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backmatter.xml
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<?xml version="1.0" encoding="UTF-8" ?>
<!-- This file is part of the book -->
<!-- -->
<!-- Logic and Proof for Teachers -->
<!-- -->
<!-- Copyright (C) 2019 Lesa L. Beverly, Kimberly M. Childs, Deborah A. Pace, Thomas W. Judson -->
<!-- -->
<!-- See the file COPYING for copying conditions. -->
<!-- See the file COPYING for copying conditions. -->
<backmatter xml:id="backmatter" xmlns:xi="http://www.w3.org/2001/XInclude">
<title>Back Matter</title> <!-- mostly for HTML sectioning -->
<appendix>
<title>More on the Integers</title>
<subsection>
<title>Strong Induction</title>
<p>We have an equivalent statement of the Principle of Mathematical Induction that is often very useful.</p>
<principle xml:id="principle-integers-second-pmi">
<title>Second Principle of Mathematical Induction</title>
<idx><h>Induction</h><h>second principle of</h></idx>
<statement>
<p>Let <m>S(n)</m> be a statement about integers for <m>n \in {\mathbb N}</m> and suppose <m>S(n_0)</m> is true for some integer <m>n_0</m>. If <m>S(n_0), S(n_0 + 1), \ldots, S(k)</m> imply that <m>S(k + 1)</m> for <m>k \geq n_0</m>, then the statement <m>S(n)</m> is true for all integers <m>n \geq n_0</m>.</p>
</statement>
</principle>
</subsection>
<subsection xml:id="subsection-backmatter-pmi">
<title>The Connection between Mathematical Induction and the Principle of Well Ordering</title>
<lemma xml:id="lemma-backmatter-smallest-number">
<statement>
<p>The Principle of Mathematical Induction implies that <m>1</m> is the least positive natural number.</p>
</statement>
<proof>
<p>Let <m>S = \{ n \in {\mathbb N} : n \geq 1 \}</m>. Then <m>1 \in S</m>. Assume that <m>n \in S</m>. Since <m>0 \lt 1</m>, it must be the case that <m>n = n + 0 \lt n + 1</m>. Therefore, <m>1 \leq n \lt n + 1</m>. Consequently, if <m>n \in S</m>, then <m>n + 1</m> must also be in <m>S</m>, and by the Principle of Mathematical Induction, and <m>S = \mathbb N</m>.</p>
</proof>
</lemma>
<theorem xml:id="theorem-backmatter-pmi-implies-pwo">
<statement>
<p>The Principle of Mathematical Induction implies the Principle of Well-Ordering. That is, every nonempty subset of <m>\mathbb N</m> contains a least element.</p>
</statement>
<proof>
<p>We must show that if <m>S</m> is a nonempty subset of the natural numbers, then <m>S</m> contains a least element. If <m>S</m> contains 1, then the theorem is true by <xref ref="lemma-backmatter-smallest-number"/>. Assume that if <m>S</m> contains an integer <m>k</m> such that <m>1 \leq k \leq n</m>, then <m>S</m> contains a least element. We will show that if a set <m>S</m> contains an integer less than or equal to <m>n + 1</m>, then <m>S</m> has a least element. If <m>S</m> does not contain an integer less than <m>n+1</m>, then <m>n+1</m> is the smallest integer in <m>S</m>. Otherwise, since <m>S</m> is nonempty, <m>S</m> must contain an integer less than or equal to <m>n</m>. In this case, by induction, <m>S</m> contains a least element.</p>
</proof>
</theorem>
</subsection>
<subsection xml:id="subsection-backmatter-fund-theorem-arithmetic">
<title>The Proof of the Fundamental Theorem of Arithmetic</title>
<theorem>
<title>Fundamental Theorem of Arithmetic</title>
<statement>
<p>Let <m>n</m> be an integer such that <m>n \gt 1</m>. Then
<me>n = p_1 p_2 \cdots p_k,</me>
where <m>p_1, \ldots, p_k</m> are primes (not necessarily distinct). Furthermore, this factorization is unique; that is, if
<me>n = q_1 q_2 \cdots q_l,</me>
then <m>k = l</m> and the <m>q_i</m>'s are just the <m>p_i</m>'s rearranged.</p>
</statement>
<proof>
<p><em>Uniqueness.</em> To show uniqueness we will use induction on <m>n</m>. The theorem is certainly true for <m>n = 2</m> since in this case <m>n</m> is prime. Now assume that the result holds for all integers <m>m</m> such that <m>1 \leq m \lt n</m>, and
<me>n = p_1 p_2 \cdots p_k = q_1 q_2 \cdots q_l,</me>
where <m>p_1 \leq p_2 \leq \cdots \leq p_k</m> and <m>q_1 \leq q_2 \leq \cdots \leq q_l</m>. By <xref ref="theorem-integers-prime-divisor"/>, <m>p_1 \mid q_i</m> for some <m>i = 1, \ldots, l</m> and <m>q_1 \mid p_j</m> for some <m>j = 1, \ldots, k</m>. Since all of the <m>p_i</m>'s and <m>q_i</m>'s are prime, <m>p_1 = q_i</m> and <m>q_1 = p_j</m>. Hence, <m>p_1 = q_1</m> since <m>p_1 \leq p_j = q_1 \leq q_i = p_1</m>. By the induction hypothesis,
<me>n' = p_2 \cdots p_k = q_2 \cdots q_l</me>
has a unique factorization. Hence, <m>k = l</m> and <m>q_i = p_i</m> for <m>i = 1, \ldots, k</m>.</p>
<p><em>Existence.</em> To show existence, suppose that there is some integer that cannot be written as the product of primes. Let <m>S</m> be the set of all such numbers. By the Principle of Well-Ordering, <m>S</m> has a smallest number, say <m>a</m>. If the only positive factors of <m>a</m> are <m>a</m> and <m>1</m>, then <m>a</m> is prime, which is a contradiction. Hence, <m>a = a_1 a_2</m> where <m>1 \lt a_1 \lt a</m> and <m>1 \lt a_2 \lt a</m>. Neither <m>a_1\in S</m> nor <m>a_2 \in S</m>, since <m>a</m> is the smallest element in <m>S</m>. So
<md>
<mrow>a_1 & = p_1 \cdots p_r</mrow>
<mrow>a_2 & = q_1 \cdots q_s</mrow>
</md>.
Therefore,
<me>a = a_1 a_2 = p_1 \cdots p_r q_1 \cdots q_s</me>.
So <m>a \notin S</m>, which is a contradiction.</p>
</proof>
</theorem>
</subsection>
</appendix>
<appendix>
<title>Notation</title>
<p>The following table defines the notation used in this book. Page numbers or references refer to the first appearance of each symbol.</p>
</appendix>
<xi:include href="./gfdl-mathbook.xml" /> <!-- GFDL License text -->
<!-- Index follows appendices, preceded only colophon -->
<index>
<index-list />
</index>
<colophon xml:id="back-colophon">
<p>This book was authored in <pretext/>.</p>
</colophon>
</backmatter>