<?xml version="1.0" encoding="utf-8" standalone="yes"?>
<rss version="2.0" xmlns:atom="http://www.w3.org/2005/Atom">
	<channel>
		<title>Proofs on tplat</title>
		<link>/docs/mathematics/pure-math/proofs/</link>
		<description>Recent content in Proofs on tplat</description>
		<generator>Hugo</generator>
		<language>en</language>
		
		
		
		
			<atom:link href="/docs/mathematics/pure-math/proofs/index.xml" rel="self" type="application/rss+xml" />
			<item>
				<title>If $n$ is even, then $n^2$ is even</title>
				<link>/docs/mathematics/pure-math/proofs/if-n-is-even-then-n2-is-even/</link>
				<pubDate>Mon, 01 Jan 0001 00:00:00 +0000</pubDate>
				<guid>/docs/mathematics/pure-math/proofs/if-n-is-even-then-n2-is-even/</guid>
				<description>&lt;h1 id=&#34;if--is-even-then--is-even&#34;&gt;If $n$ is even, then $n^2$ is even&lt;a class=&#34;anchor&#34; href=&#34;#if--is-even-then--is-even&#34;&gt;&lt;/a&gt;&lt;/h1&gt;&#xA;&lt;p&gt;&lt;strong&gt;Claim:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;If $n$ is even, then $n^2$ is even.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Assume $n$ is even. Then, by definition, there exists an integer $k$ such that:&lt;/p&gt;&#xA;$$&#xA;n=2k&#xA;$$&lt;p&gt;Then:&lt;/p&gt;&#xA;$$&#xA;n^2=(2k)^2&#xA;=4k^2&#xA;=2(2k)^2&#xA;$$&lt;p&gt;Since $2k^2$ is an integer, $n^2$ is divisible by $2$. Therefore, by direct example, $n^2$ is even.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Claim:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;If $n^2$ is even, then $n$ is even.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Assume $n$ is odd. Then, by definition, there exists integer $k$ such that:&lt;/p&gt;</description>
			</item>
			<item>
				<title>If $n$ is odd, then $n^2$ is odd</title>
				<link>/docs/mathematics/pure-math/proofs/if-n-is-odd-then-n2-is-odd/</link>
				<pubDate>Mon, 01 Jan 0001 00:00:00 +0000</pubDate>
				<guid>/docs/mathematics/pure-math/proofs/if-n-is-odd-then-n2-is-odd/</guid>
				<description>&lt;h1 id=&#34;if--is-odd-then--is-odd&#34;&gt;If $n$ is odd, then $n^2$ is odd&lt;a class=&#34;anchor&#34; href=&#34;#if--is-odd-then--is-odd&#34;&gt;&lt;/a&gt;&lt;/h1&gt;&#xA;&lt;p&gt;&lt;strong&gt;Claim:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;If $n$ is odd, then $n^2$ is odd.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Assume $n$ is odd. Then, by definition, there exists an integer $k$ such that:&lt;/p&gt;&#xA;$$&#xA;n=2k+1&#xA;$$&lt;p&gt;Then:&lt;/p&gt;&#xA;$$&#xA;n^2=(2k+1)^2&#xA;=(2k+1)(2k+1)&#xA;=4k^2+4k+1&#xA;=2(2k^2+2k)+1&#xA;$$&lt;p&gt;Let $m=2k^2+2k$, which is an integer. Then:&lt;/p&gt;&#xA;$$&#xA;n^2=2m+1&#xA;$$&lt;p&gt;Therefore, by direct example, $n^2$ is odd.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Claim:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;If $n^2$ is odd, then $n$ is odd.&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Consider the contrapositive; if $n$ is odd, then $n^2$ is odd. We have already shown this to be true. Therefore, by contrapositive, $n$ is odd.&lt;/p&gt;</description>
			</item>
			<item>
				<title>If $x^2&gt;0$, then $x&gt;0$</title>
				<link>/docs/mathematics/pure-math/proofs/if-x20-then-x0/</link>
				<pubDate>Mon, 01 Jan 0001 00:00:00 +0000</pubDate>
				<guid>/docs/mathematics/pure-math/proofs/if-x20-then-x0/</guid>
				<description>&lt;h1 id=&#34;if--then&#34;&gt;If $x^2&gt;0$, then $x&gt;0$&lt;a class=&#34;anchor&#34; href=&#34;#if--then&#34;&gt;&lt;/a&gt;&lt;/h1&gt;&#xA;&lt;p&gt;&lt;strong&gt;Claim:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;If $x^2&gt;0$, then $x&gt;0$&lt;/p&gt;&#xA;&lt;p&gt;&lt;strong&gt;Proof:&lt;/strong&gt;&lt;/p&gt;&#xA;&lt;p&gt;Let $x=-2$&lt;/p&gt;&#xA;&lt;p&gt;Then $x^2=4&gt;0$, but $x&gt;0$ is false.&lt;/p&gt;&#xA;&lt;p&gt;Therefore, by counterexample, the statement is false.&lt;/p&gt;</description>
			</item>
	</channel>
</rss>
