf{Hence show that $\exp_n(x-y)\equiv\frac{\exp_n(x)}{\exp_n(y)}\err{\frac{cX^2}{n}+\frac{gcX^2}{en}}\err{\frac{a}{n}}$.}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle a,b,p\rangle$.}
		        \end{enumerate}
		    \procedure{3.16}
			    \objective
				    Choose a rational number $X\ge 0$. The objective of the following instructions is to construct positive rational numbers $a,b$ and a procedure, $p(x,k,n)$, to show that $\exp_n(kx)\equiv\exp_n(x)^k\err{\frac{ak}{n}}$ when a complex number $x$, and non-negative integers $k,n$ such that $n>b$ and $\lVert kx\rVert^2\le X^2$ are chosen.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{3.13} on $\langle X\rangle$ and let $\langle c,d,q\rangle$ receive.
					    \item Execute \procedurehr{3.15} on $\langle X\rangle$ and let $\langle e,f,t\rangle$ receive.
					    \item Let $a=ecX^2$
					    \item Let $b=\max(d,f)$.
					    \item Let $p(x,k,n)$ be the following procedure:
					    \begin{enumerate}
					      \item If $k>0$, then for $r\in[1:k]$ do the following:
					      \begin{enumerate}
					        \item Show that $\lVert xr\rVert^2\le\lVert kx\rVert^2\le X^2$.
						      \item Hence show that $\lVert\exp_{nr}(xr)\rVert^2\le c^2$ using procedure $q$.
							    \item Hence show that $\lVert\exp_n(x)^r\rVert^2=\lVert(1+\frac{x}{n})^{nr}\rVert^2=\lVert(1+\frac{xr}{nr})^{nr}\rVert^2=\lVert\exp_{nr}(xr)\rVert^2\le c^2$
						    \end{enumerate}
						    \item Hence using procedure $t$, show that $\exp_n(kx)$
						    \begin{enumerate}
						      \item $=\exp_n(x)^0\exp_n(kx)$
						      \item $\equiv\exp_n(x)^1\exp_n((k-1)x)\err{\frac{ceX^2}{n}}$
						      \item $\equiv\exp_n(x)^2\exp_n((k-2)x)\err{\frac{ceX^2}{n}}$
						      \item $\vdots$
						      \item $\equiv\exp_n(x)^k\exp_n((k-k)x)\err{\frac{ceX^2}{n}}$
						      \item $=\exp_n(x)^k$.
						    \end{enumerate}
						    \item \textbf{Hence show that $\exp_n(kx)\equiv\exp_n(x)^k\err{\frac{kceX^2}{n}}\err{\frac{ak}{n}}$.}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,b,p\rangle$.}
				    \end{enumerate}
		    \procedure{thu2507191307}
		      \objective
		        Choose a rational number $X\ge 0$. The objective of the following instructions is to construct positive rational numbers $a,b$, and a procedure $p(x,y,n)$ to show that $\exp_n(y)\equiv\exp_n(x)\err{a(x-y)}$ when two complex numbers $x,y$ and a positive integer $n>b$ such that $\lVert x\rVert^2\le X$ and $\lVert y\rVert^2\le X$ are chosen.
		      \implementation
		        \begin{enumerate}
		          \item Execute \procedurehr{3.13} on $\langle X\rangle$ and let $\langle c,b,q\rangle$ receive.
		          \item Let $a=c^2$.
		          \item Let $p(x,y,n)$ be the following procedure:
		          \begin{enumerate}
		            \item Using procedure $q$, show that $\exp_n(x)\equiv\exp_n(y)$
		            \begin{enumerate}
		              \item $\err{\exp_n(y)-\exp_n(x)}$
		              \item $\err{(1+\frac{y}{n})^n-(1+\frac{x}{n})^n}$
		              \item $\err{(\frac{y}{n}-\frac{x}{n})\sum_r^{[0:n]}(1+\frac{y}{n})^r(1+\frac{x}{n})^{n-1-r}}$
		              \item $\err{(y-x)(\frac{1}{n}\sum_r^{[0:n]}c^2)}$
		              \item $\err{a(y-x)}$.
		            \end{enumerate}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle a,b,p\rangle$.}
		        \end{enumerate}
		    \procedure{3.21}
			    \objective
				    Choose a rational number $X\ge 0$. The objective of the following instructions is to construct two rational numbers $a,N$, and a procedure, $p(x,n)$, to show that $\exp_n(x)\equiv\sum_r^{[0:{n+1}]}\frac{x^r}{r!}\err{\frac{a}{n}}$ when a complex number $x$ and an integer $n>N$ such that $\lVert x\rVert^2\le X^2$ are chosen.
			    \implementation
				    \begin{enumerate}
					    \item Let $N=\lfloor X\rfloor+1$.
					    \item Let $a=X^2(\sum_r^{[0:N]}\frac{X^r}{r!}+\frac{X^N}{N!}\cdot\frac{1}{1-\frac{X}{N}})$.
					    \item Let $p(x,n)$ be the following procedure:
					    \begin{enumerate}
						    \item Using \procedurehr{2.06}, \procedurehr{3.04}, \procedurehr{2.05}, and \procedurehr{2.07}, show that $\exp_n(x)\equiv\sum_r^{[0:{n+1}]}\frac{x^r}{r!}$
						    \begin{enumerate}
						      \item $\err{\sum_r^{[0:{n+1}]}\frac{x^r}{r!}-\exp_n(x)}$
							    \item $\err{\sum_r^{[0:{n+1}]}\frac{x^r}{r!}-\sum_r^{[0:{n+1}]}\frac{n^{\ul{r}}}{r!}\cdot\frac{x^r}{n^r}}$
							    \item $\err{\sum_r^{[1:{n+1}]}(1-\frac{n^{\ul{r}}}{n^r})\frac{x^r}{r!}}$
							    \item $\err{\sum_r^{[1:{n+1}]}(1-\frac{n^{\ul{r}}}{n^r})\frac{X^r}{r!}}$
							    \item $\err{\sum_r^{[2:{n+1}]}(1-\frac{(n-r+1)^{r}}{n^r})\frac{X^r}{r!}}$
							    \item $\err{\sum_r^{[2:{n+1}]}(1-(1-\frac{r-1}{n})^{r})\frac{X^r}{r!}}$
							    \item $\err{\sum_r^{[2:{n+1}]}(1-(1-\frac{(r-1)r}{n}))\frac{X^r}{r!}}$
							    \item $\err{\sum_r^{[2:{n+1}]}\frac{(r-1)r}{n}\frac{X^r}{r!}}$
							    \item $\err{\frac{1}{n}\sum_r^{[2:{n+1}]}\frac{X^r}{(r-2)!}}$
							    \item $\err{\frac{X^2}{n}\sum_r^{[0:{n-1}]}\frac{X^r}{r!}}$
							    \item $\err{\frac{X^2}{n}(\sum_r^{[0:N]}\frac{X^r}{r!}+\sum_r^{[N:{n-1}]}\frac{X^r}{r!})}$
							    \item $\err{\frac{X^2}{n}(\sum_r^{[0:N]}\frac{X^r}{r!}+\sum_r^{[N:{n-1}]}\frac{X^r}{N!N^{r-N}})}$
							    \item $\err{\frac{X^2}{n}(\sum_r^{[0:N]}\frac{X^r}{r!}+\frac{X^N}{N!}\sum_r^{[N:{n-1}]}\frac{X^{r-N}}{N^{r-N}})}$
							    \item $\err{\frac{X^2}{n}(\sum_r^{[0:N]}\frac{X^r}{r!}+\frac{X^N}{N!}\sum_r^{[0:{n-N-1}]}\frac{X^{r}}{N^{r}})}$
							    \item $\err{\frac{X^2}{n}(\sum_r^{[0:N]}\frac{X^r}{r!}+\frac{X^N}{N!}\cdot\frac{1}{1-\frac{X}{N}})}$
							    \item $\err{\frac{a}{n}}$.
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,N,p\rangle$.}
				    \end{enumerate}
	    \end{multicols}
	    \mfigure
		    \begin{minipage}[c]{0.65\textwidth}
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            \draw (0.000000, 0.100000) -- (0.000000,-0.100000);
            \draw (1.000000, 0.100000) -- (1.000000,-0.100000);
            \draw (2.000000, 0.100000) -- (2.000000,-0.100000);
            \draw (0.100000, -2.000000) -- (-0.100000,-2.000000);
            \draw (0.100000, -1.000000) -- (-0.100000,-1.000000);
            \draw (0.100000, 0.000000) -- (-0.100000,0.000000);
            \draw (0.100000, 1.000000) -- (-0.100000,1.000000);
            \draw (0.100000, 2.000000) -- (-0.100000,2.000000);
            \draw (1,-0.100000) node[anchor=north,fill=white]{$1$};
            \draw (-0.100000,1) node[anchor=east,fill=white]{$i$};
            \draw (1.000000,0.000000) node[anchor=south west,fill=white]{$(1+\frac{4i}{10})^{0}$};
            \filldraw (1.000000,0.000000) circle (2pt);
            \draw (1.000000,0.000000) -- (1.000000,0.400000);
            \filldraw (1.000000,0.400000) circle (2pt);
            \draw (1.000000,0.400000) node[anchor=south west,fill=white]{$(1+\frac{4i}{10})^{1}$};
            \filldraw (1.000000,0.400000) circle (2pt);
            \draw (1.000000,0.400000) -- (0.840000,0.800000);
            \filldraw (0.840000,0.800000) circle (2pt);
            \draw (0.840000,0.800000) node[anchor=south west,fill=white]{$(1+\frac{4i}{10})^{2}$};
            \filldraw (0.840000,0.800000) circle (2pt);
            \draw (0.840000,0.800000) -- (0.520000,1.136000);
            \filldraw (0.520000,1.136000) circle (2pt);
            \draw (0.520000,1.136000) node[anchor=south west,fill=white]{$(1+\frac{4i}{10})^{3}$};
            \filldraw (0.520000,1.136000) circle (2pt);
            \draw (0.520000,1.136000) -- (0.065600,1.344000);
            \filldraw (0.065600,1.344000) circle (2pt);
            \draw (0.065600,1.344000) node[anchor=south west,fill=white]{$(1+\frac{4i}{10})^{4}$};
            \filldraw (0.065600,1.344000) circle (2pt);
            \draw (0.065600,1.344000) -- (-0.472000,1.370240);
            \filldraw (-0.472000,1.370240) circle (2pt);
            \draw (-0.472000,1.370240) node[anchor=south east,fill=white]{$(1+\frac{4i}{10})^{5}$};
            \filldraw (-0.472000,1.370240) circle (2pt);
            \draw (-0.472000,1.370240) -- (-1.020096,1.181440);
            \filldraw (-1.020096,1.181440) circle (2pt);
            \draw (-1.020096,1.181440) node[anchor=south east,fill=white]{$(1+\frac{4i}{10})^{6}$};
            \filldraw (-1.020096,1.181440) circle (2pt);
            \draw (-1.020096,1.181440) -- (-1.492672,0.773402);
            \filldraw (-1.492672,0.773402) circle (2pt);
            \draw (-1.492672,0.773402) node[anchor=south east,fill=white]{$(1+\frac{4i}{10})^{7}$};
            \filldraw (-1.492672,0.773402) circle (2pt);
            \draw (-1.492672,0.773402) -- (-1.802033,0.176333);
            \filldraw (-1.802033,0.176333) circle (2pt);
            \draw (-1.802033,0.176333) node[anchor=south east,fill=white]{$(1+\frac{4i}{10})^{8}$};
            \filldraw (-1.802033,0.176333) circle (2pt);
            \draw (-1.802033,0.176333) -- (-1.872566,-0.544480);
            \filldraw (-1.872566,-0.544480) circle (2pt);
            \draw (-1.872566,-0.544480) node[anchor=north east,fill=white]{$(1+\frac{4i}{10})^{9}$};
            \filldraw (-1.872566,-0.544480) circle (2pt);
            \draw (-1.872566,-0.544480) -- (-1.654774,-1.293507);
            \filldraw (-1.654774,-1.293507) circle (2pt);
            \draw (-1.654774,-1.293507) node[anchor=north east,fill=white]{$(1+\frac{4i}{10})^{10}$};
            \filldraw (-1.654774,-1.293507) circle (2pt);
			    \end{tikzpicture}
		    \end{minipage}
		    \begin{minipage}[c]{0.35\textwidth}
			    A plot of the list of complex numbers $(1+\frac{4\iu}{10})^{[0:11]}$. Notice that each multiplication of a complex number by $1+\frac{4\iu}{10}$ results in an anti-clockwise rotation about the origin and a small radial movement outwards. This can be seen to reflect the computation $(1+\frac{4\iu}{10})a=1a+\frac{4}{10}(a\iu)$ after one notes that $a\iu$ is perpendicular to $a$. Also note that each line segment has a length of roughly $\frac{4}{10}$ units. Hence the entire path has a length of approximately $10*\frac{4}{10}=4$ units.
		    \end{minipage}
	    \begin{multicols*}{2}
		    \declaration{3.17}
			    The notation \gls{complexcos}, where $z$ is a complex number and $n$ is a positive integer, will be used as a shorthand for $\frac{\exp_n(\iu z)+\exp_n(-\iu z)}{2}$.
		    \procedure{3.22}
			    \objective
				    Choose a rational number $x$ and a positive integer $n$. The objective of the following instructions is to show that $\Re(\exp_n(\iu x))=\cos_n(x)$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Re(\exp_n(\iu x))$
					    \begin{enumerate}
						    \item $=\frac{\exp_n(\iu x)+\conj{\exp_n(\iu x)}}{2}$
						    \item $=\frac{\exp_n(\iu x)+\exp_n(\conj{\iu x})}{2}$
						    \item $=\frac{\exp_n(\iu x)+\exp_n(-\iu x)}{2}$
						    \item $=\cos_n(x)$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{3.18}
			    The notation \gls{complexsin}, where $z$ is a complex number and $n$ is a positive integer, will be used as a shorthand for $\frac{\exp_n(\iu z)-\exp_n(-\iu z)}{2\iu}$.
		    \procedure{3.23}
			    \objective
				    Choose a rational number $x$ and a positive integer $n$. The objective of the following instructions is to show that $\Im(\exp_n(\iu x))=\sin_n(x)$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\Im(\exp_n(\iu x))$
					    \begin{enumerate}
						    \item $=\frac{\exp_n(\iu x)-\conj{\exp_n(\iu x)}}{2i}$
						    \item $=\frac{\exp_n(\iu x)-\exp_n(\conj{\iu x})}{2i}$
						    \item $=\frac{\exp_n(\iu x)-\exp_n(-\iu x)}{2i}$
						    \item $=\sin_n(x)$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{3.24}
			    \objective
				    Choose a rational number $X\ge 0$. The objective of the following instructions is to construct two rational numbers $a,b$, and a procedure, $p(x,y,n)$, to show that $\cos_n(x+y)\equiv\cos_n(x)\cos_n(y)-\sin_n(x)\sin_n(y)\err{\frac{axy}{n}}\err{\frac{aX^2}{n}}$ when two complex numbers $x,y$ and a positive integer $n>b$ such that $\lVert x\rVert^2\le X^2$ and $\lVert y\rVert^2\le X^2$ are chosen.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{3.15} on $\langle X\rangle$ and let $\langle a,b,q\rangle$ receive.
					    \item Let $p(x,y,n)$ be the following procedure:
					    \begin{enumerate}
					      \item Using procedure $q$, show that $\cos_n(x+y)$
					      \begin{enumerate}
					        \item $=\frac{1}{2}(\exp_n(\iu(x+y))+\exp_n(-\iu(x+y)))$
					        \item $\equiv\frac{1}{2}(\exp_n(\iu x)\exp_n(\iu y)+\exp_n(-\iu(x+y)))\err{\frac{a(\iu x)(\iu y)}{2n}}$
					        \item $\equiv\frac{1}{2}(\exp_n(\iu x)\exp_n(\iu y)+\exp_n(-\iu x)\exp_n(-\iu y))\allowbreak\err{\frac{a(-\iu x)(-\iu y)}{2n}}$
					        \item $=\frac{1}{4}(\exp_n(\iu x)\exp_n(\iu y)+\exp_n(-\iu x)\exp_n(-\iu y))+\frac{1}{4}(\exp_n(\iu x)\exp_n(\iu y)+\exp_n(-\iu x)\exp_n(-\iu y))$
					        \item $=\frac{1}{4}(\exp_n(\iu x)(\exp_n(\iu y)+\exp_n(-\iu y))+(\exp_n(-\iu x)-\exp_n(\iu x))\exp_n(-\iu y))+\frac{1}{4}((\exp_n(\iu x)-\exp_n(-\iu x))\exp_n(\iu y)+\exp_n(-\iu x)(\exp_n(\iu y)+\exp_n(-\iu y)))$
					        \item $=\frac{1}{2}\exp_n(\iu x)\cos_n(y)+\frac{1}{2\iu}\sin_n(x)\exp_n(-\iu y)-\frac{1}{2\iu}\sin_n(x)\exp_n(\iu y)+\frac{1}{2}\exp_n(-\iu x)\cos_n(y)$
					        \item $=\cos_n(x)\cos_n(y)-\sin_n(x)\sin_n(y)$
					      \end{enumerate}
					      \item \textbf{Hence show that $\cos_n(x+y)\equiv\cos_n(x)\cos_n(y)-\sin_n(x)\sin_n(y)\err{\frac{axy}{n}}\err{\frac{aX^2}{n}}$.}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,b,p\rangle$.}
				    \end{enumerate}
		    \procedure{3.25}
			    \objective
				    Choose a rational number $X\ge 0$. The objective of the following instructions is to construct two rational numbers $a,b$, and a procedure, $p(x,y,n)$, to show that $\sin_n(x+y)\equiv\sin_n(x)\cos_n(y)-\cos_n(x)\sin_n(y)\err{\frac{axy}{n}}\err{\frac{aX^2}{n}}$ when two complex numbers $x,y$ and a positive integer $n>b$ such that $\lVert x\rVert^2\le X^2$ and $\lVert y\rVert^2\le X^2$ are chosen.
			    \implementation
				    Implementation is analogous to that of \procedurehr{3.24}.
		    \procedure{3.26}
			    \objective
				     Choose a rational number $X\ge 0$. The objective of the following instructions is to construct two rational numbers $a,b$, and a procedure, $p(x,n)$, to show that $\cos_n(x)^2+\sin_n(x)^2\equiv 1\err{\frac{a\lVert x\rVert^2}{n}}\err{\frac{aX^2}{n}}$ when a complex number $x$ and a positive integer $n$ such that $\lVert x\rVert^2\le X^2$ and $n>b$ are chosen.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{3.15} on $\langle X\rangle$ and let $\langle a,b,q\rangle$ receive.
					    \item Let $p(x,n)$ be the following procedure:
					    \begin{enumerate}
					      \item Using procedure $q$, show that $\cos_n(x)^2+\sin_n(y)^2$
					      \begin{enumerate}
					        \item $=\frac{1}{4}(\exp_n(\iu x)+\exp_n(-\iu x))^2+\frac{1}{4\iu^2}(\exp_n(\iu x)-\exp_n(-\iu x))^2$
					        \item $=\frac{1}{4}(\exp_n(\iu x)^2+2\exp_n(\iu x)\exp_n(-\iu x)+\exp_n(-\iu x)^2-\exp_n(\iu x)^2+2\exp_n(\iu x)\exp_n(-\iu x)-\exp_n(-\iu x)^2)$
					        \item $=\exp_n(\iu x)\exp_n(-\iu x)$
					        \item $\equiv 1\err{\frac{a(-\iu x)(\iu x)}{n}}$.
					      \end{enumerate}
					      \item \textbf{Hence show that $\cos_n(x)^2+\sin_n(y)^2\equiv 1\err{\frac{a\lVert x\rVert^2}{n}}\err{\frac{aX^2}{n}}$.}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,b,p\rangle$.}
				    \end{enumerate}
		    \procedure{sat0308190647}
		      \objective
		        Choose a rational number $X\ge 0$. The objective of the following instructions is to construct two rational numbers $a,b$, and a procedure, $p(x,y,n)$, to show that $\lVert x\exp_n(\iu y)\rVert^2\equiv\lVert x\rVert^2\err{\frac{a\lVert x\rVert^2\lVert y\rVert^2}{n}}\err{\frac{a\lVert x\rVert^2X^2}{n}}$ when a complex number $x$, a rational number $y$, and a positive integer $n$ such that $\lVert y\rVert^2\le X^2$ and $n>b$ are chosen.
		      \implementation
		        \begin{enumerate}
		          \item Execute \procedurehr{3.26} on $\langle X\rangle$ and let $\langle a,b,q\rangle$ receive.
		          \item Let $p(x,y,n)$ be the following procedure:
		          \begin{enumerate}
		            \item Using procedure $q$, \procedurehr{3.22}, and \procedurehr{3.23}, show that $\lVert x\exp_n(\iu y)\rVert^2$
		            \begin{enumerate}
		              \item $=\lVert x\rVert^2\lVert\exp_n(\iu y)\rVert^2$
		              \item $=\lVert x\rVert^2\lVert\cos_n(y)+\iu\sin_n(y)\rVert^2$
		              \item $=\lVert x\rVert^2(\cos_n(y)^2+\sin_n(y)^2)$
		              \item $\equiv\lVert x\rVert^2\cdot 1\err{\lVert x\rVert^2\cdot\frac{a\lVert y\rVert^2}{n}}$.
		            \end{enumerate}
		            \item \textbf{Hence show that $\lVert x\exp_n(\iu y)\rVert^2\equiv\lVert x\rVert^2\err{\frac{a\lVert xy\rVert^2}{n}}\err{\frac{a\lVert x\rVert^2X^2}{n}}$.}
		          \end{enumerate}
		        \end{enumerate}
		    \procedure{3.29}
			    \objective
				    Choose a rational number $X\ge 0$. The objective of the following instructions is to construct two rational numbers $a,N$, and a procedure, $p(x,n)$, to show that $\cos_n(x)\equiv\sum_r^{[0:\lceil\frac{n}{2}\rceil]}\frac{(-1)^rx^{2r}}{(2r)!}\err{\frac{a}{n}}$ when a complex number $x$ and an integer $n>N$ such that $\lVert x\rVert^2\le X^2$ is chosen.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{3.21} on $\langle X\rangle$ and let $\langle a,N,q\rangle$ receive.
					    \item Let $p(x,n)$ be the following procedure:
					    \begin{enumerate}
						    \item Using procedure $q$, show that $\cos_n(x)$
						    \begin{enumerate}
						      \item $=\frac{\exp_n(\iu x)}{2}+\frac{\exp_n(-\iu x)}{2}$
						      \item $\equiv\frac{1}{2}\sum_r^{[0:{n+1}]}\frac{(\iu x)^{r}}{r!}+\frac{\exp_n(-\iu x)}{2}\err{\frac{a}{2n}}$
						      \item $\equiv\frac{1}{2}\sum_r^{[0:{n+1}]}\frac{(\iu x)^{r}}{r!}+\frac{1}{2}\sum_r^{[0:{n+1}]}\frac{(-\iu x)^{r}}{r!}\err{\frac{a}{2n}}$
						      \item $=\sum_r^{[0:{n+1}]}\frac{(\iu^r+(-\iu)^r)x^{r}}{2(r!)}$
						      \item $=\sum_r^{[0:{n+1}]}\frac{[r\bmod 2=0](-1)^{\frac{r}{2}}x^{r}}{r!}$
						      \item $=\sum_r^{[0:{\lceil\frac{n}{2}\rceil}]}\frac{(-1)^rx^{2r}}{(2r)!}$.
						    \end{enumerate}
						    \item \textbf{Hence show that $\cos_n(x)\equiv\sum_r^{[0:{\lceil\frac{n}{2}\rceil}]}\frac{(-1)^rx^{2r}}{(2r)!}\err{\frac{a}{n}}$.}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{3.30}
			    \objective
				    Choose a rational number $X\ge 0$. The objective of the following instructions is to construct two rational numbers $a,N$, and a procedure, $p(x,n)$, to show that $\sin_n(x)\equiv\sum_r^{[0:\lfloor\frac{n+1}{2}\rfloor]}\frac{(-1)^rx^{2r+1}}{(2r+1)!}\err{\frac{a}{n}}$ when a complex number $x$ and an integer $n>N$ such that $\lVert x\rVert^2\le X^2$ is chosen.
			    \implementation
				    Implementation is analogous to that of \procedurehr{3.29}.
	    \end{multicols*}
	  \chapter{Binomial and Mercator Series}
	    \begin{multicols*}{2}
		    \declaration{sun2107190610}
			    The notation \gls{complexbinomialseries}, where $x,a$ are complex numbers and $n$ is a positive integer, will be used as a shorthand for $\sum_r^{[0:n]}\binom{a}{r}x^r$.
	      \procedure{sun2107190619}
	        \objective
	          Choose a complex number $x$ and two non-negative integers $a,n$ such that $n>a$. The objective of the following instructions is to show that $(1+x)_n^a=(1+x)^a$.
	        \implementation
	          \begin{enumerate}
	            \item Using \procedurehr{sun2107190636}, show that $(1+x)^a_n=$
	            \begin{enumerate}
	              \item $=\sum_r^{[0:n]}\binom{a}{r}x^r$
	              \item $=\sum_r^{[0:n]}\frac{a^{\ul{r}}}{r!}x^r$
	              \item $=\sum_r^{[0:a+1]}\frac{a^{\ul{r}}}{r!}x^r+\sum_r^{[a+1:n]}\frac{a^{\ul{r}}}{r!}x^r$
	              \item $=\sum_r^{[0:a+1]}\frac{a^{\ul{r}}}{r!}x^r+\sum_r^{[a+1:n]}\frac{0}{r!}x^r$
	              \item $=\sum_r^{[0:a+1]}\binom{a}{r}x^r$
	              \item $=(1+x)^a$.
	            \end{enumerate}
	          \end{enumerate}
	      \procedure{sun2107190640}
	        \objective
	          Choose two complex numbers $x,y$ and a positive integer $N$. The objective of the following instructions is to show that $\binom{x+y}{N}=\sum_k^{N+1}\binom{x}{k}\binom{y}{N-k}$.
	        \implementation
	          \begin{enumerate}
	            \item If $N=0$, then do the following:
	            \begin{enumerate}
	              \item Show that $\binom{x+y}{N}=1=\sum_k^{[0:N+1]}\binom{x}{k}\binom{y}{N-k}$.
	            \end{enumerate}
	            \item Otherwise do the following:
	            \begin{enumerate}
	              \item Show that $N>0$.
	              \item Show that $\binom{x+y-1}{N-1}=\sum_k^{[0:N]}\binom{x-1}{k}\binom{y}{N-1-k}$ using \procedurehr{sun2107190640}.
	              \item Show that $\binom{x+y-1}{N-1}=\sum_k^{[0:N]}\binom{x}{k}\binom{y-1}{N-1-k}$ using \procedurehr{sun2107190640}.
	              \item Hence show that $\binom{x+y}{N}$
	              \begin{enumerate}
	                \item $=\frac{x+y}{N}\binom{x+y-1}{N-1}$
	                \item $=\frac{x}{N}\binom{x+y-1}{N-1}+\frac{y}{N}\binom{x+y-1}{N-1}$
	                \item $=\frac{x}{N}\sum_k^{[0:N]}\binom{x-1}{k}\binom{y}{N-1-k}+\frac{y}{N}\sum_k^{[0:N]}\binom{x}{k}\binom{y-1}{N-1-k}$
	                \item $=\frac{x}{N}\sum_k^{[1:N+1]}\binom{x-1}{k-1}\binom{y}{N-k}+\frac{y}{N}\sum_k^{[0:N]}\binom{x}{k}\binom{y-1}{N-1-k}$
	                \item $=\sum_k^{[0:N+1]}\frac{k}{N}\binom{x}{k}\binom{y}{N-k}+\sum_k^{[0:N+1]}\frac{N-k}{N}\binom{x}{k}\binom{y}{N-k}$
	                \item $=\sum_k^{[0:N+1]}\binom{x}{k}\binom{y}{N-k}$.
	              \end{enumerate}
	            \end{enumerate}
	          \end{enumerate}
	      \procedure{sun2107191133}
	        \objective
	          Choose complex numbers $a,b,x$ and a natural number $n$. The objective of the following instructions is to show that $(1+x)_n^a(1+x)_n^b-(1+x)_n^{a+b}=\sum_k^{[1:n]}\sum_r^{[k:n]}\binom{a}{k+n-1-r}\binom{b}{r}x^{k+n-1}$.
	        \implementation
	          \begin{enumerate}
	            \item Show that $(1+x)_n^a(1+x)_n^b-(1+x)_n^{a+b}$
	            \begin{enumerate}
	              \item $=(\sum_k^{[0:n]}\binom{a}{k}x^k)(\sum_r^{[0:n]}\binom{b}{r}x^r)-\sum_k^{[0:n]}\binom{a+b}{k}x^k$
	              \item $=\sum_k^{[0:n]}\sum_r^{[0:n]}\binom{a}{k}\binom{b}{r}x^{k+r}-\sum_k^{[0:n]}\binom{a+b}{k}x^k$
	              \item $=\sum_k^{[0:n]}\sum_r^{[0:k+1]}\binom{a}{k-r}\binom{b}{r}x^k+\sum_k^{[n:2n-1]}\sum_r^{[k-n+1:n]}\binom{a}{k-r}\binom{b}{r}x^k-\sum_k^{[0:n]}\binom{a+b}{k}x^k$
	              \item $=\sum_k^{[0:n]}\binom{a+b}{k}x^k+\sum_k^{[1:n]}\sum_r^{[k:n]}\binom{a}{k+n-1-r}\binom{b}{r}x^{k+n-1}-\sum_k^{[0:n]}\binom{a+b}{k}x^k$
	              \item $=\sum_k^{[1:n]}\sum_r^{[k:n]}\binom{a}{k+n-1-r}\binom{b}{r}x^{k+n-1}$.
	            \end{enumerate}
	          \end{enumerate}
	      \procedure{sun2107191247}
	        \objective
	          Choose two rational numbers $A>0$ and $0<X<1$. The objective of the following instructions is to construct rational numbers $Y>0$, $0<Z<1$ and a procedure $p(a,x,n)$ to show that $\lVert\binom{a}{n}x^n\rVert^2\le(YZ^n)^2$ when complex numbers $a,x$ such that $\lVert a+1\rVert^2<A^2$ and $\lVert x\rVert^2<X^2$ are chosen.
	        \implementation
	          \begin{enumerate}
	            \item Let $e=\frac{AX}{1-X}-1$.
	            \item Let $d=\lfloor\frac{AX}{1-X}\rfloor$.
	            \item Show that $d>e>-1$.
	            \item Let $Z=(1+\frac{A}{d+1})X$.
	            \item Show that $0<Z<(1+\frac{A}{e+1})X=1$.
	            \item Let $Y=Z^{-d}\prod_k^{[0:d]}\frac{(A+k+1)X}{k+1}=Z^{-d}\prod_k^{[0:d]}X(1+\frac{A}{k+1})$.
	            \item Let $p(a,x,n)$ be the following procedure:
	            \begin{enumerate}
	              \item Show that $\lvert\Re(a+1)\rvert\le A$ given that $\Re(a+1)^2\le\lVert a+1\rVert^2\le A^2$.
	              \item Hence show that $\lVert\binom{a}{n}x^n\rVert^2$
	              \begin{enumerate}
	                \item $=\lVert\frac{a^{\ul{n}}}{n!}x^n\rVert^2$
	                \item $=\lVert\prod_k^{[0:n]}(\frac{a+1-(k+1)}{k+1}\cdot x)\rVert^2$
	                \item $=\prod_k^{[0:n]}\frac{\lVert(a+1)-(k+1)\rVert^2\lVert x\rVert^2}{(k+1)^2}$
	                \item $=\prod_k^{[0:n]}\frac{(\lVert a+1\rVert^2-2\Re(a+1)(k+1)+(k+1)^2)\lVert x\rVert^2}{(k+1)^2}$
	                \item $\le\prod_k^{[0:n]}\frac{(A^2+2A(k+1)+(k+1)^2)X^2}{(k+1)^2}$
	                \item $=(\prod_k^{[0:n]}\frac{(A+k+1)X}{k+1})^2$
	                \item $=(\prod_k^{[0:n]}X(1+\frac{A}{k+1}))^2$.
	              \end{enumerate}
	              \item If $n\le d$, then do the following:
	              \begin{enumerate}
	                \item Show that $\lVert\binom{a}{n}x^n\rVert^2$
	                \begin{enumerate}
	                  \item $\le(\prod_k^{[0:n]}X(1+\frac{A}{k+1}))^2$
	                  \item $=(\prod_k^{[0:d]}X(1+\frac{A}{k+1}))^2(\prod_k^{[n:d]}X(1+\frac{A}{k+1}))^{-2}$
	                  \item $\le(\prod_k^{[0:d]}X(1+\frac{A}{k+1}))^2(X(1+\frac{A}{d+1}))^{-2(d-n)}$
	                  \item $=Y^2Z^{2n}$.
	                \end{enumerate}
	              \end{enumerate}
	              \item Otherwise do the following:
	              \begin{enumerate}
	                \item Show that $\lVert\binom{a}{n}x^n\rVert^2$
	                \begin{enumerate}
	                  \item $\le(\prod_k^{[0:n]}X(1+\frac{A}{k+1}))^2$
	                  \item $=(\prod_k^{[0:d]}X(1+\frac{A}{k+1}))^2(\prod_k^{[d:n]}X(1+\frac{A}{k+1}))^2$
	                  \item $\le(\prod_k^{[0:d]}X(1+\frac{A}{k+1}))^2(X(1+\frac{A}{d+1}))^{2(n-d)}$
	                  \item $=Y^2Z^{2n}$.
	                \end{enumerate}
	              \end{enumerate}
	            \end{enumerate}
	            \item \textbf{Yield the tuple $\langle Y,Z,p\rangle$.}
	          \end{enumerate}
	      \procedure{wed2407191422}
	        \objective
	          Choose a rational number $0<X<1$ and a positive integer $k$. The objective of the following instructions is to construct rational numbers $Y>0$, $0<Z<1$ and a procedure $p(x,n)$ to show that $\lVert n^kx^n\rVert^2\le(YZ^n)^2$ when a complex number $x$ such that $\lVert x\rVert^2\le X^2$ is chosen.
	        \implementation
	          \begin{enumerate}
	            \item Let $e=\frac{k}{1-X}-1$.
	            \item Let $d=\lfloor\frac{k}{1-X}\rfloor$.
	            \item Show that $d>e>k-1$.
	            \item Let $Z=(1+\frac{1}{d})^kX$.
	            \item Show that $Z<(1+\frac{1}{e})^kX$.
	            \item Now show that $0<Z<(1+\frac{1}{e})^kX\le\frac{1+\frac{1}{e}}{1-(k-1)\frac{1}{e}}\cdot X=1$ using \procedurehr{wed2407191348}.
	            \item Let $Y=Z^{-d}X\prod_r^{[1:d]}X(1+\frac{1}{r})^k$.
	            \item Let $p(x,n)$ be the following procedure:
	            \begin{enumerate}
	              \item Show that $\lVert n^kx^n\rVert^2$
	              \begin{enumerate}
	                \item $\le\lVert x\prod_r^{[1:n]}x\cdot\frac{(r+1)^k}{r^k}\rVert^2$
	                \item $=\lVert x\rVert^2\prod_r^{[1:n]}\lVert x\rVert^2(\frac{(r+1)^k}{r^k})^2$
	                \item $\le X^2\prod_r^{[1:n]}((1+\frac{1}{r})^kX)^2$.
	              \end{enumerate}
	              \item If $n\le d$, then do the following:
	              \begin{enumerate}
	                \item Show that $\lVert n^kx^n\rVert^2$
	                \begin{enumerate}
	                  \item $\le X^2(\prod_r^{[1:n]}X(1+\frac{1}{r})^k)^2$
	                  \item $=X^2(\prod_r^{[1:d]}X(1+\frac{1}{r})^k)^2\cdot(\prod_r^{[n:d]}X(1+\frac{1}{r})^k)^{-2}$
	                  \item $\le X^2(\prod_r^{[1:d]}X(1+\frac{1}{r})^k)^2(X(1+\frac{1}{d})^k)^{-2(d-n)}$
	                  \item $=Y^2Z^{2n}$.
	                \end{enumerate}
	              \end{enumerate}
	              \item Otherwise do the following:
	              \begin{enumerate}
	                \item Show that $\lVert n^kx^n\rVert^2$
	                \begin{enumerate}
	                  \item $\le X^2(\prod_r^{[1:n]}X(1+\frac{1}{r})^k)^2$
	                  \item $=X^2(\prod_r^{[1:d]}X(1+\frac{1}{r})^k)^2(\prod_r^{[d:n]}X(1+\frac{1}{r})^k)^2$
	                  \item $\le X^2(\prod_r^{[1:d]}X(1+\frac{1}{r})^k)^2(X(1+\frac{1}{d})^k)^{2(n-d)}$
	                  \item $=Y^2Z^{2n}$.
	                \end{enumerate}
	              \end{enumerate}
	            \end{enumerate}
	            \item \textbf{Yield the tuple $\langle Y,Z,p\rangle$.}
	          \end{enumerate}
	      \procedure{wed2407191521}
	        \objective
	          Choose two rational numbers $A>0$, $1>X>0$. The objective of the following instructions is to construct rational numbers $D>0$, $0<G<1$, and a procedure $p(x,a,b,n)$ to show that $(1+x)_n^{a+b}\equiv(1+x)_n^a(1+x)_n^b\err{DG^n}$ when $\lVert x\rVert^2\le X$, and $\lVert a\rVert^2,\lVert b\rVert^2<A$.
	        \implementation
	          \begin{enumerate}
	            \item Execute \procedurehr{sun2107191247} on $\langle A,X\rangle$ and let $\langle B,C,q\rangle$ receive.
	            \item Execute \procedurehr{wed2407191422} on $\langle C,1\rangle$ and let $\langle F,G,t\rangle$ receive.
	            \item Let $D=\frac{B^2F}{1-C}$.
	            \item Let $p(x,a,b,n)$ be the following procedure:
	            \begin{enumerate}
	              \item For each $r\in[1:n]$, do the following:
	              \begin{enumerate}
	                \item Show that $\lVert\binom{a}{r}x^r\rVert^2\le(BC^r)^2$ using procedure $q$.
	                \item Show that $\lVert\binom{b}{r}x^r\rVert^2\le(BC^r)^2$ using procedure $q$, .
	              \end{enumerate}
	              \item Show that $\lVert nC^n\rVert^2\le(FG^n)^2$ using procedure $t$.
	              \item Hence show that $(1+x)_n^{a+b}\equiv(1+x)_n^a(1+x)_n^b$
	              \begin{enumerate}
	                \item $\err{(1+x)_n^a(1+x)_n^b-(1+x)_n^{a+b}}$
	                \item $\err{\sum_k^{[1:n]}\sum_r^{[k:n]}\binom{a}{k+n-1-r}\binom{b}{r}x^{k+n-1}}$
	                \item $\err{\sum_k^{[1:n]}\sum_r^{[k:n]}\binom{a}{k+n-1-r}x^{k+n-1-r}\binom{b}{r}x^r}$
	                \item $\err{\sum_k^{[1:n]}\sum_r^{[k:n]}BC^{k+n-1-r}BC^r}$
	                \item $\err{B^2C^n\sum_k^{[1:n]}\sum_r^{[k:n]}C^{k-1}}$
	                \item $\err{B^2C^n\sum_r^{[1:n]}\sum_k^{[1:r+1]}C^{k-1}}$
	                \item $\err{B^2C^n\sum_r^{[1:n]}\frac{1}{1-C}}$
	                \item $\err{\frac{B^2}{1-C}\cdot nC^n}$
	                \item $\err{\frac{B^2F}{1-C}G^n}$
	                \item $\err{DG^n}$.
	              \end{enumerate}
	            \end{enumerate}
	            \item \textbf{Yield the tuple $\langle D,G,p\rangle$.}
	          \end{enumerate}
	      \procedure{wed2407191611}
	        \objective
	          Choose two rational numbers $A>0$, $1>X>0$. The objective of the following instructions is to construct a rational number $D$ and a procedure $p(x,n,a,k)$ to show that $\lVert((1+x)_n^a)^k\rVert^2<D^2$ when complex numbers $x,a$ and positive integers $n,k$ such that $\lVert x\rVert^2<X^2$ and $\lVert ka\rVert^2<A^2$.
	        \implementation
	          \begin{enumerate}
	            \item Execute \procedurehr{3.13} on $\langle\frac{ABX}{1-C}\rangle$ and let $\langle E,N,t\rangle$ receive.
	            \item Execute \procedurehr{sun2107191247} on $\langle A+1,X\rangle$ and let $\langle B,C,q\rangle$ receive.
	            \item Let $D=\max(E,(1+\frac{ABX}{1-C})^{\lfloor N\rfloor})$.
	            \item Let $p(x,n,a,k)$ be the following procedure:
	            \begin{enumerate}
	              \item For each $r\in[1:n]$, do the following:
	              \begin{enumerate}
	                \item Show that $\lVert a\rVert^2\le\lVert ka\rVert^2\le A^2$.
	                \item Show that $\lVert a-1\rVert^2\le(A+1)^2$.
	                \item Hence show that $\lVert\binom{a-1}{r-1}x^{r-1}\rVert^2\le(BC^r)^2$ using procedure $q$.
	              \end{enumerate}
	              \item Hence show that $\lVert k\sum_r^{[1:n]}\binom{a}{r}x^r\rVert^2$
	              \begin{enumerate}
	                \item $=\lVert k\sum_r^{[1:n]}\frac{a}{r}\binom{a-1}{r-1}x^r\rVert^2$
	                \item $=\lVert kax\sum_r^{[1:n]}\frac{1}{r}\binom{a-1}{r-1}x^{r-1}\rVert^2$
	                \item $\le(AX\sum_r^{[1:n]}BC^{r-1})^2$
	                \item $\le(\frac{ABX}{1-C})^2$.
	              \end{enumerate}
	              \item If $k>N$, then do the following:
	              \begin{enumerate}
	                \item Hence using procedure $t$, show that $\lVert((1+x)_n^a)^k\rVert^2$
	                \begin{enumerate}
	                  \item $=\lVert(\sum_r^{[0:n]}\binom{a}{r}x^r)^k\rVert^2$
	                  \item $=\lVert(1+\sum_r^{[1:n]}\binom{a}{r}x^r)^k\rVert^2$
	                  \item $=\lVert\exp_k(k\sum_r^{[1:n]}\binom{a}{r}x^r)\rVert^2$
	                  \item $\le E^2$
	                  \item $\le D^2$.
	                \end{enumerate}
	              \end{enumerate}
	              \item Otherwise do the following:
	              \begin{enumerate}
	                \item Show that $\lVert\sum_r^{[1:n]}\binom{a}{r}x^r)^k\rVert^2$
	                \begin{enumerate}
	                  \item $\le\lVert k\sum_r^{[1:n]}\binom{a}{r}x^r\rVert^2$
	                  \item $\le(\frac{ABX}{1-C})^2$.
	                \end{enumerate}
	                \item Hence show that $\lVert((1+x)_n^a)^k\rVert^2$
	                \begin{enumerate}
	                  \item $=(\lVert(1+x)_n^a\rVert^2)^k$
	                  \item $=(\lVert 1+\sum_r^{[1:n]}\binom{a}{r}x^r\rVert^2)^k$
	                  \item $\le(1+\frac{ABX}{1-C})^{2k}$
	                  \item $\le D^2$.
	                \end{enumerate}
	              \end{enumerate}
	            \end{enumerate}
	            \item \textbf{Yield $\langle D,p\rangle$.}
	          \end{enumerate}
	      \procedure{tue2008190712}
	        \objective
	          Choose two rational numbers $A>0$, $1>X>0$. The objective of the following instructions is to construct positive rational numbers $D,N$, and a procedure $p(x,a,n)$ to show that $\lVert(1+x)_n^a\rVert^2\ge D^2$ when complex numbers $x,a$ and an integer $n$ such that $\lVert x\rVert^2\le X^2$, $\lVert a\rVert\le A^2$, and $n>N$ are chosen.
	        \implementation
	          \begin{enumerate}
	            \item Execute \procedurehr{wed2407191521} on $\langle A,X\rangle$ and let $\langle a_1,b_1,p_1\rangle$ receive.
	            \item Execute \procedurehr{wed2407191422} on $\langle b_1,1\rangle$ and let $\langle a_2,b_2,p_2\rangle$ receive.
	            \item Execute \procedurehr{wed2407191611} on $\langle A,X\rangle$ and let $\langle a_3,p_3\rangle$ receive.
	            \item Let $D=\frac{1}{2a_3}$.
	            \item Let $N=2a_1a_2$.
	            \item Let $p(x,a,n)$ be the following procedure:
	            \begin{enumerate}
	              \item Show that $\lVert n{b_1}^n\rVert^2\le(a_2{b_2}^n)^2\le{a_2}^2$ using procedure $p_2$.
	              \item Hence show that $(a_1{b_1}^n)^2\le(\frac{a_1a_2}{n})^2$.
	              \item Show that $\lVert((1+x)_n^{-a})^1\rVert^2\le{a_3}^2$ using procedure $p_3$.
	              \item Using procedure $p_1$, show that $\lVert(1+x)_n^{-a}(1+x)_n^a-1\rVert^2$
	              \begin{enumerate}
	                \item $=\lVert(1+x)_n^{-a}(1+x)_n^a-(1+x)_n^{-a+a}\rVert^2$
	                \item $\le(a_1{b_1}^n)^2$.
	              \end{enumerate}
	              \item Hence using \procedurehr{3.05}, show that $\frac{1}{2}-\lVert(1+x)_n^{-a}(1+x)_n^a\rVert^2$
	              \begin{enumerate}
	                \item $=\frac{1}{2}\lVert 1\rVert^2-\lVert(1+x)_n^{-a}(1+x)_n^a\rVert^2$
	                \item $\le\lVert 1-(1+x)_n^{-a}(1+x)_n^a\rVert^2$
	                \item $\le(a_1{b_1}^n)^2$
	                \item $\le(\frac{a_1a_2}{n})^2$
	                \item $\le\frac{1}{4}$.
	              \end{enumerate}
	              \item Hence show that $(\frac{1}{2})^2$
	              \begin{enumerate}
	                \item $\le\lVert(1+x)_n^{-a}(1+x)_n^a\rVert^2$
	                \item $\le{a_3}^2\lVert(1+x)_n^a\rVert^2$.
	              \end{enumerate}
	              \item \textbf{Hence show that $D^2\le\lVert(1+x)_n^a\rVert^2$.}
	            \end{enumerate}
	            \item \textbf{Yield the tuple $\langle D,N,p\rangle$.}
	          \end{enumerate}
	      \procedure{tue2008190849}
	        \objective
	          Choose two rational numbers $A>0$ and $1>X>0$. The objective of the following instructions is to construct positive rational numbers $B,C,D$, and a procedure $p(x,a,b,n)$ to show that $(1+x)_n^{a-b}\equiv\frac{(1+x)_n^a}{(1+x)_n^b}\err{BC^n}$ when complex numbers $x,a,b$ and an integer $n$ such that $\lVert x\rVert^2\le X^2$, $\lVert a\rVert^2\le A^2$, $\lVert b\rVert^2\le A^2$, and $n>D$ are chosen.
	        \implementation
	          \begin{enumerate}
	            \item Execute \procedurehr{wed2407191521} on $\langle A,X\rangle$ and let $\langle a_1,C,p_1\rangle$ receive.
	            \item Execute \procedurehr{tue2008190712} on $\langle A,X\rangle$ and let $\langle a_2,D,p_2\rangle$ receive.
	            \item Execute \procedurehr{wed2407191611} on $\langle A,X\rangle$ and let $\langle a_3,p_3\rangle$ receive.
	            \item Let $B=(1+\frac{a_3}{a_2})a_1$.
	            \item Let $p(x,a,b,n)$ be the following procedure:
	            \begin{enumerate}
	              \item Using procedures $p_1,p_2,p_3$, show that $(1+x)_n^{a-b}$
	              \begin{enumerate}
	                \item $\equiv(1+x)_n^a(1+x)_n^{-b}\err{a_1{C}^n}$
	                \item $=(1+x)_n^a\frac{(1+x)_n^{b}(1+x)_n^{-b}}{(1+x)_n^{b}}$
	                \item $\equiv((1+x)_n^a)^1\frac{(1+x)_n^{b-b}}{(1+x)_n^{b}}\err{a_3\frac{a_1{C}^n}{a_2}}$
	                \item $=\frac{(1+x)_n^a}{(1+x)_n^b}$
	              \end{enumerate}
	              \item \textbf{Hence show that $(1+x)_n^{a-b}\equiv\frac{(1+x)_n^a}{(1+x)_n^b}\err{(1+\frac{a_3}{a_2})a_1{C}^n}\err{BC^n}$.}
	            \end{enumerate}
	            \item \textbf{Yield the tuple $\langle B,C,D,p\rangle$.}
	          \end{enumerate}
	      \procedure{wed2407191627}
	        \objective
	          Choose two rational numbers $A>0$, $1>X>0$. The objective of the following instructions is to construct rational numbers $G>0$, $0<C<1$, and a procedure $p(x,n,a,k)$ to show that $(1+x)_n^{ka}\equiv((1+x)_n^a)^k\err{GkC^n}$ when a non-negative integer $k$ and complex numbers $x,a$ such that $\lVert x\rVert^2\le X^2$ and $\lVert ka\rVert^2<A^2$ are chosen.
	        \implementation
	          \begin{enumerate}
	            \item Execute \procedurehr{wed2407191611} on $\langle A,X\rangle$ and let $\langle D,t\rangle$ receive.
	            \item Execute \procedurehr{wed2407191521} on $\langle A,X\rangle$ and let $\langle B,C,q\rangle$ receive.
	            \item Let $G=DB$.
	            \item Let $p(x,n,a,k)$ be the following procedure:
	            \begin{enumerate}
	              \item Hence using procedures $t,q$, show that $(1+x)_n^{ka}$
	              \begin{enumerate}
	                \item $=((1+x)_n^a)^0(1+x)_n^{ka}$
	                \item $\equiv((1+x)_n^a)^1(1+x)_n^{(k-1)a}\err{DBC^n}$
	                \item $\equiv((1+x)_n^a)^2(1+x)_n^{(k-2)a}\err{DBC^n}$
	                \item $\vdots$
	                \item $\equiv((1+x)_n^a)^k(1+x)_n^{(k-k)a}\err{DBC^n}$
	                \item $=((1+x)_n^a)^k$.
	              \end{enumerate}
	              \item \textbf{Hence show that $(1+x)_n^{ka}\equiv((1+x)_n^a)^k\err{kDBC^n}\err{GkC^n}$.}
	            \end{enumerate}
	            \item \textbf{Yield the tuple $\langle G,C,D,p\rangle$.}
	          \end{enumerate}
	      \procedure{sun0812190858}
	        \objective
	          Choose two non-negative rational numbers $a,b$ and two non-negative integers $r,n$ such that $b<r<n-a-1$. The objective of the following instructions is to show that $\sgn(\binom{b}{r}\binom{a}{n-r})=\sgn(\binom{b}{r+1}\binom{a}{n-r-1})$.
	        \implementation
	          \begin{enumerate}
	            \item Show that $\sgn(\frac{b-r}{r+1}\cdot\frac{n-r}{a-n+r+1})=1$
	            \begin{enumerate}
	              \item given that $\frac{b-r}{r+1}\cdot\frac{n-r}{a-n+r+1}>0$
	              \item given that $\frac{b-r}{(a+1)-(n-r)}>0$
	              \item given that $r>b$ and $n-r>a+1$.
	            \end{enumerate}
	            \item Hence show that $\sgn(\binom{b}{r+1}\binom{a}{n-r-1})$
	            \begin{enumerate}
	              \item $=\sgn(\frac{b-r}{r+1}\binom{b}{r}\cdot\frac{n-r}{a-n+r+1}\binom{a}{n-r})$
	              \item $=\sgn(\frac{b-r}{r+1}\cdot\frac{n-r}{a-n+r+1})\sgn(\binom{b}{r}\binom{a}{n-r})$
	              \item $=\sgn(\binom{b}{r}\binom{a}{n-r})$.
	            \end{enumerate}
	          \end{enumerate}
	      \procedure{sun0812190920}
	        \objective
	          Choose two non-negative rational numbers $a,b$ and an integer $n\ge\lceil a\rceil+\lceil b\rceil$. The objective of the following instructions is to show that $\sum_r^{[0:n+1]}\lVert\binom{b}{r}\binom{a}{n-r}\rVert=\sum_r^{[0:\lceil b\rceil]}\lVert\binom{b}{r}\binom{a}{n-r}\rVert+\lVert\sum_r^{[\lceil b\rceil:\lceil n-a\rceil]}\binom{b}{r}\binom{a}{n-r}\rVert+\sum_r^{[\lceil n-a\rceil:n+1]}\lVert\binom{b}{r}\binom{a}{n-r}\rVert$.
	        \implementation
	          \begin{enumerate}
	            \item Verify that $\lceil b\rceil\le\lfloor n-a\rfloor$.
	            \item For $r$ in $[\lceil b\rceil:\lfloor n-a\rceil]$, do the following:
	            \begin{enumerate}
	              \item Show that $\sgn(\binom{b}{r}\binom{a}{n-r})=\sgn(\binom{b}{r+1}\binom{a}{n-r-1})$ using \procedurehr{sun0812190858}.
	            \end{enumerate}
	            \item Hence show that $\sum_r^{[\lceil b\rceil:\lceil n-a\rceil]}\lVert\binom{b}{r}\binom{a}{n-r}\rVert=\lVert\sum_r^{[\lceil b\rceil:\lceil n-a\rceil]}\binom{b}{r}\binom{a}{n-r}\rVert$.
	            \item Hence show that $\sum_r^{[0:n+1]}\Vert\binom{b}{r}\binom{a}{n-r}\rVert$
	            \begin{enumerate}
	              \item $=\sum_r^{[0:\lceil b\rceil]}\Vert\binom{b}{r}\binom{a}{n-r}\rVert+\sum_r^{[\lceil b\rceil:\lceil n-a\rceil]}\Vert\binom{b}{r}\binom{a}{n-r}\rVert+\sum_r^{[\lceil n-a\rceil:n+1]}\Vert\binom{b}{r}\binom{a}{n-r}\rVert$
	              \item $=\sum_r^{[0:\lceil b\rceil]}\Vert\binom{b}{r}\binom{a}{n-r}\rVert+\lVert\sum_r^{[\lceil b\rceil:\lceil n-a\rceil]}\binom{b}{r}\binom{a}{n-r}\rVert+\sum_r^{[\lceil n-a\rceil:n+1]}\Vert\binom{b}{r}\binom{a}{n-r}\rVert$.
	            \end{enumerate}
	          \end{enumerate}
	      \procedure{wed2407191824}
	        \objective
	          Choose a rational number $A>0$. The objective of the following instructions is to construct rational numbers $M>1$, $N>0$, and a procedure $p(a,n)$ to show that $\lVert\binom{a}{n}\rVert^2\le(\frac{M}{n})^{2(\lceil a\rceil)}$ and $\frac{M}{n}<1$ when a rational number $-1<a<A$ and an integer $n>N$ are chosen.
	        \implementation
	          \begin{enumerate}
	            \item Let $M=2A$.
	            \item Let $N=2A$.
	            \item Let $p(a,n)$ be the following procedure:
	            \begin{enumerate}
	              \item Show that $\frac{2a}{n}<\frac{2A}{2A}=1$
	              \begin{enumerate}
	                \item given that $n>N=2A>2a$
	                \item and $-1<a<A$.
	              \end{enumerate}
	              \item Show that $n-\lfloor a\rfloor>n-a>n-\frac{n}{2}=\frac{n}{2}$
	              \begin{enumerate}
	                \item given that $\frac{n}{2}>a$
	                \item given that $n>N=2A>2a$.
	              \end{enumerate}
	              \item Hence show that $\lVert\binom{a}{n}\rVert^2$
	              \begin{enumerate}
	                \item $=\lVert\frac{a^{\ul{n}}}{n!}\rVert^2$
	                \item $=\lVert\prod_k^{[0:n]}\frac{a-k}{k+1}\rVert^2$
	                \item $=\prod_k^{[0:n]}\frac{(a-k)^2}{(k+1)^2}$
	                \item $=\prod_k^{[0:\lceil a\rceil]}(k-a)^2\cdot\prod_k^{[0:n]}\frac{(k+\lfloor a\rfloor+1-a)^2}{(k+1)^2}\cdot\prod_k^{[n-\lceil a\rceil:n]}\frac{1}{(k+1)^2}$
	                \item $\le(a^{\lceil a\rceil}\cdot 1^n\cdot(\frac{1}{n-\lfloor a\rfloor})^{\lceil a\rceil})^2$
	                \item $=(\frac{a}{n-\lfloor a\rfloor})^{2(\lceil a\rceil)}$
	                \item $\le(\frac{2a}{n})^{2(\lceil a\rceil)}$
	                \item $\le(\frac{M}{n})^{2(\lceil a\rceil)}$.
	              \end{enumerate}
	            \end{enumerate}
	            \item \textbf{Yield the tuple $\langle M,N,p\rangle$.}
	          \end{enumerate}
	      \procedure{sun0812191002}
	        \objective
	          Choose a positive integer $A$. The objective of the following instructions is to construct a rational number $B$, an integer $N$, and a procedure $p(a,b,n)$ to show that $\sum_r^{[0:n+1]}\lVert\binom{b}{r}\binom{a}{n-r}\rVert\le\frac{B}{n}$ when non-negative rational numbers $a,b$, and an integer $n$ such that $a<A$,$b<A$, and $n>N$ are chosen.
	        \implementation
	          \begin{enumerate}
	            \item Execute \procedurehr{wed2407191824} on $\langle A\rangle$ and let $\langle M,Q,q\rangle$ receive.
	            \item Let $N=\max(2A,Q+A)$.
	            \item Let $B=M^A(M^{A}+8A!A)$.
	            \item Let $p(a,b,n)$ be the following procedure:
	            \begin{enumerate}
	              \item Show that $n>N\ge Q$.
	              \item Now show that $\lVert\binom{a+b}{n}\rVert\le(\frac{M}{n})^{\lceil a+b\rceil}\le\frac{M^{\lceil A+b\rceil}}{n}\le\frac{M^{A+\lceil b\rceil}}{n}\le\frac{M^{2A}}{n}$ using procedure $q$.
	              \item For $r$ in $[0:\lceil b\rceil]$, do the following:
	              \begin{enumerate}
	                \item Show that $n-r\ge N-\lceil b\rceil\ge Q+A-A=Q$.
	                \item Now show that $\lVert\binom{a}{n-r}\rVert^2\le(\frac{M}{n-r})^{2\lceil a\rceil}\le(\frac{M^A}{r})^2\le(\frac{M^A}{n-\lfloor a\rfloor})^2\le(\frac{M^A}{n-A})^2\le(\frac{M^A}{n-\frac{1}{2}N})^2\le(\frac{M^A}{\frac{1}{2}n})^2=(\frac{2M^A}{n})^2$ using procedure $q$.
	              \end{enumerate}
	              \item For $r$ in $[\lceil n-a\rceil:n+1]$, do the following:
	              \begin{enumerate}
	                \item Show that $r\ge\lceil n-a\rceil=n-\lfloor a\rfloor\ge N-\lfloor a\rfloor\ge Q+A-A=Q$.
	                \item Now show that $\lVert\binom{b}{r}\rVert^2\le(\frac{M}{r})^{2\lceil b\rceil}\le(\frac{M^A}{r})^2\le(\frac{2M^A}{n})^2$ using procedure $q$.
	              \end{enumerate}
	              \item Hence using \procedurehr{sun0812190920}, show that $\sum_r^{[0:n+1]}\lVert\binom{b}{r}\binom{a}{n-r}\rVert$
	              \begin{enumerate}
	                \item $=\sum_r^{[0:\lceil b\rceil]}\lVert\binom{b}{r}\binom{a}{n-r}\rVert+\lVert\sum_r^{[\lceil b\rceil:\lceil n-a\rceil]}\binom{b}{r}\binom{a}{n-r}\rVert+\sum_r^{[\lceil n-a\rceil:n+1]}\lVert\binom{b}{r}\binom{a}{n-r}\rVert$
	                \item $=\sum_r^{[0:\lceil b\rceil]}\lVert\binom{b}{r}\binom{a}{n-r}\rVert+\lVert\sum_r^{[0:n+1]}\binom{b}{r}\binom{a}{n-r}-\sum_r^{[0:\lceil b\rceil]}\binom{b}{r}\binom{a}{n-r}-\sum_r^{[\lceil n-a\rceil:n+1]}\binom{b}{r}\binom{a}{n-r}\rVert+\sum_r^{[\lceil n-a\rceil:n+1]}\lVert\binom{b}{r}\binom{a}{n-r}\rVert$
	                \item $=2\sum_r^{[0:\lceil b\rceil]}\lVert\binom{b}{r}\binom{a}{n-r}\rVert+\lVert\sum_r^{[0:n+1]}\binom{b}{r}\binom{a}{n-r}\rVert+2\sum_r^{[\lceil n-a\rceil:n+1]}\lVert\binom{b}{r}\binom{a}{n-r}\rVert$
	                \item $=\lVert\binom{a+b}{n}\rVert+2(\sum_r^{[0:\lceil b\rceil]}\lVert\binom{b}{r}\binom{a}{n-r}\rVert+\sum_r^{[\lceil n-a\rceil:n+1]}\lVert\binom{b}{r}\binom{a}{n-r}\rVert)$
	                \item $\le\frac{M^{2A}}{n}+2(\sum_r^{[0:\lceil b\rceil]}A!\frac{2M^A}{n}+\sum_r^{[\lceil n-a\rceil:n+1]}\frac{2M^A}{n}A!)$
	                \item $\le\frac{M^A}{n}(M^A+8\frac{A!A}{n})$
	                \item $=\frac{B}{n}$.
	              \end{enumerate}
	            \end{enumerate}
	            \item \textbf{Yield the tuple $\langle B,N,p\rangle$.}
	          \end{enumerate}
	      \procedure{thu2507190646}
	        \objective
	          Choose a rational number $1>X\ge 0$. The objective of the following instructions is to construct rational numbers $B>0$, $N>0$, and a procedure $p(x,a,b,n)$ to show that $(1+x)_n^{a+b}\equiv(1+x)_n^a(1+x)_n^b\err{\frac{B}{n}}$ when a complex number $x$, two positive rational numbers $a,b$, and a positive integer $n$ such that $\lVert x\rVert^2\le 1$, $\Re(x)\ge-X$, $a<1$, $b<1$, and $n>N$ are chosen.
	        \implementation
	          \begin{enumerate}
	            \item Execute \procedurehr{sun0812191002} on $\langle 1\rangle$ and let $\langle M,N,q\rangle$ receive.
	            \item Let $B=\frac{2M}{1-X}$.
	            \item Let $p(x,a,b,n)$ be the following procedure:
	            \begin{enumerate}
	              \item For $r\in[1:n]$, for $k\in[0:r]$, show that $\binom{a}{k+1+n-r}(-1)^{k+1}-\binom{a}{k+n-r}(-1)^k$
	              \begin{enumerate}
	                \item $=(-1)^{k+1}(\binom{a}{k+1+n-r}+\binom{a}{k+n-r})$
	                \item $=(-1)^{k+1}\binom{a+1}{k+1+n-r}$
	                \item $=(-1)^{-(k+1)}\lVert\binom{a+1}{k+1+n-r}\rVert(-1)^{k+1+n-r}$
	                \item $=\lVert\binom{a+1}{k+1+n-r}\rVert(-1)^{n-r}$.
	              \end{enumerate}
	              \item Now show that $\sum_r^{[0:n+1]}\lVert\binom{b}{r}\binom{a}{n-r}\rVert\le\frac{M}{n}$ using procedure $q$.
	              \item Show that $\lVert x+1\rVert^2=\Re(x+1)^2+\Im(x)^2\ge(1-X)^2$.
	              \item Hence using \procedurehr{sun2107191133}, show that $(1+x)_n^{a+b}\equiv(1+x)_n^a(1+x)_n^b$
	              \begin{enumerate}
	                \item $\err{(1+x)_n^a(1+x)_n^b-(1+x)_n^{a+b}}$
	                \item $\err{\sum_k^{[1:n]}\sum_r^{[k:n]}\binom{a}{k+n-1-r}\binom{b}{r}x^{k+n-1}}$
	                \item $\err{x^n\sum_r^{[1:n]}\binom{b}{r}\sum_k^{[0:r]}\binom{a}{k+n-r}x^k}$
	                \item $\err{x^n\sum_r^{[1:n]}\binom{b}{r}\sum_k^{[0:r]}(\binom{a}{k+1+n-r}(-1)^{k+1}\cdot\frac{(-x)^{k+1}}{-x-1}-\binom{a}{k+n-r}(-1)^k\cdot\frac{(-x)^k}{-x-1}-\frac{(-x)^{k+1}}{-x-1}(\binom{a}{k+1+n-r}(-1)^{k+1}-\binom{a}{k+n-r}(-1)^k))}$
	                \item $\err{\frac{x^n}{x+1}\sum_r^{[1:n]}\binom{b}{r}(\binom{a}{n}x^r-\binom{a}{n-r}-\sum_k^{[0:r]}(-x)^{k+1}(\binom{a}{k+1+n-r}(-1)^{k+1}-\binom{a}{k+n-r}(-1)^k))}$
	                \item $\err{\frac{1}{1-X}\sum_r^{[1:n]}\lVert\binom{b}{r}\rVert(\lVert\binom{a}{n}\rVert+\lVert\binom{a}{n-r}\rVert+\sum_k^{[0:r]}\lVert\binom{a}{k+1+n-r}(-1)^{k+1}-\binom{a}{k+n-r}(-1)^k\rVert)}$
	                \item $\err{\frac{1}{1-X}\sum_r^{[1:n]}\lVert\binom{b}{r}\rVert(\lVert\binom{a}{n}\rVert+\lVert\binom{a}{n-r}\rVert+\lVert\sum_k^{[0:r]}(\binom{a}{k+1+n-r}(-1)^{k+1}-\binom{a}{k+n-r}(-1)^k)\rVert)}$
	                \item $\err{\frac{1}{1-X}\sum_r^{[1:n]}\lVert\binom{b}{r}\rVert(\lVert\binom{a}{n}\rVert+\lVert\binom{a}{n-r}\rVert+\lVert\binom{a}{n}(-1)^r-\binom{a}{n-r}\rVert)}$
	                \item $\err{\frac{2}{1-X}\sum_r^{[1:n]}\lVert\binom{b}{r}\rVert\lVert\binom{a}{n-r}\rVert}$
	                \item $\err{\frac{B}{n}}$.
	              \end{enumerate}
	            \end{enumerate}
	            \item \textbf{Yield the tuple $\langle B,N,p\rangle$.}
	          \end{enumerate}
	      \procedure{thu2507191017}
	        \objective
	          Choose a rational number $0\le X<1$. The objective of the following instructions is to construct a positive rational number $D$ such that $D>1$, and a procedure $p(x,n,a,k)$ to show that $\lVert((1+x)_n^a)^k\rVert^2<D^2$ when a complex number $x$, a rational number $a$, and positive integers $n,k$ such that $\lVert x\rVert^2\le 1$, $\Re(x)\ge-X$, and $(ka)^2<1$ are chosen.
	        \implementation
	          \begin{enumerate}
	            \item Execute \procedurehr{3.13} on $\langle\frac{2}{1-X}\rangle$ and let $\langle E,N,q\rangle$ receive.
	            \item Let $D=\max(E,(1+\frac{2}{1-X})^{\lfloor N\rfloor})$.
	            \item Let $p(x,n,a,k)$ be the following procedure:
	            \begin{enumerate}
	              \item For $t\in[1:n]$, show that $\binom{a}{t+1}(-1)^{t+1}-\binom{a}{t}(-1)^t$
	              \begin{enumerate}
	                \item $=(-1)^{t+1}(\binom{a}{t+1}+\binom{a}{t})$
	                \item $=(-1)^{t+1}\cdot\frac{(a+1)^{\ul{t+1}}}{(t+1)!}$
	                \item $>0$.
	              \end{enumerate}
	              \item Hence show that $\lVert k\sum_t^{[1:n]}\binom{a}{t}x^t\rVert^2$
	              \begin{enumerate}
	                \item $=\lVert k\sum_t^{[1:n]}(\binom{a}{t+1}(-1)^{t+1}\cdot\frac{(-x)^{t+1}}{-x-1}-\binom{a}{t}(-1)^t\cdot\frac{(-x)^t}{-x-1}-\frac{(-x)^{t+1}}{-x-1}(\binom{a}{t+1}(-1)^{t+1}-\binom{a}{t}(-1)^t))\rVert^2$
	                \item $=\frac{k^2}{\lVert x+1\rVert^2}\lVert\binom{a}{n}x^n-\binom{a}{1}x^1-\sum_t^{[1:n]}(-x)^{t+1}(\binom{a}{t+1}(-1)^{t+1}-\binom{a}{t}(-1)^t)\rVert^2$
	                \item $\le\frac{k^2}{(\Re(x)+1)^2+\Im(x)^2}(\lvert\binom{a}{n}\rvert+a+\sum_t^{[1:n]}\lvert\binom{a}{t+1}(-1)^{t+1}-\binom{a}{t}(-1)^t\rvert)^2$
	                \item $\le\frac{k^2}{(1-X)^2}(\lvert\binom{a}{n}\rvert+a+\sum_t^{[1:n]}(\binom{a}{t+1}(-1)^{t+1}-\binom{a}{t}(-1)^t))^2$
	                \item $=\frac{k^2}{(1-X)^2}(\lvert\binom{a}{n}\rvert+a+\binom{a}{n}(-1)^n-\binom{a}{1}(-1)^1)^2$
	                \item $=\frac{k^2}{(1-X)^2}(\lvert\binom{a}{n}\rvert+a-\lvert\binom{a}{n}\rvert+a)^2$
	                \item $=(\frac{2ak}{1-X})^2$
	                \item $\le(\frac{2}{1-X})^2$.
	              \end{enumerate}
	              \item If $k>N$, then do the following:
	              \begin{enumerate}
	                \item Using procedure $q$, show that $\lVert((1+x)_n^a)^k\rVert^2$
	                \begin{enumerate}
	                  \item $=\lVert(\sum_t^{[0:n]}\binom{a}{t}x^t)^k\rVert^2$
	                  \item $=\lVert(1+\sum_t^{[1:n]}\binom{a}{t}x^t)^k\rVert^2$
	                  \item $=\lVert\exp_k(k\sum_t^{[1:n]}\binom{a}{t}x^t)\rVert^2$
	                  \item $\le E^2$.
	                  \item $\le D^2$.
	                \end{enumerate}
	              \end{enumerate}
	              \item Otherwise do the following:
	              \begin{enumerate}
	                \item Show that $\lVert\sum_t^{[1:n]}\binom{a}{t}x^t\rVert^2$
	                \begin{enumerate}
	                  \item $\le\lVert k\sum_t^{[1:n]}\binom{a}{t}x^t\rVert^2$
	                  \item $\le(\frac{2}{1-X})^2$.
	                \end{enumerate}
	                \item Hence show that $\lVert((1+x)_n^a)^k\rVert^2$
	                \begin{enumerate}
	                  \item $=(\lVert(1+x)_n^a\rVert^2)^k$
	                  \item $=(\lVert 1+\sum_t^{[1:n]}\binom{a}{t}x^t\rVert^2)^k$
	                  \item $\le(1+\frac{2}{1-X})^{2k}$
	                  \item $\le D^2$.
	                \end{enumerate}
	              \end{enumerate}
	            \end{enumerate}
	            \item \textbf{Yield the tuple $\langle D,p\rangle$.}
	          \end{enumerate}
	      \procedure{thu2507190752}
	        \objective
	          Choose a rational number $0\le X<1$. The objective of the following instructions is to construct positive rational numbers $G,N$ and a procedure $p(x,n,a,k)$ to show that $(1+x)_n^{ka}\equiv((1+x)_n^a)^k\err{\frac{Gk}{n}}$ when positive integers $n,k$, a rational number $a$, and a complex number $x$ such that $\lVert x\rVert^2\le 1$, $\Re(x)\ge-X$, $k>1$, $0<ka\le 1$, and $n>N$ are chosen.
	        \implementation
	          \begin{enumerate}
	            \item Execute \procedurehr{thu2507191017} on $\langle X\rangle$ and let $\langle D,t\rangle$ receive.
	            \item Execute \procedurehr{thu2507190646} on $\langle X\rangle$ and let $\langle B,N,q\rangle$ receive.
	            \item Let $G=DB$.
	            \item Let $p(x,n,a,k)$ be the following procedure:
	            \begin{enumerate}
	              \item Using procedures $t,q$, show that $(1+x)_n^{ka}$
	              \begin{enumerate}
	                \item $=((1+x)_n^a)^0(1+x)_n^{ka}$
	                \item $\equiv((1+x)_n^a)^1(1+x)_n^{(k-1)a}\err{D\frac{B}{n}}$
	                \item $\equiv((1+x)_n^a)^2(1+x)_n^{(k-2)a}\err{D\frac{B}{n}}$
	                \item $\vdots$
	                \item $\equiv((1+x)_n^a)^k(1+x)_n^{(k-k)a}\err{D\frac{B}{n}}$
	                \item $=((1+x)_n^a)^k$.
	              \end{enumerate}
	              \item \textbf{Hence show that $(1+x)_n^{ka}\equiv((1+x)_n^a)^k\err{\frac{DBk}{n}}\err{\frac{Gk}{n}}$.}
	            \end{enumerate}
	            \item \textbf{Yield the tuple $\langle G,D,N,p\rangle$.}
	          \end{enumerate}
	      \procedure{fri2607191210}
	        \objective
	          Choose a rational number $1>X\ge 0$. The objective of the following instructions is to construct positive rational numbers $a,c$ such that $b>1$, and a procedure $p(x,n,k)$ to show that $\exp_n(n((1+x)_k^{\frac{1}{n}}-1))\equiv 1+x\err{\frac{an}{k}}$ when a complex number $x$, and positive integers $n,k$ such that $\lVert x\rVert^2\le 1$, $\Re(x)\ge-X$, $n>1$, and $k>c$ are chosen.
	        \implementation
	          \begin{enumerate}
	            \item Execute \procedurehr{thu2507190752} on $\langle X\rangle$ and let $\langle a,c,p_1\rangle$ receive.
	            \item Let $p(x,n,k)$ be the following procedure:
	            \begin{enumerate}
                \item Using procedure $p_1$ and \procedurehr{sun2107190619}, show that $\exp_n(n((1+x)_k^{\frac{1}{n}}-1))$
                \begin{enumerate}
                  \item $=(1+\frac{1}{n}(n((1+x)_k^{\frac{1}{n}}-1)))^n$
                  \item $=((1+x)_k^{\frac{1}{n}})^n$
                  \item $\equiv(1+x)_k^1\err{\frac{an}{k}}$
                  \item $=(1+x)^1$
                  \item $=1+x$.
                \end{enumerate}
                \item \textbf{Hence show that $\exp_n(n((1+x)_k^{\frac{1}{n}}-1))\equiv 1+x\err{\frac{an}{k}}$.}
	            \end{enumerate}
	            \item \textbf{Yield the tuple $\langle a,c,p\rangle$.}
	          \end{enumerate}
	       \procedure{fri2607191243}
	        \objective
	          Choose a rational number $1>X\ge 0$. The objective of the following instructions is to construct a rational number $a>0$ and a procedure $p(x,n,k)$ to show that $\lVert n((1+x)_k^{\frac{1}{n}}-1)\rVert^2\le a^2$ when positive integers $n,k$, and a complex number $x$ such that $\lVert x\rVert^2\le 1$ and $\Re(x)\ge-X$ are chosen.
	        \implementation
	          \begin{enumerate}
	            \item Let $a=\frac{2}{1-X}$.
	            \item Let $p(x,n,k)$ be the following procedure:
	            \begin{enumerate}
	              \item Show that $\lVert n((1+x)_k^{\frac{1}{n}}-1)\rVert^2$
	              \begin{enumerate}
	                \item $=\lVert n(\sum_r^{[0:k]}\binom{\frac{1}{n}}{r}x^r-1)\rVert^2$
	                \item $=\lVert n\sum_r^{[1:k]}\binom{\frac{1}{n}}{r}(-1)^r(-x)^r\rVert^2$
	                \item $=n^2\lVert\sum_r^{[1:k]}(\binom{\frac{1}{n}}{r+1}(-1)^{r+1}\cdot\frac{(-x)^{r+1}}{-x-1}-\binom{\frac{1}{n}}{r}(-1)^r\cdot\frac{(-x)^r}{-x-1}-(\binom{\frac{1}{n}}{r+1}(-1)^{r+1}-\binom{\frac{1}{n}}{r}(-1)^r)\frac{(-x)^{r+1}}{-x-1})\rVert^2$
	                \item $=\frac{n^2}{\lVert x+1\rVert^2}\lVert\binom{\frac{1}{n}}{k}x^k-\binom{\frac{1}{n}}{1}x^1-\sum_r^{[1:k]}(\binom{\frac{1}{n}}{r+1}(-1)^{r+1}-\binom{\frac{1}{n}}{r}(-1)^r)(-x)^{r+1}\rVert^2$
	                \item $\le\frac{n^2}{\lVert x+1\rVert^2}\lVert\binom{\frac{1}{n}}{k}(-1)^{k-1}+\frac{1}{n}+\sum_r^{[1:k]}(\binom{\frac{1}{n}}{r+1}(-1)^{r+1}-\binom{\frac{1}{n}}{r}(-1)^r)\rVert^2$
	                \item $=\frac{n^2}{(\Re(x)+1)^2+\Im(x)^2}(\binom{\frac{1}{n}}{k}(-1)^{k-1}+\frac{1}{n}+\binom{\frac{1}{n}}{k}(-1)^k-\binom{\frac{1}{n}}{1}(-1)^1)^2$
	                \item $\le\frac{n^2}{(1-X)^2}(\frac{2}{n})^2$
	                \item $=a^2$.
	              \end{enumerate}
	            \end{enumerate}
	            \item \textbf{Yield the tuple $\langle a,p\rangle$.}
	          \end{enumerate}
	      \declaration{fri0108191325}
	        The notation \gls{complexomega} will be used as a shorthand notation for $\frac{1}{r}(1-\prod_t^{[1:r]}(1-\frac{1}{nt}))$.
	      \procedure{thu0108191318}
	        \objective
	          Choose two positive integers $r,n$ such that $r>1$. The objective of the following instructions is to show that $\frac{\omega(r+1)}{\omega(r)}\le 1$.
	        \implementation
	          \begin{enumerate}
	            \item Using \procedurehr{2.07}, show that $\frac{\omega(r+1)}{\omega(r)}$
              \begin{enumerate}
                \item $=\frac{\frac{1}{r+1}(1-\prod_t^{[1:r+1]}(1-\frac{1}{nt}))}{\frac{1}{r}(1-\prod_t^{[1:r]}(1-\frac{1}{nt}))}$
                \item $=\frac{r}{r+1}\cdot\frac{1-(1-\frac{1}{nr})\prod_t^{[1:r]}(1-\frac{1}{nt})}{1-\prod_t^{[1:r]}(1-\frac{1}{nt})}$
                \item $=\frac{r}{r+1}\left(1+\frac{\frac{1}{nr}\prod_t^{[1:r]}(1-\frac{1}{nt})}{1-\prod_t^{[1:r]}(1-\frac{1}{nt})}\right)$
                \item $=\frac{r}{r+1}\left(1+\frac{\frac{1}{nr}}{(\prod_t^{[1:r]}(1-\frac{1}{nt}))^{-1}-1}\right)$
                \item $\le\frac{r}{r+1}\left(1+\frac{\frac{1}{nr}}{(1-\frac{1}{n(r-1)})^{-(r-1)}-1}\right)$
                \item $=\frac{r}{r+1}\left(1+\frac{\frac{1}{nr}}{(1+\frac{1}{nr-n-1})^{r-1}-1}\right)$
                \item $\le\frac{r}{r+1}\left(1+\frac{\frac{1}{nr}}{(1+\frac{1}{n(r-1)})^{r-1}-1}\right)$
                \item $\le\frac{r}{r+1}\left(1+\frac{\frac{1}{nr}}{1+\frac{r-1}{n(r-1)}-1}\right)$
                \item $=\frac{r}{r+1}(1+\frac{1}{r})$
                \item $=1$.
              \end{enumerate}
            \end{enumerate}
        \declaration{fri2607191453}
	        The notation \gls{complexnaturallogarithm} will be used as a shorthand for $\sum_r^{[1:k]}\frac{(-1)^{r-1}}{r}x^r$.
	      \procedure{fri2607191450}
	        \objective
	          Choose a rational number $1>X\ge 0$. The objective of the following instructions is to construct a positive rational number $a$ and a procedure $p(x,n,k)$ to show that $\ln_k(1+x)\equiv n((1+x)_k^{\frac{1}{n}}-1)\err{\frac{a}{n}}$ when positive integers $n,k$ and a complex number $x$ such that $\lVert x\rVert^2\le 1$ and $\Re(x)\ge-X$ are chosen.
	        \implementation
	          \begin{enumerate}
	            \item Let $a=\frac{1}{1-X}$.
	            \item Let $p(x,n,k)$ be the following procedure:
	            \begin{enumerate}
	              \item For $r\in[2:k]$, show that $\frac{\omega(r+1)}{\omega(r)}\le 1$ using \procedurehr{thu0108191318}.
	              \item Also show that $\lVert x+1\rVert^2\ge\Re(x+1)^2+\Im(x)^2\ge(1-X)^2$.
	              \item Hence show that $\ln_k(1+x)\equiv n((1+x)_k^{\frac{1}{n}}-1)$
	              \begin{enumerate}
	                \item $\err{\ln_k(1+x)-n((1+x)_k^{\frac{1}{n}}-1)}$
	                \item $\err{\sum_r^{[1:k]}\frac{(-1)^{r-1}}{r}x^r-n(\sum_r^{[0:k]}\binom{\frac{1}{n}}{r}x^r-1)}$
	                \item $\err{\sum_r^{[1:k]}\frac{(-1)^{r-1}}{r}x^r-n\sum_r^{[1:k]}\binom{\frac{1}{n}}{r}x^r}$
	                \item $\err{\sum_r^{[1:k]}\frac{(-1)^{\ul{r-1}}}{r!}x^r-\sum_r^{[1:k]}\frac{(\frac{1}{n}-1)^{\ul{r-1}}}{r!}x^r}$
	                \item $\err{\sum_r^{[1:k]}\frac{1}{r!}((-1)^{\ul{r-1}}-(\frac{1}{n}-1)^{\ul{r-1}})x^r}$
	                \item $\err{\sum_r^{[1:k]}\frac{(-1)^{\ul{r-1}}}{r!}(1-\frac{(\frac{1}{n}-1)^{\ul{r-1}}}{(-1)^{\ul{r-1}}})x^r}$
	                \item $\err{\sum_r^{[1:k]}\frac{(-1)^{r-1}}{r}(1-\prod_t^{[1:r]}\frac{\frac{1}{n}-t}{-t})x^r}$
	                \item $\err{\sum_r^{[1:k]}\omega(r)(-x)^r}$
	                \item $\err{\sum_r^{[1:k]}(\omega(r+1)\cdot\frac{(-x)^{r+1}}{-x-1}-\omega(r)\cdot\frac{(-x)^r}{-x-1}-(\omega(r+1)-\omega(r))\cdot\frac{(-x)^{r+1}}{-x-1})}$
	                \item $\err{\frac{1}{x+1}(\omega(k)(-x)^k-\omega(1)(-x)^1-\sum_r^{[1:k]}(\omega(r+1)-\omega(r))(-x)^{r+1})}$
	                \item $\err{\frac{1}{x+1}(\omega(k)+\omega(1)+\sum_r^{[2:k]}(\omega(r)-\omega(r+1))+\omega(2)-\omega(1))}$
	                \item $\err{\frac{1}{1-X}(\omega(k)-\omega(k)+\omega(2)+\omega(2)+\omega(1)-\omega(1))}$
	                \item $\err{\frac{a}{n}}$.
	              \end{enumerate}
	            \end{enumerate}
	            \item \textbf{Yield the tuple $\langle a,p\rangle$.}
	          \end{enumerate}
	      \procedure{fri2607191736}
	        \objective
	          Choose a rational number $1>X\ge 0$. The objective of the following instructions is to construct a rational number $a>0$ and a procedure $p(x,k)$ to show that $\lVert\ln_k(1+x)\rVert^2\le a^2$ when a positive integer $k$ and a complex number $x$ such that $\lVert x\rVert^2\le 1$ and $\Re(x)\ge-X$ are chosen.
	        \implementation
	          \begin{enumerate}
	            \item Let $a=\frac{2}{1-X}$.
	            \item Let $p(x,k)$ be the following procedure:
	            \begin{enumerate}
	              \item Show that $\lVert\ln_k(1+x)\rVert^2$
	              \begin{enumerate}
	                \item $=\lVert\sum_r^{[1:k]}\frac{(-1)^{r-1}}{r}x^r\rVert^2$
	                \item $=\lVert\sum_r^{[1:k]}\frac{1}{r}(-x)^r\rVert^2$
	                \item $=\lVert\sum_r^{[1:k]}(\frac{1}{r+1}\cdot\frac{(-x)^{r+1}}{-x-1}-\frac{1}{r}\cdot\frac{(-x)^r}{-x-1}-(\frac{1}{r+1}-\frac{1}{r})\cdot\frac{(-x)^{r+1}}{-x-1})\rVert^2$
	                \item $=\frac{1}{\lVert x+1\rVert^2}\lVert\frac{1}{k}(-x)^k-\frac{1}{1}(-x)^1-\sum_r^{[1:k]}(\frac{1}{r+1}-\frac{1}{r})(-x)^{r+1}\rVert^2$
	                \item $\le\frac{1}{\lVert x+1\rVert^2}(\frac{1}{k}+1+\sum_r^{[1:k]}(\frac{1}{r}-\frac{1}{r+1}))^2$
	                \item $=\frac{1}{\lVert x+1\rVert^2}(\frac{1}{k}+1-\frac{1}{k}+1)^2$
	                \item $=\frac{4}{(\Re(x)+1)^2+\Im(x)^2}$
	                \item $\le a^2$
	              \end{enumerate}
	            \end{enumerate}
	            \item \textbf{Yield the tuple $\langle a,p\rangle$.}
	          \end{enumerate}
	      \procedure{fri2607191801}
	        \objective
	          Choose a rational number $1>X\ge 0$. The objective of the following instructions is to construct positive rational numbers $a,c,d,e$ such that $b>1$, and a procedure $p(x,n,k)$ to show that $\exp_n(\ln_k(1+x))\equiv 1+x\err{\frac{an}{k}+\frac{c}{n}}$ when positive integers $n,k$, and a complex number $x$ such that $\lVert x\rVert^2\le 1$, $\Re(x)\ge-X$, $k>d$, and $n>e$ are chosen.
	        \implementation
	          \begin{enumerate}
	            \item Execute \procedurehr{fri2607191243} on $\langle X\rangle$ and let $\langle a_1,p_1\rangle$ receive.
	            \item Execute \procedurehr{fri2607191736} on $\langle X\rangle$ and let $\langle a_2,p_2\rangle$ receive.
	            \item Execute \procedurehr{thu2507191307} on $\langle\max(a_1,a_2)\rangle$ and let $\langle a_3,e,p_3\rangle$ receive.
	            \item Execute \procedurehr{fri2607191450} on $\langle X\rangle$ and let $\langle a_4,p_4\rangle$ receive.
	            \item Execute \procedurehr{fri2607191210} on $\langle X\rangle$ and let $\langle a,d,p_5\rangle$ receive.
	            \item Let $c=a_4a_3$.
	            \item Let $p(x,n,k)$ be the following procedure:
	            \begin{enumerate}
	              \item Show that $\lVert n((1+x)_k^{\frac{1}{n}}-1)\rVert^2\le {a_1}^2$ using procedure $p_1$.
	              \item Show that $\lVert\ln_k(1+x)\rVert^2\le {a_2}^2$ using procedure $p_2$.
	              \item Show that $\lVert\ln_k(1+x)-n((1+x)_k^{\frac{1}{n}}-1)\rVert^2\le(\frac{a_4}{n})^2$ using procedure $p_4$.
	              \item Now using procedures $p_3,p_5$, show that $\exp_n(\ln_k(1+x))$
	              \begin{enumerate}
	                \item $\equiv\exp_n(n((1+x)_k^{\frac{1}{n}}-1))$
	                \begin{enumerate}
	                  \item $\err{a_3(\ln_k(1+x)-n((1+x)_k^{\frac{1}{n}}-1))}$
	                  \item $\err{a_3\cdot\frac{a_4}{n}}$
	                \end{enumerate}
	                \item $\equiv 1+x\err{\frac{an}{k}}$
	              \end{enumerate}
	              \item \textbf{Hence show that $\exp_n(\ln_k(1+x))\equiv 1+x\err{\frac{a_3a_4}{n}+\frac{an}{k}}\err{\frac{c}{n}+\frac{an}{k}}$.}
	            \end{enumerate}
	            \item \textbf{Yield the tuple $\langle a,c,d,e,p\rangle$.}
	          \end{enumerate}
	    \end{multicols*}
	  \chapter{Gregory-Leibniz Series}
	    \begin{multicols}{2}
		    \declaration{3.33}
			    The notation \gls{complextau}, where $n$ is a positive integer, will be used as a shorthand for $8\Im(\ln_n(1+\iu))$.
		    \procedure{3.47}
			    \objective
				    Choose a positive integer $k$. The objective of the following instructions is to show that $\tau_k=8\sum_r^{[0:\lfloor\frac{k}{2}\rfloor]}\frac{(-1)^r}{2r+1}$.
			    \implementation
				    \begin{enumerate}
					    \item Using \declarationhr{3.33}, show that $\tau_k$
					    \begin{enumerate}
						    \item $=8\Im(\sum_r^{[1:k]}\frac{(-1)^{r-1}}{r}\iu^r)$
						    \item $=8\Im(\sum_r^{[0:\lfloor\frac{k}{2}\rfloor]}\frac{(-1)^{2r}}{2r+1}\iu^{2r+1})$
						    \item $=8\sum_r^{[0:\lfloor\frac{k}{2}\rfloor]}\frac{\iu^{2r}}{2r+1}$
						    \item $=8\sum_r^{[0:\lfloor\frac{k}{2}\rfloor]}\frac{(-1)^r}{2r+1}$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{3.49}
			    \objective
				    The objective of the following instructions is to construct positive rational numbers $a,b$ such that $a\ge 4$, and a procedure, $p(n)$, to show that $\tau_n\ge a$ when a positive integer $n\ge b$ is chosen.
			    \implementation
				    \begin{enumerate}
				      \item Let $a=\frac{16}{3}$.
					    \item \textbf{Show that $a\ge 4$.}
					    \item Let $b=4$.
					    \item Let $p(n)$ be the following procedure:
					    \begin{enumerate}
						    \item Let $d=n\div 4$.
						    \item Let $g=n\bmod 4$.
						    \item Hence show that $n=4d+g$.
						    \item If $g=0$ or $g=1$, then do the following:
						    \begin{enumerate}
						      \item Using \procedurehr{3.47}, show that $\tau_n$
						      \begin{enumerate}
						        \item $=8\sum_r^{[0:\lfloor\frac{4d+g}{2}\rfloor]}\frac{(-1)^r}{2r+1}$
						        \item $=8\sum_r^{[0:2d]}\frac{(-1)^r}{2r+1}$
						        \item $=8(1-\frac{1}{3}+\sum_r^{[2:2d]}\frac{(-1)^r}{2r+1})$
						        \item $=\frac{16}{3}+8\sum_r^{[1:d]}(\frac{1}{4r+1}-\frac{1}{4r+3})$
						        \item $\ge\frac{16}{3}$.
						      \end{enumerate}
						    \end{enumerate}
						    \item Otherwise do the following:
						    \begin{enumerate}
						      \item Show that $g=2$ or $g=3$.
						      \item Hence show that $\tau_n$
						      \begin{enumerate}
						        \item $=8\sum_r^{[0:\lfloor\frac{4d+g}{2}\rfloor]}\frac{(-1)^r}{2r+1}$
						        \item $=8\sum_r^{[0:2d+1]}\frac{(-1)^r}{2r+1}$
						        \item $=8(1-\frac{1}{3}+\sum_r^{[0:2d]}\frac{(-1)^r}{2r+1}+\frac{(-1)^{2d}}{4d+1})$
						        \item $\frac{16}{3}+8\sum_r^{[1:d]}(\frac{1}{4r+1}-\frac{1}{4r+3})+\frac{8}{4d+1}$
						        \item $\ge\frac{16}{3}$.
						      \end{enumerate}
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,b,p\rangle$.}
				    \end{enumerate}
		    \procedure{3.50}
			    \objective
				    The objective of the following instructions is to construct rational numbers $a,b$ such that $a\ge 4$ and $a^2<48$, and a procedure, $p(n)$, to show that $\tau_n\le a$ when a positive integer $n$ such that $n\ge b$ is chosen.
			    \implementation
				    \begin{enumerate}
					    \item Let $a=\frac{2104}{315}$.
					    \item \textbf{Show that $a\ge 4$.}
					    \item \textbf{Show that $a^2=\frac{4426816}{99225}<48$.}
					    \item Let $b=10$.
					    \item Let $p(n)$ be the following procedure:
					    \begin{enumerate}
						    \item Let $d=n\div 4$.
						    \item Let $g=n\bmod 4$.
						    \item Hence verify that $n=4d+g$.
						    \item If $g=0$ or $g=1$, then do the following:
						    \begin{enumerate}
						      \item Show that $\tau_n$
						      \begin{enumerate}
						        \item $=8\sum_r^{[0:\lfloor\frac{n}{2}\rfloor]}\frac{(-1)^r}{2r+1}$
						        \item $=8\sum_r^{[0:5]}\frac{(-1)^r}{2r+1}+8\sum_r^{[5:\lfloor\frac{4d+g}{2}\rfloor]}\frac{(-1)^r}{2r+1}$
						        \item $=a+8\sum_r^{[5:2d]}\frac{(-1)^r}{2r+1}$
						        \item $=a+8\sum_r^{[5:2d-1]}\frac{(-1)^r}{2r+1}+\frac{8(-1)^{2d-1}}{4d-1}$
						        \item $=a-8\sum_r^{[3:d]}(\frac{1}{4r-1}-\frac{1}{4r+1})-\frac{8}{4d-1}$
						        \item $\le a$.
						      \end{enumerate}
						    \end{enumerate}
						    \item Otherwise do the following:
						    \begin{enumerate}
						      \item Show that $g=2$ or $g=3$.
						      \item Hence show that $\tau_n$
						      \begin{enumerate}
						        \item $=8\sum_r^{[0:\lfloor\frac{n}{2}\rfloor]}\frac{(-1)^r}{2r+1}$
						        \item $=8\sum_r^{[0:5]}\frac{(-1)^r}{2r+1}+8\sum_r^{[5:\lfloor\frac{4d+g}{2}\rfloor]}\frac{(-1)^r}{2r+1}$
						        \item $=a+8\sum_r^{[5:2d+1]}\frac{(-1)^r}{2r+1}$
						        \item $=a-8\sum_r^{[2:d]}(\frac{1}{4r+3}-\frac{1}{4r+5})$
						        \item $\le a$.
						      \end{enumerate}
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,b,p\rangle$.}
				    \end{enumerate}
		    \procedure{3.53}
			    \objective
				    The objective of the following instructions is to construct positive rational numbers $a,c,d,e$, and a procedure $p(n,k)$ to show that $\exp_n(\frac{1}{4}\tau_k\iu)\equiv\iu\err{\frac{an}{k}+\frac{c}{n}}$ when integers $k,n$ such that $n>e$ and $k>d$ are chosen.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{fri2607191736} on $\langle 0\rangle$ and let $\langle a_1,p_1\rangle$ receive.
					    \item Execute \procedurehr{thu2507191359} on $\langle a_1\rangle$ and let $\langle a_2,b_2,p_2\rangle$ receive.
					    \item Execute \procedurehr{3.14} on $\langle a_1\rangle$ and let $\langle a_3,b_3,p_3\rangle$ receive.
					    \item Execute \procedurehr{fri2607191801} on $\langle 0\rangle$ and let $\langle a_4,c_4,d,e_4,p_4\rangle$ receive.
					    \item Let $a=\frac{2a_4}{a_3}$.
					    \item Let $c=\frac{2c_4}{a_3}+a_2$.
					    \item Let $e=\max(b_2,b_3,e_4)$.
					    \item Let $p(n,k)$ be the following procedure:
					    \begin{enumerate}
					      \item Show that $\lVert\ln_k(1+\iu)\rVert^2\le {a_1}^2$ using procedure $p_1$.
					      \item Hence using procedures $p_2,p_3,p_4$, show that $\exp_n(\frac{1}{4}\tau_k\iu)$
					      \begin{enumerate}
					        \item $=\exp_n(2\Im(\ln_k(1+\iu))\iu)$
					        \item $=\exp_n(\ln_k(1+\iu)-\conj{\ln_k(1+\iu)})$
					        \item $\equiv\frac{\exp_n(\ln_k(1+\iu))}{\exp_n(\conj{\ln_k(1+\iu)})}\err{\frac{a_2}{n}}$
					        \item $\equiv\frac{1+\iu}{\exp_n(\conj{\ln_k(1+\iu)})}\err{\frac{1}{a_3}(\frac{a_4n}{k}+\frac{c_4}{n})}$
					        \item $=\frac{1+\iu}{\conj{\exp_n(\ln_k(1+\iu))}}$
					        \item $\equiv\frac{1+\iu}{\conj{1+\iu}}$
					        \begin{enumerate}
					          \item $\err{\frac{(1+\iu)(\frac{a_4n}{k}+\frac{c_4}{n})}{\conj{\exp_n(\ln_k(1+\iu))}\cdot\conj{1+\iu}}}$
					          \item $\err{\frac{1}{a_3}(\frac{a_4n}{k}+\frac{c_4}{n})}$
					        \end{enumerate}
					        \item $=i$.
					      \end{enumerate}
					      \item \textbf{Hence show that $\exp_n(\frac{1}{4}\tau_k\iu)\equiv\iu\err{\frac{a_2}{n}+\frac{2}{a_3}(\frac{a_4n}{k}+\frac{c_4}{n})}\err{\frac{an}{k}+\frac{c}{n}}$.}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,c,d,e,p\rangle$.}
				    \end{enumerate}
		    \procedure{3.54}
			    \objective
				    The objective of the following instructions is to construct positive rational numbers $a,c,d,e$ such that $b>1$, and a procedure, $p(n,k)$, to show that $\exp_n(-\frac{1}{4}\tau_k\iu)\equiv -\iu\err{\frac{an}{k}+\frac{c}{n}}$ when integers $k,n$ such that $n>e$ and $k>d$ are chosen.
			    \implementation
				    Implementation is analogous to that of \procedurehr{3.53}.
		    \procedure{mon2608190753}
		      \objective
		        Choose a rational number $X\ge 0$ and an integer $K\ge 0$. The objective of the following instructions is to construct a rational number $a$, and a procedure $p(x,y,k)$ to show that $x^k\equiv y^k\err{a(x-y)}$ when two complex numbers $x,y$ and a non-negative integer $k$ such that $\lVert x\rVert^2\le X^2$, $\lVert y\rVert^2\le X^2$, and $k\le K$ are chosen.
		      \implementation
		        \begin{enumerate}
		          \item Let $a=K\max(1,X)^{K-1}$.
		          \item Let $p(x,y,k)$ be the following procedure:
		          \begin{enumerate}
		            \item Show that $x^k\equiv y^k$
		            \begin{enumerate}
		              \item $\err{y^k-x^k}$
		              \item $\err{(y-x)\sum_r^{[0:k]}x^ry^{k-1-r}}$
		              \item $\err{(y-x)\sum_r^{[0:k]}X^{k-1}}$
		              \item $\err{(y-x)KX^{k-1}}$
		              \item $\err{a(y-x)}$
		            \end{enumerate}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle a,p\rangle$.}
		        \end{enumerate}
		    \procedure{3.55}
			    \objective
				    Choose an integer $K\ge 0$. The objective of the following instructions is to construct rational numbers $a,b,c,d$, and a procedure, $p(n,m,k)$, to show that $\exp_n(\frac{k}{4}\tau_m\iu)\equiv\iu^k\err{\frac{an}{m}+\frac{b}{n}}$ when a non-negative integer $k$ and two positive integers $n,m$ such that $k\le K$, $n>c$, and $m>d$ are chosen.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{3.50} and let $\langle a_1,d,p_1\rangle$ receive.
					    \item Execute \procedurehr{3.16} on $\langle(\frac{a_1}{4})^2\rangle$ and let $\langle a_2,b_2,p_2\rangle$ receive.
					    \item Execute \procedurehr{3.53} and let $\langle a_3,b_3,c_3,p_3\rangle$ receive.
					    \item Execute \procedurehr{3.13} on $\langle(\frac{a_1}{4})^2\rangle$ and let $\langle a_4,b_4,p_4\rangle$ receive.
					    \item Execute \procedurehr{mon2608190753} on $\langle\max(1,\frac{a_1}{4}),K\rangle$ and let $\langle a_5,p_5\rangle$ receive.
					    \item Let $a=a_3a_5$.
					    \item Let $b=a_2K+b_3a_5$.
					    \item Let $c=\max(b_2,b_4,c_3)$.
					    \item Let $p(n,k,m)$ be the following procedure:
					    \begin{enumerate}
						    \item Show that $\tau_m\le a_1$ using procedure $p_1$.
						    \item Hence show that $\lVert\frac{1}{4}\tau_m\iu\rVert^2=\lVert\frac{1}{4}\tau_m\rVert^2\le(\frac{a_1}{4})^2$.
						    \item Hence show that $\lVert\exp_n(\frac{1}{4}\tau_m\iu)-\iu\rVert^2\le(\frac{a_3n}{m}+\frac{b_3}{n})^2$ using procedure $p_3$.
						    \item Hence show that $\lVert\exp_n(\frac{1}{4}\tau_m\iu)\rVert^2\le a_4$ using procedure $p_4$.
						    \item Hence using procedures $p_2,p_5$, show that $\exp_n(\frac{k}{4}\tau_m\iu)$
						    \begin{enumerate}
						      \item $\equiv\exp_n(\frac{1}{4}\tau_m\iu)^k$
						      \begin{enumerate}
						        \item $\err{\frac{a_2k}{n}}$
						        \item $\err{\frac{a_2K}{n}}$
						      \end{enumerate}
						      \item $\equiv\iu^k$
						      \begin{enumerate}
						        \item $\err{a_5(\exp_n(\frac{1}{4}\tau_m\iu)-\iu)}$
						        \item $\err{a_5(\frac{a_3n}{m}+\frac{b_3}{n})}$
						      \end{enumerate}
						    \end{enumerate}
						    \item \textbf{Hence show that $\exp_n(\frac{k}{4}\tau_m\iu)\equiv\iu^k\err{\frac{a_2K}{n}+a_5(\frac{a_3n}{m}+\frac{b_3}{n})}\err{\frac{an}{m}+\frac{b}{n}}$.}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,b,c,d,p\rangle$.}
            \end{enumerate}
		    \procedure{3.56}
			    \objective
				    Choose an integer $K\ge 0$. The objective of the following instructions is to construct rational numbers $a,b,c,d$, and a procedure, $p(n,m,k)$, to show that $\exp_n(\frac{k}{4}\tau_m i)\equiv i^k\err{\frac{an}{m}+\frac{b}{n}}$ when an integer $k$ and two positive integers $n,m$ such that $\lvert k\rvert\le K$, $n>c$, and $m>d$ are chosen.
			    \implementation
				    Implementation is an extension of that of \procedurehr{3.55} using \procedurehr{3.54}.
		    \procedure{3.57}
			    \objective
				    Choose an integer $K\ge 0$. The objective of the following instructions is to construct rational numbers $a,b,c,d$, and a procedure, $p(n,m,k)$, to show that $\cos_n(\frac{k}{4}\tau_m)\equiv\frac{\iu^k+(-\iu)^k}{2}\err{\frac{an}{m}+\frac{b}{n}}$ when an integer $k$ and two positive integers $n,m$ such that $\lvert k\rvert\le K$, $n>c$, and $m>d$ are chosen.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{3.56} on $\langle K\rangle$ and let $\langle a,b,c,d,q\rangle$ receive.
					    \item Let $p(n,m,k)$ be the following procedure:
					    \begin{enumerate}
						    \item Using procedure $q$, show that $\cos_n(\frac{k}{4}\tau_m)$
						    \begin{enumerate}
						      \item $=\frac{\exp_n(\frac{k}{4}\tau_m\iu)+\exp_n(-\frac{k}{4}\tau_m\iu)}{2}$
						      \item $=\frac{\exp_n(\frac{k}{4}\tau_m\iu)}{2}+\frac{\exp_n(-\frac{k}{4}\tau_m\iu)}{2}$
						      \item $\equiv\frac{\iu^k}{2}+\frac{\exp_n(-\frac{k}{4}\tau_m\iu)}{2}\err{\frac{1}{2}(\frac{an}{m}+\frac{b}{n})}$
						      \item $\equiv\frac{\iu^k}{2}+\frac{\iu^{-k}}{2}\err{\frac{1}{2}(\frac{an}{m}+\frac{b}{n})}$
						      \item $=\frac{\iu^k+\iu^{-k}}{2}$.
						    \end{enumerate}
						    \item \textbf{Hence show that $\cos_n(\frac{k}{4}\tau_m)\equiv\frac{\iu^k+\iu^{-k}}{2}\err{\frac{an}{m}+\frac{b}{n}}$.}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,b,c,d,p\rangle$.}
				    \end{enumerate}
		    \procedure{3.58}
			    \objective
				    Choose an integer $K\ge 0$. The objective of the following instructions is to construct rational numbers $a,b,c,d$, and a procedure, $p(n,m,k)$, to show that $\sin_n(\frac{k}{4}\tau_m)\equiv\frac{\iu^k-(-\iu)^k}{2\iu}\err{\frac{an}{m}+\frac{b}{n}}$ when an integer $k$ and two positive integers $n,m$ such that $\lvert k\rvert\le K$, $n>c$, and $m>d$ are chosen.
			    \implementation
				    Implementation is analogous to that of \procedurehr{3.57}.
		    \procedure{3.59}
			    \objective
				    Choose two integers $X\ge 0,K\ge 0$. The objective of the following instructions is to construct rational numbers $a,b,c,d$, and a procedure, $p(x,n,m,k)$, to show that $\exp_n(x+\frac{k}{4}\tau_m\iu)\equiv\iu^k\exp_n(x)\err{\frac{an}{m}+\frac{b}{n}}$ when an integer $k$ and two positive integers $n,m$ such that $\lVert x\rVert^2\le X$, $\lvert k\rvert\le K$, $n>c$, and $m>d$ are chosen.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{3.50} and let $\langle a_1,b_1,p_1\rangle$ receive.
					    \item Let $H=\max(X,\frac{K{a_1}}{4})$.
					    \item Execute \procedurehr{3.15} on $\langle H\rangle$ and let $\langle a_2,b_2,p_2\rangle$ receive.
					    \item Execute \procedurehr{3.13} on $\langle X\rangle$ and let $\langle a_3,b_3,p_3\rangle$ receive.
					    \item Execute \procedurehr{3.56} on $\langle K\rangle$ and let $\langle a_4,b_4,c_4,d_4,p_4\rangle$ receive.
					    \item Let $a=a_3a_4$.
					    \item Let $b=a_2H^2+a_3b_4$.
					    \item Let $c=\max(b_2,b_3,c_4)$.
					    \item Let $d=\max(b_1,d_4)$.
					    \item Let $p(x,n,k,m)$ be the following procedure:
					    \begin{enumerate}
						    \item Show that $\tau_m\le a_1$ using procedure $p_1$.
						    \item Hence show that $\lVert\frac{k}{4}\tau_m\iu\rVert^2=(\frac{k{\tau_m}}{4})^2$.
						    \item Hence using procedures $p_2,p_3,p_4$, show that $\exp_n(x+\frac{k}{4}\tau_m\iu)$
						    \begin{enumerate}
						      \item $\equiv\exp_n(\frac{k}{4}\tau_m\iu)\exp_n(x)\err{\frac{a_2H^2}{n}}$
						      \item $\equiv\iu^k\exp_n(x)\err{a_3(\frac{a_4n}{m}+\frac{b_4}{n})}$
						    \end{enumerate}
						    \item \textbf{Hence show that $\exp_n(x+\frac{k}{4}\tau_m\iu)\equiv\iu^k\exp_n(x)\err{\frac{a_2H^2}{n}+a_3(\frac{a_4n}{m}+\frac{b_4}{n})}\err{\frac{an}{m}+\frac{b}{n}}$.}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,b,c,d,p\rangle$.}
				    \end{enumerate}
		    \procedure{3.89}
			    \objective
				    Choose a positive integer $K$. The objective of the following instructions is to construct rational numbers $a,b,c,d$, and a procedure, $p(n,m,k)$, to show that $\exp_n(\frac{k}{K}\tau_m\iu)^K\equiv 1\err{\frac{an}{m}+\frac{b}{n}}$ when an integer $k$ and positive integers $n,m$ such that $0\le k<K$, $n\ge c$, and $m>d$ are chosen.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{3.50} and let $\langle a_1,b_1,p_1\rangle$ receive.
					    \item Execute \procedurehr{3.16} on $\langle Ka_1\rangle$ and let $\langle a_2,b_2,p_2\rangle$ receive.
					    \item Execute \procedurehr{3.56} on $\langle 4K\rangle$ and let $\langle a_3,b_3,c_3,d_3,p_3\rangle$ receive.
					    \item Let $a=a_3$.
					    \item Let $b=a_2K+b_3$.
					    \item Let $c=\max(b_2,c_3)$.
					    \item Let $d=\max(b_1,d_3)$.
					    \item Let $p(n,m,k)$ be the following procedure:
					    \begin{enumerate}
						    \item Show that $\tau_m\le a_1$ using procedure $p_1$.
						    \item Hence show that $\lVert K\frac{k}{K}\tau_m\iu\rVert=\lVert k\tau_m\rVert^2\le(Ka_1)^2$.
						    \item Now using procedures $p_2,p_3$, show that $\exp_n(\frac{k}{K}\tau_m\iu)^K$
						    \begin{enumerate}
						      \item $\equiv\exp_n(K\frac{k}{K}\tau_m\iu)\err{\frac{a_2K}{n}}$
						      \item $=\exp_n(\frac{4k}{4}\tau_m\iu)$
						      \item $\equiv\iu^{4k}\err{\frac{a_3n}{m}+\frac{b_3}{n}}$
						    \end{enumerate}
						    \item \textbf{Hence show that $\exp_n(\frac{k}{K}\tau_m\iu)^K\equiv\iu^{4k}\err{\frac{a_2K}{n}+\frac{a_3n}{m}+\frac{b_3}{n}}\err{\frac{an}{m}+\frac{b}{n}}$.}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,b,c,d,p\rangle$.}
				    \end{enumerate}
	    \end{multicols}
	    \mfigure
		    \begin{minipage}[c]{0.65\textwidth}
			    \begin{tikzpicture}[scale=1.25,every node/.style={scale=0.9}]
				    \draw[step=1,help lines] (-4.000000,-3.000000) grid (4.000000,3.000000);
            \draw[->] (-4.000000,0) -- (4.000000,0);
            \draw[->] (0,-3.000000) -- (0,3.000000);
            \draw (-4.000000, 0.100000) -- (-4.000000,-0.100000);
            \draw (-3.000000, 0.100000) -- (-3.000000,-0.100000);
            \draw (-2.000000, 0.100000) -- (-2.000000,-0.100000);
            \draw (-1.000000, 0.100000) -- (-1.000000,-0.100000);
            \draw (0.000000, 0.100000) -- (0.000000,-0.100000);
            \draw (1.000000, 0.100000) -- (1.000000,-0.100000);
            \draw (2.000000, 0.100000) -- (2.000000,-0.100000);
            \draw (3.000000, 0.100000) -- (3.000000,-0.100000);
            \draw (0.100000, -3.000000) -- (-0.100000,-3.000000);
            \draw (0.100000, -2.000000) -- (-0.100000,-2.000000);
            \draw (0.100000, -1.000000) -- (-0.100000,-1.000000);
            \draw (0.100000, 0.000000) -- (-0.100000,0.000000);
            \draw (0.100000, 1.000000) -- (-0.100000,1.000000);
            \draw (0.100000, 2.000000) -- (-0.100000,2.000000);
            \draw (1,-0.100000) node[anchor=north,fill=white]{$1$};
            \draw (-0.100000,1) node[anchor=east,fill=white]{$i$};
            \draw (1.000000,0.000000) node[anchor=south west,fill=white]{$\exp_{30}(\frac{0}{10}\tau_{100}i)$};
            \filldraw (1.000000,0.000000) circle (2pt);
            \draw (0,0) -- (1.000000,0.000000);
            \draw (0.816701,0.588280) node[anchor=south west,fill=white]{$\exp_{30}(\frac{1}{10}\tau_{100}i)$};
            \filldraw (0.816701,0.588280) circle (2pt);
            \draw (0,0) -- (0.816701,0.588280);
            \draw (0.325644,0.973269) node[anchor=south west,fill=white]{$\exp_{30}(\frac{2}{10}\tau_{100}i)$};
            \filldraw (0.325644,0.973269) circle (2pt);
            \draw (0,0) -- (0.325644,0.973269);
            \draw (-0.313007,1.012825) node[anchor=south east,fill=white]{$\exp_{30}(\frac{3}{10}\tau_{100}i)$};
            \filldraw (-0.313007,1.012825) circle (2pt);
            \draw (0,0) -- (-0.313007,1.012825);
            \draw (-0.882924,0.671288) node[anchor=south east,fill=white]{$\exp_{30}(\frac{4}{10}\tau_{100}i)$};
            \filldraw (-0.882924,0.671288) circle (2pt);
            \draw (0,0) -- (-0.882924,0.671288);
            \draw (-1.174740,0.036654) node[anchor=south east,fill=white]{$\exp_{30}(\frac{5}{10}\tau_{100}i)$};
            \filldraw (-1.174740,0.036654) circle (2pt);
            \draw (0,0) -- (-1.174740,0.036654);
            \draw (-1.051451,-0.696467) node[anchor=north east,fill=white]{$\exp_{30}(\frac{6}{10}\tau_{100}i)$};
            \filldraw (-1.051451,-0.696467) circle (2pt);
            \draw (0,0) -- (-1.051451,-0.696467);
            \draw (-0.498924,-1.276177) node[anchor=north east,fill=white]{$\exp_{30}(\frac{7}{10}\tau_{100}i)$};
            \filldraw (-0.498924,-1.276177) circle (2pt);
            \draw (0,0) -- (-0.498924,-1.276177);
            \draw (0.353481,-1.464927) node[anchor=north west,fill=white]{$\exp_{30}(\frac{8}{10}\tau_{100}i)$};
            \filldraw (0.353481,-1.464927) circle (2pt);
            \draw (0,0) -- (0.353481,-1.464927);
            \draw (1.251358,-1.116865) node[anchor=north west,fill=white]{$\exp_{30}(\frac{9}{10}\tau_{100}i)$};
            \filldraw (1.251358,-1.116865) circle (2pt);
            \draw (0,0) -- (1.251358,-1.116865);
			    \end{tikzpicture}
		    \end{minipage}
		    \begin{minipage}[c]{0.35\textwidth}
			    A plot of the list of complex numbers $\exp_{30}(\frac{[0:11]}{10}\tau_{100}i)$. Notice that when measurements are done relative to the complex number $1$, $\exp_{30}(\frac{1}{10}\tau_{100}i)$ is roughly $\frac{1}{10}$\textsuperscript{th} of a revolution, and also that each complex number has an angle that is roughly an integral multiple of that of $\exp_{30}(\frac{1}{10}\tau_{100}i)$.
		    \end{minipage}
	    \begin{multicols*}{2}
		    \procedure{3.90}
			    \objective
				    Choose a two rationals $M,N$ such that $0<M$ and $N^2<12$. The objective of the following instructions is to construct rational numbers $a,b$ such that $a>0$, and a procedure, $p(x,n)$, to show that $\lVert\cos_n(x)-1\rVert^2\ge a^2$ when a rational number $x$ and a positive integer $n$ such that $M\le\lvert x\rvert\le N$ and $n>b$ are chosen.
			    \implementation
				    \begin{enumerate}
					    \item Let $a=\frac{M^2}{4}(1-\frac{N^2}{12})$.
					    \item \textbf{Show that $a>0$.}
					    \item Let $b=4$.
					    \item Let $p(x,n)$ be the following procedure:
					    \begin{enumerate}
						    \item Using \procedurehr{3.22}, show that $(\cos_n(x)-1)^2$
						    \begin{enumerate}
							    \item $=(\frac{1}{2}((1+\frac{x\iu}{n})^n+(1-\frac{x\iu}{n})^n)-1)^2$
							    \item $=(\frac{1}{2}(\sum_r^{[0:{n+1}]}\frac{n^{\ul{r}}}{r!}(\frac{x}{n})^r\iu^r+\sum_r^{[0:{n+1}]}\frac{n^{\ul{r}}}{r!}(\frac{x}{n})^r(-\iu)^r)-1)^2$
							    \item $=(\sum_r^{[0:\lfloor\frac{n}{2}\rfloor+1]}\frac{n^{\ul{2r}}}{(2r)!}(\frac{x}{n})^{2r}(-1)^r-1)^2$
							    \item $=(\sum_r^{[1:\lfloor\frac{n}{2}\rfloor+1]}\frac{n^{\ul{2r}}}{(2r)!}(\frac{x}{n})^{2r}(-1)^r)^2$
							    \item $=(\sum_r^{[1:\lfloor\frac{\lfloor\frac{n}{2}\rfloor}{2}\rfloor+1]}(-\frac{n^{\ul{4r-2}}}{(4r-2)!}(\frac{x}{n})^{4r-2}+\frac{n^{\ul{4r}}}{(4r)!}(\frac{x}{n})^{4r})-\frac{n^{\ul{2\lfloor\frac{n}{2}\rfloor}}}{(2\lfloor\frac{n}{2}\rfloor)!}(\frac{x}{n})^{2\lfloor\frac{n}{2}\rfloor}[\lfloor\frac{n}{2}\rfloor\bmod 2=1])^2$
							    \item $\ge(\sum_r^{[1:\lfloor\frac{\lfloor\frac{n}{2}\rfloor}{2}\rfloor+1]}\frac{n^{\ul{4r-2}}}{(4r-2)!}(\frac{x}{n})^{4r-2}(-1+\frac{(n-4r+2)^{\ul{2}}}{(4r)^{\ul{2}}}(\frac{x}{n})^{2}))^2$
							    \item $\ge(\sum_r^{[1:\lfloor\frac{\lfloor\frac{n}{2}\rfloor}{2}\rfloor+1]}\frac{n^{\ul{4r-2}}}{(4r-2)!}(\frac{x}{n})^{4r-2}(-1+\frac{1}{(4r)^{\ul{2}}}(x)^{2}))^2$
							    \item $\ge(\sum_r^{[1:\lfloor\frac{\lfloor\frac{n}{2}\rfloor}{2}\rfloor+1]}\frac{n^{\ul{4r-2}}}{(4r-2)!}(\frac{x}{n})^{4r-2}(-1+\frac{1}{12}x^{2}))^2$
							    \item $\ge(\sum_r^{[1:\lfloor\frac{\lfloor\frac{n}{2}\rfloor}{2}\rfloor+1]}\frac{n^{\ul{4r-2}}}{(4r-2)!}(\frac{x}{n})^{4r-2}(-1+\frac{N^2}{12}))^2$
							    \item $\ge(\frac{n^{\ul{2}}}{2}(\frac{x}{n})^{2}(-1+\frac{N^2}{12}))^2$
							    \item $\ge(\frac{1}{4}x^{2}(-1+\frac{N^2}{12}))^2$
							    \item $\ge(\frac{M^2}{4}(-1+\frac{N^2}{12}))^2$
							    \item $=a^2$
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,b,p\rangle$.}
				    \end{enumerate}
		    \procedure{3.60}
			    \objective
				    Choose a positive integer $K$. The objective of the following instructions is to construct rational numbers $a,b,c$ such that $a>0$, and a procedure, $p(n,m,k)$, to show that $\lVert\exp_n(\frac{k}{K}\tau_m\iu)-1\rVert^2\ge a^2$ when an integer $k$ and positive integers $n,m$ such that $0<\lvert k\rvert\le\frac{K}{2}$, $n>b$, and $m>c$ are chosen.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{3.50} and let $\langle a_1,c,p_1\rangle$ receive.
					    \item Show that $(\frac{a_1}{2})^2<12$.
					    \item Execute \procedurehr{3.49} and let $\langle a_2,p_2\rangle$ receive.
					    \item Show that $a_2>0$.
					    \item Execute \procedurehr{3.90} on $\langle\frac{a_2}{K},\frac{a_1}{2}\rangle$ and let $\langle a,b,p_3\rangle$ receive.
					    \item \textbf{Show that $a>0$.}
					    \item Let $p(n,m,k)$ be the following procedure:
					    \begin{enumerate}
						    \item Show that $\frac{1}{K}\le\frac{\lvert k\rvert}{K}\le\frac{1}{2}$ given that $1\le\lvert k\rvert\le\frac{K}{2}$.
						    \item Hence show that $0<\frac{a_2}{K}\le\frac{1}{K}\tau_m\le\frac{\lvert k\rvert}{K}\tau_m\le\frac{1}{2}\tau_m\le\frac{a_1}{2}$ using procedures $p_1$ and $p_2$.
						    \item Hence show that $(\cos_n(\frac{k}{K}\tau_m)-1)^2\ge a^2$ using procedure $p_3$.
						    \item Using \procedurehr{3.22}, show that $\lVert\exp_n(\frac{k}{K}\tau_m\iu)-1\rVert^2$
						    \begin{enumerate}
							    \item $\ge\Re(\exp_n(\frac{k}{K}\tau_m\iu)-1)^2$
							    \item $=(\cos_n(\frac{k}{K}\tau_m)-1)^2$
							    \item $\ge a^2$
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,b,c,p\rangle$.}
				    \end{enumerate}
		    \procedure{3.61}
			    \objective
				    Choose a positive integer $K$. The objective of the following instructions is to construct rational numbers $a,b,c$ such that $a>0$, and a procedure, $p(n,m,j,k)$, to show that $\lVert\exp_n(\frac{k}{K}\tau_m\iu)-\exp_n(\frac{j}{K}\tau_m\iu)\rVert^2\ge a^2$ when positive integers $n,j,k,m$ such that $-K<j\le k<K$, $0<k-j\le\frac{K}{2}$, $n\ge b$, and $m\ge c$ are chosen.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{3.50} and let $\langle a_1,b_1,p_1\rangle$ receive.
					    \item Execute \procedurehr{3.14} on $\langle a_1\rangle$ and let $\langle a_2,b_2,p_2\rangle$ receive.
					    \item Execute \procedurehr{3.60} on $\langle K\rangle$ and let $\langle a_3,b_3,c_3,p_3\rangle$ receive.
					    \item Execute \procedurehr{3.15} on $\langle a_1\rangle$ and let $\langle a_4,b_4,p_4\rangle$ receive.
					    \item Let $a=\frac{1}{2}a_2a_3$.
					    \item Let $b=\max(\frac{2a_4{a_1}^2}{a_2a_3},b_3,b_4,b_2)$.
					    \item Let $c=\max(b_1,c_3)$.
					    \item Let $p(n,m,j,k)$ be the following procedure:
					    \begin{enumerate}
						    \item Show that $\lVert\frac{j}{K}\rVert^2<1$
						    \begin{enumerate}
						      \item given that $-1<\frac{j}{K}<1$
						      \item given that $-K<j<K$.
						    \end{enumerate}
						    \item Hence show that $\lVert\frac{j}{K}\tau_m\iu\rVert^2=\lVert\frac{j}{K}\rVert^2\lVert\tau_m\rVert^2\le\lVert\tau_m\rVert^2\le{a_1}^2$ using procedure $p_1$.
						    \item Hence show that $\lVert\exp_n(\frac{j}{K}\tau_m\iu)\rVert^2\ge{a_2}^2>0$ using procedure $p_2$.
						    \item Hence show that $\lVert\exp_n(\frac{k-j}{K}\tau_m\iu)-1\rVert^2\ge{a_3}^2>0$ using procedure $p_3$.
						    \item Show that $\lVert\frac{k-j}{K}\tau_m\iu\rVert^2\le\lVert\frac{k-j}{K}\rVert^2\lVert\tau_m\rVert^2\le\lVert\tau_m\rVert^2\le{a_1}^2$ given that $0<\frac{k-j}{K}\le\frac{1}{2}$.
						    \item Show that $n\ge b\ge\frac{2a_4{a_1}^2}{a_2a_3}$.
						    \item Hence show that $\lVert\exp_n(\frac{k-j}{K}\tau_m\iu)\exp_n(\frac{j}{K}\tau_m\iu)-\exp_n(\frac{k-j}{K}\tau_m\iu+\frac{j}{K}\tau_m\iu)\rVert^2\le\frac{{a_4}^2\lVert\frac{k-j}{K}\tau_m\iu\rVert^2\lVert\frac{j}{K}\tau_m\iu\rVert^2}{n^2}\le\frac{{a_4}^2{a_1}^4}{n^2}\le(\frac{a_2a_3}{2})^2$ using procedure $p_4$.
						    \item Hence using \procedurehr{sat1708191238}, show that $\lVert\exp_n(\frac{k}{K}\tau_m\iu)-\exp_n(\frac{j}{K}\tau_m\iu)\rVert^2$
						    \begin{enumerate}
							    \item $=\lVert\exp_n(\frac{k-j}{K}\tau_m\iu+\frac{j}{K}\tau_m\iu)-\exp_n(\frac{k-j}{K}\tau_m\iu)\exp_n(\frac{j}{K}\tau_m\iu)+\exp_n(\frac{k-j}{K}\tau_m\iu)\exp_n(\frac{j}{K}\tau_m\iu)-\exp_n(\frac{j}{K}\tau_m\iu)\rVert^2$
							    \item $=\lVert\exp_n(\frac{j}{K}\tau_m\iu)(\exp_n(\frac{k-j}{K}\tau_m\iu)-1)-(\exp_n(\frac{k-j}{K}\tau_m\iu+\frac{j}{K}\tau_m\iu)-\exp_n(\frac{k-j}{K}\tau_m\iu)\exp_n(\frac{j}{K}\tau_m\iu))\rVert^2$
							    \item $\ge (a_2a_3-\frac{a_2a_3}{2})^2$
							    \item $\ge a^2$.
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,b,c,p\rangle$.}
				    \end{enumerate}
		    \procedure{3.62}
			    \objective
				    Choose a positive integer $K$. The objective of the following instructions is to construct rational numbers $a,b,c$ such that $a>0$, and a procedure, $p(n,m,j,k)$, to show that $\lVert\exp_n(\frac{k}{K}\tau_m\iu)-\exp_n(\frac{j}{K}\tau_m\iu)\rVert^2\ge a^2$ when positive integers $n,j,k,m$ such that $0\le j\le k<K$, $\frac{K}{2}\le k-j<K$, $n\ge b$, and $\frac{m}{n}\ge c$ are chosen.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{3.61} on $\langle K\rangle$ and let $\langle a_1,b_1,c_1,p_1\rangle$ receive.
					    \item Execute \procedurehr{3.50} and let $\langle a_2,b_2,p_2\rangle$ receive.
					    \item Execute \procedurehr{3.59} on $\langle a_2,4\rangle$ and let $\langle a_3,b_3,c_3,d_3,p_3\rangle$ receive.
					    \item Let $a=\frac{1}{2}a_1$.
					    \item Let $b=\max(\frac{4b_3}{a_1},b_1,c_3)$.
					    \item Let $c=\max(\frac{4a_3}{a_1},\frac{c_1}{b},\frac{b_2}{b},\frac{d_3}{b})$.
					    \item Let $p(n,m,j,k)$ be the following procedure:
					    \begin{enumerate}
						    \item Show that $-\frac{K}{2}\le k-K<j<\frac{K}{2}$.
						    \item Also show that $0<j-(k-K)\le\frac{K}{2}$.
						    \item Show that $m\ge cn\ge\frac{c_1}{b}b=c_1$.
						    \item Hence show that $\lVert\exp_n(\frac{j}{K}\tau_m\iu)-\exp_n(\frac{k-K}{K}\tau_m\iu)\rVert^2\ge{a_1}^2$ using procedure $p_1$.
						    \item Show that $m\ge cn\ge\frac{b_2}{b}b=b_2$.
						    \item Hence show that $\tau_m\le a_2$ using procedure $p_2$.
						    \item Hence show that $\lVert\frac{k}{K}\tau_m\iu\rVert^2=\lVert\frac{k}{K}\rVert^2\lVert\tau_m\rVert^2\le\lVert\tau_m\rVert^2\le{a_2}^2$.
						    \item Also show that $m\ge cn\ge\frac{d_3}{b}b=d_3$.
						    \item Hence show that $\lVert\iu^{-4}\exp_n(\frac{k}{K}\tau_m\iu)-\exp_n(\frac{k}{K}\tau_m\iu-\frac{4}{4}\tau_m\iu)\rVert^2\le(\frac{a_3n}{m}+\frac{b_3}{n})^2\le(\frac{a_1}{2})^2$ using procedure $p_3$.
						    \item Now show that $\lVert\exp_n(\frac{k}{K}\tau_m\iu)-\exp_n(\frac{j}{K}\tau_m\iu)\rVert^2$
						    \begin{enumerate}
							    \item $=\lVert\exp_n(\frac{k}{K}\tau_m\iu)-\exp_n(\frac{k}{K}\tau_m\iu-\tau_m\iu)+\exp_n(\frac{k-K}{K}\tau_m\iu)-\exp_n(\frac{j}{K}\tau_m\iu)\rVert^2$
							    \item $\ge\frac{1}{2}\lVert\exp_n(\frac{k-K}{K}\tau_m\iu)-\exp_n(\frac{j}{K}\tau_m\iu)\rVert^2-\lVert\exp_n(\frac{k}{K}\tau_m\iu)-\exp_n(\frac{k}{K}\tau_m\iu-\tau_m\iu)\rVert^2$
							    \item $\ge\frac{1}{2}{a_1}^2-(\frac{a_1}{2})^2$
							    \item $\ge a^2$.
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,b,c,p\rangle$.}
				    \end{enumerate}
		    \procedure{3.63}
			    \objective
				    Choose a positive integer $K$. The objective of the following instructions is to construct rational numbers $a,b,c$ such that $a>0$, and a procedure, $p(n,m,j,k)$, to show that $\lVert\exp_n(\frac{k}{K}\tau_m\iu)-\exp_n(\frac{j}{K}\tau_m\iu)\rVert^2\ge a^2$ when positive integers $n,j,k,m$ such that $0\le j\le k<K$, $0<k-j<K$, $n\ge b$, and $\frac{m}{n}\ge c$ are chosen.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{3.61} on $\langle K\rangle$ and let $\langle a_1,b_1,c_1,p_1\rangle$ receive.
					    \item Execute \procedurehr{3.62} on $\langle K\rangle$ and let $\langle a_2,b_2,c_2,p_2\rangle$ receive.
					    \item Let $a=\min(a_1,a_2)$.
					    \item \textbf{Show that $a>0$.}
					    \item Let $b=\max(b_1,b_2)$.
					    \item Let $c=\max(\frac{c_1}{b},c_2)$.
					    \item Let $p(n,m,j,k)$ be the following procedure:
					    \begin{enumerate}
						    \item If $k-j\le\frac{K}{2}$, then do the following:
						    \begin{enumerate}
						      \item Show that $m\ge cn\ge\frac{c_1}{b}b=c_1$.
							    \item \textbf{Hence show that $\lVert\exp_n(\frac{k}{K}\tau_m\iu)-\exp_n(\frac{j}{K}\tau_m\iu)\rVert^2\ge a_1\ge a$ using procedure $p_1$.}
						    \end{enumerate}
						    \item Otherwise if $k-j>\frac{K}{2}$, then do the following:
						    \begin{enumerate}
							    \item \textbf{Show that $\lVert\exp_n(\frac{k}{K}\tau_m\iu)-\exp_n(\frac{j}{K}\tau_m\iu)\rVert^2\ge a_2\ge a$ using procedure $p_2$.}
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,b,c,p\rangle$.}
				    \end{enumerate}
		    \declaration{3.34}
			    The phrase "\gls{complexpolynomial}" will be used to indicate that the declarations and procedures pertaining to polynomials are being used but with the provison that all uses of rational numbers therein are substituted with uses of complex numbers.
		    \procedure{3.64}
			    \objective
				    Choose a positive integer $K$. The objective of the following instructions is to construct rational numbers $a,b,c,d$, and a procedure, $p(n,m)$, to construct a list of complex numbers $z$ and a list of complex polynomials $q$ such that,
				    \begin{enumerate}
					    \item $z_k=\exp_n(\frac{K-k-1}{K}\tau_m\iu)$ for $k\in[0:K]$
					    \item $q_K=\lambda^K-1$
					    \item $q_{K-1}=\sum_r^{[0:K]}\lambda^r$
					    \item $q_{k+1}=(\lambda-z_k)q_k+\Lambda(q_{k+1},z_k)$ for $k\in[0:K]$
					    \item $(q_k)_{\deg(q_k)}=1$ for $k\in[0:K+1]$
					    \item $\Lambda(q_k,z_j)\equiv 0\err{\frac{an}{m}+\frac{b}{n}}$ for $j\in[0:k]$, for $k\in[0:K+1]$
				    \end{enumerate}
				    when two positive integers $n,m$ such that $n>c$ and $\frac{m}{n}>d$ are chosen.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{3.89} on $\langle K\rangle$ and let $\langle a_1,b_1,c_1,d_1,p_1\rangle$ receive.
					    \item Execute \procedurehr{3.63} on $\langle K\rangle$ and let $\langle a_2,b_2,c_2,p_2\rangle$ receive.
					    \item Let $a=\max(1,\frac{2}{a_2})^Ka_1$.
					    \item Let $b=\max(1,\frac{2}{a_2})^Kb_1$.
					    \item Let $c=\max(c_1,b_2)$.
					    \item Let $d=\max(d_1,c_2)$.
					    \item Let $p(n,m)$ be the following procedure:
					    \begin{enumerate}
						    \item \textbf{Let $q_K=\lambda^K-1$.}
						    \item For $k\in[K:0]$, do the following:
						    \begin{enumerate}
							    \item \textbf{Let $z_k=\exp_n(\frac{K-k-1}{K}\tau_m\iu)$.}
							    \item \textbf{Now show that $\lVert\Lambda(q_K,z_k)\rVert^2\le(\frac{a_1n}{m}+\frac{b_1}{n})^2$ using procedure $p_1$.}
						    \end{enumerate}
						    \item For $k\in[K:0]$, do the following:
						    \begin{enumerate}
							    \item Let $q_k=q_{k+1}\div(\lambda-z_k)$.
							    \item Let $r_k=q_{k+1}\bmod(\lambda-z_k)$.
							    \item Show that $\deg(r_k)=0$ given that $\deg(r_k)<\deg(\lambda-z_k)=1$.
							    \item \textbf{Show that $1=(q_{k+1})_{\deg(q_{k+1})}=((\lambda-z_k)q_k+r_k)_{\deg(q_{k+1})}=(q_k)_{\deg(q_k)}$ given that $q_{k+1}=(\lambda-z_k)q_k+r_k$.}
							    \item \textbf{Show that $q_{k+1}=(\lambda-z_k)q_k+\Lambda(q_{k+1},z_k)$ given that $\Lambda(q_{k+1},z_k)=\Lambda(\lambda-z_k,z_k)\Lambda(q_k,z_k)+\Lambda(r_k,z_k)=(z_k-z_k)\Lambda(q_k,z_k)+r_k=r_k$.}
							    \item \textbf{Execute the \subprocedurehr{wed2808191630} on $\langle k,q_{k+1},z\rangle$.}
						    \end{enumerate}
						    \item Now using (cv), verify that $(\lambda-1)\sum_r^{[0:K]}\lambda^r$
						    \begin{enumerate}
							    \item $=q_K$
							    \item $=(\lambda-z_{K-1})q_{K-1}+\Lambda(q_K,z_{K-1})$
							    \item $=(\lambda-1)q_{K-1}+\Lambda(\lambda^K-1,1)$
							    \item $=(\lambda-1)q_{K-1}$.
						    \end{enumerate}
						    \item \textbf{Hence show that $\sum_r^{[0:K]}\lambda^r=q_{K-1}$.}
						    \item \textbf{Yield the tuple $\langle z,q\rangle$.}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,b,c,d,p\rangle$.}
				    \end{enumerate}
			    \subprocedure{wed2808191630}
			      \paragraph{Objective}
			        Choose a non-negative integer $k$, a complex polynomial $q_{k+1}$, and a list of complex numbers $z$ such that $z_j=\exp_n(\frac{j}{K}\tau_m\iu)$ and $\Lambda(q_{k+1},z_j)\equiv 0\err{(\frac{2}{a_2})^{K-(k+1)}(\frac{a_1n}{m}+\frac{b_1}{n})}$ for $j\in[k+1:0]$. Let $q_k=q_{k+1}\div(\lambda-z_k)$. The objective of the following instructions is to show that $\Lambda(q_{k},z_j)\equiv 0\err{(\frac{2}{a_2})^{K-k}(\frac{a_1n}{m}+\frac{b_1}{n})}\err{\frac{an}{m}+\frac{b}{n}}$ for $j\in[k:0]$.
			      \paragraph{Implementation}
			        \begin{enumerate}
			          \item For $j\in[k:0]$, do the following:
						    \begin{enumerate}
							    \item Show that $\Lambda(q_{k+1},z_j)-\Lambda(q_{k+1},z_k)=(z_j-z_k)\Lambda(q_k,z_j)$ given that $\Lambda(q_{k+1},z_j)=\Lambda(\lambda-z_k,z_j)\Lambda(q_k,z_j)+\Lambda(q_{k+1},z_k)$.
							    \item Show that $\lVert z_j-z_k\rVert^2\ge {a_2}^2$ using procedure $p_2(n,m,\min(j,k),\max(j,k))$.
							    \item Hence show that ${a_2}^2\lVert\Lambda(q_k,z_j)\rVert^2$
							    \begin{enumerate}
							      \item $\le\lVert z_j-z_k\rVert^2\lVert\Lambda(q_k,z_j)\rVert^2$
							      \item $=\lVert(z_j-z_k)\Lambda(q_k,z_j)\rVert^2$
							      \item $=\lVert\Lambda(q_{k+1},z_j)-\Lambda(q_{k+1},z_k)\rVert^2$
							      \item $\le((\frac{2}{a_2})^{K-k-1}(\frac{a_1n}{m}+\frac{b_1}{n})+(\frac{2}{a_2})^{K-k-1}(\frac{a_1n}{m}+\frac{b_1}{n}))^2$
							      \item $=(2(\frac{2}{a_2})^{K-k-1}(\frac{a_1n}{m}+\frac{b_1}{n}))^2$
							      \item $={a_2}^2((\frac{2}{a_2})^{K-k}(\frac{a_1n}{m}+\frac{b_1}{n}))^2$.
							    \end{enumerate}
							    \item \textbf{Hence show that $\lVert\Lambda(q_k,z_j)\rVert^2\le((\frac{2}{a_2})^{K-k}(\frac{a_1n}{m}+\frac{b_1}{n}))^2\le(\frac{an}{m}+\frac{b}{n})^2$.}
						    \end{enumerate}
					    \end{enumerate}
		    \procedure{3.65}
			    \objective
				    Choose a rational number $X$ and a positive integer $K$. The objective of the following instructions is to construct rational numbers $a,b,c,d$, and a procedure, $p(x,n,m)$, to show that $\sum_r^{[0:K]}x^r\equiv\prod_r^{[1:K]}(x-\exp_n(\frac{r}{K}\tau_m\iu))\err{\frac{an}{m}+\frac{b}{n}}$ when a complex number $x$ and positive integers $n,m$ such that $n>c$, $\frac{m}{n}>d$, and $\lVert x\rVert^2\le X$ are chosen.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{3.64} on $\langle K\rangle$ and let $\langle a_1,b_1,c_1,d_1,p_1\rangle$ receive.
					    \item Execute \procedurehr{3.50} and let $\langle a_2,b_2,p_2\rangle$ receive.
					    \item Execute \procedurehr{3.13} on $\langle a_2\rangle$ and let $\langle a_3,b_3,p_3\rangle$ receive.
					    \item Let $l=\sum_k^{[0:K-1]}\prod_j^{[k+1:K-1]}(X+a_3)$.
					    \item Let $a=a_1l$.
					    \item Let $b=b_1l$.
					    \item Let $c=\max(c_1,b_3)$.
					    \item Let $d=\max(d_1,b_2)$.
					    \item Let $p(x,n,m)$ be the following procedure:
					    \begin{enumerate}
						    \item Show that $\tau_m\le a_2$ using procedure $p_2$.
						    \item Execute procedure $p_1$ on $\langle n,m\rangle$ and let $\langle z,t\rangle$ receive.
						    \item For $j\in[1:K]$, do the following:
						    \begin{enumerate}
							    \item Show that $\lVert\frac{j}{K}\tau_m\iu\rVert^2=\lVert\frac{j}{K}\rVert^2\lVert\tau_m\rVert^2\le\lVert\tau_m\rVert^2\le a_2$.
							    \item Hence show that $\lVert z_j\rVert^2=\lVert\exp_n(\frac{j}{K}\tau_m\iu)\rVert^2\le a_3$ using procedure $p_3$.
						    \end{enumerate}
						    \item Hence show that $\lVert\sum_r^{[0:K]}x^r-\prod_r^{[1:K]}(x-z_r)\rVert^2$
						    \begin{enumerate}
							    \item $=\lVert\Lambda(\sum_r^{[0:K]}\lambda^r,x)-\prod_r^{[1:K]}(x-z_r)\rVert^2$
							    \item $=\lVert\Lambda(t_{K-1},x)-\prod_r^{[1:K]}(x-z_r)\rVert^2$
							    \item $=\lVert\Lambda(\prod_j^{[0:K-1]}(\lambda-z'_j)+\sum_k^{[0:K-1]}\Lambda(t_{k+1},z'_{k})\prod_j^{[k+1:K-1]}(\lambda-z'_j),x)-\prod_r^{[1:K]}(x-z_r)\rVert^2$
							    \item $=\lVert\prod_j^{[0:K-1]}(x-z'_j)+\sum_k^{[0:K-1]}\Lambda(t_{k+1},z'_{k})\prod_j^{[k+1:K-1]}(x-z'_j)-\prod_r^{[1:K]}(x-z_r)\rVert^2$
							    \item $=\lVert\sum_k^{[0:K-1]}\Lambda(t_{k+1},z'_{k})\prod_j^{[k+1:K-1]}(x-z'_j)\rVert^2$
							    \item $\le(\sum_k^{[0:K-1]}(\frac{a_1n}{m}+\frac{b_1}{n})\prod_j^{[k+1:K-1]}(X+a_3))^2$
							    \item $=((\frac{a_1n}{m}+\frac{b_1}{n})\sum_k^{[0:K-1]}\prod_j^{[k+1:K-1]}(X+a_3))^2$
							    \item $=(\frac{an}{m}+\frac{b}{n})^2$.
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,b,c,d,p\rangle$.}
				    \end{enumerate}
		    \procedure{3.66}
			    \objective
				    Choose a rational number $X$ and a positive integer $K$. The objective of the following instructions is to construct rational numbers $a,b,c,d$, and a procedure, $p(x,n,m)$, to show that $x^K-1\equiv\prod_r^{[0:K]}(x-\exp_n(\frac{r}{K}\tau_m\iu))\err{\frac{an}{m}+\frac{b}{n}}$ when a complex number $x$ and positive integers $n,m$ such that $n>c$, $\frac{m}{n}>d$, and $\lVert x\rVert^2\le X$ are chosen.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{3.65} on $\langle X,K\rangle$ and let $\langle a_1,b_1,c,d,p_1\rangle$ receive.
					    \item Let $a=(X+1)a_1$.
					    \item Let $b=(X+1)b_1$.
					    \item Let $p(x,n,m)$ be the following procedure:
					    \begin{enumerate}
						    \item Show that $\lVert\sum_r^{[0:K]}x^r-\prod_r^{[1:K]}(x-\exp_n(\frac{r}{K}\tau_m\iu))\rVert^2\le(\frac{a_1n}{m}+\frac{b_1}{n})^2$ using procedure $p_1$.
						    \item Hence show that $\lVert x^K-1-\prod_r^{[0:K]}(x-\exp_n(\frac{r}{K}\tau_m\iu))\rVert^2$
						    \begin{enumerate}
							    \item $=\lVert(x-1)\sum_r^{[0:K]}x^r-(x-1)\prod_r^{[1:K]}(x-\exp_n(\frac{r}{K}\tau_m\iu))\rVert^2$
							    \item $=\lVert x-1\rVert^2\lVert\sum_r^{[0:K]}x^r-\prod_r^{[1:K]}(x-\exp_n(\frac{r}{K}\tau_m\iu))\rVert^2$
							    \item $\le(X+1)^2(\frac{a_1n}{m}+\frac{b_1}{n})^2$
							    \item $=(\frac{an}{m}+\frac{b}{n})^2$.
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,b,c,d,p\rangle$.}
				    \end{enumerate}
		    \procedure{3.67}
			    \objective
				    Choose a rational number $X$ and a positive integer $K$. The objective of the following instructions is to construct rational numbers $a,b,c,d$, and a procedure, $p(x,n,m)$, to show that $\exp_K(x)-1\equiv x\prod_r^{[1:K]}(1-\frac{x}{K(\exp_n(\frac{r}{K}\tau_m\iu)-1)})\err{\frac{an}{m}+\frac{b}{n}}$ when a complex number $x$ and positive integers $n,m$ such that $n>c$, $\frac{m}{n}>d$, and $\lVert x\rVert^2\le X$ are chosen.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{3.66} on $\langle 1+\frac{X}{K},K\rangle$ and let $\langle a_1,b_1,c_1,d_1,p_1\rangle$ receive.
					    \item Execute \procedurehr{3.65} on $\langle 1,K\rangle$ and let $\langle a_2,b_2,c_2,d_2,p_2\rangle$ receive.
					    \item Execute \procedurehr{3.63} on $\langle K\rangle$ and let $\langle a_3,b_3,c_3,p_3\rangle$ receive.
					    \item Let $l=\frac{X}{K}(1+\frac{X}{Ka_3})^{K-1}$.
					    \item Let $a=a_1+la_2$.
					    \item Let $b=b_1+lb_2$.
					    \item Let $c=\max(c_1,c_2,b_3)$.
					    \item Let $d=\max(d_1,d_2,c_3)$.
					    \item Let $p(x,n,m)$ be the following procedure:
					    \begin{enumerate}
						    \item Show that $\lVert 1+\frac{x}{K}\rVert^2\le (1+\frac{X}{K})^2$.
						    \item Hence show that $\lVert(1+\frac{x}{K})^K-1-\prod_r^{[0:K]}(1+\frac{x}{K}-\exp_n(\frac{r}{K}\tau_m\iu))\rVert^2\le(\frac{a_1n}{m}+\frac{b_1}{n})^2$ using procedure $p_1$.
						    \item Hence show that $\lVert K-\prod_r^{[1:K]}(1-\exp_n(\frac{r}{K}\tau_m\iu))\rVert^2=\sum_r^{[0:K]}1^r-\prod_r^{[1:K]}(1-\exp_n(\frac{r}{K}\tau_m\iu))\rVert^2\le(\frac{a_2n}{m}+\frac{b_2}{n})^2$ using procedure $p_2$.
						    \item For $j\in[1:K]$, do the following:
						    \begin{enumerate}
							    \item Show that $\lVert\exp_n(\frac{j}{K}\tau_m\iu)-1\rVert^2\ge{a_3}^2$ using procedure $p_3$.
							    \item Let $z_j=K(\exp_n(\frac{j}{K}\tau_m\iu)-1)$.
						    \end{enumerate}
						    \item Hence show that $\lVert\exp_K(x)-1-x\prod_r^{[1:K]}(1-\frac{x}{z_r})\rVert^2$
						    \begin{enumerate}
							    \item $=\lVert\exp_K(x)-1-\prod_r^{[0:K]}(1+\frac{x}{K}-\exp_n(\frac{r}{K}\tau_m\iu))+\prod_r^{[0:K]}(1+\frac{x}{K}-\exp_n(\frac{r}{K}\tau_m\iu))-x\prod_r^{[1:K]}(1-\frac{x}{z_r})\rVert^2$
							    \item $=\lVert\exp_K(x)-1-\prod_r^{[0:K]}(1+\frac{x}{K}-\exp_n(\frac{r}{K}\tau_m\iu))+\frac{x}{K}\prod_r^{[1:K]}(1+\frac{x}{K}-\exp_n(\frac{r}{K}\tau_m\iu))-x\prod_r^{[1:K]}(1-\frac{x}{z_r})\rVert^2$
							    \item $=\lVert\exp_K(x)-1-\prod_r^{[0:K]}(1+\frac{x}{K}-\exp_n(\frac{r}{K}\tau_m\iu))+\frac{x}{K}\prod_r^{[1:K]}(1-\exp_n(\frac{r}{K}\tau_m\iu))\prod_r^{[1:K]}(1-\frac{x}{z_r})-x\prod_r^{[1:K]}(1-\frac{x}{z_r})\rVert^2$
							    \item $=\lVert(\exp_K(x)-1-\prod_r^{[0:K]}(1+\frac{x}{K}-\exp_n(\frac{r}{K}\tau_m\iu)))+\frac{x}{K}\prod_r^{[1:K]}(1-\frac{x}{z_r})(\prod_r^{[1:K]}(1-\exp_n(\frac{r}{K}\tau_m\iu))-K)\rVert^2$
							    \item $\le((\frac{a_1n}{m}+\frac{b_1}{n})+\frac{X}{K}(\prod_r^{[1:K]}(1+\frac{X}{Ka_3}))(\frac{a_2n}{m}+\frac{b_2}{n}))^2$
							    \item $=((\frac{a_1n}{m}+\frac{b_1}{n})+\frac{X}{K}(1+\frac{X}{Ka_3})^{K-1}(\frac{a_2n}{m}+\frac{b_2}{n}))^2$
							    \item $=(\frac{an}{m}+\frac{b}{n})^2$.
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,b,c,d,p\rangle$.}
				    \end{enumerate}
	    \end{multicols*}
	  \clearpage
	\part{Differential Arithmetic}
	  \chapter{Differential Arithmetic}
	    \begin{multicols*}{2}
	      \procedure{tue2008191129}
	        \objective
	          Choose the following:
	          \begin{enumerate}
	            \item A procedure $q_1(x,n)$ to show that $p_n(x)\equiv 0\err{a_1}$ when a complex number $x$ and a positive integer $n$ such that $P(x)$ and $n>c_1$ are chosen.
	            \item A procedure $q_2(x,n)$ to show that $t_n(x)\equiv 0\err{a_2}^2$ when a complex number $x$ and a positive integer $n$ such that $R(x)$ and $n>c_2$ are chosen.
	          \end{enumerate}
	          The objective of the following instructions is to construct the following:
	          \begin{enumerate}
	            \item Rational numbers $a_3,b_3$.
	            \item A procedure $q_3(x,n)$ to show that $p_n(x)+t_n(x)\equiv 0\err{a_3}$ when a complex number $x$ and a positive integer $n$ such that $P(x)$, $R(x)$, and $n>b_3$ are chosen.
	          \end{enumerate}
	        \implementation
	          \begin{enumerate}
	            \item Let $a_3=a_1+a_2$.
	            \item Let $b_3=\max(c_1,c_2)$.
	            \item Let $q_3(x,n)$ be the following procedure:
	            \begin{enumerate}
	              \item Show that $p_n(x)\equiv 0\err{a_1}$ using procedure $q_1$.
	              \item Show that $t_n(x)\equiv 0\err{a_2}$ using procedure $q_2$.
	              \item Hence show that $p_n(x)+t_n(x)\equiv 0\err{a_1+a_2}\err{a_3}$.
	            \end{enumerate}
	            \item \textbf{Yield the tuple $\langle a_3,b_3,q_3\rangle$.}
	          \end{enumerate}
	      \procedure{tue2008191139}
	        \objective
	          Choose the following:
	          \begin{enumerate}
	            \item A procedure $q_1(x,n)$ to show that $p_n(x)\equiv 0\err{a_1}$ when a complex number $x$ and a positive integer $n$ such that $P(x)$ and $n>c_1$ are chosen.
	            \item A procedure $q_2(x,n)$ to show that $t_n(x)\equiv 0\err{a_2}$ when a complex number $x$ and a positive integer $n$ such that $R(x)$ and $n>c_2$ are chosen.
	          \end{enumerate}
	          The objective of the following instructions is to construct the following:
	          \begin{enumerate}
	            \item Rational numbers $a_3,b_3$.
	            \item A procedure $q_3(x,n)$ to show that $p_n(x)t_n(x)\equiv 0\err{a_3}$ when a complex number $x$ and a positive integer $n$ such that $P(x)$, $R(x)$, and $n>b_3$ are chosen.
	          \end{enumerate}
	        \implementation
	          Implementation is analogous to that of \procedurehr{tue2008191129}.
	      \declaration{tue2008190516}
	        The notation \gls{differentialtaxicab}, where $x$ is a complex number, will be used as a shorthand for $\lvert\Re(x)\rvert+\lvert\Im(x)\rvert$.
	      \procedure{tue2008190655}
	        \objective
	          Choose a complex number $a$ such that $\{a\}=0$. The objective of the following instructions is to show that $a=0$.
	        \implementation
	          \begin{enumerate}
	            \item Using \declarationhr{tue2008190516}, show that $\lvert\Re(a)\rvert+\lvert\Im(a)\rvert=0$.
	            \item Hence show that $\Re(a)=0$
	            \begin{enumerate}
	              \item given that $\lvert\Re(a)\rvert=0$
	              \item given that $0\ge\lvert\Re(a)\rvert\ge 0$
	              \item given that $\lvert\Im(a)\rvert\ge 0$.
	            \end{enumerate}
	            \item Also show that $\Im(a)=0$
	            \begin{enumerate}
	              \item given that $\lvert\Im(a)\rvert=0$
	              \item given that $0\ge\lvert\Im(a)\rvert\ge 0$
	              \item given that $\lvert\Re(a)\rvert\ge 0$.
	            \end{enumerate}
	            \item \textbf{Hence show that $a=0$.}
	          \end{enumerate}
	      \procedure{tue2008190520}
	        \objective
	          Choose a complex number $a$ and a rational number $b$. The objective of the following instructions is to show that $\{ba\}=\lvert b\rvert\{a\}$.
	        \implementation
	          \begin{enumerate}
	            \item Using \declarationhr{tue2008190516}, show that $\{ba\}$
	            \begin{enumerate}
	              \item $=\lvert\Re(ba)\rvert+\lvert\Im(ba)\rvert$
	              \item $=\lvert b\Re(a)\rvert+\lvert b\Im(a)\rvert$
	              \item $=\lvert b\rvert(\lvert\Re(a)\rvert+\lvert\Im(a)\rvert)$
	              \item $=\lvert b\rvert\{a\}$.
	            \end{enumerate}
	          \end{enumerate}
	      \procedure{tue2008190540}
	        \objective
	          Choose two complex numbers $a,b$. The objective of the following instructions is to show that $\{a+b\}\le\{a\}+\{b\}$.
	        \implementation
	          \begin{enumerate}
	            \item Using \declarationhr{tue2008190516}, show that $\{a+b\}$
	            \begin{enumerate}
	              \item $=\lvert\Re(a+b)\rvert+\lvert\Im(a+b)\rvert$
	              \item $=\lvert\Re(a)+\Re(b)\rvert+\lvert\Im(a)+\Im(b)\rvert$
	              \item $\le\lvert\Re(a)\rvert+\lvert\Re(b)\rvert+\lvert\Im(a)\rvert+\lvert\Im(b)\rvert$
	              \item $=\{a\}+\{b\}$.
	            \end{enumerate}
	          \end{enumerate}
	      \procedure{tue2008190546}
	        \objective
	          Choose two complex numbers $a,b$. The objective of the following instructions is to show that $\{ab\}\le\{a\}\{b\}$.
	        \implementation
	          \begin{enumerate}
	            \item Using \procedurehr{tue2008190540}, show that $\{ab\}$
	            \begin{enumerate}
	              \item $=\{(\Re(a)+\Im(b)i)b\}$
	              \item $=\{\Re(a)b+\Im(a)bi\}$
	              \item $\le\{\Re(a)b\}+\{\Im(a)b\}$
	              \item $=(\lvert\Re(a)\rvert+\lvert\Im(a)\rvert)\{b\}$
	              \item $=\{a\}\{b\}$.
	            \end{enumerate}
	          \end{enumerate}
	      \procedure{tue2008190632}
	        \objective
	          Choose a complex number $a$. The objective of the following instructions is to show that $\lVert a\rVert^2\le\{a\}^2$.
	        \implementation
	          \begin{enumerate}
	            \item Using \procedurehr{3.04aa}, show that $\lVert a\rVert^2$
	            \begin{enumerate}
	              \item $=\lVert\Re(a)+\Im(a)i\rVert^2$
	              \item $\le(\lvert\Re(a)\rvert+\lvert\Im(a)\rvert)^2$
	              \item $=\{a\}^2$.
	            \end{enumerate}
	          \end{enumerate}
	      \procedure{tue2008190639}
	        \objective
	          Choose a complex number $a$. The objective of the following instructions is to show that $\{a\}^2\le 2\lVert a\rVert^2$.
	        \implementation
	          \begin{enumerate}
	            \item Show that $2\lVert a\rVert^2-\{a\}^2$
	            \begin{enumerate}
	              \item $=2\Re(a)^2+2\Im(a)^2-(\lvert\Re(a)\rvert+\lvert\Im(a)\rvert)^2$
	              \item $=2\Re(a)^2+2\Im(a)^2-\Re(a)^2-2\lvert\Re(a)\rvert\lvert\Im(a)\rvert-\Im(a)^2$
	              \item $=\Re(a)^2-2\lvert\Re(a)\rvert\lvert\Im(a)\rvert+\Im(a)^2$
	              \item $=(\lvert\Re(a)\rvert-\lvert\Im(a)\rvert)^2$
	              \item $\ge 0$.
	            \end{enumerate}
	            \item \textbf{Hence show that $\{a\}^2\le 2\lVert a\rVert^2$.}
	          \end{enumerate}
	      \declaration{3.29}
			    The notation \gls{differentialdiff}, where $x,z$ are complex numbers such that $z\ne 0$ and $f[x]$ is a function of $x$, will be used as a shorthand for $\frac{f(y+z)-f(y)}{z}$.
		    \procedure{3.83}
			    \objective
				    Choose two functions $f[x],g[x]$ and two complex numbers $y,z$ such that $z\ne 0$. The objective of the following instructions is to show that $\diff_{x=y}^z(f(x)+g(x))=\diff_{x=y}^zf(x)+\diff_{x=y}^zg(x)$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\diff_{x=y}^z(f(x)+g(x))$
					    \begin{enumerate}
						    \item $=\frac{(f(y+z)+g(y+z))-(f(y)+g(y))}{z}$
						    \item $=\frac{f(y+z)-f(y)}{z}+\frac{g(y+z)-g(y)}{z}$
						    \item $=\diff_{x=y}^zf(x)+\diff_{x=y}^zg(x)$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{3.84}
			    \objective
				    Choose a functions $f[x]$ and complex numbers $a,y,z$ such that $z\ne 0$. The objective of the following instructions is to show that $\diff_{x=y}^z(af(x))=a\diff_{x=y}^zf(x)$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\diff_{x=y}^z(af(x))$
					    \begin{enumerate}
						    \item $=\frac{af(y+z)-af(y)}{z}$
						    \item $=a\frac{f(y+z)-f(y)}{z}$
						    \item $=a\diff_{x=y}^zf(x)$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{mon1908191506}
		      \objective
		        Choose the following:
		        \begin{enumerate}
		          \item A procedure $q_0(x,n)$ to show that $p'_n(x)\equiv 0\err{a_0}$ when a complex number $x$ and a positive integer $n$ such that $P(x)$ and $n>b_0$ are chosen.
		          \item A procedure $q_1(x,n,\delta)$ to show that $\Delta_{y=x}^{+\delta}p_n(y)\equiv p'_n(x)\err{\frac{a_1}{n}+b_1\{\delta\}}$ when two complex numbers $x,\delta$ and a positive integer $n$ such that $P(x)$, $n>b_0$, and $0<\lVert\delta\rVert^2<{c_1}^2$ are chosen.
		          \item A procedure $q_2(x,n)$ to show that $t'_n(x)\equiv 0\err{a_2}$ when a complex number $x$ and a positive integer $n$ such that $R(x)$ and $n>b_2$ are chosen.
		          \item A procedure $q_3(x,n,\delta)$ to show that $\Delta_{y=x}^{+\delta}t_n(y)\equiv t'_n(x)\err{\frac{a_3}{n}+b_3\{\delta\}}$ when two complex numbers $x,\delta$ and a positive integer $n$ such that $R(x)$, $n>b_2$, and $0<\lVert\delta\rVert^2<{c_3}^2$ are chosen.
		        \end{enumerate}
		        The objective of the following instructions is to construct the following:
		        \begin{enumerate}
		          \item Rational numbers $a_4,b_4,a_5,b_5,c_5$.
		          \item A procedure $q_4(x,n)$ to show that $p'_n(x)+t'_n(x)\equiv 0\err{a_4}$ when a complex number $x$ and a positive integer $n$ such that $P(x)$, $R(x)$, and $n>b_4$ are chosen.
		          \item A procedure $q_5(x,n,\delta)$ to show that $\Delta_{y=x}^{+\delta}(p_n(y)+t_n(y))\equiv p'_n(x)+t'_n(x)\err{\frac{a_5}{n}+b_5\{\delta\}}$ when two complex numbers $x,\delta$ such that $P(x)$, $R(x)$, $n>b_4$, and $0<\lVert\delta\rVert^2<c_5$.
		        \end{enumerate}
		      \implementation
		        \begin{enumerate}
		          \item Let $a_5=a_1+a_3$.
		          \item Let $b_5=b_1+b_3$.
		          \item Let $q_5(x,n,\delta)$ be the following procedure:
		          \begin{enumerate}
		            \item Show that $\Delta_{y=x}^{+\delta}p_n(y)\equiv p'_n(x)\err{\frac{a_1}{n}+b_1\{\delta\}}$ using procedure $q_1$.
		            \item Show that $\Delta_{y=x}^{+\delta}t_n(y)\equiv t'_n(x)\err{\frac{a_3}{n}+b_3\{\delta\}}$ using procedure $q_3$.
		            \item Hence using \procedurehr{3.83}, show that $\Delta_{y=x}^{+\delta}(p_n(y)+t_n(y))$
		            \begin{enumerate}
		              \item $=\Delta_{y=x}^{+\delta}p_n(y)+\Delta_{y=x}^{+\delta}t_n(y)$
		              \item $\equiv p'_n(x)+\Delta_{y=x}^{+\delta}t_n(y)\err{\frac{a_1}{n}+b_1\{\delta\}}$
		              \item $\equiv p'_n(x)+t'_n(x)\err{\frac{a_3}{n}+b_3\{\delta\}}$
		            \end{enumerate}
		            \item \textbf{Hence show that $\Delta_{y=x}^{+\delta}(p_n(y)+t_n(y))\equiv p'_n(x)+t'_n(x)\err{\frac{a_5}{n}+b_5\{\delta\}}$.}
		          \end{enumerate}
		          \item Let $q_4(x,n)$ be the following procedure:
		          \begin{enumerate}
		            \item Show that $p'_n(x)\equiv 0\err{a_0}$ using procedure $q_0$.
		            \item Show that $t'_n(x)\equiv 0\err{a_2}$ using procedure $q_2$.
		            \item Hence show that $p'_n(x)+t'_n(x)\equiv 0\err{a_0+a_2}$.
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle a_4,b_4,a_5,b_5,c_5,q_4,q_5\rangle$.}
		        \end{enumerate}
		    \procedure{sat0308191134}
		      \objective
		        Choose the following:
		        \begin{enumerate}
		          \item A procedure $q_0(x,n)$ to show that $p'_n(x)\equiv 0\err{a_0}$ when a complex number $x$ and a positive integer $n$ such that $P(x)$, and $n>b_0$ are chosen
		          \item A procedure $q_1(x,n,\delta)$ to show that $\Delta_{y=x}^{+\delta}p_n(y)\equiv p'_n(x)\err{\frac{a_1}{n}+b_1\{\delta\}}$ when two complex numbers $x,\delta$ and a positive integer $n$ such that $P(x)$, $n>b_0$, and $0<\lVert\delta\rVert^2<{c_1}^2$ are chosen
		          \item A procedure $q_2(x,n)$ to show that $t'_n(x)\equiv 0\err{a_2}$ when a complex number $x$ and a positive integer $n$ such that $R(x)$, and $n>b_2$ are chosen
		          \item A procedure $q_3(x,n,\delta)$ to show that $\Delta_{y=x}^{+\delta}t_n(y)\equiv t'_n(x)\err{\frac{a_3}{n}+b_3\{\delta\}}$ when two complex numbers $x,\delta$ and a positive integer $n$ such that $R(x)$, $n>b_2$, and $0<\lVert\delta\rVert^2<{c_3}^2$ are chosen
		          \item A procedure $q_4(x,n)$ to show that $P(t_n(x))$ when a complex number $x$ and a positive integer $n$ such that $R(x)$ and $n>b_2$ are chosen
		        \end{enumerate}
		        The objective of the following instructions is to construct the following:
		        \begin{enumerate}
		          \item Rational numbers $a_5,b_5,a_6,b_6,c_6$.
		          \item A procedure $q_5(x,n)$ to show that $p'_n(t_n(x))t'_n(x)\equiv 0\err{a_5}$ when a complex number $x$ such that $R(x)$, and $n>b_5$ are chosen.
		          \item A procedure $q_6(x,n,\delta)$ to show that $\Delta_{y=x}^{x+\delta}p_n(t_n(y))\equiv p'_n(t_n(x))t'_n(x)\err{\frac{a_6}{n}+b_6\{\delta\}}$ when two complex numbers $x,dx$ such that $R(x)$, $n>b_5$, and $0<\lVert\delta\rVert^2<{c_6}^2$ are chosen.
		        \end{enumerate}
		      \implementation
		        \begin{enumerate}
		          \item Let $a_5=a_0a_2$.
		          \item Let $b_5=\max(b_0,b_2)$.
		          \item Let $a_6=a_1a_3+a_1a_2+a_0a_3$.
		          \item Let $b_6=a_1b_3+b_1a_3+2b_1b_3c_6+b_1a_2+a_0b_3$.
		          \item Let $c_6=\min(c_3,\frac{c_1}{a_3+2b_3c_3+a_2})$.
		          \item Let $q_5(x,n,\delta)$ be the following procedure:
		          \begin{enumerate}
		            \item Show that $P(t_n(x))$ using procedure $q_4$.
		            \item If $\Delta_{y=x}^{+\delta}t_n(y)=0$, then do the following:
		            \begin{enumerate}
		              \item Show that $t_n(x+\delta)=t_n(x)$ given that $t_n(x+\delta)-t_n(x)=0\delta=0$.
		              \item Hence using procedures $q_0,q_3$, show that $\Delta_{y=x}^{+\delta}p_n(t_n(y))$
		              \begin{enumerate}
		                \item $=\frac{p_n(t_n(x+\delta))-p_n(t_n(x))}{\delta}$
		                \item $=\frac{p_n(t_n(x))-p_n(t_n(x))}{\delta}$
		                \item $=0$
		                \item $=\Delta_{y=x}^{+\delta}t_n(y)p'_n(t_n(x))$
		                \item $\equiv t'_n(x)p'_n(t_n(x))\err{a_0(\frac{a_3}{n}+b_3\{\delta\})}$.
		              \end{enumerate}
		            \end{enumerate}
		            \item Otherwise do the following:
		            \begin{enumerate}
		              \item Using procedures $q_3,q_4$, show that $\Delta_{y=x}^{+\delta}t_n(y)$
		              \begin{enumerate}
		                \item $\equiv t'_n(x)\err{\frac{a_3}{n}+b_3\{\delta\}}$
		                \item $\equiv 0\err{a_2}$.
		              \end{enumerate}
		              \item Show that $\{\delta\}\le 2c_6\le 2c_3$ given that $\{\delta\}^2\le 2\lVert\delta\rVert^2\le 4{c_6}^2$.
		              \item Show that $t_n(x+\delta)-t_n(x)$
		              \begin{enumerate}
		                \item $=\Delta_{y=x}^{+\delta}t_n(y)\delta$
		                \item $\equiv 0\delta\err{(\frac{a_3}{n}+b_3\{\delta\}+a_2)c_6}\err{(a_3+2b_3c_3+a_2)c_6}\err{c_1}$.
		              \end{enumerate}
		              \item Hence using procedures $q_0,q_1$, show that $\Delta_{z=t_n(x)}^{t_n(x+\delta)-t_n(x)}p_n(z)$
		              \begin{enumerate}
		                \item $\equiv p'_n(t_n(x))\err{\frac{a_1}{n}+b_1\{\delta\}}$
		                \item $\equiv 0\err{a_0}$.
		              \end{enumerate}
		              \item Hence show that $\Delta_{y=x}^{+\delta}p_n(t_n(y))$
		              \begin{enumerate}
		                \item $=\Delta_{z=t_n(x)}^{t_n(x+\delta)-t_n(x)}p_n(z)\cdot\Delta_{y=x}^{+\delta}t_n(y)$
		                \item $\equiv p'_n(t_n(x))\Delta_{y=x}^{+\delta}t_n(y)\err{(\frac{a_1}{n}+b_1\{\delta\})(\frac{a_3}{n}+b_3\{\delta\}+a_2)}$
		                \item $\equiv p'_n(t_n(x))t'_n(x)\err{a_0(\frac{a_3}{n}+b_3\{\delta\})}$.
		              \end{enumerate}
		            \end{enumerate}
		            \item \textbf{Hence show that $\Delta_{y=x}^{+\delta}p_n(t_n(y))\equiv p'_n(t_n(x))t'_n(x)\err{\frac{a_6}{n}+b_6\{\delta\}}$.}
		          \end{enumerate}
		          \item Let $q_6(x,n)$ be the following procedure:
		          \begin{enumerate}
		            \item Show that $P(t_n(x))$ using procedure $q_4$.
		            \item Show that $p'_n(t_n(x))\equiv 0\err{a_0}$ using procedure $q_0$.
		            \item Show that $t'_n(x)\equiv 0\err{a_2}$ using procedure $q_2$.
		            \item Hence show that $p'_n(t_n(x))t'_n(x)\equiv 0\err{a_0a_2}\err{a_5}$.
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle a_5,b_5,a_6,b_6,c_6,q_5,q_6\rangle$.}
		        \end{enumerate}
		    \procedure{tue2008191001}
		      \objective
		        Choose the following:
		        \begin{enumerate}
		          \item A complex number $B$
		          \item A procedure $q_1(x,n)$ to show that $p'_n(x)\equiv 0\err{a_1}$ when a complex number $x$ and a positive integer $n$ such that $P(x)$ and $n>b_1$ are chosen.
		          \item A procedure $q_2(x,n,\delta)$ to show that $\Delta_{y=x}^{+\delta}p_n(y)\equiv p'(x)\err{\frac{a_2}{n}+b_2\{\delta\}}$ when two complex numbers $x,\delta$ and a positive integer $n$ such that $P(x)$, $n>b_1$, and $0<\lVert\delta\rVert^2\le{c_2}^2$ are chosen.
		        \end{enumerate}
		        The objective of the following instructions is to construct the following:
		        \begin{enumerate}
		          \item Rational numbers $a_3,b_3,a_4,b_4,c_4$.
		          \item A procedure $q_3(x,n)$ to show that $Bp'_n(x)\equiv 0\err{a_3}$ when a complex number $x$ and a positive integer $n$ such that $P(x)$ and $n>b_3$ are chosen.
		          \item A procedure $q_4(x,n,\delta)$ to show that $\Delta_{y=x}^{+\delta}(Bp_n(y))\equiv Bp'_n(x)\err{\frac{a_4}{n}+b_4\{\delta\}}$ when two complex numbers $x,\delta$ and a positive integer $n$ such that $P(x)$, $n>b_3$, and $0<\lVert\delta\rVert^2\le{c_4}^2$ are chosen.
		        \end{enumerate}
		      \implementation
		        \begin{enumerate}
		          \item Let $a_3=\{B\}a_1$.
		          \item Let $b_3=b_1$.
		          \item Let $a_4=\{B\}a_2$.
		          \item Let $b_4=\{B\}b_2$.
		          \item Let $c_4=c_2$.
		          \item Let $q_3(x,n)$ be the following procedure:
		          \begin{enumerate}
		            \item Show that $p'_n(x)\equiv 0\err{a_1}$ using procedure $q_1$.
		            \item Hence show that $Bp'_n(x)\equiv 0B\err{\{B\}{a_1}}\err{a_3}$.
		          \end{enumerate}
		          \item Let $q_4(x,n,\delta)$ be the following procedure:
		          \begin{enumerate}
		            \item Show that $\Delta_{y=x}^{+\delta}p_n(y)\equiv p'(x)\err{\frac{a_2}{n}+b_2\{\delta\}}$ using procedure $q_4$.
		            \item Hence show that $B\Delta_{y=x}^{+\delta}p_n(y)\equiv Bp'(x)$
		            \begin{enumerate}
		              \item $\err{\{B\}(\frac{a_2}{n}+b_2\{\delta\})}$
		              \item $\err{\frac{a_4}{n}+b_4\{\delta\}}$.
		            \end{enumerate}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle a_3,b_3,a_4,b_4,c_4,q_3,q_4\rangle$.}
		        \end{enumerate}
		    \procedure{mon1908191207}
		      \objective
		        Choose the following:
		        \begin{enumerate}
		          \item A procedure $q_0(x,n)$ to show that $p_n(x)\equiv 0\err{a_0}$ when a complex number $x$ and a positive integer $n$ such that $P(x)$ and $n>b_0$ are chosen.
		          \item A procedure $q_1(x,n)$ to show that $p'_n(x)\equiv 0\err{a_1}$ when a complex number $x$ and a positive integer $n$ such that $P(x)$ and $n>b_0$ are chosen.
		          \item A procedure $q_2(x,n,\delta)$ to show that $\Delta_{y=x}^{+\delta}p_n(y)\equiv p'_n(x)\err{\frac{a_2}{n}+b_2\{\delta\}}$ when two complex numbers $x,\delta$ and a positive integer $n$ such that $P(x)$, $n>b_0$, and $0<\lVert\delta\rVert^2<{c_2}^2$ are chosen.
		          \item A procedure $q_3(x,n)$ to show that $t_n(x)\equiv 0\err{a_3}$ when a complex number $x$ and a positive integer $n$ such that $R(x)$ and $n>b_3$ are chosen.
		          \item A procedure $q_4(x,n)$ to show that $t'_n(x)\equiv 0\err{a_4}$ when a complex number $x$ and a positive integer $n$ such that $R(x)$ and $n>b_3$ are chosen.
		          \item A procedure $q_5(x,n,\delta)$ to show that $\Delta_{y=x}^{+\delta}t_n(y)\equiv t'_n(x)\err{\frac{a_5}{n}+b_5\{\delta\}}$ when two complex numbers $x,\delta$ and a positive integer $n$ such that $R(x)$, $n>b_3$, and $0<\lVert\delta\rVert^2<{c_5}^2$ are chosen.
		        \end{enumerate}
		        The objective of the following instructions is to construct the following:
		        \begin{enumerate}
		          \item Rational numbers $a_6,b_6,a_7,b_7,c_7$.
		          \item A procedure $q_6(x,n)$ to show that $p_n(x)t'_n(x)+p'_n(x)t_n(x)\equiv 0\err{a_6}$ when a complex number $x$ and a positive integer $n$ such that $P(x)$, $R(x)$, and $n>b_6$ are chosen.
		          \item A procedure $q_7(x,n,\delta)$ to show that $\Delta_{y=x}^{+\delta}(p_n(y)t_n(y))\equiv p_n(x)t'_n(x)+p'_n(x)t_n(x)\err{\frac{a_7}{n}+b_7\{\delta\}}$ when two complex numbers $x,\delta$ such that $P(x)$, $R(x)$, $n>b_6$, and $0<\lVert\delta\rVert^2<{c_7}^2$ are chosen.
		        \end{enumerate}
		      \implementation
		        \begin{enumerate}
		          \item Let $a_6=a_0a_4+a_1a_3$.
		          \item Let $b_6=\max(b_0,b_3)$.
		          \item Let $a_7=0$.
		          \item Let $b_7=(a_5+b_5c_7+a_4)(a_2+b_2c_7+a_1)$.
		          \item Let $c_7=\min(c_2,c_5)$.
		          \item Let $q_7(x,n,\delta)$ be the following procedure:
		          \begin{enumerate}
		            \item Show that $\{\delta\}\le 2c_7$ given that $\{\delta\}^2\le 2\lVert\delta\rVert^2\le 4{c_7}^2$.
		            \item Hence using procedures $q_2,q_1$, show that $\Delta_{y=x}^{+\delta}p_n(y)$
		            \begin{enumerate}
		              \item $\equiv p'_n(x)\err{\frac{a_2}{n}+b_2\{\delta\}}$
		              \item $\equiv 0\err{a_1}$.
		            \end{enumerate}
		            \item Hence using procedures $q_5,q_4$, show that $\Delta_{y=x}^{+\delta}t_n(y)$
		            \begin{enumerate}
		              \item $\equiv t'_n(x)\err{\frac{a_5}{n}+b_5\{\delta\}}$
		              \item $\equiv 0\err{a_4}$.
		            \end{enumerate}
		            \item Show that $p_n(x)\equiv 0\err{a_0}$ using procedure $q_0$.
		            \item Show that $t_n(x)\equiv 0\err{a_6}$ using procedure $q_3$.
		            \item Hence show that $\Delta_{y=x}^{+\delta}(p_n(y)t_n(y))$
		            \begin{enumerate}
		              \item $=p_n(x+\delta)\Delta_{y=x}^{+\delta}t_n(y)+t_n(x)\Delta_{y=x}^{+\delta}p_n(y)$
		              \item $=(p_n(x)+\delta\Delta_{y=x}^{+\delta}p_n(y))\Delta_{y=x}^{+\delta}t_n(y)+t_n(x)\Delta_{y=x}^{+\delta}p_n(y)$
		              \item $=p_n(x)\Delta_{y=x}^{+\delta}t_n(y)+\delta\Delta_{y=x}^{+\delta}p_n(y)\Delta_{y=x}^{+\delta}t_n(y)+t_n(x)\Delta_{y=x}^{+\delta}p_n(y)$
		              \item $\equiv p_n(x)\Delta_{y=x}^{+\delta}t_n(y)+0\delta\Delta_{y=x}^{+\delta}p_n(y)+t_n(x)\Delta_{y=x}^{+\delta}p_n(y)\err{(\frac{a_5}{n}+b_5\{\delta\}+a_4)\{\delta\}(\frac{a_2}{n}+b_2\{\delta\}+a_1)}$
		            \end{enumerate}
		            \item \textbf{Hence show that $\Delta_{y=x}^{+\delta}(p_n(y)t_n(y))\equiv p_n(x)\Delta_{y=x}^{+\delta}t_n(y)+t_n(x)\Delta_{y=x}^{+\delta}p_n(y)\err{\frac{a_7}{n}+b_7\{\delta\}}$.}
		          \end{enumerate}
		          \item Let $q_6(x,n)$ be the following procedure:
		          \begin{enumerate}
		            \item Show that $p'_n(x)\equiv 0\err{a_1}$ using procedure $q_1$.
		            \item Show that $t'_n(x)\equiv 0\err{a_4}$ using procedure $q_4$.
		            \item Show that $p_n(x)\equiv 0\err{a_0}$ using procedure $q_0$.
		            \item Show that $t_n(x)\equiv 0\err{a_3}$ using procedure $q_3$.
		            \item \textbf{Hence show that $p_n(x)t'_n(x)+p'_n(x)t_n(x)\equiv 0\err{a_0a_4+a_1a_3}\err{a_6}$.}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle a_6,b_6,a_7,b_7,c_7,q_6,q_7\rangle$.}
		        \end{enumerate}
		    \procedure{fri2308191803}
		      \objective
		        Choose the following:
		        \begin{enumerate}
		          \item A procedure $q_0(x,n)$ to show that $p'_n(x)\equiv q'_n(x)\err{\frac{a_0}{n}}$ when a complex number $x$ and a positive integer $n$ such that $P(x)$ and $n>b_0$ are chosen.
		          \item A procedure $q_1(x,n)$ to show that $p'_n(x)\equiv 0\err{a_1}$ when a complex number $x$ and a positive integer $n$ such that $P(x)$ and $n>b_1$ are chosen.
		          \item A procedure $q_2(x,n,\delta)$ to show that $\Delta_{y=x}^{+\delta}p_n(y)\equiv p'_n(x)\err{\frac{a_2}{n}+b_2\{\delta\}}$ when two complex numbers $x,\delta$ and a positive integer $n$ such that $P(x)$, $n>b_1$, and $0<\lVert\delta\rVert^2\le{c_2}^2$ are chosen.
		        \end{enumerate}
		        The objective of the following instructions is to construct the following:
		        \begin{enumerate}
		          \item Rational numbers $a_3,b_3,a_4,b_4,c_4$.
		          \item A procedure $q_3(x,n)$ to show that $q'_n(x)\equiv 0\err{a_3}$ when a complex number $x$ and a positive integer $n$ such that $P(x)$ and $n>b_3$ are chosen.
		          \item A procedure $q_4(x,n,\delta)$ to show that $\Delta_{y=x}^{+\delta}p_n(y)\equiv q'_n(x)\err{\frac{a_4}{n}+b_4\{\delta\}}$ when two complex numbers $x,\delta$ and a positive integer $n$ such that $P(x)$, $n>b_3$, and $0<\lVert\delta\rVert^2\le{c_4}^2$ are chosen.
		        \end{enumerate}
		      \implementation
		        \begin{enumerate}
		          \item Let $a_3=a_0+a_1$.
		          \item Let $b_3=\max(b_0,b_1)$.
		          \item Let $a_4=a_0+a_2$.
		          \item Let $b_4=b_2$.
		          \item Let $c_4=c_2$.
		          \item Let $q_3(x,n)$ be the following procedure:
		          \begin{enumerate}
		            \item Show that $p'_n(x)\equiv q'_n(x)\err{\frac{a_0}{n}}$ using procedure $q_0$.
		            \item Show that $p'_n(x)\equiv 0\err{a_1}$ using procedure $q_1$.
		            \item \textbf{Hence show that $q'_n(x)\equiv 0\err{a_3}$.}
		          \end{enumerate}
		          \item Let $q_4(x,n,\delta)$ be the following procedure:
		          \begin{enumerate}
		            \item Using procedures $q_0,q_2$, show that $\Delta_{y=x}^{+\delta}p_n(y)$
		            \begin{enumerate}
		              \item $\equiv p'_n(x)\err{\frac{a_2}{n}+b_2\{\delta\}}$
		              \item $\equiv q'_n(x)\err{\frac{a_0}{n}}$.
		            \end{enumerate}
		            \item Hence show that $\Delta_{y=x}^{+\delta}p_n(y)\equiv q'_n(x)$
		            \begin{enumerate}
		              \item $\err{\frac{a_2}{n}+b_2\{\delta\}+\frac{a_0}{n}}$
		              \item $\err{\frac{a_4}{n}+b_4\{\delta\}}$.
		            \end{enumerate}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle a_3,b_3,a_4,b_4,c_4,q_3,q_4\rangle$.}
		        \end{enumerate}
      \end{multicols*}
    \chapter{Common Derivatives}
      \begin{multicols}{2}
		    \procedure{tue2008191151}
		      \objective
		        Choose a complex number $B$ and a rational number $D>0$. The objective of the following instructions is to construct rational numbers $a,b,c,d$, a procedure $p(x,n)$ to show that $0\equiv 0\err{a}$ when a complex number $x$ and a positive integer $n$ such that $n>d$ are chosen, and a procedure $q(x,n,\delta)$ to show that $\Delta_{y=x}^{+\delta}B\equiv 0\err{\frac{b}{n}+c\{\delta\}}$ when in addition a complex number $\delta$ such that $0<\lVert\delta\rVert^2\le D^2$ is chosen.
		      \implementation
		        \begin{enumerate}
		          \item Let $a=b=c=d=0$.
		          \item Let $p(x,n)$ be the following procedure:
		          \begin{enumerate}
		            \item \textbf{Show that $0\equiv 0\err{a}$.}
		          \end{enumerate}
		          \item Let $q(x,n,\delta)$ be the following procedure:
		          \begin{enumerate}
		            \item \textbf{Show that $\Delta_{y=x}^{+\delta}B\equiv 0\err{\frac{b}{n}+c\{\delta\}}$.}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle a,b,c,d,p,q\rangle$.}
		        \end{enumerate}
		    \procedure{tue2008191209}
		      \objective
		        Choose a positive integer $N$ and positive rational numbers $X,D$. The objective of the following instructions is to construct rational numbers $a,b,c,d$, a procedure $p(x,n)$ to show that $Nx^{N-1}\equiv 0\err{a}$ when a complex number $x$ and a positive integer $n$ such that $\lVert x\rVert^2\le X^2$ and $n>d$ are chosen, and a procedure $q(x,n,\delta)$ to show that $\Delta_{y=x}^{+\delta}y^N\equiv Nx^{N-1}\err{\frac{b}{n}+c\{\delta\}}$ when in addition a complex number $\delta$ such that $0<\lVert\delta\rVert^2\le D^2$ is chosen.
		      \implementation
		        \begin{enumerate}
		          \item Let $a=NX^{N-1}$.
		          \item Let $b=d=0$.
		          \item Let $c=\sum_r^{[0:N-1]}\binom{N}{r}X^rD^{N-r-2}$.
		          \item Let $p(x,n)$ be the following procedure:
		          \begin{enumerate}
		            \item \textbf{Show that $Nx^{N-1}\equiv 0\allowbreak\err{Nx^{N-1}}\allowbreak\err{NX^{N-1}}\allowbreak\err{a}$.}
		          \end{enumerate}
		          \item Let $q(x,n,\delta)$ be the following procedure:
		          \begin{enumerate}
		            \item Show that $\Delta_{y=x}^{+\delta}y^N\equiv Nx^{N-1}$
		            \begin{enumerate}
		              \item $\err{\frac{(x+\delta)^N-x^N}{\delta}-Nx^{N-1}}$
		              \item $\err{\frac{1}{\delta}(\sum_r^{[0:N+1]}\binom{N}{r}x^r\delta^{N-r}-x^N)-Nx^{N-1}}$
		              \item $\err{\sum_r^{[0:N]}\binom{N}{r}x^r\delta^{N-r-1}-Nx^{N-1}}$
		              \item $\err{\delta(\sum_r^{[0:N-1]}\binom{N}{r}x^r\delta^{N-r-2})}$
		              \item $\err{\frac{b}{n}+c\{\delta\}}$.
		            \end{enumerate}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle a,b,c,d,p,q\rangle$.}
		        \end{enumerate}
		    \procedure{tue2008191254}
		      \objective
		        Choose two rational numbers $X>D>0$. The objective of the following instructions is to construct rational numbers $a,b,c,d$, a procedure $p(x,n)$ to show that $-\frac{1}{x^2}\equiv 0\err{a}$ when a complex number $x$ and a positive integer $n$ such that $\lVert x\rVert^2\ge X^2$ and $n>b$ are chosen, and a procedure $q(x,n,\delta)$ to show that $\Delta_{y=x}^{+\delta}\frac{1}{y}\equiv-\frac{1}{x^2}\err{\frac{c}{n}+d\{\delta\}}$ when in addition a complex number $\delta$ such that $0<\lVert\delta\rVert^2\le D^2$ is chosen.
		      \implementation
		        \begin{enumerate}
		          \item Let $a=\frac{1}{X^2}$.
		          \item Let $b=c=0$.
		          \item Let $d=\frac{1}{X^2(X-D)}$.
		          \item Let $p(x,n)$ be the following procedure:
		          \begin{enumerate}
		            \item \textbf{Show that $\frac{1}{x^2}\equiv 0\allowbreak\err{\frac{1}{x^2}}\allowbreak\err{\frac{1}{X^2}}\allowbreak\err{a}$.}
		          \end{enumerate}
		          \item Let $q(x,n,\delta)$ be the following procedure:
		          \begin{enumerate}
		            \item Show that $\Delta_{y=x}^{+\delta}\frac{1}{y}\equiv-\frac{1}{x^2}$
		            \begin{enumerate}
		              \item $\err{\frac{1}{\delta}(\frac{1}{x+\delta}-\frac{1}{x})+\frac{1}{x^2}}$
		              \item $\err{\frac{1}{\delta}\cdot\frac{-\delta}{x(x+\delta)}+\frac{1}{x^2}}$
		              \item $\err{\frac{1}{x^2}-\frac{1}{x(x+\delta)}}$
		              \item $\err{\frac{\delta}{x^2(x+\delta)}}$
		              \item $\err{\frac{\{\delta\}}{X^2(X-D)}}$
		              \item $\err{\frac{c}{n}+d\{\delta\}}$.
		            \end{enumerate}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle a,b,c,d,p,q\rangle$.}
		        \end{enumerate}
		    \procedure{tue2008191341}
		      \objective
		        Choose a positive integer $N$ and positive rational numbers $X<Y$. The objective of the following instructions is to construct positive rational numbers $a,b,c,d,e$, a procedure $p(x,n)$ to show that $-Nx^{-N-1}\equiv 0\err{a}$ when a complex number $x$ and a positive integer $n$ such that $X^2\le\lVert x\rVert^2\le Y^2$ and $n>b$ are chosen, and a procedure $q(x,n,\delta)$ to show that $\Delta_{y=x}^{+\delta}y^{-N}\equiv-Nx^{-N-1}\err{\frac{c}{n}+d\{\delta\}}$ when in addition a complex number $\delta$ such that $0<\lVert\delta\rVert^2\le e^2$ is chosen.
		      \implementation
		        \begin{enumerate}
		          \item Execute the following in post-order:
		          \begin{enumerate}
		            \item Execute \procedurehr{sat0308191134} on $\langle q_2,q_3,q_4,q_5,q_6\rangle$ and let $\langle a,b,c,d,e,q_0,q_1\rangle$ receive.
		            \begin{enumerate}
		              \item Execute \procedurehr{tue2008191254} on $\langle X^N,\frac{X^N}{2}\rangle$ and let $\langle\cdots,q_2,q_3\rangle$ receive.
		              \item Execute \procedurehr{tue2008191209} on $\langle N,Y,Y\rangle$ and let $\langle\cdots,q_4,q_5\rangle$ receive.
		              \item Let $q_6(x,n)$ be the following procedure:
		              \begin{enumerate}
		                \item Show that $\lVert x^N\rVert^2=(\lVert x\rVert^2)^N\ge(X^2)^N=(X^N)^2$.
		              \end{enumerate}
		            \end{enumerate}
		          \end{enumerate}
		          \item Let $p(x,n)$ be the following procedure:
		          \begin{enumerate}
		            \item \textbf{Show that $-Nx^{-N-1}=-\frac{1}{(x^N)^2}\cdot Nx^{N-1}\equiv 0\err{a}$ using procedure $q_0$.}
		          \end{enumerate}
		          \item Let $q(x,n,\delta)$ be the following procedure:
		          \begin{enumerate}
		            \item Using procedure $q_1$, show that $\Delta_{y=x}^{+\delta}y^{-N}$
		            \begin{enumerate}
		              \item $=\frac{1}{\delta}(((x+\delta)^N)^{-1}-(x^N)^{-1})$
		              \item $\equiv-\frac{1}{(x^N)^2}\cdot Nx^{N-1}\err{\frac{c}{n}+d\{\delta\}}$
		              \item $=-Nx^{-N-1}$
		            \end{enumerate}
		            \item \textbf{Hence show that $\Delta_{y=x}^{+\delta}y^{-N}\equiv-Nx^{-N-1}\err{\frac{c}{n}+d\{\delta\}}$.}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle a,b,c,d,e,p,q\rangle$.}
		        \end{enumerate}
		    \procedure{3.18}
			    \objective
				    Choose a rational number $D\ge 0$. The objective of the following instructions is to construct two rational numbers $a,c$ and a procedure, $p(n,\delta)$, to show that $\diff_{x=0}^{+\delta}\exp_n(x)\equiv 1\err{a\delta}\err{a\{\delta\}}$ when a complex number $\delta$ and a positive integer $n$ such that $0<\lVert\delta\rVert^2\le D^2$ and $n>c$ are chosen.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{3.13} on $\langle D\rangle$ and let $\langle a,c,q\rangle$ receive.
					    \item Let $p(\delta,n)$ be the following procedure:
					    \begin{enumerate}
						    \item Now using \procedurehr{2.04}, and procedure $q$, show that $\exp_n(\delta)-1\equiv\delta$
						    \begin{enumerate}
						      \item $\err{\exp_n(\delta)-1-\delta}$
							    \item $\err{(1+\frac{\delta}{n})^n-1-\delta}$
							    \item $\err{\frac{\delta}{n}\sum_r^{[0:n]}(1+\frac{\delta}{n})^r-n\frac{\delta}{n}}$
							    \item $\err{\frac{\delta}{n}\sum_r^{[0:n]}((1+\frac{\delta}{n})^r-1)}$
							    \item $\err{\frac{\delta}{n}\sum_r^{[0:n]}\frac{\delta}{n}\sum_k^{[0:r]}(1+\frac{\delta}{n})^k}$
							    \item $\err{\frac{\delta^2}{n^2}\sum_r^{[0:n]}\sum_k^{[0:r]}(1+\frac{\delta}{n})^k}$
							    \item $\err{\frac{\delta^2}{n^2}\sum_r^{[0:n]}\sum_k^{[0:r]}a}$
							    \item $\err{\delta^2a}$.
						    \end{enumerate}
						    \item \textbf{Therefore show that $\diff_{x=0}^{+\delta}\exp_n(x)\equiv 1\err{a\delta}\err{a\{\delta\}}$.}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,c,p\rangle$.}
				    \end{enumerate}
		    \procedure{3.19}
			    \objective
				    Choose two rational numbers $X\ge 0$, $D\ge 0$. The objective of the following instructions is to construct rational numbers $l,a,b,d$, a procedure $t(x,n)$ to show that $\exp_n(x)\equiv 0\err{l}$ when a complex number $x$ and a positive integer $n$ such that $\lVert x\rVert^2\le X^2$ and $n>d$ are chosen, and a procedure, $q(x,n,\delta)$, to show that $\diff_{y=x}^{+\delta}\exp_n(y)\equiv\exp_n(x)\err{\frac{a}{n}+b\{\delta\}}$ when in addition a complex number $\delta$ such that $0<\lVert\delta\rVert^2\le D^2$ is chosen. 
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{3.15} on $\langle\max(X,D)\rangle$ and let $\langle e,f,u\rangle$ receive.
					    \item Execute \procedurehr{3.18} on $\langle X\rangle$ and let $\langle h,j,r\rangle$ receive.
					    \item Execute \procedurehr{3.13} on $\langle X\rangle$ and let $\langle l,m,t\rangle$ receive.
					    \item Let $a=eX$.
					    \item Let $b=lh$.
					    \item Let $d=\max(f,j,m)$.
					    \item Let $p(x,n,\delta)$ be the following procedure:
					    \begin{enumerate}
						    \item Using procedures $u,r,t$, show that $\Delta_{y=x}^{+\delta}\exp_n(y)$
						    \begin{enumerate}
						      \item $=\frac{\exp_n(x+\delta)-\exp_n(x)}{\delta}$
						      \item $=\frac{\exp_n(x)\exp_n(\delta)-\exp_n(x)}{\delta}$
						      \begin{enumerate}
						        \item $\err{\frac{ex\delta}{n\delta}}$
						        \item $\err{\frac{eX}{n}}$
						      \end{enumerate}
						      \item $=\exp_n(x)\Delta_{y=0}^{\delta}\exp_n(y)$
						      \item $\equiv\exp_n(x)\cdot 1\err{lh\{\delta\}}$.
						    \end{enumerate}
						    \item \textbf{Hence show that $\diff_{y=x}^{+\delta}\exp_n(y)\equiv\exp_n(x)\err{\frac{ex}{n}+lh\{\delta\}}\err{\frac{a}{n}+b\{\delta\}}$.}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle a,b,d,p\rangle$.}
				    \end{enumerate}
		     \procedure{3.27}
			    \objective
				    Choose non-negative rational numbers $X,D$. The objective of the following instructions is to construct rational numbers $l,d,a,b$, a procedure $q(x,n)$ to show that $\cos_n(x)\equiv 0\err{l}$ when a complex number $x$ and a positive integer $n$ such that $\lVert x\rVert^2\le X^2$ and $n>d$ are chosen, and a procedure $p(x,n,\delta)$ to show that $\diff_{y=x}^{+\delta}\sin_n(y)\equiv\cos_n(x)\err{\frac{a}{n}+b\{\delta\}}$ when in addition a complex number $\delta$ such that $0<\lVert\delta\rVert^2\le D^2$ is chosen.
			    \implementation
				    \begin{instructions}
					    \item Execute the following in post-order:
					    \begin{instructions}
					      \item Execute \procedurehr{tue2008191001} on $\langle\frac{1}{2\iu},q_2,q_3\rangle$ and let $\langle l,d,a,b,D,q_0,q_1\rangle$ receive.
					      \begin{instructions}
					        \item Execute \procedurehr{mon1908191506} on $\langle q_4,q_5,q_6,q_7\rangle$ and let $\langle q_2,q_3\rangle$ receive.
					        \begin{instructions}
					          \item Execute \procedurehr{sat0308191134} on $\langle q_8,q_9,q_{10},q_{11},q_{12}\rangle$ and let $\langle q_4,q_5\rangle$ receive.
					          \begin{instructions}
					            \item Execute \procedurehr{3.19} on $\langle X,D\rangle$ and let $\langle q_8,q_9\rangle$ receive.
					            \item Execute \procedurehr{tue2008191001} on $\langle\iu,q_{13},q_{14}\rangle$ and let $\langle q_{10},q_{11}\rangle$ receive.
					            \begin{instructions}
					              \item Execute \procedurehr{tue2008191209} on $\langle 1,X,D\rangle$ and let $\langle q_{13},q_{14}\rangle$ receive.
					            \end{instructions}
					            \item Let $q_{12}(x,n)$ be the following procedure:
					            \begin{instructions}
					              \item Show that $\lVert\iu x\rVert^2=\lVert x\rVert^2\le X^2$.
					            \end{instructions}
					          \end{instructions}
					          \item Execute \procedurehr{tue2008191001} on $\langle-1,q_{15},q_{16}\rangle$ and let $\langle q_6,q_7\rangle$.
					          \begin{instructions}
					            \item Execute \procedurehr{sat0308191134} on $\langle q_{17},q_{18},q_{19},q_{20},q_{21}\rangle$ and let $\langle q_{15},q_{16}\rangle$ receive.
					            \begin{instructions}
					              \item Execute \procedurehr{3.19} on $\langle X,D\rangle$ and let $\langle q_{17},q_{18}\rangle$ receive.
					              \item Execute \procedurehr{tue2008191001} on $\langle-\iu,q_{22},q_{23}\rangle$ and let $\langle q_{19},q_{20}\rangle$ receive.
					              \begin{instructions}
					                \item Execute \procedurehr{tue2008191209} on $\langle 1,X,D\rangle$ and let $\langle q_{22},q_{23}\rangle$ receive.
					              \end{instructions}
					              \item Let $q_{21}(x,n)$ be the following procedure:
					              \begin{instructions}
					                \item Show that $\lVert -\iu x\rVert^2=\lVert x\rVert^2\le X^2$.
					              \end{instructions}
					            \end{instructions}
					          \end{instructions}
					        \end{instructions}
					      \end{instructions}
					    \end{instructions}
					    \item Let $p(x,n)$ be the following procedure:
					    \begin{enumerate}
					      \item Using procedure $q_0$, show that $\cos_n(x)$
					      \begin{enumerate}
					        \item $=\frac{1}{2}(\exp_n(\iu x)+\exp_n(-\iu x))$
					        \item $=\frac{1}{2\iu}(\exp_n(\iu x^1)\cdot\iu\cdot 1x^0+(-1)\exp_n(-\iu x^1)\cdot(-\iu)\cdot 1\cdot x^0)$
					        \item $\equiv 0\err{l}$.
					      \end{enumerate}
					    \end{enumerate}
					    \item Let $q(x,n,\delta)$ be the following procedure:
					    \begin{enumerate}
					      \item Using procedure $q_1$, show that $\Delta_{y=x}^{+\delta}\sin_n(y)$
					      \begin{enumerate}
					        \item $=\Delta_{y=x}^{+\delta}(\frac{\exp_n(\iu x)-\exp_n(-\iu x)}{2\iu})$
					        \item $=\Delta_{y=x}^{+\delta}(\frac{1}{2\iu}(\exp_n(\iu x^1)+(-1)\exp_n((-\iu)x^1)))$
					        \item $\equiv\frac{1}{2\iu}(\exp_n(\iu x^1)\cdot\iu\cdot 1x^0+(-1)\exp_n(-\iu x^1)\cdot(-\iu)\cdot 1\cdot x^0)\err{\frac{a}{n}+b\{\delta\}}$
					        \item $=\frac{\exp_n(\iu x)+\exp_n(-\iu x)}{2}$
					        \item $=\cos_n(x)$.
					      \end{enumerate}
					      \item \textbf{Hence show that $\Delta_{y=x}^{+\delta}\sin_n(y)\equiv\cos_n(x)\err{\frac{a}{n}+b\{\delta\}}$.}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle l,d,a,b,D,p,q\rangle$.}
				    \end{instructions}
		    \procedure{3.28}
			    \objective
				    Choose non-negative rational numbers $X,D$. The objective of the following instructions is to construct rational numbers $l,d,a,b$, a procedure $q(x,n)$ to show that $-\sin_n(x)\equiv 0\err{l}$ when a complex number $x$ and a positive integer $n$ such that $\lVert x\rVert^2\le X^2$ and $n>d$ are chosen, and a procedure $p(x,n,\delta)$ to show that $\diff_{y=x}^{+\delta}\cos_n(y)\equiv-\sin_n(x)\err{\frac{a}{n}+b\{\delta\}}$ when in addition a complex number $\delta$ such that $0<\lVert\delta\rVert^2\le D^2$ is chosen.
			    \implementation
				    Implementation is analogous to that of \procedurehr{3.27}.
		    \procedure{wed2108191034}
		      \objective
		        Choose non-negative rational numbers $X,D$ such that $X+D<1$. The objective of the following instructions is to construct rational numbers $l,d,a,b$, a procedure $p(x,n)$ to show that $(1+x)_{n-1}^{-1}\equiv 0\err{l}$ when a complex number $x$ and a positive integer $n$ such that $\lVert x\rVert^2\le X^2$ and $n>d$ are chosen, and a procedure $q(x,n,\delta)$ to show that $\Delta_{y=x}^{+\delta}\ln_n(1+y)\equiv(1+x)_{n-1}^{-1}\err{\frac{a}{n}+b\{\delta\}}$ when in addition a complex number $\delta$ such that $0<\lVert\delta\rVert^2\le D^2$ is chosen.
		      \implementation
		        \begin{enumerate}
		          \item Execute \procedurehr{wed2407191611} on $\langle 1,X\rangle$ and let $\langle a_1,p_1\rangle$ receive.
		          \item Let $l=a_1$.
		          \item Let $d=1$.
		          \item Let $a=0$.
		          \item Let $b=\frac{1}{D^2((X+D)^{-1}-1)}$.
		          \item Let $p(x,n)$ be the following procedure:
		          \begin{enumerate}
		            \item Using procedure $p_1$, show that $(1+x)_{n-1}^{-1}$
		            \begin{enumerate}
		              \item $=((1+x)_{n-1}^{-1})^1$
		              \item $\equiv 0\err{a_1}$.
		            \end{enumerate}
		          \end{enumerate}
		          \item Let $q(x,n,\delta)$ be the following procedure:
		          \begin{enumerate}
		            \item Using \procedurehr{2.05}, show that $\Delta_{y=x}^{+\delta}\ln_n(1+y)\equiv(1+x)_{n-1}^{-1}$
		            \begin{enumerate}
		              \item $\err{\Delta_{y=x}^{+\delta}\ln_n(1+y)-(1+x)_{n-1}^{-1}}$
		              \item $\err{\frac{1}{\delta}(\sum_r^{[1:n]}\frac{(-1)^{r-1}}{r}(x+\delta)^r-\sum_r^{[1:n]}\frac{(-1)^{r-1}}{r}x^r)-\sum_r^{[0:n-1]}\binom{-1}{r}x^r}$
		              \item $\err{\frac{1}{\delta}\sum_r^{[1:n]}\frac{(-1)^{r-1}}{r}(\sum_k^{[0:r+1]}\binom{r}{k}x^k\delta^{r-k}-x^r)-\sum_r^{[0:n-1]}(-1)^rx^r}$
		              \item $\err{\sum_r^{[1:n]}\frac{(-1)^{r-1}}{r}\sum_k^{[0:r]}\binom{r}{k}x^k\delta^{r-1-k}-\sum_r^{[0:n-1]}(-1)^rx^r}$
		              \item $\err{\sum_r^{[1:n]}\frac{(-1)^{r-1}}{r}(\sum_k^{[0:r]}\binom{r}{k}x^k\delta^{r-1-k}-rx^{r-1})}$
		              \item $\err{\delta(\sum_r^{[1:n]}\frac{(-1)^{r-1}}{r}\sum_k^{[0:r-1]}\binom{r}{k}x^k\delta^{r-2-k})}$
		              \item $\err{\delta(\sum_r^{[1:n]}\sum_k^{[0:r-1]}\binom{r}{k}X^kD^{r-2-k})}$
		              \item $\err{\delta(\frac{1}{D^2}\sum_r^{[1:n]}\sum_k^{[0:r-1]}\binom{r}{k}X^kD^{r-k})}$
		              \item $\err{\delta(\frac{1}{D^2}\sum_r^{[1:n]}(X+D)^r)}$
		              \item $\err{\delta(\frac{1}{D^2((X+D)^{-1}-1)})}$
		              \item $\err{\frac{a}{n}+b\{\delta\}}$.
		            \end{enumerate}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle l,d,a,b,p,q\rangle$.}
		        \end{enumerate}
		    \procedure{wed2108191140}
		      \objective
		        Choose non-negative rational numbers $X,D$ such that $X+D<1$. The objective of the following instructions is to construct rational numbers $l,d,a,b$, a procedure $p(x,n)$ to show that $\frac{1}{1+x}\equiv 0\err{l}$ when a complex number $x$ and a positive integer $n$ such that $\lVert x\rVert^2\le X^2$ and $n>d$ are chosen, and a procedure $q(x,n,\delta)$ to show that $\Delta_{y=x}^{+\delta}\ln_n(1+y)\equiv\frac{1}{1+x}\err{\frac{a}{n}+b\{\delta\}}$ when in addition a complex number $\delta$ such that $0<\lVert\delta\rVert^2\le D^2$ is chosen.
		      \implementation
		        \begin{enumerate}
		          \item Execute \procedurehr{wed2407191422} on $\langle X,1\rangle$ and let $\langle a_1,b_1,p_1\rangle$ receive.
		          \item Execute \procedurehr{wed2108191034} on $\langle X,D\rangle$ and let $\langle a_2,b_2,c_2,b,p_2,q_2\rangle$ receive.
		          \item Let $l=\frac{1}{1-X}$.
		          \item Let $d=\max(1,b_2)$.
		          \item Let $a=2a_1l+c_2$.
		          \item Let $p(x,n)$ be the following procedure:
		          \begin{enumerate}
		            \item Show that $\lvert\Re(x)\rvert\le X$ given that $\Re(x)^2\le\lVert x\rVert^2\le X^2$.
		            \item Hence show that $\lVert 1+x\rVert^2$
		            \begin{enumerate}
		              \item $\ge\Re(1+x)^2$
		              \item $=(1+\Re(x))^2$
		              \item $\le(1-X)^2$.
		            \end{enumerate}
		            \item Hence show that $\frac{1}{1+x}\equiv 0\allowbreak\err{\frac{1}{1+x}}\allowbreak\err{\frac{1}{1-X}}\allowbreak\err{l}$.
		          \end{enumerate}
		          \item Let $q(x,n,\delta)$ be the following procedure:
		          \begin{enumerate}
		            \item Show that $\lVert\frac{1}{1+x}\rVert^2\le l^2$ using procedure $p$.
		            \item Show that $\lVert(n-1)(-x)^{n-1}\rVert^2\le(a_1{b_1}^{n-1})^2\le{a_1}^2$ using procedure $p_1$.
		            \item Hence show that $\lVert(-x)^{n-1}\rVert^2\le(\frac{a_1}{n-1})^2\le(\frac{2a_1}{n})^2$.
		            \item Now using procedure $q_2$, show that $\Delta_{y=x}^{+\delta}\ln_n(1+y)$
		            \begin{enumerate}
		              \item $\equiv(1+x)_{n-1}^{-1}\err{\frac{c_2}{n}+b\{\delta\}}$
		              \item $=\frac{1-(-x)^{n-1}}{1-(-x)}$
		              \item $\equiv\frac{1}{1+x}\err{\frac{(-x)^{n-1}}{1+x}}\err{\frac{2a_1l}{n}}$.
		            \end{enumerate}
		            \item \textbf{Hence show that $\Delta_{y=x}^{+\delta}\ln_n(1+y)\equiv\frac{1}{1+x}\err{\frac{c_2}{n}+b\{\delta\}+\frac{2a_1l}{n}}\err{\frac{a}{n}+b\{\delta\}}$.}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle l,d,a,b,p,q\rangle$.}
		        \end{enumerate}
		    \procedure{sun0812191401}
		      \objective
		        Choose a rational number $X$ such that $0<X\le 1$. The objective of the following instructions is to construct a rational number $f$ such that $0\le f<1$, and a procedure $p(x)$ to show that $\Re(\frac{x-1}{x+1})\ge -f$ and $\lVert\frac{x-1}{x+1}\rVert^2\le 1$ when a complex number $x$ such that $\Re(x)\ge 0$ and $\lVert x\rVert^2\ge X^2$ is chosen.
		      \implementation
		        \begin{enumerate}
		          \item Let $f=\frac{1-X^2}{1+X^2}$.
		          \item Verify that $0\le f<1$.
		          \item Let $p(x)$ be the following procedure:
		          \begin{enumerate}
		            \item Show that $\lVert\frac{x-1}{x+1}\rVert^2\le 1$
		            \begin{enumerate}
		              \item given that $\lVert x-1\rVert^2\le\lVert x+1\rVert^2$
		              \item given that $(\Re(x)-1)^2\le(\Re(x)+1)^2$
		              \item given that $0\le\Re(x)$.
		            \end{enumerate}
		            \item Show that $\Re(\frac{x-1}{x+1})$
		            \begin{enumerate}
		              \item $=\Re(\frac{(x-1)(\conj{x+1})}{(x+1)(\conj{x+1})})$
		              \item $=\Re(\frac{\lVert x\rVert^2+x-\conj{x}-1}{\lVert x\rVert^2+x+\conj{x}+1})$
		              \item $=\frac{\lVert x\rVert^2-1}{\lVert x\rVert^2+2\Re(x)+1}$
		              \item $\ge\frac{X^2-1}{\lVert x\rVert^2+2\Re(x)+1}$
		              \item $\ge\frac{X^2-1}{X^2+1}=-f$.
		            \end{enumerate}
		          \end{enumerate}
		        \end{enumerate}
		    \declaration{wed2108191408}
		      The notation \gls{differentialln}, where $x$ is a complex number, will be used as a shorthand for $\ln_n(1+\frac{x-1}{x+1})-\ln_n(1-\frac{x-1}{x+1})$ when $\Re(x)\ge 0$, $\ln_n(\frac{x}{i})+\frac{\tau_n}{4}\iu$ when $\Im(x)\ge 0$, and $\ln_n(xi)-\frac{\tau_n}{4}\iu$ if otherwise.
		    \procedure{thu2208191250}
		      \objective
		        Choose a rational number $X$ such that $1\ge X> 0$. The objective of the following instructions is to construct a positive rational number $a$, and a procedure $p(x,k)$ to show that $\lVert\ln_k(x)\rVert^2\le a^2$ when a positive integer $k$ and a complex number $x$ such that $\lVert x\rVert^2\ge X^2$ and $\Re(x)\ge 0$ are chosen.
		      \implementation
		        \begin{enumerate}
		          \item Execute \procedurehr{sun0812191401} on $\langle X\rangle$ and let $\langle a_1,p_1\rangle$ receive.
		          \item Verify that $0<a_1<1$.
		          \item Execute \procedurehr{fri2607191736} on $\langle a_1\rangle$ and let $\langle a_2,p_2\rangle$ receive.
		          \item Let $a=2a_2$.
		          \item Let $p(x,k)$ be the following procedure:
		          \begin{enumerate}
		            \item Show that $\lVert\frac{x-1}{x+1}\rVert^2\le{a_1}^2$ using procedure $p_1$.
		            \item Show that $\lVert\ln_k(1+\frac{x-1}{x+1})\rVert^2\le{a_2}^2$ using procedure $p_2$.
		            \item Show that $\lVert\ln_k(1-\frac{x-1}{x+1})\rVert^2\le{a_2}^2$ using procedure $p_2$.
		            \item Hence show that $\lVert\ln_k(x)\rVert^2$
		            \begin{enumerate}
		              \item $=\lVert\ln_k(1+\frac{x-1}{x+1})-\ln_k(1-\frac{x-1}{x+1})\rVert^2$
		              \item $\le a^2$.
		            \end{enumerate}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle a,p\rangle$.}
		        \end{enumerate}
		    \procedure{sun0812191440}
		      \objective
		        Choose a rational number $X$ such that $0<X\le 1$. The objective of the following instructions is to construct positive rational numbers $a,b$, and a procedure $p(x,k)$ to show that $\lVert\ln_k(x)\rVert^2\le a^2$ when a positive integer $k$ and a complex number $x$ such that $\lVert x\rVert^2\ge X^2$ and $k>b$ are chosen.
		      \implementation
		        \begin{enumerate}
		          \item Execute \procedurehr{thu2208191250} on $\langle X\rangle$ and let $\langle a_1,p_1\rangle$ receive.
		          \item Execute \procedurehr{3.50} and let $\langle a_2,b_2,p_2\rangle$ receive.
		          \item Verify that $a_2>0$.
		          \item Let $a=a_1+\frac{a_2}{4}$.
		          \item Let $p(x,k)$ be the following procedure:
		          \begin{enumerate}
		            \item Show that $\frac{1}{4}\tau_k\le\frac{a_2}{4}$ using procedure $p_2$.
		            \item If $\Re(x)\ge 0$, then do the following:
		            \begin{enumerate}
		              \item \textbf{Show that $\lVert\ln_k(x)\rVert^2\le{a_1}^2\le a^2$ using procedure $p_1$.}
		            \end{enumerate}
		            \item Otherwise if $\Im(x)\ge 0$, then do the following:
		            \begin{enumerate}
		              \item Show that $\lVert\frac{x}{\iu}\rVert^2=\lVert x\rVert^2\ge X^2$.
		              \item Show that $\Re(\frac{x}{\iu})=\Re(\Im(x)-\Re(x)\iu)=\Im(x)\ge 0$.
		              \item Hence show that $\lVert\ln_k(\frac{x}{\iu})\rVert^2\le{a_1}^2$ using procedure $p_1$.
		              \item \textbf{Hence show that $\lVert\ln_k(x)\rVert^2=\lVert\ln_k(\frac{x}{\iu})+\frac{\tau_k}{4}\iu\rVert^2\le(a_1+\frac{a_2}{4})^2=a^2$.}
		            \end{enumerate}
		            \item Otherwise do the following:
		            \begin{enumerate}
		              \item Show that $\lVert x\iu\rVert^2=\lVert x\rVert^2\ge X^2$.
		              \item Show that $\Re(x\iu)=\Re(-\Im(x)+\Re(x)\iu)=-\Im(x)>0$.
		              \item Hence show that $\lVert\ln_k(x\iu)\rVert^2\le{a_1}^2$ using procedure $p_1$.
		              \item \textbf{Hence show that $\lVert\ln_k(x)\rVert^2=\lVert\ln_k(x\iu)-\frac{\tau_k}{4}\iu\rVert^2\le(a_1+\frac{a_2}{4})^2=a^2$.}
		            \end{enumerate}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle a,b,p\rangle$.}
		        \end{enumerate}
		    \procedure{wed2108191401}
		      \objective
		        Choose a rational number $X$ such that $1\ge X> 0$. The objective of the following instructions is to construct positive rational numbers $a,c,d,e$ and a procedure $p(x,n,k)$ to show that $\exp_n(\ln_k(x))\equiv x\err{\frac{an}{k}+\frac{c}{n}}$ when a complex number $x$ and integers $k,n$ such that $\Re(x)\ge 0$, $\lVert x\rVert^2\ge X^2$, $n>e$, and $k>d$ are chosen.
		      \implementation
		        \begin{enumerate}
		          \item Execute \procedurehr{sun0812191401} on $\langle X\rangle$ and let $\langle f,p_0\rangle$ receive.
		          \item Execute \procedurehr{fri2607191736} on $\langle f\rangle$ and let $\langle a_1,p_1\rangle$ receive.
		          \item Execute \procedurehr{thu2507191359} on $\langle a_1\rangle$ and let $\langle a_2,b_2,p_2\rangle$ receive.
		          \item Execute \procedurehr{3.14} on $\langle a_1\rangle$ and let $\langle a_3,b_3,p_3\rangle$ receive.
		          \item Execute \procedurehr{fri2607191801} on $\langle f\rangle$ and let $\langle a_4,c_4,d,e_4,p_4\rangle$ receive.
		          \item Let $a=\frac{a_4}{a_3}+\frac{a_4(1+f)}{a_3(1-f)}$.
		          \item Let $c=a_2+\frac{c_4}{a_3}+\frac{c_4(1+f)}{a_3(1-f)}$.
		          \item Let $e=\max(b_2,b_3,e_4)$.
		          \item Let $p(x,n,k)$ be the following procedure:
		          \begin{enumerate}
		            \item Show that $\Re(\frac{x-1}{x+1})^2\le\lVert\frac{x-1}{x+1}\rVert^2\le f^2$ using procedure $p_0$.
		            \item Hence show that $\lvert\Re(\frac{x-1}{x+1})\rvert\le f$.
		            \item Show that $\lVert\ln_k(1+\frac{x-1}{x+1})\rVert^2\le{a_1}^2$ using procedure $p_1$.
		            \item Show that $\lVert\ln_k(1-\frac{x-1}{x+1})\rVert^2\le{a_1}^2$ using procedure $p_1$.
		            \item Hence using procedures $p_0,p_2,p_3,p_4$, show that $\exp_n(\ln_k(x))$
		            \begin{enumerate}
		              \item $=\exp_n(\ln_k(1+\frac{x-1}{x+1})-\ln_k(1-\frac{x-1}{x+1}))$
		              \item $\equiv\frac{\exp_n(\ln_k(1+\frac{x-1}{x+1}))}{\exp_n(\ln_k(1-\frac{x-1}{x+1}))}\err{\frac{a_2}{n}}$
		              \item $\equiv\frac{1+\frac{x-1}{x+1}}{\exp_n(\ln_k(1-\frac{x-1}{x+1}))}\err{\frac{1}{a_3}(\frac{a_4n}{k}+\frac{c_4}{n})}$
		              \item $\equiv\frac{1+\frac{x-1}{x+1}}{1-\frac{x-1}{x+1}}\err{(\frac{1+f}{a_3(1-f)})(\frac{a_4n}{k}+\frac{c_4}{n})}$
		              \item $=x$.
		            \end{enumerate}
		            \item Hence show that $\exp_n(\ln_k(x))\equiv x$
		            \begin{enumerate}
		              \item $\err{\frac{a_2}{n}+\frac{1}{a_3}(\frac{a_4n}{k}+\frac{c_4}{n})+(\frac{1+f}{a_3(1-f)})(\frac{a_4n}{k}+\frac{c_4}{n})}$
		              \item $\err{\frac{an}{k}+\frac{c}{n}}$.
		            \end{enumerate}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle a,c,d,e,p\rangle$.}
		        \end{enumerate}
		    \procedure{sun0812191512}
		      \objective
		        Choose a rational number $X$ such that $0<X\le 1$. The objective of the following instructions is to construct positive rational numbers $a,c,d,e$ and a procedure $p(x,n,k)$ to show that $\exp_n(\ln_k(x))\equiv x\err{\frac{an}{k}+\frac{c}{n}}$ when a complex number $x$ and integers $k,n$ such that $\lVert x\rVert^2\ge X^2$, $n>e$, and $k>d$ are chosen.
		      \implementation
		        \begin{enumerate}
		          \item Execute \procedurehr{wed2108191401} on $\langle X\rangle$ and let $\langle a_1,b_1,c_1,d_1,p_1\rangle$ receive.
		          \item Execute \procedurehr{sun0812191440} on $\langle X\rangle$ and let $\langle a_2,b_2,p_2\rangle$ receive.
		          \item Execute \procedurehr{3.59} on $\langle a_2,1\rangle$ and let $\langle a_3,b_3,c_3,d_3,p_3\rangle$ receive.
		          \item Let $a=a_2+a_3$.
		          \item Let $c=b_1+b_3$.
		          \item Let $d=\max(c_1,b_2,d_3)$.
		          \item Let $e=\max(c_3,d_1)$.
		          \item Let $p(x,n,k)$ be the following procedure:
		          \begin{enumerate}
		            \item If $\Re(x)\ge 0$, then do the following:
		            \begin{enumerate}
		              \item \textbf{Show that $\exp_n(\ln_k(x))\equiv x\err{\frac{a_2n}{k}+\frac{b_1}{n}}\err{\frac{an}{k}+\frac{c}{n}}$ using procedure $p_1$.}
		            \end{enumerate}
		            \item Otherwise if $\Im(x)\ge 0$, then do the following:
		            \begin{enumerate}
		              \item Show that $\lVert\frac{x}{\iu}\rVert^2=\lVert x\rVert^2\ge X^2$.
		              \item Show that $\Re(\frac{x}{\iu})=\Re(\Im(x)-\Re(x)\iu)=\Im(x)\ge 0$.
		              \item Hence show that $\exp_n(\ln_k(\frac{x}{\iu}))\equiv\frac{x}{\iu}\err{\frac{a_2n}{k}+\frac{b_1}{n}}$ using procedure $p_1$.
		              \item Hence show that $\lVert\ln_k(\frac{x}{\iu})\rVert^2\le{a_2}^2$ using procedure $p_2$.
		              \item Hence using procedure $p_3$, show that $\exp_n(\ln_k(x))$
		              \begin{enumerate}
		                \item $=\exp_n(\ln_k(\frac{x}{\iu})+\frac{\tau_k}{4}\iu)$
		                \item $\equiv\iu^1\exp_n(\ln_k(\frac{x}{\iu}))\err{\frac{a_3n}{k}+\frac{b_3}{n}}$
		                \item $\equiv\iu\cdot\frac{x}{\iu}\err{\frac{a_2n}{k}+\frac{b_1}{n}}$
		                \item $=x$.
		              \end{enumerate}
		              \item \textbf{Hence show that $\exp_n(\ln_k(x))\equiv x\err{\frac{an}{k}+\frac{c}{n}}$.}
		            \end{enumerate}
		            \item Otherwise do the following:
		            \begin{enumerate}
		              \item Show that $\lVert x\iu\rVert^2=\lVert x\rVert^2\ge X^2$.
		              \item Show that $\Re(x\iu)=\Re(-\Im(x)+\Re(x)\iu)=-\Im(x)>0$.
		              \item Hence show that $\exp_n(\ln_k(x\iu))\equiv x\iu\err{\frac{a_2n}{k}+\frac{b_1}{n}}$ using procedure $p_1$.
		              \item Hence show that $\lVert\ln_k(x\iu)\rVert^2\le{a_2}^2$ using procedure $p_2$.
		              \item Hence using procedure $p_3$, show that $\exp_n(\ln_k(x))$
		              \begin{enumerate}
		                \item $=\exp_n(\ln_k(x\iu)-\frac{\tau_k}{4}\iu)$
		                \item $\equiv\iu^{-1}\exp_n(\ln_k(x\iu))\err{\frac{a_3n}{k}+\frac{b_3}{n}}$
		                \item $\equiv\iu^{-1}x\iu\err{\frac{a_2n}{k}+\frac{b_1}{n}}$
		                \item $=x$.
		              \end{enumerate}
		              \item \textbf{Hence show that $\exp_n(\ln_k(x))\equiv x\err{\frac{an}{k}+\frac{c}{n}}$.}
		            \end{enumerate}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle a,c,d,e,p\rangle$.}
		        \end{enumerate}
		    \procedure{wed2108191324}
		      \objective
		        Choose two rational numbers $X,\epsilon$ such that $0<\epsilon<1$ and $X\ge 0$. The objective of the following instructions is to construct a rational number $f$ such that $0<f<1$, and a procedure $p(x)$ to show that $\lVert\frac{x-1}{x+1}\rVert^2\le f^2$ when a complex number $x$ such that $\Re(x)\ge\epsilon$ and $\lVert x\rVert^2\le X^2$ is chosen.
		      \implementation
		        \begin{enumerate}
		          \item Let $f=1-\frac{2\epsilon}{X^2+1+2\epsilon}$.
		          \item Show that $0<f<1$.
		          \item Let $p(x)$ be the following procedure:
		          \begin{enumerate}
		            \item Show that $X^2+1+2\epsilon\le\frac{4\epsilon}{1-f^2}$
		            \begin{enumerate}
		              \item given that $\frac{4\epsilon}{X^2+1+2\epsilon}\ge 1-f^2$
		              \item given that $f^2=1-\frac{4\epsilon}{X^2+1+2\epsilon}+(\frac{2\epsilon}{X^2+1+2\epsilon})^2\ge 1-\frac{4\epsilon}{X^2+1+2\epsilon}$.
		            \end{enumerate}
		            \item Hence show that $\lVert x\rVert^2$
		            \begin{enumerate}
		              \item $\le X^2$
		              \item $\le\frac{4\epsilon}{1-f^2}-(1+2\epsilon)$
		              \item $=\frac{4\epsilon-1+f^2-2\epsilon+2\epsilon f^2}{1-f^2}$
		              \item $=\frac{f^2-1+2\epsilon(1+f^2)}{1-f^2}$
		              \item $\le\frac{f^2-1+2\Re(x)(1+f^2)}{1-f^2}$.
		            \end{enumerate}
		            \item \textbf{Hence show that $\lVert\frac{x-1}{x+1}\rVert^2\le f^2$}
		            \begin{enumerate}
		              \item given that $(\Re(x)-1)^2+\Im(x)^2\le f^2((\Re(x)+1)^2+\Im(x)^2)$
		              \item given that $(1-f^2)(\Re(x)^2+\Im(x)^2)\le f^2-1+2\Re(x)(1+f^2)$.
		            \end{enumerate}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle f,p\rangle$.}
		        \end{enumerate}
		    \procedure{wed2108191603}
		      \objective
		        Choose two rational numbers $X,\epsilon$ such that $0<\epsilon<1$ and $X\ge 0$. The objective of the following instructions is to construct rational numbers $c,d,a,b,e$, a procedure $p(x,n)$ to show that $\lVert\frac{1}{x}\rVert^2\le c^2$ when a complex number $x$ and a positive integer $n$ such that $\Re(x)\ge\epsilon$, $\lVert x\rVert^2\le X^2$, and $n>d$ are chosen, and a procedure $q(x,n,\delta)$ to show that $\Delta_{y=x}^{+\delta}\ln_n(y)\equiv\frac{1}{x}\err{\frac{a}{n}+b\{\delta\}}$ when in addition a complex number $\delta$ such that $0<\lVert\delta\rVert^2\le e^2$ is chosen.
		      \implementation
		        \begin{instructions}
		          \item Execute the following in post-order:
		          \begin{instructions}
		            \item Execute \procedurehr{mon1908191506} on $\langle q_3,q_4,q_5,q_6\rangle$ and let $\langle c,d,a,b,e,q_1,q_2\rangle$ receive.
		            \begin{instructions}
		              \item Execute \procedurehr{sat0308191134} on $\langle q_7,q_8,q_9,q_{10},q_{11}\rangle$ and let $\langle q_3,q_4\rangle$ receive.
		              \begin{instructions}
		                \item Execute \procedurehr{wed2108191140} on $\langle e_1,\frac{1-e_1}{2}\rangle$ and let $\langle q_7,q_8\rangle$ receive.
		                \begin{instructions}
		                  \item Execute \procedurehr{wed2108191324} on $\langle X,\epsilon\rangle$ and let $\langle e_1,q_{12}\rangle$ receive.
		                \end{instructions}
		                \item Execute \procedurehr{mon1908191506} on $\langle q_{13},q_{14},q_{15},q_{16}\rangle$ and let $\langle q_9,q_{10}\rangle$ receive.
		                \begin{instructions}
		                  \item Execute \procedurehr{tue2008191151} on $\langle 1,1\rangle$ and let $\langle q_{13},q_{14}\rangle$ receive.
		                  \item Execute \procedurehr{tue2008191001} on $\langle-2,q_{17},q_{18}\rangle$ and let $\langle q_{15},q_{16}\rangle$ receive.
		                  \begin{instructions}
		                    \item Execute \procedurehr{sat0308191134} on $\langle q_{19},q_{20},q_{21},q_{22},q_{23}\rangle$ and let $\langle q_{17},q_{18}\rangle$ receive.
		                    \begin{instructions}
		                      \item Execute \procedurehr{tue2008191341} on $\langle 1,1+\epsilon,X+1\rangle$ and let $\langle q_{19},q_{20}\rangle$ receive.
		                      \item Execute \procedurehr{mon1908191506} on $\langle q_{24},q_{25},q_{26},q_{27}\rangle$ and let $\langle q_{21},q_{22}\rangle$ receive.
		                      \begin{instructions}
		                        \item Execute \procedurehr{tue2008191151} on $\langle 1,1\rangle$ and let $\langle q_{24},q_{25}\rangle$ receive.
		                        \item Execute \procedurehr{tue2008191209} on $\langle 1,X,1\rangle$ and let $\langle q_{26},q_{27}\rangle$ receive.
		                      \end{instructions}
		                      \item Let $q_{24}(x,n)$ be the following procedure:
		                      \begin{instructions}
		                        \item Show that $(1+\epsilon)^2\le(1+\Re(x))^2=\Re(x+1)^2\le\lVert x+1\rVert^2\le(X+1)^2$.
		                      \end{instructions}
		                    \end{instructions}
		                  \end{instructions}
		                \end{instructions}
		                \item Let $q_{11}(x,n)$ be the following procedure:
		                \begin{instructions}
		                  \item Show that $\lVert 1-2(x+1)^{-1}\rVert^2=\lVert\frac{x-1}{x+1}\rVert^2\le{e_1}^2$ using procedure $q_{12}$.
		                \end{instructions}
		              \end{instructions}
		              \item Execute \procedurehr{tue2008191001} on $\langle-1,q_{28},q_{29}\rangle$ and let $\langle q_5,q_6\rangle$ receive.
		              \begin{instructions}
		                \item Execute \procedurehr{sat0308191134} on $\langle q_{30},q_{31},q_{32},q_{33},q_{34}\rangle$ and let $\langle q_{28},q_{29}\rangle$ receive.
		                \begin{instructions}
		                  \item Execute \procedurehr{wed2108191140} on $\langle e_1,\frac{1-e_1}{2}\rangle$ and let $\langle q_{30},q_{31}\rangle$ receive.
		                  \item Execute \procedurehr{tue2008191001} on $\langle-1,q_9,q_{10}\rangle$ and let $\langle q_{32},q_{33}\rangle$ receive.
		                  \item Let $q_{34}(x,n)$ be the following procedure:
		                  \begin{instructions}
		                    \item Show that $\lVert-(1+(-2)(x+1)^{-1})\rVert^2=\lVert\frac{x-1}{x+1}\rVert^2\le{e_1}^2$ using procedure $q_{12}$.
		                  \end{instructions}
		                \end{instructions}
		              \end{instructions}
		            \end{instructions}
		          \end{instructions}
		          \item Let $p(x,n)$ be the following procedure:
		          \begin{instructions}
		            \item \textbf{Show that $\frac{1}{x}\equiv 0\allowbreak\err{\frac{1}{x}}\allowbreak\err{\frac{1}{\Re(x)}}\allowbreak\err{\frac{1}{\epsilon}}\allowbreak\err{c}$.}
		          \end{instructions}
		          \item Let $q(x,n,\delta)$ be the following procedure:
		          \begin{instructions}
		            \item Using procedure $q_2$, show that $\Delta_{y=x}^{+\delta}\ln_n(y)$
		            \begin{enumerate}
		              \item $=\Delta_{y=x}^{+\delta}(\ln_n(1+\frac{y-1}{y+1})-\ln_n(1-\frac{y-1}{y+1}))$
		              \item $=\Delta_{y=x}^{+\delta}(\ln_n(1+(1+(-2)(y+1)^{-1}))+(-1)\ln_n(1+(-1)(1+(-2)(y+1)^{-1})))$
		              \item $\equiv((1+(1+(-2)(x+1)^{-1}))^{-1}(0+(-2)(-1)(x+1)^{-2}(1+0)+(-1)((1+(-1)(1+(-2)(x+1)^{-1}))^{-1}\cdot(0+(-1)(0+(-2)(-1)(x+1)^{-2}(1+0))))))\err{\frac{a}{n}+b\{\delta\}}$
		              \item $=\frac{1}{x}$
		            \end{enumerate}
		            \item \textbf{Hence show that $\Delta_{y=x}^{+\delta}\ln_n(y)\equiv\frac{1}{x}\err{\frac{a}{n}+b\{\delta\}}$.}
		          \end{instructions}
		          \item \textbf{Yield the tuple $\langle c,d,a,b,e,p,q\rangle$.}
		        \end{instructions}
		    \procedure{thu2208191330}
		      \objective
		        Choose a complex number $A$ and non-negative rational numbers $X,D$ such that $X+D<1$. The objective of the following instructions is to construct rational numbers $l,d,a,b$, a procedure $p(x,n)$ to show that $\lVert A(1+x)_{n-1}^{A-1}\rVert^2\le l^2$ when a complex number $x$ and a positive integer $n$ such that $\lVert x\rVert^2\le X^2$ and $n>d$ are chosen, and a procedure $q(x,n,\delta)$ to show that $\Delta_{y=x}^{+\delta}(1+y)_n^A\equiv A(1+x)_{n-1}^{A-1}\err{\frac{a}{n}+b\{\delta\}}$ when in addition a complex number $\delta$ such that $0<\lVert\delta\rVert^2\le D^2$ is chosen.
		      \implementation
		        \begin{enumerate}
		          \item Execute \procedurehr{sun2107191247} on $\langle\{A+1\},X+D\rangle$ and let $\langle a_1,b_1,p_1\rangle$ receive.
		          \item Execute \procedurehr{wed2407191611} on $\langle\{A-1\},X\rangle$ and let $\langle a_2,p_2\rangle$ receive.
		          \item Let $l=\{A\}a_2$.
		          \item Let $d=1$.
		          \item Let $a=0$.
		          \item Let $b=\frac{a_1b_1}{D^2(1-b_1)}$.
		          \item Let $p(x,n)$ be the following procedure:
		          \begin{enumerate}
		            \item Using procedure $p_2$, show that $A(1+x)_{n-1}^{A-1}\equiv 0$
		            \begin{enumerate}
		              \item $\err{A(1+x)_{n-1}^{A-1}}$
		              \item $\err{\{A\}a_2}$
		              \item $\err{l}$.
		            \end{enumerate}
		          \end{enumerate}
		          \item Let $q(x,n,\delta)$ be the following procedure:
		          \begin{enumerate}
		            \item For $r\in[1:n]$, do the following:
		            \begin{enumerate}
		              \item Show that $\lVert A+1\rVert^2\le\{A+1\}^2$.
		              \item Hence show that $\lVert\binom{A}{r}(X+D)^r\rVert^2\le(a_1{b_1}^r)^2$ using procedure $p_1$.
		              \item Hence show that $\lVert\binom{A}{r}\sum_k^{[0:r-1]}\binom{r}{k}x^k\delta^{r-2-k}\rVert^2$
		              \begin{enumerate}
		                \item $\le\lVert\binom{A}{r}\rVert^2(\sum_k^{[0:r-1]}\binom{r}{k}X^kD^{r-2-k})^2$
		                \item $=\lVert\binom{A}{r}\rVert^2(\frac{1}{D^2}\sum_k^{[0:r-1]}\binom{r}{k}X^kD^{r-k})^2$
		                \item $\le\lVert\frac{1}{D^2}\binom{A}{r}(X+D)^r\rVert^2$
		                \item $\le(\frac{a_1{b_1}^r}{D^2})^2$.
		              \end{enumerate}
		            \end{enumerate}
		            \item Now show that $\Delta_{y=x}^{+\delta}(1+y)_{n}^A\equiv A(1+x)_{n-1}^{A-1}$
		            \begin{enumerate}
		              \item $\err{\frac{1}{\delta}((1+x+\delta)_n^A-(1+x)_n^A)-A(1+x)_{n-1}^{A-1}}$
		              \item $\err{\frac{1}{\delta}(\sum_r^{[0:n]}\binom{A}{r}(x+\delta)^r-\sum_r^{[0:n]}\binom{A}{r}x^r)-A\sum_r^{[0:n-1]}\binom{A-1}{r}x^r}$
		              \item $\err{\frac{1}{\delta}(\sum_r^{[0:n]}\binom{A}{r}(\sum_k^{[0:r+1]}\binom{r}{k}x^k\delta^{r-k}-x^r))-A\sum_r^{[0:n-1]}\binom{A-1}{r}x^r}$
		              \item $\err{\sum_r^{[0:n]}\binom{A}{r}\sum_k^{[0:r]}\binom{r}{k}x^k\delta^{r-1-k}-\sum_r^{[0:n-1]}(r+1)\binom{A}{r+1}x^r}$
		              \item $\err{\sum_r^{[1:n]}\binom{A}{r}\sum_k^{[0:r]}\binom{r}{k}x^k\delta^{r-1-k}-\sum_r^{[1:n]}r\binom{A}{r}x^{r-1}}$
		              \item $\err{\delta(\sum_r^{[1:n]}\binom{A}{r}\sum_k^{[0:r-1]}\binom{r}{k}x^k\delta^{r-2-k})}$
		              \item $\err{\delta(\sum_r^{[1:n]}\frac{a_1{b_1}^r}{D^2})}$
		              \item $\err{\delta(\frac{a_1b_1}{D^2(1-b_1)})}$
		              \item $\err{\frac{a}{n}+b\{\delta\}}$.
		            \end{enumerate}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle l,d,a,b,p,q\rangle$.}
		        \end{enumerate}
		    \procedure{thu2208191432}
		      \objective
		        Choose a complex number $A$ and non-negative rational numbers $X,D$ such that $X+D<1$. The objective of the following instructions is to construct rational numbers $l,d,a,b$, a procedure $p(x,n)$ to show that $\lVert A(1+x)_{n}^{A-1}\rVert^2\le l^2$ when a complex number $x$ and a positive integer $n$ such that $\lVert x\rVert^2\le X^2$ and $n>d$ are chosen, and a procedure $q(x,n,\delta)$ to show that $\Delta_{y=x}^{+\delta}(1+y)_n^A\equiv A(1+x)_{n}^{A-1}\err{\frac{a}{n}+b\{\delta\}}$ when in addition a complex number $\delta$ such that $0<\lVert\delta\rVert^2\le D^2$ is chosen.
		      \implementation
		        \begin{enumerate}
		          \item Execute \procedurehr{sun2107191247} on $\langle\{A\},X\rangle$ and let $\langle a_1,b_1,p_1\rangle$ receive.
		          \item Execute \procedurehr{wed2407191611} on $\langle\{A-1\},X\rangle$ and let $\langle a_2,p_2\rangle$ receive.
		          \item Execute \procedurehr{wed2407191422} on $\langle b_1,1\rangle$ and let $\langle a_3,b_3,p_3\rangle$ receive.
		          \item Execute \procedurehr{thu2208191330} on $\langle X,D\rangle$ and let $\langle a_4,b_4,c_4,d_4,p_4,q_4\rangle$ receive.
		          \item Let $l=\{a\}a_2$.
		          \item Let $d=\max(1,b_4)$.
		          \item Let $a=c_4+2\{A\}a_1a_3$.
		          \item Let $b=d_4$.
		          \item Let $p(x,n)$ be the following procedure:
		          \begin{enumerate}
		            \item Using procedure $p_2$, show that $A(1+x)_n^{A-1}\equiv 0$
		            \begin{enumerate}
		              \item $\err{A(1+x)_n^{A-1}}$
		              \item $\err{\{A\}a_2}$
		              \item $\err{l}$.
		            \end{enumerate}
		          \end{enumerate}
		          \item Let $q(x,n,\delta)$ be the following procedure:
		          \begin{enumerate}
		            \item Show that $\lVert\binom{A-1}{n-1}x^{n-1}\rVert^2\le(a_1{b_1}^{n-1})^2$ using procedure $p_1$.
		            \item Show that $\lVert(n-1){b_1}^{n-1}\rVert^2\le(a_3{b_3}^{n-1})^2\le{a_3}^2$ using procedure $p_3$.
		            \item Hence show that $\lVert{b_1}^{n-1}\rVert^2\le(\frac{a_3}{n-1})^2\le(\frac{2a_3}{n})^2$.
		            \item Now using procedure $q_4$, show that $\Delta_{y=x}^{+\delta}(1+y)_n^A$
		            \begin{enumerate}
		              \item $\equiv A(1+x)_{n-1}^{A-1}\err{\frac{c_4}{n}+d_4\{\delta\}}$
		              \item $\equiv A(1+x)_{n}^{A-1}$
		              \begin{enumerate}
		                \item $\err{A\binom{A-1}{n-1}x^{n-1}}$
		                \item $\err{\{A\}a_1{b_1}^{n-1}}$
		                \item $\err{\frac{2\{A\}a_1a_3}{n}}$.
		              \end{enumerate}
		            \end{enumerate}
		            \item Hence show that $\Delta_{y=x}^{+\delta}(1+y)_n^A\equiv A(1+x)_{n}^{A-1}$
		            \begin{enumerate}
		              \item $\err{\frac{c_4}{n}+d_4\{\delta\}+\frac{2\{A\}a_1a_3}{n}}$
		              \item $\err{\frac{a}{n}+b\{\delta\}}$.
		            \end{enumerate}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle l,d,a,b,p,q\rangle$.}
		        \end{enumerate}
		    \declaration{thu2208191619}
		      The notation \gls{differentialpow}, where $x,a$ are complex numbers and $n$ is a positive integer, will be used as a shorthand for $(1+\frac{x-1}{x+1})_n^a(1-\frac{x-1}{x+1})_n^{-a}$.
		    \procedure{sat2408190819}
		      \objective
		        Choose three non-negative rational numbers $A,X,\epsilon$ such that $0<\epsilon<1$. The objective of the following instructions is to construct  positive rational numbers $B,C,D$, and a procedure $p(x,a,n)$ to show that $x_n^a\equiv x^a\err{BC^n}$ when a complex number $x$, and integers $a,n$ such that $\lVert x\rVert^2\le X^2$, $\lVert a\rVert^2\le A^2$, $\Re(x)\ge\epsilon$, and $n>D$ are chosen.
		      \implementation
		        \begin{enumerate}
		          \item Execute \procedurehr{wed2108191324} on $\langle X,\epsilon\rangle$ and let $\langle a_1,p_1\rangle$ receive.
		          \item Show that $0<a_1<1$.
		          \item Execute \procedurehr{tue2008190849} on $\langle A,a_1\rangle$ and let $\langle a_2,b_2,c_2,p_2\rangle$ receive.
		          \item Execute \procedurehr{wed2407191611} on $\langle A,a_1\rangle$ and let $\langle a_3,p_3\rangle$ receive.
		          \item Let $B=a_3a_2$.
		          \item Let $C=b_2$.
		          \item Let $D=\max(c_2,A)$.
		          \item Let $p(x,a,n)$ be the following procedure:
		          \begin{enumerate}
		            \item Show that $\lVert\frac{x-1}{x+1}\rVert^2\le{a_1}^2$ using procedure $p_1$.
		            \item If $a\ge 0$, then do the following:
		            \begin{enumerate}
		              \item Using \procedurehr{sun2107190619} and procedures $p_2,p_3$, show that $x_n^a$
		              \begin{enumerate}
		                \item $=(1+\frac{x-1}{x+1})_n^a(1-\frac{x-1}{x+1})_n^{0-a}$
		                \item $\equiv(1+\frac{x-1}{x+1})_n^a\frac{(1-\frac{x-1}{x+1})_n^{0}}{(1-\frac{x-1}{x+1})_n^{a}}\err{a_3a_2{b_2}^n}$
		                \item $=\frac{(1+\frac{x-1}{x+1})^a}{(1-\frac{x-1}{x+1})^{a}}$
		                \item $=x^a$
		              \end{enumerate}
		            \end{enumerate}
		            \item Otherwise do the following:
		            \begin{enumerate}
		              \item Using \procedurehr{sun2107190619} and procedures $p_2,p_3$, show that $x_n^a$
		              \begin{enumerate}
		                \item $=(1+\frac{x-1}{x+1})_n^{0-(-a)}(1-\frac{x-1}{x+1})_n^{-a}$
		                \item $\equiv\frac{(1+\frac{x-1}{x+1})_n^{0}}{(1+\frac{x-1}{x+1})_n^{-a}}(1-\frac{x-1}{x+1})_n^{-a}\err{a_3a_2{b_2}^n}$
		                \item $=\frac{(1-\frac{x-1}{x+1})_n^{-a}}{(1+\frac{x-1}{x+1})_n^{-a}}$
		                \item $=(\frac{1}{x})^{-a}$
		                \item $=x^a$.
		              \end{enumerate}
		            \end{enumerate}
		            \item \textbf{Hence show that $x_n^a\equiv x^a\allowbreak\err{a_3a_2{b_2}^n}\allowbreak\err{BC^n}$.}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle B,C,D,p\rangle$.}
		        \end{enumerate}
		    \procedure{sat2408191109}
		      \objective
		        Choose three non-negative rational numbers $A,X,\epsilon$ such that $0<\epsilon<1$. The objective of the following instructions is to construct positive rational numbers $B,C$, and a procedure $p(x,a,b,n)$ to show that $x_n^{a+b}\equiv x_n^ax_n^b\err{BC^n}$ when complex numbers $x,a,b$, and a positive integer $n$ such that $\lVert x\rVert^2\le X^2$, $\lVert a\rVert^2\le A^2$, $\lVert b\rVert^2\le A^2$, and $\Re(x)\ge\epsilon$ are chosen.
		      \implementation
		        \begin{enumerate}
		          \item Execute \procedurehr{wed2108191324} on $\langle X,\epsilon\rangle$ and let $\langle a_1,p_1\rangle$ receive.
		          \item Execute \procedurehr{wed2407191521} on $\langle A,a_1\rangle$ and let $\langle a_2,b_2,p_2\rangle$ receive.
		          \item Execute \procedurehr{wed2407191611} on $\langle 2A,a_1\rangle$ and let $\langle a_3,p_3\rangle$ receive.
		          \item Let $B=a_2a_3(1+a_3)$.
		          \item Let $C=b_2$.
		          \item Let $p(x,a,b,n)$ be the following procedure:
		          \begin{enumerate}
		            \item Using procedures $p_1,p_2,p_3$, show that $x_n^{a+b}$
		            \begin{enumerate}
		              \item $=(1+\frac{x-1}{x+1})_n^{a+b}(1-\frac{x-1}{x+1})_n^{-(a+b)}$
		              \item $\equiv(1+\frac{x-1}{x+1})_n^{a}(1+\frac{x-1}{x+1})_n^{b}(1-\frac{x-1}{x+1})_n^{(-a)+(-b)}\err{a_2{b_2}^na_3}$
		              \item $\equiv(1+\frac{x-1}{x+1})_n^{a}(1+\frac{x-1}{x+1})_n^{b}(1-\frac{x-1}{x+1})_n^{-a}(1-\frac{x-1}{x+1})_n^{-b}\err{{a_3}^2a_2{b_2}^n}$
		              \item $=x_n^ax_n^b$.
		            \end{enumerate}
		            \item \textbf{Hence show that $x_n^{a+b}\equiv x_n^ax_n^b\err{BC^n}$.}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle B,C,p\rangle$.}
		        \end{enumerate}
		    \procedure{sat2408191137}
		      \objective
		        Choose three non-negative rational numbers $A,X,\epsilon$ such that $0<\epsilon<1$. The objective of the following instructions is to construct a positive rational number $D$, and a procedure $p(x,n,a,k)$ to show that $(x_n^a)^k\equiv 0\err{D}$ when complex numbers $x,a,k$, and a positive integer $n$ such that $\lVert x\rVert^2\le X^2$, $\lVert ka\rVert^2\le A^2$, and $\Re(x)\ge\epsilon$ are chosen.
		      \implementation
		        \begin{enumerate}
		          \item Execute \procedurehr{wed2108191324} on $\langle X,\epsilon\rangle$ and let $\langle a_1,p_1\rangle$ receive.
		          \item Execute \procedurehr{wed2407191611} on $\langle A,a_1\rangle$ and let $\langle a_2,p_2\rangle$ receive.
		          \item Let $D={a_2}^2$.
		          \item Let $p(x,n,a,k)$ be the following procedure:
		          \begin{enumerate}
		            \item Using procedures $p_1,p_2$, show that $(x_n^a)^k$
		            \begin{enumerate}
		              \item $=((1+\frac{x-1}{x+1})_n^a(1-\frac{x-1}{x+1})_n^{-a})^k$
		              \item $=((1+\frac{x-1}{x+1})_n^a)^k((1-\frac{x-1}{x+1})_n^{-a})^k$
		              \item $\equiv 0((1-\frac{x-1}{x+1})_n^{-a})^k\err{{a_2}^2}$
		              \item $=0$.
		            \end{enumerate}
		            \item \textbf{Hence show that $(x_n^a)^k\equiv 0\err{D}$.}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle D,p\rangle$.}
		        \end{enumerate}
		    \procedure{thu2908190744}
		      \objective
		        Choose three non-negative rational numbers $A,X,\epsilon$ such that $0<\epsilon<1$. The objective of the following instructions is to construct a positive rational number $D$, and a procedure $p(x,n,a)$ to show that $\lVert x_n^a\rVert^2\ge D^2$ when complex numbers $x,a$, and a positive integer $n$ such that $\lVert x\rVert^2\le X^2$, $\lVert a\rVert^2\le A^2$, and $\Re(x)\ge\epsilon$ are chosen.
		      \implementation
		        \begin{enumerate}
		          \item Execute \procedurehr{wed2108191324} on $\langle X,\epsilon\rangle$ and let $\langle a_1,p_1\rangle$ receive.
		          \item Execute \procedurehr{tue2008190712} on $\langle A,a_1\rangle$ and let $\langle a_2,b_2,p_2\rangle$ receive.
		          \item Let $D={a_2}^2$.
		          \item Let $p(x,n,a)$ be the following procedure:
		          \begin{enumerate}
		            \item Show that $\lVert\frac{x-1}{x+1}\rVert^2\le{a_1}^2$ using procedure $p_1$.
		            \item Show that $\lVert(1+\frac{x-1}{x+1})_n^a\rVert^2\ge{a_2}^2$ using procedure $p_2$.
		            \item Show that $\lVert(1-\frac{x-1}{x+1})_n^{-a}\rVert^2\ge{a_2}^2$ using procedure $p_2$.
		            \item Hence using \declarationhr{thu2208191619}, show that $\lVert x_n^a\rVert^2$
		            \begin{enumerate}
		              \item $=\lVert(1+\frac{x-1}{x+1})_n^a\rVert^2\lVert(1-\frac{x-1}{x+1})_n^{-a}\rVert^2$
		              \item $\ge{a_2}^2{a_2}^2$
		              \item $=D^2$.
		            \end{enumerate}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle D,p\rangle$.}
		        \end{enumerate}
		    \procedure{thu2908190802}
		      \objective
		        Choose three non-negative rational numbers $A,X,\epsilon$ such that $0<\epsilon<1$. The objective of the following instructions is to construct positive rational numbers $B,C,D$, and a procedure $p(x,a,b,n)$ to show that $x_n^{a-b}\equiv\frac{x_n^a}{x_n^b}\err{BC^n}$ when complex numbers $x,a,b$, and a positive integer $n$ such that $\lVert x\rVert^2\le X^2$, $\lVert a\rVert^2\le A^2$, $\lVert b\rVert^2\le A^2$, $\Re(x)\ge\epsilon$, and $n>D$ are chosen.
		      \implementation
		        \begin{enumerate}
		          \item Execute \procedurehr{wed2108191324} on $\langle X,\epsilon\rangle$ and let $\langle a_1,p_1\rangle$ receive.
		          \item Execute \procedurehr{tue2008190849} on $\langle A,a_1\rangle$ and let $\langle a_2,b_2,c_2,p_2\rangle$ receive.
		          \item Execute \procedurehr{tue2008190712} on $\langle A,a_1\rangle$ and let $\langle a_3,b_3,p_3\rangle$ receive.
		          \item Execute \procedurehr{wed2407191611} on $\langle 2A,a_1\rangle$ and let $\langle a_4,p_4\rangle$ receive.
		          \item Let $B=a_2a_4(1+\frac{1}{a_3})$.
		          \item Let $C=b_2$.
		          \item Let $D=\max(c_2,b_3)$.
		          \item Let $p(x,a,b,n)$ be the following procedure:
		          \begin{enumerate}
		            \item Using procedures $p_1,p_2,p_3,p_4$, show that $x_n^{a-b}$
		            \begin{enumerate}
		              \item $=(1+\frac{x-1}{x+1})_n^{a-b}(1-\frac{x-1}{x+1})_n^{(-a)-(-b)}$
		              \item $\equiv(1+\frac{x-1}{x+1})_n^{a-b}\frac{(1-\frac{x-1}{x+1})_n^{-a}}{(1-\frac{x-1}{x+1})_n^{-b}}\err{a_4a_2{b_2}^n}$
		              \item $\equiv\frac{(1+\frac{x-1}{x+1})_n^{a}}{(1+\frac{x-1}{x+1})_n^{b}}\frac{(1-\frac{x-1}{x+1})_n^{-a}}{(1-\frac{x-1}{x+1})_n^{-b}}\err{a_2{b_2}^n\frac{a_4}{a_3}}$
		              \item $=\frac{x_n^a}{x_n^b}$.
		            \end{enumerate}
		            \item \textbf{Hence show that $x_n^{a-b}\equiv\frac{x_n^a}{x_n^b}\err{a_4a_2{b_2}^n+a_2{b_2}^n\frac{a_4}{a_3}}\err{BC^n}$.}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle B,C,D,p\rangle$.}
		        \end{enumerate}
		    \procedure{sat2408191538}
		      \objective
		        Choose three non-negative rational numbers $A,X,\epsilon$ such that $0<\epsilon<1$. The objective of the following instructions is to construct a positive rational number $B,C$, and a procedure $p(x,n,a,k)$ to show that $(x_n^a)^k\equiv x_n^{ak}\err{BC^n}$ when complex numbers $x,a,k$, and a positive integer $n$ such that $\lVert x\rVert^2\le X^2$, $\lVert ka\rVert^2\le A^2$, and $\Re(x)\ge\epsilon$ are chosen.
		      \implementation
		        \begin{enumerate}
		          \item Execute \procedurehr{wed2108191324} on $\langle X,\epsilon\rangle$ and let $\langle a_1,p_1\rangle$ receive.
		          \item Execute \procedurehr{wed2407191627} on $\langle A,a_1\rangle$ and let $\langle a_2,b_2,p_2\rangle$ receive.
		          \item Execute \procedurehr{wed2407191611} on $\langle A,a_1\rangle$ and let $\langle a_3,p_3\rangle$ receive.
		          \item Let $B=2a_2a_3$.
		          \item Let $C=b_2$.
		          \item Let $p(x,n,a,k)$ be the following procedure:
		          \begin{enumerate}
		            \item Using procedures $p_1,p_2,p_3$, show that $(x_n^a)^k$
		            \begin{enumerate}
		              \item $=((1+\frac{x-1}{x+1})_n^a(1-\frac{x-1}{x+1})_n^{-a})^k$
		              \item $=((1+\frac{x-1}{x+1})_n^a)^k((1-\frac{x-1}{x+1})_n^{-a})^k$
		              \item $\equiv(1+\frac{x-1}{x+1})_n^{ak}((1-\frac{x-1}{x+1})_n^{-a})^k\err{a_2{b_2}^na_3}$
		              \item $\equiv((1+\frac{x-1}{x+1})_n^{ak})^1(1-\frac{x-1}{x+1})_n^{-ak}\err{a_3a_2{b_2}^n}$
		              \item $=x_n^{ak}$.
		            \end{enumerate}
		            \item \textbf{Hence show that $(x_n^a)^k\equiv x_n^{ak}\err{BC^n}$.}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle B,C,p\rangle$.}
		        \end{enumerate}
		    \procedure{thu2208191610}
		      \objective
		        Choose a complex number $f$ and two rational numbers $X,\epsilon$ such that $0<\epsilon<1$ and $X\ge 0$. Let $g(f,x,n)$ be a shorthand for $[2f(1+\frac{x-1}{x+1})_n^f\cdot(1-\frac{x-1}{x+1})_n^{-f-1}(x+1)^{-2}]+[2f(1+\frac{x-1}{x+1})_n^{f-1}\cdot(1-\frac{x-1}{x+1})_n^{-f}(x+1)^{-2}]$. The objective of the following instructions is to construct rational numbers $c,d,a,b,e$, a procedure $p(x,n)$ to show that $\lVert g(f,x,n)\rVert^2\le c^2$ when a complex number $x$ and a positive integer $n$ such that $\Re(x)\ge\epsilon$, $\lVert x\rVert^2$, and $n>d$ are chosen, and a procedure $q(x,n,\delta)$ to show that $\Delta_{y=x}^{+\delta}x_n^f\equiv g(f,x,n)\err{\frac{a}{n}+b\{\delta\}}$ when in addition a complex number $\delta$ such that $0<\lVert\delta\rVert^2\le e^2$ is chosen.
		      \implementation
		        \begin{instructions}
		          \item Execute the following in post-order:
		          \begin{instructions}
		            \item Execute \procedurehr{mon1908191207} on $\langle q_7,q_3,q_4,q_8,q_5,q_6\rangle$ and let $\langle c,d,a,b,e,p,q\rangle$ receive.
		            \begin{instructions}
		              \item Execute \procedurehr{wed2108191324} on $\langle X,\epsilon\rangle$ and let $\langle e_1,q_9\rangle$ receive.
		              \item Execute \procedurehr{wed2407191611} on $\langle \{f\},e_1\rangle$ and let $\langle e_2,q_{10}\rangle$ receive.
		              \item Let $q_7(x,n)$ be the following procedure:
		              \begin{instructions}
		                \item Show that $\lVert\frac{x-1}{x+1}\rVert^2\le{e_1}^2$ using procedure $q_9$.
		                \item Using procedure $q_{10}$, show that $\lVert(1+(1+(-2)(x+1)^{-1}))_n^f\rVert^2$
		                \begin{instructions}
		                  \item $=\lVert(1+\frac{x-1}{x+1})_n^f\rVert^2$
		                  \item $\le{e_2}^2$.
		                \end{instructions}
		              \end{instructions}
		              \item Let $q_8(x,n)$ be the following procedure:
		              \begin{instructions}
		                \item Show that $\lVert\frac{x-1}{x+1}\rVert^2\le{e_1}^2$ using procedure $q_9$.
		                \item Using procedure $q_{10}$, show that $\lVert(1-(1+(-2)(x+1)^{-1}))_n^{-f}\rVert^2$
		                \begin{instructions}
		                  \item $=\lVert(1-\frac{x-1}{x+1})_n^{-f}\rVert^2$
		                  \item $\le{e_2}^2$.
		                \end{instructions}
		              \end{instructions}
		              \item Execute \procedurehr{sat0308191134} on $\langle q_{11},q_{12},q_{13},q_{14},q_{15}\rangle$ and let $\langle q_3,q_4\rangle$ receive.
		              \begin{instructions}
		                \item Execute \procedurehr{thu2208191432} on $\langle f,e_1,\frac{1-e_1}{2}\rangle$ and let $\langle q_{11},q_{12}\rangle$ receive.
		                \item Execute \procedurehr{mon1908191506} on $\langle q_{16},q_{17},q_{18},q_{19}\rangle$ and let $\langle q_{13},q_{14}\rangle$ receive.
		                \begin{instructions}
		                  \item Execute \procedurehr{tue2008191151} on $\langle 1,1\rangle$ and let $\langle q_{16},q_{17}\rangle$ receive.
		                  \item Execute \procedurehr{tue2008191001} on $\langle-2,q_{20},q_{21}\rangle$ and let $\langle q_{18},q_{19}\rangle$ receive.
		                  \begin{instructions}
		                    \item Execute \procedurehr{sat0308191134} on $\langle q_{22},q_{23},q_{24},q_{25},q_{26}\rangle$ and let $\langle q_{20},q_{21}\rangle$ receive.
		                    \begin{instructions}
		                      \item Execute \procedurehr{tue2008191341} on $\langle 1,1+\epsilon,1+X\rangle$ and let $\langle q_{22},q_{23}\rangle$ receive.
		                      \item Execute \procedurehr{mon1908191506} on $\langle q_{27},q_{28},q_{29},q_{30}\rangle$ and let $\langle q_{24},q_{25}\rangle$ receive.
		                      \begin{instructions}
		                        \item Execute \procedurehr{tue2008191151} on $\langle 1,1\rangle$ and let $\langle q_{27},q_{28}\rangle$ receive.
		                        \item Execute \procedurehr{tue2008191209} on $\langle 1,X,1\rangle$ and let $\langle q_{29},q_{30}\rangle$ receive.
		                      \end{instructions}
		                      \item Let $q_{26}(x,n)$ be the following procedure:
		                      \begin{instructions}
		                        \item Show that $(1+\epsilon)^2$
		                        \begin{instructions}
		                          \item $\le(1+\Re(x))^2$
		                          \item $\le\Re(x+1)^2$
		                          \item $\le\lVert x+1\rVert^2$
		                          \item $\le (X+1)^2$.
		                        \end{instructions}
		                      \end{instructions}
		                    \end{instructions}
		                  \end{instructions}
		                \end{instructions}
		                \item Let $q_{15}(x,n)$ be the following procedure:
		                \begin{instructions}
		                  \item \textbf{Hence show that $\lVert 1+(-2)(x+1)^{-1}\rVert^2=\lVert\frac{x-1}{x+1}\rVert^2\le{e_1}^2$ using procedure $q_9$.}
		                \end{instructions}
		              \end{instructions}
		              \item Execute \procedurehr{sat0308191134} on $\langle q_{31},q_{32},q_{33},q_{34},q_{35}\rangle$ and let $\langle q_5,q_6\rangle$ receive.
		              \begin{instructions}
		                \item Execute \procedurehr{thu2208191432} on $\langle-f,e_1,\frac{1-e_1}{2}\rangle$ and let $\langle q_{31},q_{32}\rangle$ receive.
		                \item Execute \procedurehr{tue2008191001} on $\langle-1,q_{13},q_{14}\rangle$ and let $\langle q_{33},q_{34}\rangle$ receive.
		                \item Let $q_{35}(x,n)$ be the following procedure:
		                \begin{instructions}
		                  \item \textbf{Hence show that $\lVert-(1+(-2)(x+1)^{-1})\rVert^2=\lVert\frac{x-1}{x+1}\rVert^2\le{e_1}^2$ using procedure $q_9$.}
		                \end{instructions}
		              \end{instructions}
		            \end{instructions}
		          \end{instructions}
		        \end{instructions}
		    \procedure{thu2208191859}
		      \objective
		        Choose a complex number $f$ and two rational numbers $X,\epsilon$ such that $0<\epsilon<1$ and $X\ge 0$. The objective of the following instructions is to construct rational numbers $c,d,a,b,e$, a procedure $p(x,n)$ to show that $\lVert fx_n^{f-1}\rVert^2\le c^2$ when a complex number $x$ and a positive integer $n$ such that $\Re(x)\ge\epsilon$, $\lVert x\rVert^2\le X^2$, and $n>d$ are chosen, and a procedure $q(x,n,\delta)$ to show that $\Delta_{y=x}^{+\delta}x_n^f\equiv fx_n^{f-1}\err{\frac{a}{n}+b\{\delta\}}$ when in addition a complex number $\delta$ such that $0<\lVert\delta\rVert^2\le e^2$ is chosen.
		      \implementation
		        \begin{instructions}
		          \item Execute \procedurehr{wed2108191324} on $\langle X,\epsilon\rangle$ and let $\langle a_1,p_1\rangle$ receive.
		          \item Execute \procedurehr{wed2407191521} on $\langle\{f\}+1,a_1\rangle$ and let $\langle a_2,b_2,p_2\rangle$ receive.
		          \item Execute \procedurehr{wed2407191611} on $\langle\{f\}+1,a_1\rangle$ and let $\langle a_3,p_3\rangle$ receive.
		          \item Execute \procedurehr{wed2407191422} on $\langle b_2,1\rangle$ and let $\langle a_4,b_4,p_4\rangle$ receive.
		          \item Execute \procedurehr{thu2208191610} on $\langle f,X,\epsilon\rangle$ and let $\langle p_5,p_6\rangle$ receive.
		          \item Let $t$ be \subprocedurehr{wed2808191509}.
		          \item \textbf{Execute \procedurehr{fri2308191803} on $\langle t,p_5,p_6\rangle$ and let $\langle c,d,a,b,e,p,q\rangle$ receive.}
		        \end{instructions}
		      \subprocedure{wed2808191509}
		        \subobjective
		          Choose a complex number $f$ and two rational numbers $X,\epsilon$ such that $0<\epsilon<1$ and $X\ge 0$. Let $g(f,x,n)$ be a shorthand for $[2f(1+\frac{x-1}{x+1})_n^f\cdot(1-\frac{x-1}{x+1})_n^{-f-1}(x+1)^{-2}]+[2f(1+\frac{x-1}{x+1})_n^{f-1}\cdot(1-\frac{x-1}{x+1})_n^{-f}(x+1)^{-2}]$. The objective of the following instructions is to construct a rational number $h$, and a procedure $t(x,n)$ to show that $g(f,x,n)\equiv fx_n^{f-1}\err{\frac{h}{n}}$ when a complex number $x$ and a positive integer $n$ such that $\Re(x)\ge\epsilon$, $\lVert x\rVert^2\le X^2$, and $n>d$ are chosen.
		        \subimplementation
		          \begin{enumerate}
		            \item Let $h=\{f\}a_2a_3a_4((\frac{2}{1+\epsilon})^2+1)$.
		            \item Let $t(x,n)$ be the following procedure:
		            \begin{enumerate}
		              \item Show that $\lVert\frac{1}{(x+1)^2}\rVert^2$
		              \begin{enumerate}
		                \item $=\frac{1}{\lVert 1+x\rVert^4}$
		                \item $=\frac{1}{(\Re(1+x)^2+\Im(x)^2)^2}$
		                \item $\le(\frac{1}{1+\epsilon})^4$.
		              \end{enumerate}
		              \item Using procedures $p_1$,$p_2$,$p_3$,$p_4$, show that $[2f(1+\frac{x-1}{x+1})_n^f\cdot(1-\frac{x-1}{x+1})_n^{-f-1}(x+1)^{-2}]+[2f(1+\frac{x-1}{x+1})_n^{f-1}\cdot(1-\frac{x-1}{x+1})_n^{-f}(x+1)^{-2}]$
		              \begin{enumerate}
		                \item $\equiv[2f(1+\frac{x-1}{x+1})_n^{f-1}(1+\frac{x-1}{x+1})_n^1\cdot(1-\frac{x-1}{x+1})_n^{-f-1}(x+1)^{-2}]+[2f(1+\frac{x-1}{x+1})_n^{f-1}\cdot(1-\frac{x-1}{x+1})_n^{-f}(x+1)^{-2}]$
		                \begin{enumerate}
		                  \item $\err{2\{f\}a_2{b_2}^na_3(\frac{1}{1+\epsilon})^2}$
		                  \item $\err{\frac{2\{f\}a_2a_3a_4}{n}(\frac{1}{1+\epsilon})^2}$
		                \end{enumerate}
		                \item $\equiv[2f(1+\frac{x-1}{x+1})_n^{f-1}(1+\frac{x-1}{x+1})_n^1\cdot(1-\frac{x-1}{x+1})_n^{-f-1}(x+1)^{-2}]+[2f(1+\frac{x-1}{x+1})_n^{f-1}\cdot(1-\frac{x-1}{x+1})_n^{-f-1}(1-\frac{x-1}{x+1})_n^{1}(x+1)^{-2}]$
		                \begin{enumerate}
		                  \item $\err{2\{f\}a_3a_2{b_2}^n(\frac{1}{1+\epsilon})^2}$
		                  \item $\err{\frac{2\{f\}a_3a_2a_4}{n}(\frac{1}{1+\epsilon})^2}$
		                \end{enumerate}
		                \item $=2f(1+\frac{x-1}{x+1})_n^{f-1}(1-\frac{x-1}{x+1})_n^{-f-1}[(1+\frac{x-1}{x+1})^1\cdot(x+1)^{-2}+(1-\frac{x-1}{x+1})^{1}(x+1)^{-2}]$
		                \item $=4f(1+\frac{x-1}{x+1})_n^{f-1}(1-\frac{x-1}{x+1})_n^{-f-1}[(x+1)^{-2}]$
		                \item $=f(1+\frac{x-1}{x+1})_n^{f-1}(1-\frac{x-1}{x+1})_n^{-f-1}(1-\frac{x-1}{x+1})_n^{2}$
		                \item $\equiv f(1+\frac{x-1}{x+1})_n^{f-1}(1-\frac{x-1}{x+1})_n^{-(f-1)}$
		                \begin{enumerate}
		                  \item $\err{\{f\}a_3a_2{b_2}^n}$
		                  \item $\err{\frac{\{f\}a_3a_2a_4}{n}}$
		                \end{enumerate}
		                \item $=fx_n^{f-1}$
		              \end{enumerate}
		              \item \textbf{Hence show that $[2f(1+\frac{x-1}{x+1})_n^f\cdot(1-\frac{x-1}{x+1})_n^{-f-1}(x+1)^{-2}]+[2f(1+\frac{x-1}{x+1})_n^{f-1}\cdot(1-\frac{x-1}{x+1})_n^{-f}(x+1)^{-2}]\equiv fx_n^{f-1}\err{\frac{h}{n}}$.}
		            \end{enumerate}
		            \item \textbf{Yield the tuple $\langle h,t\rangle$.}
		          \end{enumerate}
		    \end{multicols}
	    \mfigure
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            \filldraw (3.100000,1.760666) circle (1pt);
            \filldraw (3.000000,0.288676) circle (1pt);
            \draw (3.000000,0.288676) -- (3.100000,0.283983);
            \filldraw (3.100000,0.283983) circle (1pt);
            \filldraw (3.100000,1.760666) circle (1pt);
            \draw (3.100000,1.760666) -- (3.200000,1.788832);
            \filldraw (3.200000,1.788832) circle (1pt);
            \filldraw (3.100000,0.283983) circle (1pt);
            \draw (3.100000,0.283983) -- (3.200000,0.279511);
            \filldraw (3.200000,0.279511) circle (1pt);
            \filldraw (3.200000,1.788832) circle (1pt);
            \draw (3.200000,1.788832) -- (3.300000,1.816559);
            \filldraw (3.300000,1.816559) circle (1pt);
            \filldraw (3.200000,0.279511) circle (1pt);
            \draw (3.200000,0.279511) -- (3.300000,0.275244);
            \filldraw (3.300000,0.275244) circle (1pt);
            \filldraw (3.300000,1.816559) circle (1pt);
            \draw (3.300000,1.816559) -- (3.400000,1.843865);
            \filldraw (3.400000,1.843865) circle (1pt);
            \filldraw (3.300000,0.275244) circle (1pt);
            \draw (3.300000,0.275244) -- (3.400000,0.271167);
            \filldraw (3.400000,0.271167) circle (1pt);
            \filldraw (3.400000,1.843865) circle (1pt);
            \draw (3.400000,1.843865) -- (3.500000,1.870770);
            \filldraw (3.500000,1.870770) circle (1pt);
            \filldraw (3.400000,0.271167) circle (1pt);
            \draw (3.400000,0.271167) -- (3.500000,0.267266);
            \filldraw (3.500000,0.267266) circle (1pt);
            \filldraw (3.500000,1.870770) circle (1pt);
            \draw (3.500000,1.870770) -- (3.600000,1.897289);
            \filldraw (3.600000,1.897289) circle (1pt);
            \filldraw (3.500000,0.267266) circle (1pt);
            \draw (3.500000,0.267266) -- (3.600000,0.263530);
            \filldraw (3.600000,0.263530) circle (1pt);
            \filldraw (3.600000,1.897289) circle (1pt);
            \draw (3.600000,1.897289) -- (3.700000,1.923437);
            \filldraw (3.700000,1.923437) circle (1pt);
            \filldraw (3.600000,0.263530) circle (1pt);
            \draw (3.600000,0.263530) -- (3.700000,0.259946);
            \filldraw (3.700000,0.259946) circle (1pt);
            \filldraw (3.700000,1.923437) circle (1pt);
            \draw (3.700000,1.923437) -- (3.800000,1.949228);
            \filldraw (3.800000,1.949228) circle (1pt);
            \filldraw (3.700000,0.259946) circle (1pt);
            \draw (3.700000,0.259946) -- (3.800000,0.256505);
            \filldraw (3.800000,0.256505) circle (1pt);
            \filldraw (3.800000,1.949228) circle (1pt);
            \draw (3.800000,1.949228) -- (3.900000,1.974676);
            \filldraw (3.900000,1.974676) circle (1pt);
            \filldraw (3.800000,0.256505) circle (1pt);
            \draw (3.800000,0.256505) -- (3.900000,0.253198);
            \filldraw (3.900000,0.253198) circle (1pt);
            \filldraw (3.900000,1.974676) circle (1pt);
            \draw (3.900000,1.974676) -- (4.000000,1.999792);
            \filldraw (4.000000,1.999792) circle (1pt);
            \filldraw (3.900000,0.253198) circle (1pt);
            \draw (3.900000,0.253198) -- (4.000000,0.250016);
            \filldraw (4.000000,0.250016) circle (1pt);
			    \end{tikzpicture}
		    \end{minipage}
		    \begin{minipage}[c]{0.35\textwidth}
			    A plot of the lists of complex numbers $(\frac{[0:40]}{10})_{15}^{\frac{1}{2}}$ and $\frac{1}{2}(\frac{[0:40]}{10})_{15}^{\frac{1}{2}-1}$. Note that the steepness of $(\frac{[0:40]}{10})_{15}^{\frac{1}{2}}$ is approximately given by the y-coordinates of $\frac{1}{2}(\frac{[0:40]}{10})_{15}^{\frac{1}{2}-1}$. That is, where the graph of $(\frac{[0:40]}{10})_{15}^{\frac{1}{2}}$ is rapidly increasing, the graph of $\frac{1}{2}(\frac{[0:40]}{10})_{15}^{\frac{1}{2}-1}$ has a relatively large positive value, and where the graph of $(\frac{[0:40]}{10})_{15}^{\frac{1}{2}}$ flattens out, the graph of $\frac{1}{2}(\frac{[0:40]}{10})_{15}^{\frac{1}{2}-1}$ has a relatively small positive value.
		    \end{minipage}
	  \chapter{Integral Arithmetic}
	    \begin{multicols*}{2}
	      \declaration{3.30}
			    The notation \gls{differentialintegral}, where:
			    \begin{enumerate}
			      \item $f(r,\delta_r)$ is a procedure to construct a complex number when complex numbers $r,\delta_r$ such that $P(r,\delta_r)$ are chosen
			      \item $R$ is a non-empty list of complex numbers such that $P(R_t,R_{t+1}-R_t)$ for $t\in[0:\lvert R\rvert-1]$
			    \end{enumerate}
			    will be used as a shorthand for $\sum_t^{[0:\lvert R\rvert-1]}f(R_t,R_{t+1}-R_{t})$.
		    \procedure{3.86}
			    \objective
				    Choose the following:
				    \begin{enumerate}
				      \item A procedure $f(r,\delta)$ to construct a complex number when complex numbers $r,\delta_r$ such that $P(r,\delta_r)$ are chosen.
				      \item A procedure $g(r,\delta)$ to construct a complex number when complex numbers $r,\delta_r$ such that $Q(r,\delta_r)$ are chosen.
				      \item A non-empty list of complex numbers $R$ such that $P(R_t,R_{t+1}-R_t)$ and $Q(R_t,R_{t+1}-R_t)$ for $t\in[0:\lvert R\rvert-1]$.
				    \end{enumerate}
				    The objective of the following instructions is to show that $\int_r^R(f(r,\delta_r)+g(r,\delta_r))=\int_r^Rf(r,\delta_r)+\int_r^Rg(r,\delta_r)$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\int_r^R(f(r,\delta_r)+g(r,\delta_r))$
					    \begin{enumerate}
						    \item $=\sum_t^{[0:\lvert R\rvert-1]}(f(R_t,R_{t+1}-R_t)+g(R_t,R_{t+1}-R_t))$
						    \item $=\sum_t^{[0:\lvert R\rvert-1]}f(R_t,R_{t+1}-R_t)+\sum_t^{[0:\lvert R\rvert-1]}g(R_t,R_{t+1}-R_t)$
						    \item $=\int_r^Rf(r,\delta_r)+\int_r^Rg(r,\delta_r)$
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{3.87}
			    \objective
				    Choose the following:
				    \begin{enumerate}
				      \item A complex number $a$.
				      \item A procedure $f(r,\delta)$ to construct a complex number when complex numbers $r,\delta_r$ such that $P(r,\delta_r)$ are chosen.
				      \item A non-empty list of complex numbers $R$ such that $P(R_t, R_{t+1}-R_t)$ for $t\in[0:\lvert R\rvert-1]$.
				    \end{enumerate}
				     The objective of the following instructions is to show that $\int_r^Raf(r,\delta_r)=a\int_r^Rf(r,\delta_r)$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\int_r^Raf(r,\delta_r)$
					    \begin{enumerate}
						    \item $=\sum_t^{[0:\lvert R\rvert-1]}af(R_t,R_{t+1}-R_t)$
						    \item $=a\sum_t^{[0:\lvert R\rvert-1]}f(R_t,R_{t+1}-R_t)$
						    \item $=a\int_r^Rf(r,\delta_r)$
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{3.88}
			    \objective
				    Choose the following:
				    \begin{enumerate}
				      \item A procedure $f(r)$ to construct a complex number when a complex number $r$ such that $P(r)$ is chosen.
				      \item A non-empty list of complex numbers $R$ such that $P(R_t)$ for $t\in[0:\lvert R\rvert-1]$.
				      \item A non-empty list of complex numbers $S$ such that $P(S_t)$ for $t\in[0:\lvert R\rvert-1]$ and $R_{\lvert R\rvert-1}=S_0$.
				    \end{enumerate}
				    The objective of the following instructions is to show that $\int_r^{R^\frown S}f(r)\delta_r=\int_r^Rf(r)\delta_r+\int_r^Sf(r)\delta_r$.
			    \implementation
				    \begin{enumerate}
					    \item Let $T=R^\frown S$.
					    \item Show that $\int_r^{T}f(r)\delta_r$
					    \begin{enumerate}
						    \item $=\sum_t^{[0:\lvert T\rvert-1]}f(T_t)(T_{t+1}-T_t)$
						    \item $=\sum_t^{[0:\lvert R\rvert-1]}f(T_t)(T_{t+1}-T_t)+\sum_t^{[\lvert R\rvert-1:\lvert R\rvert]}f(T_t)(T_{t+1}-T_t)+\sum_t^{[\lvert R\rvert:\lvert T\rvert-1]}f(T_t)(T_{t+1}-T_t)$
						    \item $=\sum_t^{[0:\lvert R\rvert-1]}f(R_t)(R_{t+1}-R_t)+f(T_{\lvert R\rvert-1})(T_{\lvert R\rvert}-T_{\lvert R\rvert-1})+\sum_t^{[\lvert R\rvert:\lvert T\rvert-1]}f(S_{t-\lvert R\rvert})\cdot(S_{t+1-\lvert R\rvert}-S_{t-\lvert R\rvert})$
						    \item $=\sum_t^{[0:\lvert R\rvert-1]}f(R_t)(R_{t+1}-R_t)+f(T_{\lvert R\rvert-1})(S_0-R_{\lvert R\rvert-1})+\sum_t^{[0:\lvert S\rvert-1]}f(S_{t})(S_{t+1}-S_{t})$
						    \item $=\int_r^Rf(r)\delta_r+\int_r^Sf(r)\delta_r$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{3.34}
			    \objective
				    Choose the following:
				    \begin{enumerate}
				      \item A procedure $f(r)$ to construct a complex number when a complex number $r$ such that $P(r)$ is chosen.
				      \item A non-empty list of complex numbers $R$ such that $P(R_t)$ for $t\in[0:\lvert R\rvert-1]$.
				    \end{enumerate}
				    The objective of the following instructions is to show that $\int_r^R\delta_r\diff_{z=r}^{+\delta_r}f(z)=f(R_{\lvert R\rvert-1})-f(R_0)$.
			    \implementation
				    \begin{enumerate}
					    \item Show that $\int_r^R\delta_r\diff_{z=r}^{\delta_r}f(z)$
					    \begin{enumerate}
						    \item $=\int_r^R\delta_r(\frac{f(r+\delta_r)-f(r)}{\delta_r})$
						    \item $=\int_r^R(f(r+\delta_r)-f(r))$
						    \item $=\sum_k^{[0:\lvert R\rvert-1]}(f(R_{k+1})-f(R_k))$
						    \item $=f(R_{\lvert R\rvert-1})-f(R_0)$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{3.31}
			    The notation \gls{differentialDeltaX}, where $X$ is a list, will be used as a shorthand for $\langle X_1-X_0,X_2-X_1,\cdots,X_{\lvert X\rvert-1}-X_{\lvert X\rvert-2}\rangle$.
		    \procedure{fri3008190328}
		      \objective
		        Choose the following:
		        \begin{enumerate}
		          \item A non-negative rational number $A$.
				      \item A procedure $q_1(x,n,\delta)$ to show that $\Delta_{y=x}^{+\delta}f_n(y)\equiv f'_n(x)\err{\frac{a_1}{n}+b_1\{\delta\}}$ when two complex numbers $x,\delta$ and a positive integer $n$ such that $P(x)$, $n>c_1$, and $0<\lVert\delta\rVert^2<{d_1}^2$ are chosen.
				    \end{enumerate}
				    The objective of the following instructions is to construct the following:
				    \begin{enumerate}
				      \item Non-negative rational numbers $a,b,c,d$.
				      \item A procedure $p(R,n)$ to show that $\int_r^Rf'_n(r)\delta_r\equiv f_n(R_{\lvert R\rvert-1})-f_n(R_0)\err{\frac{a}{n}+b\max(\{\Delta R\})}$ when an integer $n$ and a non-empty list of complex numbers $R$ such that $P(R_t)$ and $0<\lVert R_{t+1}-R_t\rVert^2<{d}^2$ for $t\in[0:\lvert R\rvert-1]$, $\int_r^R\{\delta_r\}\le A$, and $n>c$ are chosen.
				    \end{enumerate}
		      \implementation
		        \begin{enumerate}
		          \item Let $a=a_1A$.
		          \item Let $b=b_1A$.
		          \item Let $c=c_1$.
		          \item Let $d=d_1$.
		          \item Let $p(R,n)$ be the following procedure:
		          \begin{enumerate}
		            \item Using procedure $q_1$, show that $\int_r^Rf'_n(r)\delta_r$
		            \begin{enumerate}
		              \item $=\sum_k^{[0:\lvert R\rvert-1]}f'_n(R_k)(R_{k+1}-R_k)$
		              \item $\equiv\sum_k^{0:\lvert R\rvert-1}\Delta_{y=R_k}^{R_{k+1}-R_k}f_n(y)(R_{k+1}-R_k)$
		              \begin{enumerate}
		                \item $\err{\sum_k^{0:\lvert R\rvert-1}(\frac{a_1}{n}+b_1\{R_{k+1}-R_k\})\{R_{k+1}-R_k\}}$
		                \item $\err{(\frac{a_1}{n}+b_1\max(\{\Delta R\}))\sum_k^{0:\lvert R\rvert-1}\{R_{k+1}-R_k\}}$
		                \item $\err{(\frac{a_1}{n}+b_1\max(\{\Delta R\}))\int_r^R\{\delta_r\}}$
		                \item $\err{(\frac{a_1}{n}+b_1\max(\{\Delta R\}))A}$
		                \item $\err{\frac{a}{n}+b\max(\{\Delta R\})}$
		              \end{enumerate}
		              \item $=\sum_k^{0:\lvert R\rvert-1}\frac{f_n(R_k+(R_{k+1}-R_k))-f_n(R_k)}{R_{k+1}-R_k}\cdot(R_{k+1}-R_k)$
		              \item $=\sum_k^{0:\lvert R\rvert-1}(f_n(R_{k+1})-f_n(R_k))$
		              \item $=f_n(R_{\lvert R\rvert-1})-f_n(R_0)$.
		            \end{enumerate}
		            \item \textbf{Therefore show that $\int_r^Rf'_n(r)\delta_r\equiv f_n(R_{\lvert R\rvert-1})-f_n(R_0)\err{\frac{a}{n}+b\max(\{\Delta R\})}$.}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle a,b,c,d,p\rangle$.}
		        \end{enumerate}
		    \procedure{fri3008190457}
		      \objective
		        Choose the following:
		        \begin{enumerate}
		          \item A non-negative rational number $A$.
				      \item A procedure $q_1(x,n,\delta)$ to show that $\Delta_{y=x}^{+\delta}g_n(y)\equiv g'_n(x)\err{\frac{a_1}{n}+b_1\{\delta\}}$ when two complex numbers $x,\delta$ and a positive integer $n$ such that $P(x)$, $n>c_1$, and $0<\lVert\delta\rVert^2<{d_1}^2$ are chosen.
				      \item A procedure $q_2(x,n)$ to show that $f_n(x)\equiv 0\err{a_2}$ when a complex number $x$ and a positive integer $n$ such that $Q(x)$ and $n>b_2$ are chosen.
				      \item A procedure $q_3(x,n)$ to show that $Q(g_n(x))$ when a complex number $x$ and a positive integer $n$ such that $P(x)$ and $n>c_1$ are chosen
				    \end{enumerate}
				    The objective of the following instructions is to construct the following:
				    \begin{enumerate}
				      \item Non-negative rational numbers $a,b,c,d$.
				      \item A procedure $p(R,n)$ to show that $\int_r^{g(R)}f_n(r)\delta_r\equiv\int_r^Rf_n(g_n(r))g'_n(r)\delta_r\err{\frac{a}{n}+b\max(\{\Delta R\})}$ when an integer $n$ and a non-empty list of complex numbers $R$ such that $P(R_t)$ and $0<\lVert R_{t+1}-R_t\rVert^2<{d}^2$ for $t\in[0:\lvert R\rvert-1]$, $\int_r^R\{\delta_r\}\le A$, and $n>c$ are chosen.
				    \end{enumerate}
			    \implementation
			      \begin{enumerate}
			        \item Let $a=a_1a_2A$.
			        \item Let $b=b_1a_2A$.
			        \item Let $c=\max(c_1,b_2)$.
			        \item Let $d=d_1$.
			        \item Let $p(R,n)$ be the following procedure:
			        \begin{enumerate}
			          \item Using procedures $q_1,q_2,q_3$, show that $\int_r^{g_n(R)}f_n(r)\delta_r$
			          \begin{enumerate}
			            \item $=\sum_k^{[0:\lvert R\rvert-1]}f_n(g_n(R_k))(g_n(R_{k+1})-g_n(R_k))$
			            \item $=\sum_k^{[0:\lvert R\rvert-1]}f_n(g_n(R_k))\Delta_{y=R_k}^{R_{k+1}-R_k}g_n(y)(R_{k+1}-R_k)$
			            \item $\equiv\sum_k^{[0:\lvert R\rvert-1]}f_n(g_n(R_k))g'_n(R_k)(R_{k+1}-R_k)$
			            \begin{enumerate}
			              \item $\err{\sum_k^{[0:\lvert R\rvert-1]}a_2(\frac{a_1}{n}+b_1\{R_{k+1}-R_k\})\{R_{k+1}-R_k\}}$
			              \item $\err{a_2(\frac{a_1}{n}+b_1\max(\{\Delta R\}))\sum_k^{[0:\lvert R\rvert-1]}\{R_{k+1}-R_k\}}$
			              \item $\err{a_2(\frac{a_1}{n}+b_1\max(\{\Delta R\}))\int_r^R\{\delta_r\}}$
			              \item $\err{a_2(\frac{a_1}{n}+b_1\max(\{\Delta R\}))A}$
			              \item $\err{\frac{a}{n}+b\max(\{\Delta R\})}$
			            \end{enumerate}
			            \item $=\int_r^Rf_n(g_n(r))g'_n(r)\delta_r$.
			          \end{enumerate}
			          \item \textbf{Hence show that $\int_r^{g_n(R)}f_n(r)\delta_r\equiv\int_r^Rf_n(g_n(r))g'_n(r)\delta_r\err{\frac{a}{n}+b\max(\{\Delta R\})}$.}
			        \end{enumerate}
			        \item \textbf{Yield the tuple $\langle a,b,c,d,p\rangle$.}
			      \end{enumerate}
		    \procedure{fri3008190709}
		      \objective
		        Choose three rational numbers $A,X,\epsilon$ such that $0<\epsilon<1$ and $X\ge 0$. The objective of the following instructions is to construct the following:
				    \begin{enumerate}
				      \item Non-negative rational numbers $a,b,c,d$.
				      \item A procedure $p(R,n)$ to show that $\int_r^R\frac{\delta_r}{r}\equiv\ln_n(R_{\lvert R\rvert-1})\err{\frac{a}{n}+b\max(\{\Delta R\})}$ when an integer $n$ and a non-empty list of complex numbers $R$ such that $\Re(R_t)\ge\epsilon$, $\lVert R_t\rVert^2\le X^2$ and $0<\lVert R_{t+1}-R_t\rVert^2<{d}^2$ for $t\in[0:\lvert R\rvert-1]$, $R_0=1$, $\int_r^R\{\delta_r\}\le A$, and $n>c$ are chosen.
				    \end{enumerate}
		      \implementation
		        \begin{enumerate}
		          \item Execute \procedurehr{wed2108191603} on $\langle X,\epsilon\rangle$ and let $\langle\cdots,q\rangle$ receive.
		          \item Hence execute \procedurehr{fri3008190328} on $\langle A,q\rangle$ and let $\langle a,b,c,d,t\rangle$ receive.
		          \item Let $p(R,n)$ be the following procedure:
		          \begin{enumerate}
		            \item Using procedure $t$, show that $\int_r^R\frac{\delta_r}{r}$
		            \begin{enumerate}
		              \item $\equiv\ln_n(R_{\lvert R\rvert-1})-\ln_n(R_0)\err{\frac{a}{n}+b\max(\{\Delta R\})}$
		              \item $=\ln_n(R_{\lvert R\rvert-1})-\ln_n(1)$
		              \item $=\ln_n(R_{\lvert R\rvert-1})$.
		            \end{enumerate}
		            \item \textbf{Hence show that $\int_r^R\frac{\delta_r}{r}\equiv\ln_n(R_{\lvert R\rvert-1})\err{\frac{a}{n}+b\max(\{\Delta R\})}$.}
		          \end{enumerate}
		          \item \textbf{Yield the tuple $\langle a,b,c,d,p\rangle$.}
		        \end{enumerate}
	    \end{multicols*}
	  \clearpage
	\part{Matrix Arithmetic}
	  \chapter{Matrix Arithmetic}
	    \begin{multicols*}{2}
		    \declaration{4.28}
			    The phrase "\gls{matrix}" will be used as a shorthand for a list of equally lengthed lists of polynomials. In particular, the phrase "$m\times n$ matrix" will be used as a shorthand for a length-$m$ list of length-$n$ lists of polynomials.
		    \declaration{4.29}
			    The notation \gls{matrixentry}, where $A$ is a matrix and $I,J$ are lists of indicies, will be used as a shorthand for $\langle(A_j)_J\for j\in I\rangle$.
		    \declaration{4.30}
			    The phrase "\gls{matrixA=B}", where $A,B$ are $m\times n$ matrices, will be used as a shorthand for "$A_{i,j}=B_{i,j}$ for $j\in[0:n]$, for $i\in[0:m]$".
		    \procedure{4.73}
			    \objective
				    Choose an $m\times n$ matrix $A$. The objective of the following instructions is to show that $A=A$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that $A_{i,j}=A_{i,j}$ for $j\in[0:n]$, for $i\in[0:m]$.
					    \item \textbf{Hence verify that $A=A$.}
				    \end{enumerate}
		    \procedure{4.74}
			    \objective
				    Choose two $m\times n$ matrices $A,B$ such that $A=B$. The objective of the following instructions is to show that $B=A$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that $A_{i,j}=B_{i,j}$ for $j\in[0:n]$, for $i\in[0:m]$.
					    \item Hence verify that $B_{i,j}=A_{i,j}$ for $j\in[0:n]$, for $i\in[0:m]$.
					    \item \textbf{Hence verify that $B=A$.}
				    \end{enumerate}
		    \procedure{4.75}
			    \objective
				    Choose three $m\times n$ matrices $A,B,C$ such that $A=B$ and $B=C$. The objective of the following instructions is to show that $A=C$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that $A_{i,j}=B_{i,j}$ for $j\in[0:n]$, for $i\in[0:m]$.
					    \item Verify that $B_{i,j}=C_{i,j}$ for $j\in[0:n]$, for $i\in[0:m]$.
					    \item Hence verify that $A_{i,j}=C_{i,j}$ for $j\in[0:n]$, for $i\in[0:m]$.
					    \item \textbf{Hence verify that $A=C$.}
				    \end{enumerate}
		    \declaration{4.31}
			    The notation \gls{matrixA+B}, where $A,B$ are $m\times n$ matrices, will be used as a shorthand for the list $\langle\langle A_{i,j}+B_{i,j}\for j\in[0:n]\rangle\for i\in[0:m]\rangle$.
		    \procedure{4.76}
			    \objective
				    Choose four $m\times n$ matrices $A,B,C,D$ such that $A=C$ and $B=D$. The objective of the following instructions is to show that $A+B=C+D$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that $A_{i,j}=C_{i,j}$ for $j\in[0:n]$, for $i\in[0:m]$.
					    \item Verify that $B_{i,j}=D_{i,j}$ for $j\in[0:n]$, for $i\in[0:m]$.
					    \item Hence verify that $A+B$
					    \begin{enumerate}
						    \item $=\langle\langle A_{i,j}+B_{i,j}\for j\in[0:n]\rangle\for i\in[0:m]\rangle$
						    \item $=\langle\langle C_{i,j}+D_{i,j}\for j\in[0:n]\rangle\for i\in[0:m]\rangle$
						    \item $=C+D$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{4.77}
			    \objective
				    Choose three $m\times n$ matrices $A,B,C$. The objective of the following instructions is to show that $(A+B)+C=A+(B+C)$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that $(A+B)+C$
					    \begin{enumerate}
						    \item $=\langle\langle (A+B)_{i,j}+C_{i,j}\for j\in[0:n]\rangle\for i\in[0:m]\rangle$
						    \item $=\langle\langle (A_{i,j}+B_{i,j})+C_{i,j}\for j\in[0:n]\rangle\for i\in[0:m]\rangle$
						    \item $=\langle\langle A_{i,j}+(B_{i,j}+C_{i,j})\for j\in[0:n]\rangle\for i\in[0:m]\rangle$
						    \item $=\langle\langle A_{i,j}+(B+C)_{i,j}\for j\in[0:n]\rangle\for i\in[0:m]\rangle$
						    \item $=A+(B+C)$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{4.78}
			    \objective
				    Choose two $m\times n$ matrices $A,B$. The objective of the following instructions is to show that $A+B=B+A$.
			    \implementation
				    \begin{enumerate}
					    \item $A+B$
					    \begin{enumerate}
						    \item $=\langle\langle A_{i,j}+B_{i,j}\for j\in[0:n]\rangle\for i\in[0:m]\rangle$
						    \item $=\langle\langle B_{i,j}+A_{i,j}\for j\in[0:n]\rangle\for i\in[0:m]\rangle$
						    \item $=B+A$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{4.32}
			    The notation \gls{matrix0mn} will contextually be used as a shorthand for the list $\langle\langle 0\for j\in[0:n]\rangle\for i\in[0:m]\rangle$ where the natural numbers $m,n$ are determined by the context.
		    \procedure{4.79}
			    \objective
				    Choose an $m\times n$ matrix $A$. The objective of the following instructions is to show that $0+A=A$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that $0+A$
					    \begin{enumerate}
						    \item $=0_{m\times n}+A$
						    \item $=\langle\langle 0_{i,j}+A_{i,j}\for j\in[0:n]\rangle\for i\in[0:m]\rangle$
						    \item $=\langle\langle 0+A_{i,j}\for j\in[0:n]\rangle\for i\in[0:m]\rangle$
						    \item $=\langle\langle A_{i,j}\for j\in[0:n]\rangle\for i\in[0:m]\rangle$
						    \item $=A$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{4.33}
			    The notation \gls{matrix-A}, where $A$ is an $m\times n$ matrix, will be used as a shorthand for the list $\langle\langle -A_{i,j}\for j\in[0:n]\rangle\for i\in[0:m]\rangle$.
		    \procedure{4.80}
			    \objective
				    Choose two $m\times n$ matrices $A,B$ such that $A=B$. The objective of the following instructions is to show that $-A=-B$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that $A_{i,j}=B_{i,j}$ for $j\in[0:n]$, for $i\in[0:m]$.
					    \item Hence verify that $-A$
					    \begin{enumerate}
						    \item $=\langle\langle -A_{i,j}\for j\in[0:n]\rangle\for i\in[0:m]\rangle$
						    \item $=\langle\langle -B_{i,j}\for j\in[0:n]\rangle\for i\in[0:m]\rangle$
						    \item $=-B$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{4.81}
			    \objective
				    Choose a $m\times n$ matrix $A$. The objective of the following instructions is to show that $-A+A=0$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that $-A+A$
					    \begin{enumerate}
						    \item $\langle\langle (-A)_{i,j}+A_{i,j}\for j\in[0:n]\rangle\for i\in[0:m]\rangle$
						    \item $\langle\langle -(A_{i,j})+A_{i,j}\for j\in[0:n]\rangle\for i\in[0:m]\rangle$
						    \item $\langle\langle 0\for j\in[0:n]\rangle\for i\in[0:m]\rangle$
						    \item $=0$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{4.34}
			    The notation \gls{matrixAB}, where $A$ is an $m\times n$ matrix and $B$ is an $n\times k$ matrix, will be used as a shorthand for the list $\langle\langle\sum_r^{[0:n]}A_{i,r}B_{r,j}\for j\in[0:k]\rangle\for i\in[0:m]\rangle$.
		    \procedure{4.82}
			    \objective
				    Choose two $m\times n$ matrices $A,C$ and two $n\times k$ matrices $B,D$ such that $A=C$ and $B=D$. The objective of the following instructions is to show that $AB=CD$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that $A_{i,j}=C_{i,j}$ for $j\in[0:n]$, for $i\in[0:m]$.
					    \item Verify that $B_{i,j}=D_{i,j}$ for $j\in[0:k]$, for $i\in[0:n]$.
					    \item Hence verify that $AB$
					    \begin{enumerate}
						    \item $=\langle\langle\sum_r^{[0:n]}A_{i,r}B_{r,j}\for j\in[0:k]\rangle\for i\in[0:m]\rangle$
						    \item $=\langle\langle\sum_r^{[0:n]}C_{i,r}D_{r,j}\for j\in[0:k]\rangle\for i\in[0:m]\rangle$
						    \item $=CD$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{4.02}
			    \objective
				    Choose an $m\times n$ matrix, A, an $n\times p$ matrix, B, and a $p\times q$ matrix, C. The objective of the following instructions is to show that $(AB)C=A(BC)$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that $(AB)C$
					    \begin{enumerate}
						    \item $=\langle\langle\sum_r^{[0:p]}(AB)_{i,r}C_{r,j}\for j\in[0:q]\rangle\for i\in[0:m]\rangle$
						    \item $=\langle\langle\sum_r^{[0:p]}(\sum_l^{[0:n]}A_{i,l}B_{l,r})C_{r,j}\for j\in[0:q]\rangle\for i\in[0:m]\rangle$
						    \item $=\langle\langle\sum_r^{[0:p]}\sum_l^{[0:n]}A_{i,l}B_{l,r}C_{r,j}\for j\in[0:q]\rangle\for i\in[0:m]\rangle$
						    \item $=\langle\langle\sum_l^{[0:n]}\sum_r^{[0:p]}A_{i,l}B_{l,r}C_{r,j}\for j\in[0:q]\rangle\for i\in[0:m]\rangle$
						    \item $=\langle\langle\sum_l^{[0:n]}A_{i,l}\sum_r^{[0:p]}B_{l,r}C_{r,j}\for j\in[0:q]\rangle\for i\in[0:m]\rangle$
						    \item $=\langle\langle\sum_l^{[0:n]}A_{i,l}(BC)_{l,j}\for j\in[0:q]\rangle\for i\in[0:m]\rangle$
						    \item $=A(BC)$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{4.35}
			    The notation \gls{matrixamm}, where $a\ne 0$ is a polynomial, will contextually be used as a shorthand for the list $\langle\langle a[i=j]\for j\in[0:m]\rangle\for i\in[0:m]\rangle$.
		    \procedure{4.84}
			    \objective
				    Choose an $m\times n$ matrix, $A$. The objective of the following instructions is to show that $1A=A$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that $1A$
					    \begin{enumerate}
						    \item $=1_mA$
						    \item $=\langle\langle\sum_r^{[0:m]}1_{i,r}A_{r,j}\for j\in[0:n]\rangle\for i\in[0:m]\rangle$
						    \item $=\langle\langle\sum_r^{[0:m]}[i=r]A_{r,j}\for j\in[0:n]\rangle\for i\in[0:m]\rangle$
						    \item $=\langle\langle A_{i,j}\for j\in[0:n]\rangle\for i\in[0:m]\rangle$
						    \item $=A$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{4.85}
			    \objective
				    Choose an $m\times n$ matrix $A$, and two $n\times k$ matrices $B,C$. The objective of the following instructions is to show that $A(B+C)=AB+AC$.
			    \implementation
				    \begin{enumerate}
					    \item $A(B+C)$
					    \begin{enumerate}
						    \item $=\langle\langle\sum_r^{[0:n]}A_{i,r}(B+C)_{r,j}\for j\in[0:k]\rangle\for i\in[0:m]\rangle$
						    \item $=\langle\langle\sum_r^{[0:n]}A_{i,r}(B_{r,j}+C_{r,j})\for j\in[0:k]\rangle\for i\in[0:m]\rangle$
						    \item $=\langle\langle\sum_r^{[0:n]}(A_{i,r}B_{r,j}+A_{i,r}C_{r,j})\for j\in[0:k]\rangle\for i\in[0:m]\rangle$
						    \item $=\langle\langle\sum_r^{[0:n]}A_{i,r}B_{r,j}+\sum_r^{[0:n]}A_{i,r}C_{r,j}\for j\in[0:k]\rangle\for i\in[0:m]\rangle$
						    \item $=\langle\langle\sum_r^{[0:n]}A_{i,r}B_{r,j}\for j\in[0:k]\rangle\for i\in[0:m]\rangle+\langle\langle\sum_r^{[0:n]}\sum_r^{[0:n]}A_{i,r}C_{r,j}\for j\in[0:k]\rangle\for i\in[0:m]\rangle$
						    \item $=AB+AC$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{4.36}
			    The phrase "row $i$ of $A$" and the notation \gls{matrixrow}, where $A$ is an $m\times n$ matrix and $0\le i<m$, will be used as a shorthand for $A_{i,[0:n]}$.
		    \declaration{4.37}
			    The phrase "column $i$ of $A$" and the notation \gls{matrixcol}, where $A$ is an $m\times n$ matrix and $0\le i<n$, will be used as a shorthand for $A_{[0:m],i}$.
		    \procedure{4.00}
			    \objective
				    Choose an $m\times 2$ matrix, $A$. Let $\deg(0)=\infty$. Let $k=\min(\deg(A_{0,0}),\deg(A_{0,1}))$ and $q=\deg(A_{0,0})$. The objective of the following instructions is to make $A_{0,1}=0$, $\deg(A_{0,0})\le k$, and either leave $A_{*,0}$ unchanged or make $\deg(A_{0,0})<q$ by a sequence of operations whereby, in each step a polynomial times either of the columns is added to the other.
			    \implementation
				    \begin{enumerate}
					    \item Let $A$ be our working matrix.
					    \item While $A_{0,1}\ne 0$, do the following:
					    \begin{enumerate}
						    \item If $\deg(A_{0,0})\le\deg(A_{0,1})$, then:
						    \begin{enumerate}
							    \item Subtract $\frac{(A_{0,1})_{\deg(A_{0,1})}}{(A_{0,0})_{\deg(A_{0,0})}}\lambda^{\deg(A_{0,1})-\deg(A_{0,0})}$ times $A_{0,0}$ from $A_{0,1}$.
							    \item Now verify that either $A_{0,1}$'s degree has decreased or $A_{0,1}=0$.
						    \end{enumerate}
						    \item Otherwise, do the following:
						    \begin{enumerate}
							    \item Let $p=\frac{(A_{0,0})_{\deg(A_{0,0})}}{(A_{0,1})_{\deg(A_{0,1})}}\lambda^{\deg(A_{0,0})-\deg(A_{0,1})}$.
							    \item If $A_{0,0}=pA_{0,1}$, then do the following:
							    \begin{enumerate}
								    \item Add $1-p$ times $A_{0,1}$ to $A_{0,0}$.
								    \item Verify that now $A_{0,0}=A_{0,1}$.
							    \end{enumerate}
							    \item Otherwise, do the following:
							    \begin{enumerate}
								    \item Verify that $A_{0,0}\ne pA_{0,1}$.
								    \item Add $-p$ times $A_{0,1}$ to $A_{0,0}$.
							    \end{enumerate}
							    \item Therefore verify that $A_{0,0}\ne 0$.
							    \item Also verify that $A_{0,0}$'s degree has decreased.
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Verify that $A_{0,1}=0$.}
					    \item Verify that the changes to $A_{0,0}$, if any, have decreased its degree.
					    \item If both operations are well-defined, then do the following:
					    \begin{enumerate}
						    \item Verify that all changes to $A_{0,1}$ but the last have decreased its degree.
						    \item Verify that $\deg(A_{0,0})\le$ the degree of the penultimate value of $A_{0,1}$.
					    \end{enumerate}
					    \item \textbf{Therefore verify that $\deg(A_{0,0})\le k$.}
					    \item If $A_{*,0}$ was changed, then do the following:
					    \begin{enumerate}
						    \item Verify that $A_{0,0}$ was also changed.
						    \item \textbf{Therefore verify that $\deg(A_{0,0})<q$.}
					    \end{enumerate}
					    \item \textbf{Yield the tuple $\langle A\rangle$.}
				    \end{enumerate}
		    \declaration{4.01}
			    The phrase "\gls{matrixdiag}" will be used as a shorthand for matrix positions such that the row index equals the column index.
		    \declaration{4.02}
			    The phrase "\gls{matrixdiagmatrix}" will be used to refer to matrices with $0$s in all off-diagonal positions.
		    \procedure{4.01}
			    \objective
				    Choose a $m\times n$ matrix, $A$. The objective of the following instructions is to transform $A$ into an $m\times n$ diagonal matrix by a sequence of operations whereby either a polynomial times any of the columns is added to a different column, or a polynomial times any of the rows is added to a different row.
			    \implementation
				    \begin{enumerate}
					    \item If $m=0$ or $n=0$, then do the following:
					    \begin{enumerate}
						    \item \textbf{Verify that $A$ is an $m\times n$ diagonal matrix.}
						    \item \textbf{Yield the tuple $\langle A\rangle$.}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \item Verify that $m>0$ and $n>0$.
					    \item Let $A$ be our working matrix.
					    \item Now do the following:
						    \begin{enumerate}
						    \item While $A_{0,[1:n]}\ne 0$, do the following:
						    \begin{enumerate}
							    \item Select the $m\times 2$ matrix whose top-right entry coincides with the last non-zero entry of the first row
							    \item Apply \procedurehr{4.00} on this submatrix.
							    \item Verify that the top-left and top-right entries of the submatrix are now non-zero and zero respectively.
							    \item If $A_{*,0}$ was modified by (5aii), then do the following:
							    \begin{enumerate}
								    \item Verify that $\deg(A_{0,0})$ decreased.
								    \item Go back to (5).
							    \end{enumerate}
						    \end{enumerate}
						    \item Now do the same operations as in (a), but this time with the operations themselves reflected across the matrix's diagonal.
					    \end{enumerate}
					    \item Verify that $A_{0,[1:n]}=0$.
					    \item Also verify that $A_{[1:m],0}=0$.
					    \item Apply \procedurehr{4.01} on the submatrix $A_{[1:m],[1:n]}$.
					    \item Verify that (8)'s execution leaves the first row and column unchanged.
					    \item Also verify that $A_{[1:m],[1:n]}$ is now a $(m-1)\times(n-1)$ diagonal matrix.
					    \item \textbf{Therefore verify that $A$ is now an $m\times n$ diagonal matrix.}
					    \item \textbf{Yield the tuple $\langle A\rangle$.}
				    \end{enumerate}
		    \declaration{4.04}
			    The phrase "\gls{matrixtilt}" will be used to refer to square matrices with only $1$s on the diagonal, a single polynomial off the diagonal, and $0$s everywhere else.
		    \procedure{4.03}
			    \objective
				    Choose a procedure, $A$, and two non-negative integers $m,n$. The objective of the following instructions is, once $A$ has been executed, to construct a list of $m\times m$ tilts, $M$, and a list of $n\times n$ tilts, $N$ such that $M_{\lvert M\rvert-1-i}$ equals $1_m$ after applying the $i^{th}$ row operation carried out by $A$ also on it, and $N_i$ equals $1_n$ after applying the $i^{th}$ row operation carried out by $A$ also on it.
			    \implementation
				    \begin{enumerate}
					    \item Make an empty list, $N$.
					    \item Augment procedure $A$ so that each time a polynomial $x$ times a column $i$ is added onto column $j$, an $n\times n$ matrix that only has $1$s on its diagonal, and such that the only non-zero entry off its diagonal is $x$ at position $(i,j)$, is appended onto $N$.
					    \item Make an empty list, $M$.
					    \item Also augment procedure $A$ so that each time a polynomial $x$ times a row $i$ is added onto row $j$, an $n\times n$ matrix that only has $1$s on its diagonal, and such that the only non-zero entry off its diagonal is $x$ at position $(j,i)$, is prepended onto $M$.
					    \item Now run procedure $A$.
					    \item \textbf{Yield the tuple $\langle M,N\rangle$.}
				    \end{enumerate}
		    \procedure{4.04}
			    \objective
				    Choose a $m\times n$ matrix, $A$. The objective of the following instructions is to show that $1_mA=A=A1_n$.
			    \implementation
				    \begin{enumerate}
					    \item For $0\le r<m$, do the following:
					    \begin{enumerate}
						    \item For $0\le t<n$, do the following:
						    \begin{enumerate}
							    \item Verify that $(1_mA)_{r,t}=\sum_u^{[0:m]} (1_m)_{r,u}A_{u,t}=(1_m)_{r,r}A_{r,t}=1*A_{r,t}=A_{r,t}$.
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Therefore verify that $1_mA=A$.}
					    \item For $0\le r<m$, do the following:
					    \begin{enumerate}
						    \item For $0\le t<n$, do the following:
						    \begin{enumerate}
							    \item Verify that $(A1_n)_{r,t}=\sum_u^{[0:m]} A_{r,u}(1_n)_{u,t}=A_{r,t}(1_n)_{t,t}=A_{r,t}*1=A_{r,t}$.
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Therefore verify that $A1_n=A$.}
				    \end{enumerate}
		    \declaration{4.05}
			    The notation \gls{matrixA^-1}, where $A$ is a list of $m\times m$ tilts, will be used to refer to the result yielded by executing the following instructions:
			    \begin{enumerate}
				    \item Let $A^{-1}$ be $\langle\rangle$.
				    \item For $i$ in $[0:\lvert A\rvert]$, do the following:
				    \begin{enumerate}
					    \item Let $(j,k)$ be the position of the off diagonal entry of $A_i$.
					    \item Let $B$ equal $A_i$ but with entry $(j,k)$ negated.
					    \item Now prepend $B$ onto $A^{-1}$.
				    \end{enumerate}
				    \item \textbf{Yield $\langle A^{-1}\rangle$.}
			    \end{enumerate}
		    \procedure{4.05}
			    \objective
				    Choose a list of $m\times m$ tilts, $A$. The objective of the following instructions is to show that $A_*{A^{-1}}_*=1_m$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that $\lvert A\rvert=\lvert A^{-1}\rvert$.
					    \item For $i$ in $[0:\lvert A\rvert]$, do the following:
					    \begin{enumerate}
						    \item Let $(j,k)$ be the position of the off diagonal entry of $A_i$.
						    \item Let $B={A^{-1}}_{\lvert A\rvert-1-i}$.
						    \item For $r$ in $[0:m]$ and $r\ne j$, do the following:
						    \begin{enumerate}
							    \item For $t$ in $[0:m]$, do the following:
							    \begin{enumerate}
								    \item Verify that $(A_iB)_{r,t}=\sum_u^{[0:m]} (A_i)_{r,u}B_{u,t}=(A_i)_{r,r}B_{r,t}=1*B_{r,t}=[r=t]$.
							    \end{enumerate}
						    \end{enumerate}
						    \item For $t$ in $[0:m]$ and $t\ne k$, do the following:
						    \begin{enumerate}
							    \item Verify that $(A_iB)_{j,t}=\sum_u^{[0:m]} (A_i)_{j,u}B_{u,t}=(A_i)_{j,t}B_{t,t}=(A_i)_{j,t}*1=[j=t]$.
						    \end{enumerate}
						    \item Verify that $(A_iB)_{j,k}=\sum_u^{[0:m]} (A_i)_{j,u}B_{u,k}=(A_i)_{j,j}B_{j,k}+(A_i)_{j,k}B_{k,k}=1*B_{j,k}+(A_i)_{j,k}*1=B_{j,k}+(A_i)_{j,k}=0$.
						    \item Therefore verify that $A_iB=1_m$.
					    \end{enumerate}
					    \item Therefore using \procedurehr{4.02} and \procedurehr{4.04}, verify that $A_*{A^{-1}}_*$
					    \begin{enumerate}
						    \item $=A_0\cdots A_{\lvert A\rvert-2}A_{\lvert A\rvert-1}{A^{-1}}_0{A^{-1}}_1\cdots {A^{-1}}_{\lvert A\rvert-1}$
						    \item $=A_0\cdots A_{\lvert A\rvert-3}A_{\lvert A\rvert-2}1_m{A^{-1}}_1{A^{-1}}_2\cdots {A^{-1}}_{\lvert A\rvert-1}$
						    \item $=A_0\cdots A_{\lvert A\rvert-3}A_{\lvert A\rvert-2}{A^{-1}}_1{A^{-1}}_2\cdots {A^{-1}}_{\lvert A\rvert-1}$
						    \item $\vdots$
						    \item $=A_01_m{A^{-1}}_{\lvert A\rvert-1}$
						    \item $=A_0{A^{-1}}_{\lvert A\rvert-1}$
						    \item $=1_m$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{4.06}
			    \objective
				    Choose a list of $m\times m$ tilts, $A$. The objective of the following instructions is to show that $(A^{-1})^{-1}=A$ and ${A^{-1}}_*A_*=1_m$.
			    \implementation
				    \begin{enumerate}
					    \item \textbf{Verify that $(A^{-1})^{-1}=A$.}
					    \item \textbf{Therefore using \procedurehr{4.05}, verify that ${A^{-1}}_*A_*={A^{-1}}_*{(A^{-1})^{-1}}_*=1_m$.}
				    \end{enumerate}
		    \procedure{4.07}
			    \objective
				    Choose a $2\times 2$ diagonal matrix, $A$. The objective of the following instructions is to construct polynomials $u,v$ and transform $A$ into a $2\times 2$ diagonal matrix, $A'$, such that $A'_{1,1}=uA'_{0,0}$ and $A_{0,0}=vA'_{0,0}$ by a sequence of operations whereby either a polynomial times any of the columns is added to a different column, or a polynomial times any of the rows is added to a different row.
			    \implementation
				    \begin{enumerate}
					    \item Add row $1$ to row $0$.
					    \item Now verify that $A_{0,1}=A_{1,1}$.
					    \item Set $A'=A$ and let $A'$ be our working matrix.
					    \item Let $\langle M,N\rangle$ receive the results of executing \procedurehr{4.03} on the pair $\langle 2,2\rangle$ and the following procedure:
					    \begin{enumerate}
						    \item Execute \procedurehr{4.00} on $A'$.
					    \end{enumerate}
					
					    \item Using (4), verify that $M$ is empty.
					    \item Using (4) and (5), verify that $AN_*=M_*AN_*=A'$.
					    \item Using (6), verify that $A=A1_n=AN_*{N^{-1}}_*=A'{N^{-1}}_*$.
					    \item Using (4), verify that $A'_{0,1}=0$.
					    \item \textbf{Using (4) and (7), verify that $A_{0,0}=A'_{0,0}{{N^{-1}}_*}_{0,0}+A'_{0,1}{{N^{-1}}_*}_{1,0}=A'_{0,0}{{N^{-1}}_*}_{0,0}$.}
					    \item Using (4) and (7), verify that $A_{1,1}=A_{0,1}=A'_{0,0}{{N^{-1}}_*}_{0,1}+A'_{0,1}{{N^{-1}}_*}_{1,1}=A'_{0,0}{{N^{-1}}_*}_{0,1}$.
					    \item Using (2), verify that $A_{1,0}=0$.
					    \item Using (6) and (11), verify that $A'_{1,0}=A_{1,0}{N_*}_{0,0}+A_{1,1}{N_*}_{1,0}=A_{1,1}{N_*}_{1,0}=A'_{0,0}{{N^{-1}}_*}_{0,1}{N_*}_{1,0}$.
					    \item \textbf{Using (6) and (11), verify that $A'_{1,1}=A_{1,0}{N_*}_{0,1}+A_{1,1}{N_*}_{1,1}=A_{1,1}{N_*}_{1,1}=A'_{0,0}{{N^{-1}}_*}_{0,1}{N_*}_{1,1}$.}
					    \item Subtract ${{N^{-1}}_*}_{0,1}{N_*}_{1,0}$ times row $0$ from row $1$.
					    \item Now using (14) and (12), verify that $A'_{1,0}=0$.
					    \item \textbf{Therefore verify that $A'$ is a $2\times 2$ diagonal matrix.}
					    \item \textbf{Let $A=A'$.}
					    \item \textbf{Yield $\langle{{N^{-1}}_*}_{0,1}{N_*}_{1,1},{{N^{-1}}_*}_{0,0}\rangle$.}
				    \end{enumerate}
		    \procedure{4.08}
			    \objective
				    Choose a $m\times n$ matrix, $A$ such that $\min(m,n)>0$. The objective of the following instructions is to define a list of polynomials $u$ and transform $A$ into an $m\times n$ diagonal matrix such that $A_{k,k}=u_kA_{0,0}$ for $k$ in $[0:\min(m,n)]$ by a sequence of operations whereby either a polynomial times any of the columns is added to a different column, or a polynomial times any of the rows is added to a different row.
			    \implementation
				    \begin{enumerate}
					    \item Let $u=\langle 1\rangle$.
					    \item Execute \procedurehr{4.01} on $A$.
					    \item Verify that $A$ is an $m\times n$ diagonal matrix.
					    \item For $j$ in $[1:\min(m,n)]$, do the following:
					    \begin{enumerate}
						    \item Using (h), verify that $A_{k,k}=u_kA_{0,0}$ for $k$ in $[0:j]$.
						    \item Set $A'=A$.
						    \item Execute \procedurehr{4.07} on $A'_{\langle 0,j\rangle,\langle 0,j\rangle}$ and let $\langle u_j,v\rangle$ receive.
						    \item Using (c), verify that $A$ and $A'$ are the same modulo positions $\langle 0,0\rangle$ and $\langle j,j\rangle$.
						    \item Therefore verify that $A'$ is an $m\times n$ diagonal matrix.
						    \item Also, using (c), verify that $A'_{j,j}=u_jA'_{0,0}$.
						    \item Also, for $k$ in $[1:j]$, do the following:
						    \begin{enumerate}
							    \item Using (a), (c), and (d), verify that $A'_{k,k}=A_{k,k}=u_kA_{0,0}=u_kA'_{0,0}v$.
							    \item Set $u_k=u_kv$.
							    \item Hence verify that $A'_{k,k}=u_kA'_{0,0}$.
						    \end{enumerate}
						    \item Therefore verify that $A_{k,k}=u_kA_{0,0}$ for $k$ in $[0:j+1]$.
						    \item Now let $A=A'$.
					    \end{enumerate}
					    \item \textbf{Hence using (4h), verify that $A_{k,k}=u_kA_{0,0}$ for $k$ in $[0:\min(m,n)]$.}
					    \item \textbf{Also, using (4e), verify that $A$ is an $m\times n$ diagonal matrix.}
					    \item \textbf{Yield $\langle u\rangle$.}
				    \end{enumerate}
		    \procedure{4.09}
			    \objective
				    Choose a $m\times n$ matrix, $A$, and a $n\times k$ matrix, $B$. Choose integers $0\le a<m$, $0\le b<n$, and $0\le c<k$. The objective of the following instructions is to show that
				    \begin{enumerate}
					    \item $(AB)_{[0:a],[0:c]}=A_{[0:a],[0:b]}B_{[0:b],[0:c]}+A_{[0:a],[b:n]}B_{[b:n],[0:c]}$
					    \item $(AB)_{[0:a],[c:k]}=A_{[0:a],[0:b]}B_{[0:b],[c:k]}+A_{[0:a],[b:n]}B_{[b:n],[c:k]}$
					    \item $(AB)_{[a:m],[0:c]}=A_{[a:m],[0:b]}B_{[0:b],[0:c]}+A_{[a:m],[b:n]}B_{[b:n],[0:c]}$
					    \item $(AB)_{[a:m],[c:k]}=A_{[a:m],[0:b]}B_{[0:b],[c:k]}+A_{[a:m],[b:n]}B_{[b:n],[c:k]}$.
				    \end{enumerate}
			    \implementation
				    \begin{enumerate}
					    \item For each $0\le i<a$, do the following:
					    \begin{enumerate}
						    \item For each $0\le j<c$, do the following:
							    \begin{enumerate}
								    \item Verify that $(AB)_{i,j}=\sum_{p}^{[0:n]} A_{i,p}B_{p,j}=\sum_{p}^{[0:b]} A_{i,p}B_{p,j}+\sum_p^{[b:n]} A_{i,p}B_{p,j}=\sum_p^{[0:{b}]} (A_{[0:a],[0:b]})_{i,p}(B_{[0:b],[0:c]})_{p,j}+\sum_p^{[0:{n-b}]} (A_{[0:a],[b:n]})_{i,p}(B_{[b:n],[0:c]})_{p,j}=(A_{[0:a],[0:b]}B_{[0:b],[0:c]})_{i,j}+(A_{[0:a],[b:n]}B_{[b:n],[0:c]})_{i,j}$.
							    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Therefore verify that $(AB)_{[0:a],[0:c]}=A_{[0:a],[0:b]}B_{[0:b],[0:c]}+A_{[0:a],[b:n]}B_{[b:n],[0:c]}$.}
					    \item \textbf{Using computations analogous to (1) and (2), show items (2), (3), and (4) of the objective.}
				    \end{enumerate}
		    \declaration{4.06}
			    The phrase "number of rows of $A$" and the notation \gls{matrixrows}, where $A$ is an $m\times n$ matrix, will be used as a shorthand for $m$.
		    \declaration{4.07}
			    The phrase "number of columns of $A$" and the notation \gls{matrixcols}, where $A$ is an $m\times n$ matrix, will be used as a shorthand for $n$.
		    \declaration{4.08}
			    The notation \gls{matrixbdiag}, where $C$ is a list of rational square matrices, will be used to refer to the result yielded by executing the following instructions:
			    \begin{enumerate}
				    \item Let $E$ be a $0\times 0$ matrices.
				    \item Now for $i$ in $[0:\lvert C\rvert]$:
				    \begin{enumerate}
					    \item Add $\cols(C_i)$ columns filled with zeros to the right end of $E$.
					    \item Add $\rows(C_i)$ rows filled with zeros to the bottom end of $E$.
					    \item Set the bottom-right corner of $E$ equal to $C_i$.
				    \end{enumerate}
				    \item \textbf{Yield the tuple $\langle E\rangle$.}
			    \end{enumerate}
		    \procedure{4.10}
			    \objective
				    Choose a $m\times n$ matrix, $A$. Let $A_{-1,-1}=1$. The objective of the following instructions is to construct the list of polynomials $v$ and transform $A$ into an $m\times n$ diagonal matrix such that $A_{k,k}=v_kA_{k-1,k-1}$ for $k$ in $[0:\min(m,n)]$ by a sequence of operations whereby either a polynomial times any of the columns is added to a different column, or a polynomial times any of the rows is added to a different row.
			    \implementation
				    \begin{enumerate}
					    \item If $\min(m,n)=0$, then do the following:
					    \begin{enumerate}
						    \item \textbf{Verify that $A$ is an $m\times n$ diagonal matrix.}
						    \item \textbf{Yield $\langle\rangle$.}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
						    \item Apply \procedurehr{4.08} on $A$, and let $\langle u\rangle$ receive.
						    \item Verify that $A$ is an $m\times n$ diagonal matrix.
						    \item Verify that $A_{k,k}=u_{k}A_{0,0}$ for $k$ in $[0:\min(m,n)]$.
						    \item Let $B,C$ be an $(m-1)\times(n-1)$ diagonal matrix with $u_{1:\lvert u\rvert}$ on the diagonal.
						    \item Let $\langle M,N\rangle$ receive the results of executing \procedurehr{4.03} on the pair $\langle m-1,n-1\rangle$ and the following procedure:
						    \begin{enumerate}
							    \item Execute \procedurehr{4.10} on $C$ and let $\langle w\rangle$ receive.
						    \end{enumerate}
						    \item Therefore verify that $C$ is an $(m-1)\times(n-1)$ diagonal matrix.
						    \item Also verify that $C=M_*BN_*$.
						    \item Let $C_{-1,-1}=1$.
						    \item Now using (ei), verify that $C_{k,k}=w_kC_{k-1,k-1}$ for $k$ in $[0:\min(m,n)-1]$.
						    \item Therefore using (c), verify that $A_{0,0}C=M_*(A_{0,0}B)N_*=M_*A_{[1:m],[1:n]}N_*$.
						    \item Premultiply $A$ by $\diag(1,M_k)$ for $k$ in $[\lvert M\rvert:0]$.
						    \item Postmultiply $A$ by $\diag(1,N_k)$ for $k$ in $[0:\lvert N\rvert]$.
						    \item Now verify that $A_{[1:m],[1:n]}=A_{0,0}C$.
						    \item Now let $u=\langle A_{0,0}\rangle^{\frown}w$.
						    \item \textbf{Therefore verify that $A_{k,k}=u_kA_{k-1,k-1}$ for $k$ in $[0:\min(m,n)]$.}
						    \item \textbf{Yield the tuple $\langle u\rangle$.}
					    \end{enumerate}
				    \end{enumerate}
	    \end{multicols*}
	  \chapter{Compound Matrices}
	    \begin{multicols*}{2}
		    \declaration{4.09}
			    The notation \gls{matrixdet}, where $A$ is a $m\times m$ matrix, will be used to refer to the result yielded by executing the following instructions:
			    \begin{enumerate}
				    \item If $m=0$, then do the following:
				    \begin{enumerate}
					    \item \textbf{Yield the tuple $\langle 1\rangle$.}
				    \end{enumerate}
				    \item Otherwise, do the following:
				    \begin{enumerate}
					    \item Let $h_r=A_{[0:r]^\frown[r+1,m],[1:m]}$ for $r$ in $[0:m]$.
					    \item \textbf{Yield the tuple $\langle\sum_r^{[0:m]} (-1)^{r}A_{r,0}\det(h_r)\rangle$.}
				    \end{enumerate}
			    \end{enumerate}
		    \procedure{4.11}
			    \objective
				    Choose a polynomial $p$. Choose two $1\times m$ matrices, $B$ and $C$. Choose an integer $0\le i<m$. Choose a $m\times m$ matrix, $A$, such that its $i^{th}$ row is $B+pC$. Let $A'$ be $A$ but with the $i^{th}$ row replaced by $B$ and let $A''$ be $A$ but with the $i^{th}$ row replaced by $C$. The objective of the following instructions is to show that $\det(A)=\det(A')+p\det(A'')$.
			    \implementation
				    \begin{enumerate}
					    \item If $m=1$, then do the following:
					    \begin{enumerate}
						    \item Verify that $i=0$.
						    \item \textbf{Therefore verify that $\det(A)=A_{0,0}=B_{0,0}+pC_{0,0}=\det(A')+p\det(A'')$.}
					    \end{enumerate}
					    \item Otherwise, do the following:
					    \begin{enumerate}
						    \item For $r$ in $[0:i]$, do the following:
						    \begin{enumerate}
							    \item Verify that $(A_{[0:r]^\frown[r+1:m],[1:m]})_{i-1,*}=B+pC$.
							    \item Verify that $A'_{[0:r]^\frown[r+1:m],[1:m]}$ is $A_{[0:r]^\frown[r+1:m],[1:m]}$ with row $i-1$ replaced by $B$.
							    \item Verify that $A''_{[0:r]^\frown[r+1:m],[1:m]}$ is $A_{[0:r]^\frown[r+1:m],[1:m]}$ with row $i-1$ replaced by $C$.
							    \item Execute \procedurehr{4.11} on $\langle p,B,C,i-1,A_{[0:r]^\frown[r+1:m],[1:m]}\rangle$.
							    \item Therefore verify that $\det(A_{[0:r]^\frown[r+1:m],[1:m]})=\det(A'_{[0:r]^\frown[r+1:m],[1:m]})+p\det(A''_{[0:r]^\frown[r+1:m],[1:m]})$.
						    \end{enumerate}
						    \item For $r$ in $[i+1:m]$, do the following:
						    \begin{enumerate}
							    \item Verify that $(A_{[0:r]^\frown[r+1:m],[1:m]})_{i,*}=B+pC$.
							    \item Verify that $A'_{[0:r]^\frown[r+1:m],[1:m]}$ is $A_{[0:r]^\frown[r+1:m],[1:m]}$ with row $i$ replaced by $B$.
							    \item Verify that $A''_{[0:r]^\frown[r+1:m],[1:m]}$ is $A_{[0:r]^\frown[r+1:m],[1:m]}$ with row $i$ replaced by $C$.
							    \item Execute \procedurehr{4.11} on $\langle p,B,C,i,A_{[0:r]^\frown[r+1:m],[1:m]}\rangle$.
							    \item Therefore verify that $\det(A_{[0:r]^\frown[r+1:m],[1:m]})=\det(A'_{[0:r]^\frown[r+1:m],[1:m]})+p\det(A''_{[0:r]^\frown[r+1:m],[1:m]})$.
						    \end{enumerate}
						    \item Therefore using (av) and (bv), verify that $\det(A)$
						    \begin{enumerate}
							    \item $=\sum_r^{[0:m]} (-1)^rA_{r,0}\det(A_{[0:r]^\frown[r+1:m],[1:m]})$
							    \item $=\sum_r^{[0:i]} (-1)^rA_{r,0}\det(A_{[0:r]^\frown[r+1:m],[1:m]})+(-1)^iA_{i,0}\det(A_{[0:i]^\frown[i+1:m],[1:m]})+\sum_r^{[i+1:m]} (-1)^rA_{r,0}\det(A_{[0:r]^\frown[r+1:m],[1:m]})$
							    \item $=\sum_r^{[0:i]} (-1)^rA_{r,0}(\det(A'_{[0:r]^\frown[r+1:m],[1:m]})+p\det(A''_{[0:r]^\frown[r+1:m],[1:m]}))+(-1)^i(A'_{i,0}+pA''_{i,0})\det(A_{[0:i]^\frown[i+1:m],[1:m]})+\\\sum_r^{[i+1:m]} (-1)^rA_{r,0}(\det(A'_{[0:r]^\frown[r+1:m],[1:m]})+p\det(A''_{[0:r]^\frown[r+1:m],[1:m]}))$
							    \item $=\sum_r^{[0:i]} (-1)^rA_{r,0}\det(A'_{[0:r]^\frown[r+1:m],[1:m]})+(-1)^iA'_{i,0}\det(A_{[0:i]^\frown[i+1:m],[1:m]})+\sum_r^{[i+1:m]} (-1)^rA_{r,0}\det(A'_{[0:r]^\frown[r+1:m],[1:m]})
								    +\sum_r^{[0:i]} (-1)^rA_{r,0}p\det(A''_{[0:r]^\frown[r+1:m],[1:m]})+(-1)^ipA''_{i,0}\det(A_{[0:i]^\frown[i+1:m],[1:m]})+\sum_r^{[i+1:m]} (-1)^rA_{r,0}p\det(A''_{[0:r]^\frown[r+1:m],[1:m]})$
							    \item $=\sum_r^{[0:m]} (-1)^rA'_{r,0}\det(A'_{[0:r]^\frown[r+1:m],[1:m]})+p\sum_r^{[0:m]} (-1)^rA''_{r,0}\det(A''_{[0:r]^\frown[r+1:m],[1:m]})$
							    \item $=\det(A')+p\det(A'')$.
						    \end{enumerate}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{4.12}
			    \objective
				    Choose a polynomial $p$. Choose two $m\times 1$ matrices, $B$ and $C$. Choose an integer $0\le i<m$. Choose a $m\times m$ matrix, $A$, such that its $i^{th}$ column is $B+pC$. Let $A'$ be $A$ but with the $i^{th}$ column replaced by $B$ and let $A''$ be $A$ but with the $i^{th}$ column replaced by $C$. The objective of the following instructions is to show that $\det(A)=\det(A')+p\det(A'')$.
			    \implementation
				    \begin{enumerate}
					    \item If $i=0$, then verify that $\det(A)$
					    \begin{enumerate}
						    \item $=\sum_r^{[0:m]} (-1)^{r}A_{r,0}\det(A_{[0:r]^\frown[r+1:m],[1:m]})$
						    \item $=\sum_r^{[0:m]} (-1)^{r}(B+pC)_{r,0}\det(A_{[0:r]^\frown[r+1:m],[1:m]})$
						    \item $=\sum_r^{[0:m]} (-1)^{r}(B)_{r,0}\det(A_{[0:r]^\frown[r+1:m],[1:m]})+\sum_r^{[0:m]} (-1)^{r}(pC)_{r,0}\det(A_{[0:r]^\frown[r+1:m],[1:m]})$
						    \item $=\sum_r^{[0:m]} (-1)^{r}(B)_{r,0}\det(A_{[0:r]^\frown[r+1:m],[1:m]})+p\sum_r^{[0:m]} (-1)^{r}(C)_{r,0}\det(A_{[0:r]^\frown[r+1:m],[1:m]})$
						    \item $=\sum_r^{[0:m]} (-1)^{r}(A')_{r,0}\det(A'_{[0:r]^\frown[r+1:m],[1:m]})+p\sum_r^{[0:m]} (-1)^{r}(A'')_{r,0}\det(A''_{[0:r]^\frown[r+1:m],[1:m]})$
						    \item $=\det(A')+p\det(A'')$
					    \end{enumerate}
					    \item Otherwise, do the following:
					    \begin{enumerate}
						    \item For $r$ in $[0:m]$, do the following:
						    \begin{enumerate}
							    \item Execute \procedurehr{4.12} on $\langle p,B_{[0:r]^\frown[r+1:m],0},C_{[0:r]^\frown[r+1:m],0},i-1,A_{[0:r]^\frown[r+1:m],[1:m]}\rangle$.
							    \item Therefore verify that $\det(A_{[0:r]^\frown[r+1:m],[1:m]})=\det(A'_{[0:r]^\frown[r+1:m],[1:m]})+p\det(A''_{[0:r]^\frown[r+1:m],[1:m]})$.
						    \end{enumerate}
						    \item Therefore using (a), verify that $\det(A)$
						    \begin{enumerate}
							    \item $=\sum_r^{[0:m]} (-1)^{r}A_{r,0}\cdot\det(A_{[0:r]^\frown[r+1:m],[1:m]})$
							    \item $=\sum_r^{[0:m]} (-1)^{r}A_{r,0}(\det(A'_{[0:r]^\frown[r+1:m],[1:m]})+p\det(A''_{[0:r]^\frown[r+1:m],[1:m]}))$
							    \item $=\sum_r^{[0:m]} (-1)^{r}A'_{r,0}\det(A'_{[0:r]^\frown[r+1:m],[1:m]})+\sum_r^{[0:m]} (-1)^{r}A''_{r,0}p\det(A''_{[0:r]^\frown[r+1:m],[1:m]})$
							    \item $=\det(A')+p\det(A'')$.
						    \end{enumerate}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{4.13}
			    \objective
				    Choose a $m\times m$ matrix, $A$. Choose an integer $0<i<m$. Let $A'$ be $A$ with rows $i-1$ and $i$ swapped. The objective of the following instructions is to show that $\det(A')=-\det(A)$.
			    \implementation
				    \begin{enumerate}
					    \item If $m=2$, then do the following:
					    \begin{enumerate}
						    \item Verify that $i=1$.
						    \item Therefore verify that $\det(A')=A'_{0,0}A'_{1,1}-A'_{1,0}A'_{0,1}=A_{1,0}A_{0,1}-A_{0,0}A_{1,1}=-\det(A)$.
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
						    \item For $r$ in $[0:i-1]$, do the following:
						    \begin{enumerate}
							    \item Verify that $A_{[0:r]^\frown[r+1:m],[1:m]}$ is the same as $A'_{[0:r]^\frown[r+1:m],[1:m]}$ but with rows $i-2$ and $i-1$ swapped.
							    \item Execute \procedurehr{4.13} on $\langle A_{[0:r]^\frown[r+1:m],[1:m]},i-1\rangle$.
							    \item Hence verify that $\det(A'_{[0:r]^\frown[r+1:m],[1:m]})=-\det(A_{[0:r]^\frown[r+1:m],[1:m]})$.
						    \end{enumerate}
						    \item For $r$ in $[i+1:m]$, do the following:
						    \begin{enumerate}
							    \item Verify that $A_{[0:r]^\frown[r+1:m],[1:m]}$ is the same as $A'_{[0:r]^\frown[r+1:m],[1:m]}$ but with rows $i-1$ and $i$ swapped.
							    \item Execute \procedurehr{4.13} on $\langle A_{[0:r]^\frown[r+1:m],[1:m]},i\rangle$.
							    \item Hence verify that $\det(A'_{[0:r]^\frown[r+1:m],[1:m]})=-\det(A_{[0:r]^\frown[r+1:m],[1:m]})$.
						    \end{enumerate}
						    \item Verify that $\det(A)$
						    \begin{enumerate}
							    \item $=\sum_r^{[0:m]} (-1)^rA_{r,0}\det(A_{[0:r]^\frown[r+1:m],[1:m]})$
							    \item $=\sum_r^{[0:{i-1}]} (-1)^rA_{r,0}\det(A_{[0:r]^\frown[r+1:m],[1:m]})+(-1)^{i-1}A_{i-1,0}\det(A_{[0:i-1]^\frown[i:m],[1:m]})+(-1)^iA_{i,0}\det(A_{[0:i]^\frown[i+1:m],[1:m]})+\sum_r^{[i+1:m]} (-1)^rA_{r,0}\det(A_{[0:r]^\frown[r+1:m],[1:m]})$
							    \item $=-\sum_r^{[0:{i-1}]} (-1)^rA'_{r,0}\det(A'_{[0:r]^\frown[r+1:m],[1:m]})-(-1)^{i}A'_{i,0}\det(A'_{[0:i]^\frown[i+1:m],[1:m]})-(-1)^{i-1}A'_{i-1,0}\det(A'_{[0:i-1]^\frown[i:m],[1:m]})-\sum_r^{[i+1:m]} (-1)^rA'_{r,0}\det(A'_{[0:r]^\frown[r+1:m],[1:m]})$
							    \item $=-\sum_r^{[0:m]} (-1)^rA'_{r,0}\det(A'_{[0:r]^\frown[r+1:m],[1:m]})$
							    \item $=-\det(A')$.
						    \end{enumerate}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{4.14}
			    \objective
				    Choose a $m\times m$ matrix, $A$. Choose an integer $0<i<m$. Let $A'$ be $A$ with columns $i-1$ and $i$ swapped. The objective of the following instructions is to show that $\det(A')=-\det(A)$.
			    \implementation
				    \begin{enumerate}
					    \item If $i=1$, then verify that $\det(A)$
					    \begin{enumerate}
						    \item $=\sum_r^{[0:m]} (-1)^{r}A_{r,0}\det(A_{[0:r]^\frown[r+1:m],[1:m]})$
						    \item $=\sum_r^{[0:m]} (-1)^{r}A_{r,0}\sum_t^{[r+1:m]} (-1)^{t-1}A_{t,1}*\\\det(A_{[0:r]^\frown[r+1:t]^\frown[t+1:m],[2:m]})+\\\sum_t^{[0:m]} (-1)^{t}A_{t,0}\sum_r^{[0:{t}]} (-1)^{r}A_{r,1}*\\\det(A_{[0:r]^\frown[r+1:t]^\frown[t+1:m],[2:m+1]})$
						    \item $=\sum_t^{[0:m]} (-1)^{t-1}A_{t,1}\sum_r^{[0:{t}]} (-1)^{r}A_{r,0}*\\\det(A_{[0:r]^\frown[r+1:t]^\frown[t+1:m],[2:m+1]})+\\\sum_r^{[0:m]} (-1)^{r}A_{r,1}\sum_t^{[r+1:m]} (-1)^{t}A_{t,0}*\\\det(A_{[0:r]^\frown[r+1:t]^\frown[t+1:m],[2:m+1]})$
						    \item $=\sum_t^{[0:m]} (-1)^{t-1}A'_{t,0}\sum_r^{[0:{t}]} (-1)^{r}A'_{r,1}*\\\det(A'_{[0:r]^\frown[r+1:t]^\frown[t+1:m],[2:m+1]})+\\\sum_r^{[0:m]} (-1)^{r}A'_{r,0}\sum_t^{[r+1:m]} (-1)^{t}A'_{t,1}*\\\det(A'_{[0:r]^\frown[r+1:t]^\frown[t+1:m],[2:m]})$
						    \item $=-(\sum_r^{[0:m]} (-1)^{r}A'_{r,0}\sum_t^{[r+1:m]} (-1)^{t-1}A'_{t,1}*\\\det(A'_{[0:r]^\frown[r+1:t]^\frown[t+1:m],[2:m]})+\\\sum_t^{[0:m]} (-1)^{t}A'_{t,0}\sum_r^{[0:{t}]} (-1)^{r}A'_{r,1}*\\\det(A'_{[0:r]^\frown[r+1:t]^\frown[t+1:m],[2:m]}))$
						    \item $=-\det(A')$.
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
						    \item Verify that $i>1$.
						    \item For $r$ in $[0:m]$, do the following:
						    \begin{enumerate}
							    \item Execute \procedurehr{4.14} on $\langle i-1,A_{[0:r]^\frown[r+1:m],[1:m]}\rangle$.
							    \item Therefore verify that $\det(A_{[0:r]^\frown[r+1:m],[1:m]})=-\det(A'_{[0:r]^\frown[r+1:m],[1:m]})$.
						    \end{enumerate}
						    \item Therefore using (bii), verify that $\det(A)=\sum_r^{[0:m]} (-1)^{r}A_{r,0}\cdot\det(A_{[0:r]^\frown[r+1:m],[1:m]})=\sum_r^{[0:m]} (-1)^{r}A'_{r,0}\cdot(-\det(A'_{[0:r]^\frown[r+1:m],[1:m]}))=-\det(A')$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{4.15}
			    \objective
				    Choose integers $0<i<m$. Choose a $m\times m$ matrix, $A$, such that columns $i-1$ and $i$ are the same. The objective of the following instructions is to show that $\det(A)=0$.
			    \implementation
				    \begin{enumerate}
					    \item Let $A'$ be $A$ with columns $i-1$ and $i$ swapped.
					    \item Execute \procedurehr{4.14} on $\langle A, i\rangle$.
					    \item Also, verify that $A'=A$.
					    \item Therefore verify that $\det(A)=\det(A')=-\det(A)$.
					    \item \textbf{Therefore verify that $\det(A)=0$.}
				    \end{enumerate}
		    \procedure{4.16}
			    \objective
				    Choose integers $0<i<m$. Choose a $m\times m$ matrix, $A$, such that rows $i-1$ and $i$ are the same. The objective of the following instructions is to show that $\det(A)=0$.
			    \implementation
				    Instructions are analogous to those of \procedurehr{4.15}.
		    \procedure{4.17}
			    \objective
				    Choose integers $0\le i<m$. Choose an integer $-i\le j<m-i$. Choose a $m\times m$ matrix, $A$. Let $A'$ be $A$ but with column $i$ moved $j$ places. The objective of the following instructions is to show that $\det(A')=(-1)^j\det(A)$.
			    \implementation
				    \begin{enumerate}
					    \item Let $B=\langle A\rangle$.
					    \item For $k$ in $[i:i+j]$, do the following:
					    \begin{enumerate}
						    \item Let $B_{\lvert B\rvert}$ be the result of swapping columns $k$ and $k+1$ of $B_{\lvert B\rvert-1}$.
						    \item Using \procedurehr{4.14}, verify that $\det(B_{\lvert B\rvert-1})=-\det(B_{\lvert B\rvert-2})$.
					    \end{enumerate}
					    \item Verify that $A'=B_{\lvert B\rvert-1}$.
					    \item \textbf{Therefore verify that $\det(A')=\det(B_{\lvert B\rvert-1})=(-1)^1\det(B_{\lvert B\rvert-2})=\cdots=(-1)^j\det(B_{0})=(-1)^j\det(A)$.}
				    \end{enumerate}
		    \procedure{4.18}
			    \objective
				    Choose integers $0\le i<m$. Choose an integer $-i\le j<m-i$. Choose a $m\times m$ matrix, $A$. Let $A'$ be $A$ but with row $i$ moved $j$ places. The objective of the following instructions is to show that $\det(A')=(-1)^j\det(A)$.
			    \implementation
				    Instructions are analogous to those of \procedurehr{4.17}.
		    \declaration{4.10}
			    The notation \gls{matrixC_k(A)}, where $A$ is a $m\times n$ matrix and $k$ is an integer such that $0\le k\le\min(m,n)$, will be used to refer to the $\binom{m}{k}\times\binom{n}{k}$ matrix with the following specification:
			    \begin{enumerate}
				    \item The rows are labeled by the colexicographically sorted list of increasing length-$k$ sequences whose elements are picked from $[0:m]$.
				    \item The columns are labeled by the colexicographically sorted list of increasing length-$k$ sequences whose elements are picked from $[0:n]$.
				    \item For each row label $I$: For each column label $J$: The entry at position $(I,J)$ is $\det(A_{I,J})$.
			    \end{enumerate}
		    \declaration{4.11}
			    The notation \gls{matrixlabelledentry} will be used to refer to the entry of $A$ with row label $I$ and column label $J$.
		    \procedure{4.19}
			    \objective
				    Choose two integers $0\le k\le m$. The objective of the following instructions is to show that $C_k(1_m)=1_{\binom{m}{k}}$.
			    \implementation
				    \begin{enumerate}
					    \item For each row label $I$ of $C_k(1_m)$, for each column label $J$ of $C_k(1_m)$, do the following:
					    \begin{enumerate}
						    \item If $I=J$, then do the following:
						    \begin{enumerate}
							    \item Verify that $((1_m)_{I,J})_{i,j}=((1_m)_{J,J})_{i,j}=(1_m)_{J_i,J_j}=[J_i=J_j]=[i=j]$ for $0\le i<k$, for $0\le j<k$.
							    \item Therefore verify that $(C_k(1_m))_{\ul{I},\ul{J}}=1_k$.
							    \item \textbf{Therefore verify that $(C_k(1_m))_{\ul{I},\ul{J}}=\det((1_m)_{I,J})=\det(1_k)=1$.}
						    \end{enumerate}
						    \item Otherwise, do the following:
						    \begin{enumerate}
							    \item Verify that $I\ne J$.
							    \item Let $i$ be the index of an element of $I$ that is not an element of $J$.
							    \item Now verify that $(1_m)_{I_i,j}=[I_i=j]=0$, for each $j$ in $J$.
							    \item Therefore verify that $((1_m)_{I,J})_{i,*}=0_{1\times k}$.
							    \item \textbf{Therefore verify that $(C_k(1_m))_{\ul{I},\ul{J}}=\det((1_m)_{I,J})=0$.}
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Therefore verify that $C_k(1_m)=1_{\binom{m}{k}}$.}	
				    \end{enumerate}
		    \procedure{4.20}
			    \objective
				    Choose an integer $0\le k\le\min(m,n)$. Choose a $m\times m$ tilt, $A$, such that the off diagonal entry is the polynomial $p$ at $(i,j)$. Also choose a $m\times n$ matrix, $B$. The objective of the following instructions is to construct a $\binom{m}{k}\times\binom{m}{k}$ matrix $D$ such that $C_k(AB)=DC_k(B)$.
			    \implementation
				    \begin{enumerate}
					    \item Let $D=C_k(1_m)=1_{\binom{m}{k}}$.
					    \item Verify that $AB$ equals $B$, but with its row $i$ having $p$ times $B$'s row $j$ added to it.
					    \item Go through the row labels, $I$, of $C_k(AB)$ and do the following:
					    \begin{enumerate}
						    \item If $i\notin I$, then do the following:
						    \begin{enumerate}
							    \item Verify that $(AB)_{I,*}=B_{I,*}$.
							    \item Therefore for each column label $J$, verify that ${C_k(AB)}_{\ul{I},\ul{J}}=\det((AB)_{I,J})=\det(B_{I,J})={C_k(B)}_{\ul{I},\ul{J}}$.
							    \item \textbf{Therefore verify that $(C_k(AB))_{\ul{I},*}=(C_k(B))_{\ul{I},*}$.}
						    \end{enumerate}
						    \item Otherwise, if $i\in I$, then:
						    \begin{enumerate}
							    \item Let $I'$ be $I$ but with an in-place replacement of $i$ by $j$.
							    \item For each column label $J$: Using \procedurehr{4.12}, verify that ${C_k(AB)}_{\ul{I},\ul{J}}=\det((AB)_{I,J})=\det(B_{I,J})+p*\det(B_{I',J})$.
							    \item If $j\in I$, then do the following:
							    \begin{enumerate}
								    \item Verify that the sequence $I'$ contains two $j$s.
								    \item For each column label $J$: Using \procedurehr{4.16} verify that $\det(B_{I',J})=0$.
								    \item Therefore for each column label $J$: verify that ${C_k(AB)}_{\ul{I},\ul{J}}=\det(B_{I,J})={C_k(B)}_{\ul{I},\ul{J}}$.
								    \item \textbf{Therefore verify that ${C_k(AB)}_{\ul{I},*}={C_k(B)}_{\ul{I},*}$.}
							    \end{enumerate}
							    \item Otherwise if $j\notin I$, do the following:
							    \begin{enumerate}
								    \item Let $l$ be the signed number of places that the $j$ introduced above needs to be moved in order to make $I'$ an increasing sequence.
								    \item Let $I''$ be obtained from $I'$ by moving the integer $j$ in $I'$ by $l$ places.
								    \item For each column label $J$: Using \procedurehr{4.18}, verify that $\det(B_{I',J})=(-1)^l\det(B_{I'',J})$.
								    \item Therefore for each column label $J$: Verify that ${C_k(AB)}_{\ul{I},\ul{J}}=\det(B_{I,J})+p*\det(B_{I',J})=\det(B_{I,J})+(-1)^lp*\det(B_{I'',J})$.
								    \item Verify that $I''$ is a row label of $C_k(B)$.
								    \item Therefore for each column label $J$: Verify that ${C_k(AB)}_{\ul{I},\ul{J}}=\det(B_{I,J})+(-1)^lp*\det(B_{I'',J})={C_k(B)}_{\ul{I},\ul{J}}+(-1)^lp*{C_k(B)}_{\ul{I''},\ul{J}}$.
								    \item \textbf{Therefore verify that $(C_k(AB))_{\ul{I},*}=(C_k(B))_{\ul{I},*}+(-1)^lp(C_k(B))_{\ul{I''},*}$.}
								    \item \textbf{Set $D_{\ul{I},\ul{I''}}$ to $(-1)^lp$.}
							    \end{enumerate}
						    \end{enumerate}
						    \item \textbf{Therefore verify that ${C_k(AB)}_{\ul{I},*}=D_{\ul{I},*}C_k(B)$.}
					    \end{enumerate}
					    \item \textbf{Therefore verify that $C_k(AB)=DC_k(B)$.}
					    \item \textbf{Yield $\langle D\rangle$.}
				    \end{enumerate}
		    \procedure{4.21}
			    \objective
				    Choose an $m\times n$ diagonal matrix, $A$. Also choose an $n\times n$ matrix, $B$. Also choose an integer $0\le k\le\min(m,n)$. The objective of the following instructions is to construct an $\binom{m}{k}\times\binom{n}{k}$ diagonal matrix $D$ such that $C_k(AB)=DC_k(B)$.
			    \implementation
				    \begin{enumerate}
					    \item Let $D=C_k(0_{m\times n})=0_{\binom{m}{k}\times\binom{n}{k}}$.
					    \item Verify that $AB$ equals $B_{[0:\min(m,n)],*}$ with each row $i$ multiplied by $A_{i,i}$.
					    \item Go through the row labels, $I$, of $C_k(AB)$ and do the following:
					    \begin{enumerate}
						    \item If $I_k<\min(m,n)$, then do the following:
						    \begin{enumerate}
							    \item Verify that every element of $I$ is less than $\min(m,n)$.
							    \item Let $A_0=A$.
							    \item For $i$ in $[0:k]$: Let $A_{i+1}$ equal $A_{i}$ but with position $(I_i,I_i)$ set to $1$.
							    \item For each column label $J$: Repeatedly using \procedurehr{4.12}, verify that ${C_k(AB)}_{I,J}$
							    \begin{enumerate}
								    \item $=\det((AB)_{I,J})$
								    \item $=\det((A_0B)_{I,J})$
								    \item $=A_{I_0,I_0}\det((A_1B)_{I,J})$
								    \item $=A_{I_0,I_0}A_{I_1,I_1}\det((A_2B)_{I,J})$
								    \item $\vdots$
								    \item $=A_{I_0,I_0}A_{I_1,I_1}\cdots A_{I_{k-1},I_{k-1}}\det((A_kB)_{I,J})$
								    \item $=A_{I_0,I_0}A_{I_1,I_1}\cdots A_{I_{k-1},I_{k-1}}\det(B_{I,J})$
								    \item $=A_{I_0,I_0}A_{I_1,I_1}\cdots A_{I_{k-1},I_{k-1}}{C_k(B)}_{\ul{I},\ul{J}}$.
							    \end{enumerate}
							    \item \textbf{Therefore verify that $(C_k(AB))_{\ul{I},*}=A_{I_1,I_1}A_{I_1,I_1}\cdots A_{I_k,I_k}*(C_k(B))_{\ul{I},*}$.}
							    \item \textbf{Set $D_{\ul{I},\ul{I}}$ to $A_{I_0,I_0}A_{I_1,I_1}\cdots A_{I_{k-1},I_{k-1}}$.}
						    \end{enumerate}
						    \item Otherwise if $I_k\ge\min(m,n)$, then do the following:
						    \begin{enumerate}
							    \item Using the precondition, verify that $A_{I_k,*}=0_{1\times n}$.
							    \item Therefore verify that $(AB)_{I_k,*}=0_{1\times n}$.
							    \item Therefore verify that $((AB)_{I,*})_{k,*}=0_{1\times n}$.
							    \item Therefore for each column label $J$: verify that ${C_k(AB)}_{\ul{I},\ul{J}}=\det((AB)_{I,J})=0$.
							    \item \textbf{Therefore verify that $(C_k(AB))_{\ul{I},*}$ is zero.}
						    \end{enumerate}
						    \item \textbf{Therefore verify that ${C_k(AB)}_{\ul{I},*}=D_{\ul{I},*}C_k(B)$.}
					    \end{enumerate}
					    \item \textbf{Verify that $D$ is diagonal.}
					    \item \textbf{Verify that $C_k(AB)=DC_k(B)$.}
					    \item \textbf{Yield $\langle D\rangle$.}
				    \end{enumerate}
		    \procedure{4.22}
			    \objective
				    Choose an integer $0\le k\le\min(m,n)$. Choose a $m\times m$ tilt, $A$. Also choose a $m\times n$ matrix, $B$. The objective of the following instructions is to show that $C_k(AB)=C_k(A)C_k(B)$.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{4.20} on matrices $A$ and $1_m$ and let $\langle D\rangle$ receive.
					    \item Using \procedurehr{4.19}, verify that $C_k(A)=C_k(A1_m)=DC_k(1_m)=D1_{\binom{m}{k}}=D$.
					    \item Execute \procedurehr{4.20} on $\langle A,B\rangle$ and let $\langle D'\rangle$ receive.
					    \item Verify that $D'=D=C_k(A)$.
					    \item \textbf{Therefore verify that $C_k(AB)=D'C_k(B)=C_k(A)C_k(B)$.}
				    \end{enumerate}
		    \procedure{4.23}
			    \objective
				    Choose an integer $0\le k\le\min(m,n)$. Choose an $n\times n$ tilt, $A$. Also choose a $m\times n$ matrix, $B$. The objective of the following instructions is to show that $C_k(BA)=C_k(B)C_k(A)$.
			    \implementation
				    Instructions are analogous to those of \procedurehr{4.22}.
		    \procedure{4.24}
			    \objective
				    Choose an integer $0\le k\le\min(m,n)$. Choose an $m\times n$ diagonal matrix, $A$. Also choose a $n\times n$ matrix, $B$. The objective of the following instructions is to show that $C_k(AB)=C_k(A)C_k(B)$.
			    \implementation
				    Instructions are analogous to those of \procedurehr{4.22}.
		    \procedure{4.25}
			    \objective
				    Choose a $m\times n$ matrix, $A$. Let $D_{-1,-1}=1$. The objective of the following instructions is to construct a list of $m\times m$ tilts, $M$, an $m\times n$ diagonal matrix, $D$, a list of polynomials, $v$, and a list of $n\times n$ tiltss, $N$, such that $M_*AN_*=D$, $A={M^{-1}}_*D{N^{-1}}_*$, and $D_{i,i}=v_iD_{i-1,i-1}$ for $i$ in $[0:\min(m,n)]$.
			    \implementation
				    \begin{enumerate}
					    \item Let $D$ be a copy of $A$.
					    \item Let $\langle M,N\rangle$ receive the results of executing \procedurehr{4.03} on the pair $\langle m,n\rangle$ and the following procedure:
						    \begin{enumerate}
							    \item Execute \procedurehr{4.10} on the matrix $D$ and let $\langle v\rangle$ receive.
						    \end{enumerate}
					    \item \textbf{Verify that $D_{i,i}=v_iD_{i-1,i-1}$ for $i$ in $[0:\min(m,n)]$.}
					    \item \textbf{Verify that $M_*AN_*=D$.}
					    \item Hence verify that $A=1_mA1_n={M^{-1}}_*M_*AN_*{N^{-1}}_*={M^{-1}}_*D{N^{-1}}_*$.
					    \item \textbf{Yield the tuple $\langle M,D,v,N\rangle$.}
				    \end{enumerate}
		    \procedure{4.26}
			    \objective
				    Choose integers $0\le k\le\min(m,n,p)$. Choose a $m\times n$ matrix, $A$. Also choose a $n\times p$ matrix, $B$. The objective of the following instructions is to show that $C_k(AB)=C_k(A)C_k(B)$.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{4.25} on $A$ and let $\langle M,D,,N\rangle$ receive.
					    \item Using repeated applications of \procedurehr{4.24}, verify that $C_k(AB)$
					    \begin{enumerate}
						    \item $=C_k({M^{-1}}_0\cdots {M^{-1}}_{\lvert M\rvert-1}D{N^{-1}}_0\cdots {N^{-1}}_{\lvert N\rvert-1}B)$
						    \item $=C_k({M^{-1}}_0)\cdots C_k({M^{-1}}_{\lvert M\rvert-1})*C_k(D)*C_k({N^{-1}}_0)\cdots C_k({N^{-1}}_{\lvert N\rvert-1})C_k(B)$
						    \item $=C_k({M^{-1}}_0\cdots {M^{-1}}_{\lvert M\rvert-1}D{N^{-1}}_0\cdots {N^{-1}}_{\lvert N\rvert-1})C_k(B)$
						    \item $=C_k(A)C_k(B)$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{4.27}
			    \objective
				    Choose a $m\times m$ matrix, $A$. Let $D$ be a copy of $A$. Execute \procedurehr{4.10} on $D$. The objective of the following instructions is to show that $\det(A)$ is the product of the diagonal entries of $D$.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{4.25} on $A$ and let $\langle M,D,,N\rangle$ receive.
					    \item Using \procedurehr{4.26}, verify that $\det(A)$
					    \begin{enumerate}
						    \item $=C_m(A)$
						    \item $=C_m({M^{-1}}_0\cdots {M^{-1}}_{\lvert M\rvert-1}D{N^{-1}}_0\cdots {N^{-1}}_{\lvert N\rvert-1})$
						    \item $=C_m({M^{-1}}_0)\cdots C_m({M^{-1}}_{\lvert M\rvert-1})C_m(D)C_m({N^{-1}}_0)\allowbreak\cdots C_m({N^{-1}}_{\lvert N\rvert-1})$
						    \item $=1\cdots 1C_m(D)1\cdots 1=C_m(D)$
						    \item $=\det(D)$
						    \item $=\prod_r^{[0:m]} D_{r,r}$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{4.12}
			    The notation \gls{matrixtranspose}, where $A$ is a $m\times n$ matrix, will be used to refer to the $n\times m$ matrix such that ${A^T}_{i,j}=A_{j,i}$ for $i$ in $[0:n]$, for $j$ in $[0:m]$.
		    \procedure{4.28}
			    \objective
				    Choose a $m\times n$ matrix, $A$, and a $n\times k$ matrix, $B$. The objective of the following instructions is to show that $B^TA^T=(AB)^T$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that $B^TA^T$ and $(AB)^T$ have dimensions $k\times m$.
					    \item For $i$ in $[0:k]$: For $j$ in $[0:m]$:
					    \begin{enumerate}
						    \item Verify that $(B^TA^T)_{i,j}=\sum_l^{[0:n]} B_{l,i}A_{j,l}=\sum_l^{[0:n]} A_{j,l}B_{l,i}=(AB)_{j,i}=((AB)^T)_{i,j}$.
					    \end{enumerate}
					    \item \textbf{Therefore verify that $B^TA^T=(AB)^T$.}
				    \end{enumerate}
		    \procedure{4.29}
			    \objective
				    Choose a $m\times m$ matrix, $A$. The objective of the following instructions is to show that $\det(A^T)=\det(A)$.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{4.25} on $A$ and let $\langle M,D,,N\rangle$ receive.
					    \item Therefore using procedures \procedurehr{4.27} and \procedurehr{4.28}, verify that $\det(A^T)$
					    \begin{enumerate}
						    \item $=\det(({M^{-1}}_0\cdots {M^{-1}}_{\lvert M\rvert-1}D{N^{-1}}_0\cdots {N^{-1}}_{\lvert N\rvert-1})^T)$
						    \item $=\det(({N^{-1}}_{\lvert N\rvert-1})^T\cdots({N^{-1}}_0)^TD^T({M^{-1}}_{\lvert M\rvert-1})^T\allowbreak\cdots({M^{-1}}_0)^T)$
						    \item $=\det(D^T)$
						    \item $=\det(D)$
						    \item $=\det({M^{-1}}_0\cdots {M^{-1}}_{\lvert M\rvert-1}D{N^{-1}}_0\cdots {N^{-1}}_{\lvert N\rvert-1})$
						    \item $=\det(A)$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{4.30}
			    \objective
				    Choose a $m\times n$ matrix, $A$, and an integer $0\le k\le\min(m,n)$. The objective of the following instructions is to show that $C_k(A)^T=C_k(A^T)$.
			    \implementation
				    \begin{enumerate}
					    \item For each row label $I$ of $C_k(A^T)$, do the following:
					    \begin{enumerate}
						    \item For each column label $J$ of $C_k(A^T)$, do the following:
						    \begin{enumerate}
							    \item Using \procedurehr{4.29}, verify that $(C_k(A^T))_{\ul{I},\ul{J}}=\det((A^T)_{I,J})=\det(A_{J,I})=(C_k(A))_{\ul{J},\ul{I}}$.
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Therefore verify that $(C_k(A))^T=(C_k(A^T))$.}
				    \end{enumerate}
	    \end{multicols*}
	  \chapter{Polynomials and Normal Forms}
	    \begin{multicols*}{2}
		    \procedure{4.31}
			    \objective
				    Choose a $m\times n$ rational matrix, $A$, and a $m\times p$ rational matrix, $B$. Execute \procedurehr{4.25} on $A$ and let $\langle M,D,,N\rangle$ receive the result. If the indices of the rows of $D$ that are entirely zero are also the indices of the rows of $M_*B$ that are entirely zero, then the objective of the following instructions is to construct a $n\times p$ rational matrix $E$ such that $AE=B$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that $A={M^{-1}}_*D{N^{-1}}_*$.
					    \item Verify that ${M^{-1}}_*$, $D$, and ${N^{-1}}_*$ are rational matrices.
					    \item Let $C$ be an $n\times p$ matrix with its $i^{th}$ row given as follows:
					    \begin{enumerate}
						    \item If $D_{i,i}\ne 0$, then do the following:
						    \begin{enumerate}
							    \item Let row $i$ be row $i$ of $M_*B$ divided by $D_{i,i}$.
						    \end{enumerate}
						    \item Otherwise, do the following:
						    \begin{enumerate}
							    \item \textbf{Choose $p$ rational numbers to fill up the row.}
						    \end{enumerate}
					    \end{enumerate}
					    \item Verify that $DC=M_*B$.
					    \item Let $E$ be $N_*C$.
					    \item \textbf{Therefore using \procedurehr{4.05}, verify that $AE={M^{-1}}_*D{N^{-1}}_*E={M^{-1}}_*D{N^{-1}}_*N_*C={M^{-1}}_*D1_nC={M^{-1}}_*DC={M^{-1}}_*M_*B=1_mB=B$.}
					    \item \textbf{Yield the tuple $\langle E\rangle$.}
				    \end{enumerate}
		    \declaration{4.13}
			    The notation \gls{matrixleftdiv} will be used to refer to the result yielded by executing \procedurehr{4.31} on $\langle A,B\rangle$.
		    \procedure{4.32}
			    \objective
				    Choose a $m\times n$ rational matrix, $A$, and a $p\times n$ rational matrix, $B$. Execute \procedurehr{4.25} on $A$ and let $\langle M,D,,N\rangle$ receive the result. If the indices of the columns of $D$ that are entirely zero are also the indices of the columns of $BN_*$ that are entirely zero, then the objective of the following instructions is to construct a $p\times m$ rational matrix $E$ such that $EA=B$.
			    \implementation
				    Instructions are analogous to those of \procedurehr{4.31}.
		    \declaration{4.14}
			    The notation \gls{matrixrightdiv} will be used to refer to the result yielded by executing \procedurehr{4.32} on $\langle A,B\rangle$.
		    \procedure{4.33}
			    \objective
				    Choose a $m\times n$ rational matrix, $A$, a $n\times p$ rational matrix, $E$, and a $m\times p$ rational matrix, $B$ such that $AE=B$. Execute \procedurehr{4.25} on $A$ and let $\langle M,D,,N\rangle$ receive the result. If the indices of the rows of $D$ that are entirely zero are not also the indices of the rows of $M_*B$ that are entirely zero, then the objective of the following instructions is to show that $0\ne 0$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that ${M^{-1}}_*D{N^{-1}}_*E=AE=B$.
					    \item Therefore verify that $D{N^{-1}}_*E=M_*B$.
					    \item Let $i$ be an integer such that $D_{i,*}$ is zero and yet $(M_*B)_{i,*}$ is not zero.
					    \item Verify that $D_{i,*}=D_{i,*}{N^{-1}}_*E=(D{N^{-1}}_*E)_{i,*}=(M_*B)_{i,*}$.
					    \item Let $j$ be an integer such that $(M_*B)_{i,j}\ne 0$.
					    \item \textbf{Now verify that $0=D_{i,j}=(M_*B)_{i,j}\ne 0$.}
				    \end{enumerate}
		    \procedure{4.34}
			    \objective
				    Choose a $p\times m$ rational matrix, $E$, a $m\times n$ rational matrix, $A$, and a $p\times n$ rational matrix, $B$ such that $EA=B$. Execute \procedurehr{4.25} on $A$ and let $\langle M,D,,N\rangle$ receive the result. If the indices of the columns of $D$ that are entirely zero are not also the indices of the columns of $BN_*$ that are entirely zero, then the objective of the following instructions is to show that $0\ne 0$.
			    \implementation
				    Instructions are analogous to those of \procedurehr{4.33}.
		    \procedure{4.35}
			    \objective
				    Choose two $m\times m$ rational matrices, $A$ and $B$, such that $AB=1_m$. The objective of the following instructions is to show that either $0=1$ or $BA=1_m$.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{4.25} on $B$ and let $\langle M,D,,N\rangle$ receive the result.
					    \item Verify that $B={M^{-1}}_*D{N^{-1}}_*$.
					    \item If $D$ has a zero on its diagonal, then do the following:
					    \begin{enumerate}
						    \item Using \procedurehr{4.27}, verify that $\det(1_m)=\det(AB)=\det(A)\det(B)=\det(A)\det(D)=\det(A)*0=0$.
						    \item Also verify that $\det(1_m)=1^m=1$.
						    \item Therefore verify that $0=1$.
						    \item \textbf{Abort procedure.}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
						    \item Verify that $D$ does not have a zero on its diagonal.
						    \item Verify that $B\backslash 1_m=1_m(B\backslash 1_m)=AB(B\backslash 1_m)=A(B(B\backslash 1_m))=A1_m=A$.
						    \item \textbf{Therefore verify that $BA=B(B\backslash 1_m)=1_m$.}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{4.36}
			    \objective
				    Choose an $m\times m$ matrix, $M$, and an $m\times m$ rational matrix, $B$. The objective of the following instructions is to construct a $m\times m$ matrix, $Q$, and a $m\times m$ rational matrix, $R$, such that $M=(\lambda 1_m-B)Q+R$.
			    \implementation
				    \begin{enumerate}
					    \item Let $M_0\lambda^b+M_1\lambda^{b-1}+\cdots+M_b\lambda^0=M$, where the $M_i$ are $m\times m$ rational matrices.
					    \item Now let $R=B^bM_0+B^{b-1}M_1+\cdots+B^0M_b$.
					    \item Let $Q=\sum_k^{[1:b]} (\lambda^{k-1}1_mB^0+\lambda^{k-2}1_mB^1+\cdots+\lambda^01_mB^{k-1})M_k$.
					    \item Verify that $M-R=(\lambda 1_m-B)\sum_k^{[1:b]} (\lambda^{k-1}1_mB^0+\lambda^{k-2}1_mB^1+\cdots+\lambda^01_mB^{k-1})M_k=(\lambda 1_m-B)Q$.
					    \item \textbf{Verify that $M=(\lambda 1_m-B)Q+R$.}
					    \item \textbf{Yield the tuple $\langle Q,R\rangle$.}
				    \end{enumerate}
		    \procedure{4.37}
			    \objective
				    Choose an $m\times m$ matrix, $M$, and an $m\times m$ rational matrix, $B$. The objective of the following instructions is to construct a $m\times m$ matrix, $Q$, and a $m\times m$ rational matrix, $R$, such that $M=Q(\lambda 1_m-B)+R$.
			    \implementation
				    The instructions are analogous to those of \procedurehr{4.36}.
		    \procedure{4.38}
			    \objective
				    Choose two $m\times m$ rational matrices, $B,A$, and two lists of $m\times m$ tilts such that $\lambda 1_m-B=M(\lambda 1_m-A)N$. The objective of the following instructions is to either show that $0=1$ or to construct $m\times m$ rational matrices $R_1$ and $R_3$ such that $1_m=R_1R_3$ and $B=R_1AR_3$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that $(\lambda 1_m-B)N^{-1}=M(\lambda 1_m-A)NN^{-1}=M(\lambda 1_m-A)1_m=M(\lambda 1_m-A)$.
					    \item Execute \procedurehr{4.37} on $\langle M,B\rangle$ and let $\langle Q_1,R_1\rangle$ receive.
					    \item Verify that $M=(\lambda 1_m-B)Q_1+R_1$.
					    \item Execute \procedurehr{4.37} on $\langle N^{-1},A\rangle$ and let $\langle Q_2,R_2\rangle$ receive.
					    \item Verify that $N^{-1}=Q_2(\lambda 1_m-A)+R_2$.
					    \item By substituting $M$ and $N^{-1}$ into (2), verify that $(\lambda 1_m-B)(Q_2(\lambda 1_m-A)+R_2)=((\lambda 1_m-B)Q_1+R_1)(\lambda 1_m-A)$.
					    \item By rearranging both sides, verify that $(\lambda 1_m-B)(Q_2-Q_1)(\lambda 1_m-A)=R_1(\lambda 1_m-A)-(\lambda 1_m-B)R_2$.
					    \item By equating the coefficients of different powers of $\lambda$ both sides, verify that $Q_2-Q_1=0_{m\times m}$.
					    \item Verify that $R_1(\lambda 1_m-A)-(\lambda 1_m-B)R_2=(\lambda 1_m-B)(Q_2-Q_1)(\lambda 1_m-A)=(\lambda 1_m-B)0_{m\times m}(\lambda 1_m-A)=0_{m\times m}$.
					    \item Therefore by adding $(\lambda 1_m-B)R_2$ to both sides, verify that $\lambda R_1-R_1A=R_1(\lambda 1_m-A)=(\lambda 1_m-B)R_2=\lambda R_2-BR_2$.
					    \item By equating the coefficients of $\lambda$ on both sides, verify that $R_1=R_2$.
					    \item Therefore verify that $R_1A=BR_1$.
					    \item Execute \procedurehr{4.37} on $\langle M^{-1},A\rangle$ and let $\langle Q_3,R_3\rangle$ receive.
					    \item Verify that $M^{-1}=(\lambda 1_m-A)Q_3+R_3$.
					    \item Verify that $1_m=MM^{-1}=((\lambda 1_m-B)Q_1+R_1)M^{-1}=(\lambda 1_m-B)Q_1M^{-1}+R_1M^{-1}=(\lambda 1_m-B)Q_1M^{-1}+R_1(\lambda I-A)Q_3+R_1R_3=(\lambda 1_m-B)Q_1M^{-1}+(\lambda I-B)R_1Q_3+R_1R_3=(\lambda 1_m-B)(Q_1M^{-1}+R_1Q_3)+R_1R_3$.
					    \item By equating the powers of $\lambda$ on both sides, verify that $Q_1M^{-1}+R_1Q_3=0$.
					    \item By substituting zero for $Q_1M^{-1}+R_1Q_3$, \textbf{verify that $1_m=(\lambda 1_m-B)0_{m\times m}+R_1R_3=R_1R_3$.}
					    \item \textbf{Therefore using \procedurehr{4.35}, verify that $R_3R_1=1_m$.}
					    \item \textbf{Also, verify that $B=B1_m=BR_1R_3=R_1AR_3$.}
					    \item \textbf{Yield the pair $(R_1,R_3)$.}
				    \end{enumerate}
		    \procedure{4.39}
			    \objective
				    Choose a $m\times n$ matrix, $A$. Choose two integers $0\le i,j<m$ such that $i\ne j$. The objective of the following instructions is to negate row $i$ and swap it with row $j$ using only elementary row operations.
			    \implementation
				    \begin{enumerate}
					    \item Let $A$ be our working matrix.
					    \item Subtract row $j$ from row $i$.
					    \item Add row $i$ to row $j$.
					    \item Subtract row $j$ from row $i$.
					    \item \textbf{Verify that the $i^{th}$ row has been negated and swapped with the $j^{th}$ row.}
				    \end{enumerate}
		    \procedure{4.40}
			    \objective
				    Choose a $m\times n$ matrix, $A$. Choose two integers $0\le i,j<n$ such that $i\ne j$. The objective of the following instructions is to negate column $i$ and swap it with row $j$ using only elementary column operations.
			    \implementation
				    The instructions are analogous to those of \procedurehr{4.39}.
		    \procedure{4.41}
			    \objective
				    Choose an $m\times n$ diagonal matrix, $A$. Choose two integers $0\le i,j<\min(m,n)$ such that $i\ne j$. The objective of the following instructions is to swap $B_{i,i}$ and $B_{j,j}$ using only elementary row and column operations.
			    \implementation
				    \begin{enumerate}
					    \item Let $A$ be our working matrix.
					    \item Use \procedurehr{4.40} to negate the $i^{th}$ row and swap it with the $j^{th}$ row.
					    \item Use \procedurehr{4.40} to negate the $i^{th}$ column and swap it with the $j^{th}$ column.
					    \item \textbf{Therefore, overall verify that $B_{i,i}$ and $B_{j,j}$ have been swapped.}
				    \end{enumerate}
		    \procedure{4.42}
			    \objective
				    Choose an $m\times n$ diagonal matrix, $A$. Choose two integers $0\le i,j<\min(m,n)$ such that $i\ne j$. Choose a rational $k\ne 0$. The objective of the following instructions is to multiply $B_{i,i}$ by $k$ and $B_{j,j}$ by $\frac{1}{k}$ using only elementary row and column operations.
			    \implementation
				    \begin{enumerate}
					    \item Let $A$ be our working matrix.
					    \item Add $k$ times row $i$ to row $j$.
					    \item Subtract $\frac{1}{k}$ times row $j$ from row $i$.
					    \item Add $k$ times row $i$ to row $j$.
					    \item Verify that the $i^{th}$ row has been scaled by $k$, the $j^{th}$ row by $-\frac{1}{k}$, and that both these rows are swapped.
					    \item Use \procedurehr{4.40} to negate the $i^{th}$ row and swap it with the $j^{th}$ row.
					    \item \textbf{Therefore, overall verify that $B_{i,i}$ has been multiplied by $k$, and $B_{j,j}$ by $\frac{1}{k}$.}
				    \end{enumerate}
		    \procedure{4.43}
			    \objective
				    Choose a $m\times m$ rational matrix, $A$. Execute \procedurehr{4.10} on the polynomial matrix $\lambda I-A$ and let $\langle B\rangle$ be the result. The objective of the following instructions is to show that either none of the diagonal entries of $B$ are equal to zero, or $1=0$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that $\det(\lambda I-A)$ is a monic polynomial of degree $m$.
					    \item Therefore using \procedurehr{4.27}, verify that $\det(B)=\det(\lambda I-A)$.
					    \item Therefore verify that $\det(B)$ is a monic polynomial of degree $m$.
					    \item If any of the diagonal entries of $B$ equal zero, then do the following:
					    \begin{enumerate}
						    \item Verify that $\det(B)=B_{0,0}B_{1,1}\cdots B_{m-1,m-1}=0$.
						    \item Therefore using (3) and (4a), verify that $1=0$.
						    \item \textbf{Abort procedure.}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
						    \item \textbf{Verify that none of the diagonal entries of $B$ equal zero.}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{4.44}
			    \objective
				    Choose a positive integer $m$ and an $m\times m$ rational matrix, $A$. Execute \procedurehr{4.25} on the polynomial matrix $\lambda 1_m-A$ and let $\langle ,B,v,\rangle$ be the result. The objective of the following instructions is to either show that $0<0$ or to construct an integer $a$ such that $\sum_i^{[a:m]}\deg(B_{i,i})=m$, $\deg(B_{i,i})>0$ for $i$ in $[a:m]$, and $\deg(B_{i,i})=0$ for $i$ in $[0:a]$.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{4.43} on $A$.
					    \item If $\deg(B_{i,i})=0$ for $i$ in $[0:m]$, then do the following:
					    \begin{enumerate}
						    \item Verify that $\det(\lambda 1_m-A)=\det(B)=B_{0,0}B_{1,1}\cdots B_{m-1,m-1}$.
						    \item \textbf{Therefore verify that $0<m=\deg(\det(\lambda 1_m-A))=\deg(B_{0,0}B_{1,1}\cdots B_{m-1,m-1})=0+0+\cdots+0=0$.}
						    \item \textbf{Abort procedure.}
					    \end{enumerate}
					    \item Otherwise do the following:
					    \begin{enumerate}
						    \item Let $0\le a<m$ be the least integer such that $\deg(B_{a,a})>0$.
						    \item \textbf{Verify that $\deg(B_{i,i})=0$ for $i$ in $[0:a]$.}
						    \item \textbf{Verify that $\sum_i^{[a:m]}\deg(B_{i,i})=\sum_i^{[0:m]}\deg(B_{i,i})=\deg(B_{0,0}B_{1,1}\cdots B_{m-1,m-1})=\deg(\det(B))=\deg(\lambda 1_m-A)=m$.}
						    \item For $i$ in $[a+1:m]$, do the following:
						    \begin{enumerate}
							    \item Verify that $B_{i,i}=u_iB_{i-1,i-1}$.
							    \item Verify that $B_{i,i}\ne 0$.
							    \item Therefore verify that $u_i\ne 0$.
							    \item \textbf{Therefore verify that $\deg(B_{i,i})=\deg(u_iB_{i-1,i-1})\ge\deg(B_{i-1,i-1})>0$.}
						    \end{enumerate}
						    \item \textbf{Yield the tuple $\langle a\rangle$.}
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{4.19}
			    The notation \gls{matrixe} will be used to refer to the $k\times 1$ rational matrix such that its $i^{th}$ entry, $1$, is the only non-zero entry.
		    \declaration{4.22}
			    The notation \gls{matrixmattp} will be used as a shorthand for $\sum_j^{[0:t]}p_je_j$.
		    \declaration{4.16}
			    The notation \gls{matrixcompmatrix}, where $p\ne 0$ is a monic polynomial such that $\deg(p)>0$, will be used as a shorthand for the $\deg(p)\times\deg(p)$ rational matrix of the following constitution:
			    \begin{enumerate}
				    \item Its first $\deg(p)-1$ columns equal the last $\deg(p)-1$ columns of $1_k$.
				    \item Its last column is $-\mat_{\deg(p)}(p)$.
			    \end{enumerate}
		    \procedure{4.45}
			    \objective
				    Choose a monic polynomial, $p$ such that $\deg(p)>0$. Let $k=\deg(p)$. Choose a $k\times k$ matrix, $D$, such that $D=\lambda 1_k-\comp(p)$. The objective of the following instructions is to transform $D$ into $\diag(1,\cdots,1,p)$ by a sequence of elementary operations.
			    \implementation
				    \begin{enumerate}
					    \item Let the matrix $D$ be our working matrix.
					    \item For $i$ in $[k:1]$, add $\lambda$ times row $i$ to row $i-1$.
					    \item Verify that $D$'s first $k-1$ columns are now the last $k-1$ columns of $-1_k$.
					    \item Verify that $D$'s last column is $p$ followed by some other polynomials.
					    \item For $i$ in $[1:k]$, subtract $D_{i,k-1}$ times column $i-1$ from column $k-1$.
					    \item Verify that $D$'s last column is now $p$ followed by zeros.
					    \item For $i$ in $[1:k]$, negate row $i-1$ and exchange it with row $i$ using \procedurehr{4.40}.
					    \item \textbf{Therefore verify that $D=\diag(1,\cdots,1,p)$.}
				    \end{enumerate}
		    \procedure{4.46}
			    \objective
				    Choose a positive integer $m$ and an $m\times m$ rational matrix, $A$. Execute \procedurehr{4.03} on the polynomial matrix $\lambda 1_m-A$ and let $\langle,B,,\rangle$ receive the result. Execute \procedurehr{4.44} on $A$ and let $\langle a\rangle$ receive the result. Let $E_i=\comp(\mon(B_{a+i,a+i}))$ for $i$ in $[0:m-a]$. The objective of the following instructions is to first show that $\cols(\diag(E))=m$, and second to apply a sequence of elementary operations on $\lambda 1_m-\diag(E)$ to obtain the matrix $B$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that the diagonal of $B$ comprises $a$ rationals followed by $B_{a,a},B_{a+1,a+1},\cdots,B_{m-1,m-1}$.
					    \item \textbf{Using \procedurehr{4.45}, verify that $\cols(\diag(E))=\sum_i^{[0:{\lvert E\rvert}]}\cols(E_i)=\sum_i^{[0:{\lvert E\rvert}]}\cols(\comp(\mon(B_{a+i,a+i})))=\sum_i^{[0:{\lvert E\rvert}]}\deg(\mon(B_{a+i,a+i}))=\sum_i^{[0:{m-a}]}\deg(B_{a+i,a+i})=\sum_i^{[a:m]}\deg(B_{i,i})=m$.}
					    \item Let $F=\lambda 1_m-\diag(E)$.
					    \item Now for $i$ in $[0:\lvert E\rvert]$:
					    \begin{enumerate}
						    \item Let $j=\sum_r^{[0:{i}]}\cols(E_r)$.
						    \item Let $k=j+\cols(E_i)$.
						    \item Apply \procedurehr{4.45} on the tuple $\langle\mon(B_{a+i,a+i}),F_{[j:k],[j:k]}\rangle$.
					    \end{enumerate}
					    \item Now verify that $F$ is an $m\times m$ diagonal rational matrix.
					    \item Also verify that the diagonal of $F$ comprises $\mon(B_{a,a}),\mon(B_{a+1,a+1}),\cdots,\mon(B_{m-1,m-1})$ and $a$ $1$s.
					    \item Rearrange the diagonal of $F$ so that $\mon(B_{i,i})$ is at the $i^{th}$ position on the diagonal for $i$ in $[a:m]$ by doing pairwise swaps. In general, swap the $i^{th}$ and $j^{th}$ diagonal entries using \procedurehr{4.41}.
					    \item For $i$ in $[0:m-1]$, do the following:
					    \begin{enumerate}
						    \item Let $k=\frac{(B_{i,i})_{\deg(B_{i,i})}}{(F_{i,i})_{\deg(F_{i,i})}}$.
						    \item Scale $B_{i,i}$ by $k$ and $B_{i+1,i+1}$ by $\frac{1}{k}$ using \procedurehr{4.42}.
						    \item Now verify that $F_{i,i}=B_{i,i}$.
					    \end{enumerate}
					    \item Now verify that $\det(F)_{m}=\det(\lambda 1_m-\diag(E))_m=1=\det(\lambda 1_m-A)_m=\det(B)_m$.
					    \item Therefore verify that $(F_{m,m})_{\deg(F_{m,m})}$
					    \begin{enumerate}
						    \item $=\frac{\det(F)_m}{(\det(F_{[1:m],[1:m]}))_{m-\deg(F_{m,m})}}$
						    \item $=\frac{\det(B)_m}{(\det(B_{[1:m],[1:m]}))_{m-\deg(B_{m,m})}}$
						    \item $=(B_{m,m})_{\deg(B_{m,m})}$.
					    \end{enumerate}
					    \item Therefore verify that $F_{m,m}=B_{m,m}$.
					    \item \textbf{Therefore verify that $F=B$.}
				    \end{enumerate}
		    \procedure{4.47}
			    \objective
				    Choose a $m\times m$ rational matrix, $A$. Execute \procedurehr{4.44} on $A$ and let $\langle a\rangle$ receive the result. Let $E_i=\comp(\mon(B_{a+i,a+i}))$ for $i$ in $[0:m-a]$. The objective of the following instructions is to either show that $0=1$ or to construct $m\times m$ rational matrices $R,T$ such that $A=R\diag(E)T$, $RT=1_m$, and $TR=1_m$.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{4.25} on the polynomial matrix $\lambda 1_m-A$ and let $\langle P,B,,Q\rangle$ be the result.
					    \item Verify that $P_*(\lambda 1_m-A)Q_*=B$.
					    \item Verify that $\lambda 1_m-A={P^{-1}}_*B{Q^{-1}}_*$.
					    \item Let $Z$ be a variant of \procedurehr{4.25} where every occurence of \procedurehr{4.10} in its instructions is replaced with \procedurehr{4.46}, and where every mention of $v$ is ignored.
					    \item Execute procedure $Z$ on the matrix $\lambda 1_m-\diag(E)$ and let $\langle M,,,N\rangle$ receive the result.
					    \item Verify that $M_*(\lambda 1_m-\diag(E))N_*=B$.
					    \item Verify that $\lambda 1_m-A={P^{-1}}_*B{Q^{-1}}_*={P^{-1}}_*M(\lambda 1_m-\diag(E))N{Q^{-1}}_*$.
					    \item Execute \procedurehr{4.38} on the matrices $\langle A,{P}^{-1}M,\diag(E),N{Q}^{-1}\rangle$. Let the tuple $\langle R,T\rangle$ be the result.
					    \item \textbf{Verify that $A=R\diag(E)T$.}
					    \item \textbf{Verify that $RT=1_m$.}
					    \item \textbf{Verify that $TR=1_m$.}
					    \item \textbf{Yield the tuple $\langle R,E,T\rangle$.}
				    \end{enumerate}
		    \procedure{4.86}
			    \objective
				    Choose two polynomials $a,b$ and an $m\times m$ matrix $C$ such that $a=b$. The objective of the following instructions is to show that $\Lambda(a,C)=\Lambda(b,C)$.
			    \implementation
				    Implementation is analogous to that of \procedurehr{2.51}.
		    \procedure{4.87}
			    \objective
				    Choose two polynomials $a,b$ and an $m\times n$ matrix $C$. The objective of the following instructions is to show that $\Lambda(a+b,C)=\Lambda(a,C)+\Lambda(b,C)$.
			    \implementation
				    Implementation is analogous to that of \procedurehr{2.56}.
		    \procedure{4.88}
			    \objective
				    Choose a polynomial $a$ and an $m\times m$ matrix $B$. The objective of the following instructions is to show that $\Lambda(-a,B)=-\Lambda(a,B)$.
			    \implementation
				    Implementation is analogous to that of \procedurehr{2.00}.
		    \procedure{4.89}
			    \objective
				    Choose two polynomials $a,b$ and an $m\times m$ matrix $C$. The objective of the following instructions is to show that $\Lambda(ab,C)=\Lambda(a,C)\Lambda(b,C)$.
			    \implementation
				    Implementation is analogous to that of \procedurehr{2.64}.
		    \procedure{4.48}
			    \objective
				    Choose a polynomial, $r$, and $m\times m$ rational matrices, $R,A,S$ such that $SR=1_m$. The objective of the following instructions is to show that $\Lambda(r,RAS)=R\Lambda(r,A)S$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that $\Lambda(r,RAS)$
					    \begin{enumerate}
						    \item $=\sum_j^{[0:\lvert r\rvert]}r_j(RAS)^j$
						    \item $=\sum_j^{[0:\lvert r\rvert]}r_jRA^jS$
						    \item $=R(\sum_j^{[0:\lvert r\rvert]}r_jA^j)S$
						    \item $=R\Lambda(r,A)S$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{4.49}
			    \objective
				    Choose a list of $m\times m$ rational matrices, $A$, and a polynomial, $r$. The objective of the following instructions is to show that $\Lambda(r,\diag(A))=\diag(\Lambda(r,A))$.
			    \implementation
				    \begin{enumerate}
					    \item For $i=0$ up to $i=t$, by repeated applications of \procedurehr{4.09}, verify that $\diag(A)^i$ evaluates to $\diag(A^i)$.
					    \item Therefore verify that $\Lambda(r,\diag(A))$
					    \begin{enumerate}
						    \item $=\sum_j^{[0:\lvert r\rvert]}r_j\diag(A)^j$
						    \item $=\sum_j^{[0:\lvert r\rvert]}r_j\diag(A^j)$
						    \item $=\sum_j^{[0:\lvert r\rvert]}\diag(r_jA^j)$
						    \item $=\diag(\sum_j^{[0:\lvert r\rvert]}r_jA^j)$
						    \item \textbf{$=\diag(\Lambda(r,A))$.}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{4.50}
			    \objective
				    Choose a $m\times m$ rational matrix, $A$, and a polynomial, $r$. Execute \procedurehr{4.47} on the matrix $A$ and let the tuple $\langle R_1,E,R_3\rangle$ receive the result. The objective of the following instructions is to show that $\Lambda(r,A)=R_1\diag(\Lambda(r,E))R_3$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that $R_3R_1=1_m$.
					    \item Using \procedurehr{4.48}, verify that $\Lambda(r,A)=\Lambda(r,R_1\diag(E)R_3)=R_1\Lambda(r,\diag(E))R_3$.
					    \item Using \procedurehr{4.49}, verify that $\Lambda(r,\diag(E))=\diag(\Lambda(r,E))$.
					    \item \textbf{Therefore verify that $\Lambda(r,A)=R_1\diag(\Lambda(r,E))R_3$.}
				    \end{enumerate}
		    \procedure{4.51}
			    \objective
				    Choose a monic polynomial $p\ne 0$ such that $\deg(p)>0$. The objective of the following instructions is to show that $\Lambda(p,\comp(p))=0_{\deg(p)\times\deg(p)}$.
			    \implementation
				    \begin{enumerate}
					    \item Let $G=\comp(p)$.
					    \item For $i$ in $[0:\deg(p)]$, verify that $G^{i}e_0=G^{i-1}e_1=\cdots=G^{0}e_i=e_i$.
					    \item Therefore, for $i\in[0:\deg(p)]$, do the following:
					    \begin{enumerate}
						    \item Using (1), verify that $\Lambda(p,G)e_i$
						    \begin{enumerate}
							    \item $=(\sum_j^{[0:\lvert p\rvert]}p_{j}G^{j})e_i$
							    \item $=(\sum_j^{[0:\lvert p\rvert]}p_{j}G^{j})G^{i}e_0$
							    \item $=G^{i}(GG^{\deg(p)-1}+\sum_j^{[0:\deg(p)]}p_{j}G^{j})e_0$
							    \item $=G^{i}(Ge_{\deg(p)-1}+\sum_j^{[0:\deg(p)]}p_{j}e_j)$
							    \item $=G^{i}0_{\deg(p)\times 1}$
							    \item $=0_{\deg(p)\times 1}$.
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Therefore verify that $\Lambda(p,\comp(p))=\Lambda(p,G)=0_{\deg(p)\times\deg(p)}$.}
				    \end{enumerate}
		    \declaration{4.20}
			    The notation \gls{matrixlast}, where $A$ is an $m\times m$ rational matrix, will be used as a shorthand for the polynomial yielded by executing the following instructions:
			    \begin{enumerate}
				    \item Execute \procedurehr{4.25} on the polynomial matrix $\lambda 1_m-A$ and let the tuple $\langle,B,,\rangle$ receive the result.
				    \item Yield $\langle B_{m-1,m-1}\rangle$.
			    \end{enumerate}
		    \procedure{4.52}
			    \objective
				    Choose a $m\times m$ rational matrix, $A$. The objective of the following instructions is to show that either $1=0$ or $\last_A\ne 0$.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{4.43} on $A$.
					    \item \textbf{Therefore verify that $\last_A\ne 0$.}
				    \end{enumerate}
		    \procedure{4.53}
			    \objective
				    Choose a $m\times m$ rational matrix, $A$. The objective of the following instructions is to either show that $0<0$ or to show that $\Lambda(\last_A,A)=0_{m\times m}$.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{4.25} on the matrix $A$ and let the tuple $\langle M,B,v,N\rangle$ receive the result.
					    \item Execute \procedurehr{4.44} on $A$ and let $\langle a\rangle$ receive.
					    \item Execute \procedurehr{4.47} on $A$ and let $\langle R,E,T\rangle$ receive.
					    \item For $j$ in $[0:\lvert E\rvert]$:
					    \begin{enumerate}
						    \item Verify that $E_j=\comp(\mon(B_{a+j,a+j}))$.
						    \item Verify that $\last_A=B_{m-1,m-1}=B_{a+j,a+j}\prod_r^{[a+j+1:m]} v_r$.
						    \item Let $k=\deg(\mon(B_{a+j,a+j}))$.
						    \item Therefore using \procedurehr{4.51} verify that $\Lambda(\last_A,E_j)=\Lambda(B_{m-1,m-1},E_j)=\Lambda(B_{a+j,a+j},\comp(\mon(B_{a+j,a+j})))\prod_r^{[a+j+1:m]}\Lambda(v_r,E_j)=0_{k\times k}\prod_r^{[a+j+1:m]}\Lambda(v_r,E_j)=0_{k\times k}$.
					    \end{enumerate}
					    \item \textbf{Therefore using \procedurehr{4.50} verify that $\Lambda(\last_A,A)=R\diag(\Lambda(\last_A,E))T=R\diag(\Lambda(B_{m-1,m-1},E))T=R0_{m\times m}T=0_{m\times m}$.}
				    \end{enumerate}
		    \procedure{4.54}
			    \objective
				    Choose a monic polynomial $p$ such that $\deg(p)>0$. Choose a polynomial $g\ne 0$ such that $\deg(g)<\deg(p)$. The objective of the following instructions is to show that $\Lambda(g,\comp(p))\ne 0_{\deg(p)\times\deg(p)}$.
			    \implementation
				    \begin{enumerate}
					    \item Let $G=\comp(p)$.
					    \item Therefore using \declarationhr{4.16}, verify that $\Lambda(g,G)e_0=(\sum_j^{[0:\deg(g)+1]}g_jG^j)e_0=\sum_j^{[0:\deg(g)+1]}g_je_j\ne 0_{\deg(p)\times 1}$.
					    \item \textbf{Therefore verify that $\Lambda(g,G)\ne 0_{\deg(p)\times\deg(p)}$.}
				    \end{enumerate}
		    \procedure{4.55}
			    \objective
				    Choose a polynomial $g$ and a monic polynomial $p$ such that $\deg(p)=\deg(g)>0$ and $\Lambda(g,\comp(p))=0_{\deg(g)\times\deg(g)}$. The objective of the following instructions is to show that $g=g_{\deg(g)}p$.
			    \implementation
				    \begin{enumerate}
					    \item Let $G=\comp(p)$.
					    \item Using \declarationhr{4.16}, verify that $0_{\deg(g)\times 1}=\Lambda(g,G)e_0=(\sum_j^{[0:\lvert g\rvert]}g_jG^j)e_0=g_{\deg(g)}Ge_{\deg(g)-1}+\sum_j^{[0:\deg(g)]}g_je_j$.
					    \item Therefore for $i$ in $[0:\deg(g)]$, do the following:
					    \begin{enumerate}
						    \item Verify that $0=(g_{\deg(g)}Ge_{\deg(g)-1}+\sum_j^{[0:\deg(g)]}g_je_j)_{i,0}$.
						    \item Therefore using \declarationhr{4.16}, verify that $-g_{\deg(g)}p_{i}+g_{i}=0$.
						    \item Therefore verify that $g_{i}=g_{\deg(g)}p_{i}$.
					    \end{enumerate}
					    \item \textbf{Therefore verify that $g=g_{\deg(g)}p$.}
				    \end{enumerate}
		    \procedure{4.56}
			    \objective
				    Choose a $m\times m$ rational matrix, $A$. Choose a polynomial $p\ne 0$, such that $\Lambda(p,A)=0_{m\times m}$. The objective of the following instructions is to either show that $0\ne 0$ or to construct a polynomial $f$ such that $p=f\last_A$.
			    \implementation
				    \begin{enumerate}
					    \item Let $F$ be the $1\times 2$ matrix $\langle\langle p,\last_A\rangle\rangle$.
					    \item Execute \procedurehr{4.25} on $F$ and let $\langle M,D,,N\rangle$ receive the result.
					    \item Verify that $D_{0,0}\ne 0$.
					    \item Let $g=D_{0,0}$.
					    \item Verify that $F={M^{-1}}_*D{N^{-1}}_*=D{N^{-1}}_*$.
					    \item Verify that $\last_A=F_{0,1}=D_{0,0}{{N^{-1}}_*}_{0,1}+D_{0,1}{{N^{-1}}_*}_{1,1}=D_{0,0}{{N^{-1}}_*}_{0,1}=g{{N^{-1}}_*}_{0,1}$.
					    \item Therefore verify that ${{N^{-1}}_*}_{0,1}\ne 0$.
					    \item Let $u=\deg(\last_A)$.
					    \item Now verify that $u=\deg(\last_A)=\deg(D_{0,0}{{N^{-1}}_*}_{0,1})\ge\deg(D_{0,0})=\deg(g)$.
					    \item Verify that $D=M_*FN_*=FN_*$.
					    \item Therefore verify that $g=D_{0,0}={N_*}_{0,0}p+{N_*}_{1,0}\last_A$.
					    \item Therefore using \procedurehr{4.51}, verify that $\Lambda(g,A)=\Lambda({N_*}_{0,0},A)\Lambda(p,A)+\Lambda({N_*}_{1,0},A)\Lambda(\last_A,A)=\Lambda({N_*}_{0,0},A)0_{m\times m}+\Lambda({N_*}_{1,0},A)0_{m\times m}=0_{m\times m}$.
					    \item Execute \procedurehr{4.47} on the matrix $A$ and let the tuple $\langle R_1,E,R_3\rangle$ receive the result.
					    \item Using \procedurehr{4.50}, and \procedurehr{4.47}, verify that $\diag(\Lambda(g,E))=1_m\diag(\Lambda(g,E))1_m=R_3R_1\diag(\Lambda(g,E))R_3R_1=R_3\Lambda(g,A)R_1=R_30_{m\times m}R_1=0_{m\times m}$.
					    \item Let $G=\comp(\mon(\last_A))$.
					    \item Verify that $\Lambda(g,G)=\Lambda(g,E_{\lvert E\rvert-1})=\diag(\Lambda(g,E))_{[m-u:m],[m-u:m]}=0_{u\times u}$.
					    \item If $\deg(g)<u$, then:
					    \begin{enumerate}
						    \item Using \procedurehr{4.54}, verify that $\Lambda(g,G)\ne 0_{u\times u}$.
						    \item \textbf{Therefore using (16), verify that $0_{u\times u}=\Lambda(g,G)\ne 0_{u\times u}$.}
						    \item \textbf{Abort procedure.}
					    \end{enumerate}
					    \item Otherwise, do the following:
					    \begin{enumerate}
						    \item Verify that $\deg(g)=u$.
						    \item Using \procedurehr{4.55}, verify that $g=g_{\deg(g)}\last_A$.
						    \item \textbf{Therefore verify that $p=F_{0,0}=D_{0,0}{{N^{-1}}_*}_{0,0}+D_{0,1}{{N^{-1}}_*}_{1,0}={{N^{-1}}_*}_{0,0}g+{{N^{-1}}_*}_{1,0}*0={{N^{-1}}_*}_{0,0}g={{N^{-1}}_*}_{0,0}g_{\deg(g)}\last_A$.}
						    \item \textbf{Yield the tuple $\langle {{N^{-1}}_*}_{0,0}g_{\deg(g)}\rangle$.}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{4.57}
			    \objective
				    Choose an $m\times n$ rational matrix, $A$, and an $n\times m$ rational matrix, $B$, such that $AB=1_m$. The objective of the following instructions is to show that either $0=1$ or every column of $B$ is non-zero.
			    \implementation
				    \begin{enumerate}
					    \item If any column $i$ of $B$, $Be_i$, is equal to zero, then:
					    \begin{enumerate}
						    \item Verify that $0_{n\times 1}=A0_{n\times 1}=A(Be_i)=(AB)e_i=1_me_i=e_i$.
						    \item \textbf{Therefore verify that 0=1.}
						    \item \textbf{Abort procedure.}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{4.58}
			    \objective
				    Choose a $m\times m$ rational matrix, $A$. Choose a polynomial $p$ such that $p\ne 0$, $\Lambda(p,A)=0$, and $\deg(p)<\deg(\last_A)$. The objective of the following instructions is to show that $0<0$.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{4.56} on $A$ and $p$ and let $f$ receive.
					    \item Now verify that $p=f\last_A$.
					    \item Now using the precondition and (2), verify that $f\ne 0$ and $\last_A\ne 0$.
					    \item \textbf{Therefore using the precondition, (2), and (3), verify that $\deg(\last_A)>\deg(p)=\deg(f\last_A)\ge\deg(\last_A)$.}
					    \item \textbf{Abort procedure.}
				    \end{enumerate}
		    \declaration{4.21}
			    The notation \gls{matrixpows}, where $A$ is a $m\times m$ rational matrix, will be used as a shorthand for the result yielded by executing the following instructions:
			    \begin{enumerate}
				    \item Let $t=\deg(\last_A)$.
				    \item Make an $m^2\times t$ matrix, $B$, whose $i^{th}$ column is the sequential concatenation of the columns of $A^i$.
				    \item \textbf{Yield $\langle B\rangle$.}
			    \end{enumerate}
		    \procedure{4.59}
			    \objective
				    Choose a $m\times m$ rational matrix, $A$. Execute \procedurehr{4.25} on $\pows(A)$ and let the tuple $\langle M,D,,N\rangle$ receive the result. Let $t=\cols(\pows(A))$. The objective of the following instructions is to show that either $0<0$ or to show that ${C_t(D)}={C_t(D)}_{0,0}e_0\ne 0$.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{4.25} on $\pows(A)$ and let the tuple $\langle M,D,,N\rangle$ receive the result.
					    \item Verify that $M_*\pows(A)N_*=D$.
					    \item Using \procedurehr{4.05}, verify that ${M^{-1}}_*M_*\pows(A)N_*=1_{m^2}\pows(A)N_*=\pows(A)N_*={M^{-1}}_*D$.
					    \item If ${C_t(D)}_{0,0}=0$, then:
					    \begin{enumerate}
						    \item Verify that for some $0\le i<t$, $D_{i,i}=0$.
						    \item Therefore verify that $De_i=0_{m^2\times 1}$.
						    \item Therefore verify that $\pows(A)(Ne_i)=(\pows(A)N)e_i=(M^{-1}D)e_i=M^{-1}(De_i)=0_{m^2\times 1}$.
						    \item Let $p=N_{0,i}\lambda^0+N_{1,i}\lambda^1+\cdots+N_{t-1,i}\lambda^{t-1}$.
						    \item Therefore verify that $\Lambda(p,A)=0_{m\times m}$.
						    \item Execute \procedurehr{4.57} on ${N^{-1}}_*$ and $N_*$.
						    \item Therefore verify that $p\ne 0$.
						    \item Execute \procedurehr{4.58} on $A$ and $p$.
						    \item \textbf{Abort procedure.}
					    \end{enumerate}
					    \item Otherwise, do the following:
					    \begin{enumerate}
						    \item Execute \procedurehr{4.21} on $\langle D,1_t,t\rangle$ and let $E$ receive.
						    \item Verify that $C_t(D)=C_t(D1_t)=EC_t(1_t)=E*1=E$.
						    \item Verify that $E$ is a $\binom{m^2}{t}\times\binom{t}{t}$ diagonal matrix.
						    \item Therefore verify that $C_t(D)$ is a $\binom{m^2}{t}\times 1$ diagonal matrix.
						    \item \textbf{Therefore verify that $C_t(D)={C_t(D)}_{0,0}e_0\ne 0$.}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{4.60}
			    \objective
				    Choose a $m\times m$ rational matrix, $A$. Let $t=\cols(\pows(A))$. The objective of the following instructions is to show that either $0<0$ or to show that ${C_t(\pows(A))}\ne 0$.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{4.25} on $\pows(A)$ and let the tuple $\langle M,D,,N\rangle$ receive the result.
					    \item Verify that $\pows(A)={M^{-1}}_*D{N^{-1}}_*$.
					    \item Execute \procedurehr{4.57} on $C_t(M_*),C_t({M^{-1}}_*)$.
					    \item Hence verify that all columns of $C_t({M^{-1}}_*)$ are non-zero.
					    \item Execute \procedurehr{4.59} on $A$.
					    \item Verify that $C_t(D)={C_t(D)}_{0,0}e_0\ne 0$.
					    \item Therefore verify that ${C_t(D)}_{0,0}\ne 0$.
					    \item Execute \procedurehr{4.57} on $C_t(N_*),C_t({N^{-1}}_*)$.
					    \item Hence verify that $C_t(N^{-1})\ne 0$.
					    \item \textbf{Verify that $C_t(\pows(A))=C_t({M^{-1}}_*D{N^{-1}}_*)=C_t({M^{-1}}_*)C_t(D)C_t({N^{-1}}_*)=C_t({M^{-1}}_*){C_t(D)}_{0,0}e_0C_t({N^{-1}}_*)={C_t(D)}_{0,0}C_t({N^{-1}}_*)C_t({M^{-1}}_*)e_0\ne 0_{\binom{m^2}{t}\times 1}$.}
				    \end{enumerate}
		    \declaration{4.26}
			    The notation \gls{matrixtr}, where $A$ is a square matrix, will be used as a shorthand for the sum of its diagonal entries.
		    \procedure{4.68}
			    \objective
				    Choose two $m\times m$ matrices $A,B$. The objective of the following instructions is to show that $\tr(A+B)=\tr(A)+\tr(B)$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that $\tr(A+B)$
					    \begin{enumerate}
						    \item $=\sum_r^{[0:m]}(A+B)_{r,r}$
						    \item $=\sum_r^{[0:m]}(A_r+B_r)_{r,r}$
						    \item $=\sum_r^{[0:m]}A_{r,r}+\sum_r^{[0:m]}B_{r,r}$
						    \item $=\tr(A)+\tr(B)$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{4.69}
			    \objective
				    Choose a polynomial $b$ and an $m\times m$ matrix $A$. The objective of the following instructions is to show that $\tr(bA)=b\tr(A)$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that $\tr(bA)$
					    \begin{enumerate}
						    \item $=\tr(b_{m\times m}A)$
						    \item $=\sum_r^{[0:m]}(b_{m\times m}A)_{r,r}$
						    \item $=\sum_r^{[0:m]}\sum_t^{[0:m]}(b_{m\times m})_{r,t}A_{t,r}$
						    \item $=\sum_r^{[0:m]}(b_{m\times m})_{r,r}A_{r,r}$
						    \item $=\sum_r^{[0:m]}bA_{r,r}$
						    \item $=b\sum_r^{[0:m]}A_{r,r}$
						    \item $=b\tr(A)$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{4.70}
			    \objective
				    Choose an $m\times n$ matrix $A$ and an $n\times m$ matrix $B$. The objective of the following instructions is to show that $\tr(AB)=\tr(BA)$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that $\tr(AB)$
					    \begin{enumerate}
						    \item $=\sum_r^{[0:m]}(AB)_{r,r}$
						    \item $=\sum_r^{[0:m]}\sum_t^{[0:n]}A_{r,t}B_{t,r}$
						    \item $=\sum_t^{[0:n]}\sum_r^{[0:m]}B_{t,r}A_{r,t}$
						    \item $=\sum_t^{[0:n]}(BA)_{t,t}$
						    \item $=\tr(BA)$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{4.71}
			    \objective
				    Choose an $m\times n$ matrix $A$ such that $A\ne 0$. The objective of the following instructions is to show that $\tr(A^TA)>0$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that $\tr(A^TA)$
					    \begin{enumerate}
						    \item $=\sum_r^{[0:n]}(A^TA)_{r,r}$
						    \item $=\sum_r^{[0:n]}\sum_t^{[0:m]}(A^T)_{r,t}A_{t,r}$
						    \item $=\sum_r^{[0:n]}\sum_t^{[0:m]}A_{t,r}A_{t,r}$
						    \item $=\sum_r^{[0:n]}\sum_t^{[0:m]}(A_{t,r})^2$
						    \item $>0$.
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{4.27}
			    The phrase "\gls{matrixsymmetric}" will be used to refer to matrices $A$ such that "$A^T=A$".
		    \procedure{4.61}
			    \objective
				    Choose a symmetric $m\times m$ rational matrix, $A$. Let $t=\deg(\last_A)$. Choose two polynomials $u,w$ such that $\deg(u)<t$ and $\deg(w)<t$. The objective of the following instructions is to show that $\tr(\Lambda(uw,A))=\mat(u)^T\pows(A)^T\pows(A)\mat_t(w)$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that $\tr(\Lambda(uw,A))$
					    \begin{enumerate}
						    \item $=\tr(\Lambda(u,A)\Lambda(w,A))$
						    \item $=\tr((\sum_p^{[0:t]} u_pA^{p})(\sum_q^{[0:t]} w_qA^{q}))$
						    \item $=\tr(\sum_p^{[0:t]}\sum_q^{[0:t]} u_pw_qA^{p}A^{q})$
						    \item $=\sum_p^{[0:t]}\sum_q^{[0:t]} u_pw_q\tr(A^{p}A^{q})$
						    \item $=\sum_p^{[0:t]}\sum_q^{[0:t]} u_pw_q\sum_e^{[0:m]}\sum_f^{[0:m]}{A^{p}}_{e,f}\cdot{A^{q}}_{f,e}$
						    \item $=\sum_p^{[0:t]}\sum_q^{[0:t]} u_pw_q\sum_e^{[0:m]}\sum_f^{[0:m]}{A^{p}}_{f,e}\cdot{A^{q}}_{f,e}$
						    \item $=\sum_p^{[0:t]}\sum_q^{[0:t]} u_pw_q\sum_g^{[0:{m^2}]}{\pows(A)}_{g,p}{\pows(A)}_{g,q}$
						    \item $=\sum_p^{[0:t]}\sum_q^{[0:t]} u_pw_q(\pows(A)^T\pows(A))_{p,q}$
						    \item $=\sum_p^{[0:t]} u_p(\pows(A)^T\pows(A)\mat_t(w))_{p}$
						    \item $=\mat_t(u)^T\pows(A)^T\pows(A)\mat_t(w)$
					    \end{enumerate}
				    \end{enumerate}
		    \declaration{4.25}
			    The notation \gls{matrixsel}, where $A$ is an $m\times m$ rational matrix, will be used as a shorthand for the result yielded by executing the following instructions:
			    \begin{enumerate}
				    \item Using \procedurehr{4.30}, \procedurehr{4.60}, and \procedurehr{4.71}, verify that $C_t(\pows(A)^T\pows(A))=C_t(\pows(A)^T)C_t(\pows(A))={C_t(\pows(A))}^TC_t(\pows(A))=\tr({C_t(\pows(A))}^TC_t(\pows(A)))>0$.
				    \item Let $t=\deg(\last_A)$.
				    \item Let $H=(\pows(A)^T\pows(A))\backslash e_{t-1}$.
				    \item \textbf{Yield $\langle\frac{\sum_j^{[0:t]}H_{j,0}\lambda^j}{(\last_A)_t}\rangle$.}
			    \end{enumerate}
		    \procedure{4.62}
			    \objective
				    Choose a symmetric $m\times m$ rational matrix, $A$. Let $t=\deg(\last_A)$. Choose a polynomial $u$ such that $\deg(u)<t$. The objective of the following instructions is to show that $\tr(\Lambda(u\sel_A,A))=\frac{u_{t-1}}{(\last_A)_t}$.
			    \implementation
				    \begin{enumerate}
					    \item Using \procedurehr{4.61}, verify that $\tr(\Lambda(u\sel_A,A))$
					    \begin{enumerate}
						    \item $=\mat(u)^T\pows(A)^T\pows(A)\mat_t(\sel_A)$
						    \item $=\frac{\mat(u)^T\pows(A)^T\pows(A)((\pows(A)^T\pows(A))\backslash e_{t-1})}{(\last_A)_t}$
						    \item $=\frac{\mat(u)^Te_{t-1}}{(\last_A)_t}$
						    \item $=\frac{\mat(u)_{t-1,0}}{(\last_A)_t}$
						    \item $=\frac{u_{t-1}}{(\last_A)_t}$.
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{4.63}
			    \objective
				    Choose a symmetric $m\times m$ rational matrix, $A$. The objective of the following instructions is to either show that $0\ne 0$ or construct polynomials $u,v$ such that $u\last_A+v\sel_A=1$.
			    \implementation
				    \begin{enumerate}
					    \item Let $t=\deg(\last_A)$.
					    \item Let $G$ be the $1\times 2$ matrix $\langle\langle\last_A,\sel_A\rangle\rangle$.
					    \item Execute \procedurehr{4.25} on $G$ and let the tuple $\langle M,D,,N\rangle$ receive.
					    \item Verify that $G={M^{-1}}_*D{N^{-1}}_*$.
					    \item Verify that $\last_A\ne 0$.
					    \item Therefore verify that $D_{0,0}\ne 0$.
					    \item If $\deg(D_{0,0})>0$, then do the following:
					    \begin{enumerate}
						    \item Let $b={{N^{-1}}_*}_{0,0}$.
						    \item Verify that $\last_A=bD_{0,0}$.
						    \item Therefore verify that $b\ne 0$.
						    \item Let $z=\deg(b)$.
						    \item Verify that $t=\deg(\last_A)=\deg(bD_{0,0})=\deg(b)+\deg(D_{0,0})>\deg(b)=z$.
						    \item Let $c={{N^{-1}}_*}_{0,1}$.
						    \item Verify that $\sel_A=cD_{0,0}$.
						    \item Let $u=\lambda^{t-z-1}b$.
						    \item Execute \procedurehr{4.62} on $A$ and $u$.
						    \item Hence verify that $(\last_A)_t\tr(\Lambda(u\sel_A,A))=u_{t-1}=b_z\ne 0$.
						    \item Also verify that $\tr(\Lambda(u\sel_A,A))$
						    \begin{enumerate}
							    \item $=\tr(\Lambda(\lambda^{t-z-1}bcD_{0,0},A))$
							    \item $=\tr(\Lambda(\lambda^{t-z-1}c\last_A,A))$
							    \item $=\tr(\Lambda(\lambda^{t-z-1}c,A)\Lambda(\last_A,A))$
							    \item $=\tr(\Lambda(\lambda^{t-z-1}c,A)0_{m\times m})$
							    \item $=\tr(0_{m\times m})$
							    \item $=0$.
						    \end{enumerate}
						    \item \textbf{Therefore verify that $0\ne 0$.}
						    \item \textbf{Abort procedure.}
					    \end{enumerate}
					    \item Otherwise, do the following:
					    \begin{enumerate}
						    \item Verify that $\deg(D_{0,0})=0$.
						    \item Let $u=\frac{N_{0,0}}{D_{0,0}}$.
						    \item Let $v=\frac{N_{1,0}}{D_{0,0}}$.
						    \item \textbf{Verify that $u\last_A+v\sel_A=1$.}
						    \item \textbf{Yield the tuple $\langle u,v\rangle$.}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{4.64}
			    \objective
				    Choose a symmetric $m\times m$ rational matrix $A$, where $m>0$. Let $t=\deg(\last_A)$. The objective of the following instructions is to either show that $0\ne 0$ or to construct lists of polynomials $s,q$ such that
				    \begin{enumerate}
					    \item For $i=0$ to $i=t$, $\deg(s_i)=i$.
					    \item For $i=0$ to $i=t$, $\sgn((s_i)_i)=\sgn((s_t)_t)$.
					    \item For $i=1$ to $i=t-1$, $s_{i-1}+s_{i+1}=q_is_i$.
					    \item $s_t=\last_A$.
				    \end{enumerate}
			    \implementation
				    \begin{enumerate}
					    \item Let $s_t=\last_A$.
					    \item Execute \procedurehr{4.63} on $A$ and let $\langle u,s_{t+1}\rangle$ receive the result.
					    \item Hence verify that $us_t+s_{t+1}\sel_A=1$.
					    \item Let $q_t=s_{t+1}\div s_t$.
					    \item Let $s_{t-1}=s_{t+1}\bmod s_t$.
					    \item Verify that $s_{t+1}=q_ts_t+s_{t-1}$, where $\deg(s_{t-1})<\deg(s_t)=t$.
					    \item Therefore verify that $us_t+(q_ts_t+s_{t-1})\sel_A=1$.
					    \item Therefore verify that $\Lambda(s_{t-1}\sel_A,A)=\Lambda(us_t+(q_ts_t+s_{t-1})\sel_A,A)=\Lambda(1,A)=1_{m}$.
					    \item Therefore using \procedurehr{4.62}, verify that $\frac{(s_{t-1})_{t-1}}{(s_t)_t}=\tr(\Lambda(s_{t-1}\sel_A,A)=\tr(1_{m})=m>0$.
					    \item For $i\in[t:1]$, do the following:
					    \begin{enumerate}
						    \item Let $q_i=(-s_{i+1})\div(-s_i)$.
						    \item Let $s_{i-1}=(-s_{i+1})\bmod(-s_i)$.
						    \item Verify that $\deg(q_i)=1$.
						    \item Verify that $(q_i)_1=\frac{(s_{i+1})_{i+1}}{(s_i)_i}$.
						    \item Also verify that $-s_{i+1}=-q_is_i+s_{i-1}$.
						    \item Therefore verify that $q_is_i=s_{i+1}+s_{i-1}$.
						    \item Therefore verify that $q_is_i-s_{i+1}=s_{i-1}$.
						    \item Execute \procedurehr{2.26} on the tuple $\langle s,q,i-1\rangle$ and let $\langle p,j\rangle$ receive.
						    \item Verify that $s_{i-1}=ps_{t-1}+js_t$.
						    \item Verify that $\deg(p)=t-1-(i-1)=t-i$.
						    \item Verify that $\deg(j)=t-2-(i-1)=t-1-i$
						    \item Therefore verify that $\Lambda(s_{i-1},A)=\Lambda(ps_{t-1}+js_t,A)=\Lambda(ps_{t-1},A)+\Lambda(j,A)\Lambda(s_t,A)=\Lambda(ps_{t-1},A)+\Lambda(j,A)0_{m\times m}=\Lambda(ps_{t-1},A)$.
						    \item If $\Lambda(p,A)=0$, then do the following:
						    \begin{enumerate}
							    \item Execute \procedurehr{4.58} on $A$ and $p$.
							    \item \textbf{Abort procedure.}
						    \end{enumerate}
						    \item Otherwise, if $\Lambda(s_{i-1},A)=0_{m\times m}$, then do the following:
						    \begin{enumerate}
							    \item Verify that $\Lambda(ps_{t-1}\sel_A,A)=\Lambda(ps_{t-1},A)\Lambda(\sel_A,A)=\Lambda(s_{i-1},A)\Lambda(\sel_A,A)=0_{m\times m}\Lambda(\sel_A,A)=0_{m\times m}$.
							    \item Verify that $\Lambda(ps_{t-1}\sel_A,A)=\Lambda(p,A)\Lambda(s_{t-1}\sel_A,A)=\Lambda(p,A)1_{m}=\Lambda(p,A)\ne0_{m\times m}$.
							    \item Therefore verify that $0\ne 0$.
							    \item \textbf{Abort procedure.}
						    \end{enumerate}
						    \item Otherwise if $\Lambda(s_{i-1}\sel_A,A)=0_{m\times m}$, then do the following:
						    \begin{enumerate}
							    \item Verify that $\Lambda(s_{i-1}\sel_As_{t-1},A)=\Lambda(s_{i-1}\sel_A,A)\Lambda(s_{t-1},A)=0_{m\times m}\Lambda(s_{t-1},A)=0_{m\times m}$.
							    \item Verify that $\Lambda(s_{i-1}\sel_As_{t-1},A)=\Lambda(s_{i-1},A)\Lambda(\sel_As_{t-1},A)=\Lambda(s_{i-1},A)1_{m}=\Lambda(s_{i-1},A)\ne 0_{m\times m}$.
							    \item Therefore verify that $0_{m\times m}\ne 0_{m\times m}$.
							    \item \textbf{Abort procedure.}
						    \end{enumerate}
						    \item Otherwise, do the following:
						    \begin{enumerate}
							    \item Verify that $\deg(s_{i-1})<i$.
							    \item Verify that $\Lambda(s_{i-1}\sel_A,A)\ne 0_{m\times m}$.
							    \item Execute the \subprocedurehr{wed2808191621} on the tuple $(i-1, s_{i-1})$.
							    \item Hence using \procedurehr{4.71}, verify that $\frac{(s_{i-1})_{i-1}}{(s_i)_i}=\tr(\Lambda({s_{i-1}}^2{\sel_A}^2,A))=\tr((\Lambda(s_{i-1}\sel_A,A))^2)=\tr((\Lambda(s_{i-1}\sel_A,A))^T(\Lambda(s_{i-1}\sel_A,A)))>0$.
							    \item \textbf{Therefore verify that $\sgn((s_{i-1})_{i-1})=\sgn((s_i)_i)$.}
						    \end{enumerate}
					    \end{enumerate}
					    \item Yield the tuple $\langle s_{[0:t+1]}, q_{[0:t]}\rangle$.
				    \end{enumerate}
			    \subprocedure{wed2808191621}
				    \subobjective
					    Choose an integer $0\le k\le t$ such that polynomial $s_k$ is defined. Choose a polynomial $g$ such that $\deg(g)\le\min(k,t-1)$. The objective of the following instructions is to show that $\tr(\Lambda(gs_k{\sel_A}^2,A))=\frac{g_k}{(s_{k+1})_{k+1}}$.
				    \subimplementation
					    \begin{enumerate}
						    \item If $k=t$, then verify that $\tr(\Lambda(gs_k{\sel_A}^2,A))$
						    \begin{enumerate}
							    \item $=\tr(\Lambda(gs_t{\sel_A}^2,A))$
							    \item $=\tr(\Lambda(g{\sel_A}^2,A)\Lambda(s_t,A))$
							    \item $=\tr(\Lambda(g{\sel_A}^2,A)0_{m\times m})$
							    \item $=0$
							    \item $=\frac{g_k}{(s_{k+1})_{k+1}}$.
						    \end{enumerate}
						    \item Otherwise if $k=t-1$, then verify that $\tr(\Lambda(gs_k{\sel_A}^2,A))$
						    \begin{enumerate}
							    \item $=\tr(\Lambda(gs_{t-1}{\sel_A}^2,A))$
							    \item $=\tr(\Lambda(g{\sel_A},A)\Lambda(s_{t-1}{\sel_A},A))$
							    \item $=\tr(\Lambda(g\sel_A,A)1_{m})$
							    \item $=\tr(\Lambda(g\sel_A,A))$
							    \item \textbf{=$\frac{g_k}{(s_{k+1})_{k+1}}$.}
						    \end{enumerate}
						    \item Otherwise if $k<t-1$, then do the following:
						    \begin{enumerate}
							    \item Verify that $\deg(gq_{k+1})=k+1\le t-1$.
							    \item Execute the \subprocedurehr{wed2808191621} on the tuple $\langle k+1, gq_{k+1}\rangle$.
							    \item Now verify that $\tr(\Lambda((gq_{k+1})s_{k+1}{\sel_A}^2,A))=\frac{\frac{(s_{k+2})_{k+2}}{(s_{k+1})_{k+1}}g_k}{(s_{k+2})_{k+2}}=\frac{g_k}{(s_{k+1})_{k+1}}$.
							    \item Verify that $\deg(g)\le k\le t-2$.
							    \item Execute the \subprocedurehr{wed2808191621} on the tuple $\langle k+2,g\rangle$.
							    \item Now verify that $\tr(\Lambda(gs_{k+2}{\sel_A}^2,A))=\frac{g_{k+2}}{(s_{k+3})_{k+3}}=\frac{0}{(s_{k+3})_{k+3}}=0$.
							    \item Therefore verify that $\tr(\Lambda(gs_k{\sel_A}^2,A))$
							    \begin{enumerate}
								    \item $=\tr(\Lambda(g(q_{k+1}s_{k+1}+s_{k+2}){\sel_A}^2,A))$
								    \item $=\tr(\Lambda(gq_{k+1}s_{k+1}{\sel_A}^2+gs_{k+2}{\sel_A}^2,A))$
								    \item $=\tr(\Lambda(gq_{k+1}s_{k+1}{\sel_A}^2,A)+\Lambda(gs_{k+2}{\sel_A}^2,A))$
								    \item $=\tr(\Lambda(gq_{k+1}s_{k+1}{\sel_A}^2,A))+\tr(\Lambda(gs_{k+2}{\sel_A}^2,A))$
								    \item $=\frac{g_k}{(s_{k+1})_{k+1}}+0$
								    \item $=\frac{g_k}{(s_{k+1})_{k+1}}$.
							    \end{enumerate}
						    \end{enumerate}
					    \end{enumerate}
		    \procedure{4.65}
			    \objective
				    Choose a symmetric $m\times m$ rational matrix, $A$. Let $t=\deg(\last_A)$. The objective of the following instructions is to either show that $0<0$ or to construct two lists of rational numbers $c,d$ such that $c_0<d_0\le c_1<d_1\le\cdots\le c_{t-1}<d_{t-1}$ and $0\ne\sgn(\Lambda(\last_A,c_i))=-\sgn(\Lambda(\last_A,d_i))$ for $i$ in $[0:t]$.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{4.64} on the matrix $A$ and let the tuple $\langle s,q\rangle$ receive the result.
					    \item Execute \procedurehr{2.24} supplying the tuple $\langle s,q\rangle$. Let the tuple $\langle c,d\rangle$ receive the result.
					    \item \textbf{Verify that $c_0<d_0\le c_1<d_1\le\cdots\le c_{t-1}<d_{t-1}$.}
					    \item \textbf{Verify that $\sgn(\Lambda(\last_A,c_i))=-\sgn(\Lambda(\last_A,d_i))$ for $i$ in $[0:t]$.}
					    \item \textbf{Yield $\langle c,d\rangle$.}
				    \end{enumerate}
		    \procedure{4.66}
			    \objective
				    Choose a symmetric $m\times m$ rational matrix, $A$. Let $t=\deg(\last_A)$. Execute \procedurehr{4.65} on $A$ and let the tuple $\langle c,d\rangle$ receive the result. Execute \procedurehr{4.25} on $A$ and let the tuple $\langle,,u,\rangle$ receive the result. The objective of the following instructions is to either show that $1=-1$ or to construct a list of non-negative integers $k$ such that $0\ne\sgn(\Lambda(u_{k_i},c_i))=-\sgn(\Lambda(u_{k_i},d_i))$ for $i$ in $[0:t]$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that $\last_A=u_0u_1\cdots u_{m-1}$.
					    \item For $i$ in $[0:t]$, do the following:
					    \begin{enumerate}
						    \item Using the precondition, verify that $0\ne\sgn(\Lambda(\last_A,c_i))=-\sgn(\Lambda(\last_A,d_i))$.
						    \item If $0\in\sgn(\Lambda(u,c_i))$, then do the following:
						    \begin{enumerate}
							    \item Verify that $0$
							    \begin{enumerate}
								    \item $=\sgn(\Lambda(u_0,c_i))\sgn(\Lambda(u_1,c_i))\cdots\sgn(\Lambda(u_{m-1},c_i))$
								    \item $=\sgn(\Lambda(u_0,c_i)\Lambda(u_1,c_i)\cdots\Lambda(u_{m-1},c_i))$
								    \item $=\sgn(\Lambda(u_0u_1\cdots u_{m-1},c_i))$
								    \item $=\sgn(\Lambda(\last_A,c_i))$
								    \item $\ne 0$.
							    \end{enumerate}
						    \end{enumerate}
						    \item If $0\in\sgn(\Lambda(u,d_i))$, then do the following:
						    \begin{enumerate}
							    \item Verify that $0$
							    \begin{enumerate}
								    \item $=\sgn(\Lambda(u_0,d_i))\sgn(\Lambda(u_1,d_i))\cdots\sgn(\Lambda(u_{m-1},d_i))$
								    \item $=\sgn(\Lambda(u_0,d_i)\Lambda(u_1,d_i)\cdots\Lambda(u_{m-1},d_i))$
								    \item $=\sgn(\Lambda(u_0u_1\cdots u_{m-1},d_i))$
								    \item $=\sgn(\Lambda(\last_A,d_i))$
								    \item $\ne 0$.
							    \end{enumerate}
						    \end{enumerate}
						    \item If $\sgn(\Lambda(u_j,c_i))=\sgn(\Lambda(u_j,d_i))\for j\in[0:m]$, then do the following:
						    \begin{enumerate}
							    \item Verify that $\sgn(\Lambda(\last_A,c_i))$
							    \begin{enumerate}
								    \item $=\sgn(\Lambda(u_0u_1\cdots u_{m-1},c_i))$
								    \item $=\sgn(\Lambda(u_0,c_i))\sgn(\Lambda(u_1,c_i))\cdots\sgn(\Lambda(u_{m-1},c_i))$
								    \item $=\sgn(\Lambda(u_0,d_i))\sgn(\Lambda(u_1,d_i))\cdots\sgn(\Lambda(u_{m-1},d_i))$
								    \item $=\sgn(\Lambda(u_0u_1\cdots u_{m-1},d_i))$
								    \item $=\sgn(\Lambda(\last_A,d_i))$.
							    \end{enumerate}
							    \item \textbf{Therefore verify that $1=-1$.}
							    \item \textbf{Abort procedure.}
						    \end{enumerate}
						    \item Otherwise do the following:
						    \begin{enumerate}
							    \item \textbf{Let $k_i$ be the least integer such that $0\ne\sgn(\Lambda(u_{k_i},c_i))=-\sgn(\Lambda(u_{k_i},d_i))$.}
						    \end{enumerate}
					    \end{enumerate}
					    \item \textbf{Yield $\langle k\rangle$.}
				    \end{enumerate}
		    \procedure{4.67}
			    \objective
				    Choose a symmetric $m\times m$ rational matrix, $A$. Execute \procedurehr{4.25} on $A$ and let the tuple $\langle,,u,\rangle$ receive the result. Execute \procedurehr{2.18} on $A$ and let $k$ receive. Let $t=\deg(\last_A)$. Let $n_j=\sum_i^{[0:t]} [k_i=j]$ for $j$ in $[0:m]$. The objective of the following instructions is to either show that $0<0$, or to show that $n_i=\deg(u_i)$ for $i$ in $[0:m]$.
			    \implementation
				    \begin{enumerate}
					    \item Verify that $\sum_j^{[0:m]} n_j=\sum_j^{[0:m]}\sum_i^{[0:t]} [k_i=j]=\sum_i^{[0:t]}\sum_j^{[0:m]} [k_i=j]=\sum_i^{[0:t]} 1=t$.
					    \item If for any $i$ in $[0:m]$, $n_i>\deg(u_i)$, then do the following:
					    \begin{enumerate}
						    \item Execute \procedurehr{2.18} on the polynomial $u_i$ along with $\deg(u_i)+1$ of the distinct pairs $\langle c_l,d_l\rangle$ such that $k_l=i$.
						    \item \textbf{Abort procedure.}
					    \end{enumerate}
					    \item Otherwise if for any $i$ in $[0:m]$, $n_i<\deg(u_i)$, then do the following:
					    \begin{enumerate}
						    \item Verify that $\sum_i^{[0:m]} n_j<\sum_i^{[0:m]} \deg(u_j)=t$.
						    \item Therefore using (1) and (a), verify that $\sum_i^{[0:m]} n_j<\sum_i^{[0:m]} n_j$.
						    \item \textbf{Abort procedure.}
					    \end{enumerate}
					    \item Otherwise, do the following:
					    \begin{enumerate}
						    \item \textbf{For all $i$ in $[0:m]$, verify that $n_i=\deg(u_i)$.}
					    \end{enumerate}
				    \end{enumerate}
		    \procedure{4.72}
			    \objective
				    Choose a symmetric $m\times m$ rational matrix, $A$. Let $t=\deg(\last_A)$. Execute \procedurehr{4.66} on the matrix $A$ and let the tuple $\langle k\rangle$ receive the result. The objective of the following instructions is to either show that $0<0$ or to show that $\sum_i^{[0:t]}(m-k_i)=m$.
			    \implementation
				    \begin{enumerate}
					    \item Execute \procedurehr{4.25} on the matrix $A$ and let the tuple $\langle,D,u,\rangle$.
					    \item Using \procedurehr{4.67}, verify that $\sum_i^{[0:t]}(m-k_i)$
					    \begin{enumerate}
						    \item $=\sum_i^{[0:t]}\sum_j^{[0:m]} [k_i\le j]$
						    \item $=\sum_j^{[0:m]}\sum_i^{[0:t]} [k_i\le j]$
						    \item $=\sum_j^{[0:m]}\sum_i^{[0:t]} [k_i\le j]\sum_l^{[0:m]} [k_i=l]$
						    \item $=\sum_j^{[0:m]}\sum_l^{[0:m]}\sum_i^{[0:t]} [k_i\le j][k_i=l]$
						    \item $=\sum_j^{[0:m]}\sum_l^{[0:m]}\sum_i^{[0:t]} [l\le j][k_i=l]$
						    \item $=\sum_j^{[0:m]}\sum_l^{[0:m]} [l\le j]\sum_i^{[0:t]} [k_i=l]$
						    \item $=\sum_j^{[0:m]}\sum_l^{[0:m]} [l\le j]\deg u_l$
						    \item $=\sum_j^{[0:m]}\sum_l^{[0:{j+1}]} \deg u_l$
						    \item $=\sum_j^{[0:m]} \deg D_{j,j}$
						    \item $=m$
					    \end{enumerate}
				    \end{enumerate}
	    \end{multicols*}
	  \clearpage
	\begin{multicols*}{2}
		\begingroup
			\renewcommand{\addcontentsline}[3]{}
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					Harold Edwards.
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				  Darij Grinberg.
				  \textit{Notes on the Combinatorial Fundamentals of Algebra}.
				  Ludwig Maximilian University of Munich, 2020.
				\bibitem{introductiontheoryofnumbers}
					Hugh L. Montgomery, Ivan Niven, Herbert S. Zuckerman.
					\textit{An Introduction to the Theory of Numbers}.
					John Wiley \& Sons, 1991.
				\bibitem{calculusofapproximations}
				  Mieczyslaw Warmus.
				  \textit{Calculus of Approximations}.
				  Polish Scientific Publishers PWN, 1958.
				\bibitem{philosophicalgrammar}
					Ludwig Wittgenstein.
					\textit{Philosophical Grammar}.
					Edited by Rush Rhees.
					Translated by Anthony Kenny.
					Basil Blackwell, Oxford, 1974.
			\end{thebibliography}
		\endgroup
	\end{multicols*}
\end{document}
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