<?xml version="1.0"?>
<feed xmlns="http://www.w3.org/2005/Atom" xml:lang="en">
	<id>https://mindpowe.red/wiki/index.php?action=history&amp;feed=atom&amp;title=User%3AKarlhahn%2FUser_e-irrational</id>
	<title>User:Karlhahn/User e-irrational - Revision history</title>
	<link rel="self" type="application/atom+xml" href="https://mindpowe.red/wiki/index.php?action=history&amp;feed=atom&amp;title=User%3AKarlhahn%2FUser_e-irrational"/>
	<link rel="alternate" type="text/html" href="https://mindpowe.red/wiki/index.php?title=User:Karlhahn/User_e-irrational&amp;action=history"/>
	<updated>2026-04-07T04:52:39Z</updated>
	<subtitle>Revision history for this page on the wiki</subtitle>
	<generator>MediaWiki 1.34.2</generator>
	<entry>
		<id>https://mindpowe.red/wiki/index.php?title=User:Karlhahn/User_e-irrational&amp;diff=13805&amp;oldid=prev</id>
		<title>imported&gt;Voidxor: No need to subst the {{Userbox}} template (doing so makes life difficult). Add sentence period.</title>
		<link rel="alternate" type="text/html" href="https://mindpowe.red/wiki/index.php?title=User:Karlhahn/User_e-irrational&amp;diff=13805&amp;oldid=prev"/>
		<updated>2018-02-03T00:55:09Z</updated>

		<summary type="html">&lt;p&gt;No need to subst the {{Userbox}} template (doing so makes life difficult). Add sentence period.&lt;/p&gt;
&lt;p&gt;&lt;b&gt;New page&lt;/b&gt;&lt;/p&gt;&lt;div&gt;{{Userbox&lt;br /&gt;
| id = &amp;lt;math&amp;gt;e \ne \frac {p} {q}&amp;lt;/math&amp;gt;&lt;br /&gt;
| id-s = 16&lt;br /&gt;
| id-c = #ffffff&lt;br /&gt;
| info = This user can prove that the number, &amp;lt;big&amp;gt;'''''e'''''&amp;lt;/big&amp;gt;, is irrational.&lt;br /&gt;
| info-a = center&lt;br /&gt;
| info-c = #F3F3F3&lt;br /&gt;
| info-fc = #505080&lt;br /&gt;
| info-s = 10&lt;br /&gt;
| border-c = black&lt;br /&gt;
}}&lt;br /&gt;
&lt;br /&gt;
&amp;lt;noinclude&amp;gt;&lt;br /&gt;
{{Clear}}&lt;br /&gt;
&lt;br /&gt;
'''Usage: &amp;lt;nowiki&amp;gt;{{User:Karlhahn/User e-irrational}}&amp;lt;/nowiki&amp;gt;'''&lt;br /&gt;
&lt;br /&gt;
'''PROOF:'''&lt;br /&gt;
&lt;br /&gt;
If &amp;lt;math&amp;gt;\scriptstyle e&amp;lt;/math&amp;gt; were rational, then&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;math&amp;gt;e = \frac {p} {q}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
where &amp;lt;math&amp;gt;\scriptstyle p&amp;lt;/math&amp;gt; and &amp;lt;math&amp;gt;\scriptstyle q&amp;lt;/math&amp;gt; are both positive integers. Hence&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;math&amp;gt;q e = p&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
making &amp;lt;math&amp;gt;\scriptstyle qe&amp;lt;/math&amp;gt; an integer. Multiplying both sides by &amp;lt;math&amp;gt;\scriptstyle (q-1)!&amp;lt;/math&amp;gt;,&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;math&amp;gt;q! e = p(q-1)!&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
so clearly &amp;lt;math&amp;gt;\scriptstyle q! e&amp;lt;/math&amp;gt; is also an integer. By [[Maclaurin series]]&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;math&amp;gt;e = \sum_{k=0}^\infty \frac {1} {k!}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
Multiplying both sides by &amp;lt;math&amp;gt;\scriptstyle q!&amp;lt;/math&amp;gt;:&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;math&amp;gt;q! e = \sum_{k=0}^\infty \frac {q!} {k!}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
The first &amp;lt;math&amp;gt;\scriptstyle q+1&amp;lt;/math&amp;gt; terms of this sum are integers. It follows that the sum of the remaining terms must also be an integer. The sum of those remaining terms is&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;math&amp;gt;r = \sum_{k=1}^\infty \frac {q!} {(q+k)!}&amp;lt;/math&amp;gt;&lt;br /&gt;
making &amp;lt;math&amp;gt;\scriptstyle r&amp;lt;/math&amp;gt; an integer. Observe that&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;math&amp;gt;\frac {q!} {(q+k)!} &amp;lt; \frac {q!} {q! q^k} = \frac {1} {q^k}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
&lt;br /&gt;
So&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;math&amp;gt;r = \sum_{k=1}^\infty \frac {q!} {(q+k)!} &amp;lt; \sum_{k=1}^\infty \frac {1} {q^k}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
But&lt;br /&gt;
&lt;br /&gt;
:&amp;lt;math&amp;gt;\sum_{k=1}^\infty \frac {1} {q^k} = \frac {1} {q-1}&amp;lt;/math&amp;gt;&lt;br /&gt;
&lt;br /&gt;
This means that &amp;lt;math&amp;gt;\scriptstyle 0 &amp;lt; r &amp;lt; \frac {1} {q-1} \le 1&amp;lt;/math&amp;gt;, which requires that &amp;lt;math&amp;gt;\scriptstyle r&amp;lt;/math&amp;gt; be an integer between zero and one. That is clearly impossible, hence &amp;lt;math&amp;gt;\scriptstyle e&amp;lt;/math&amp;gt; is irrational&lt;br /&gt;
[[Category:Mathematics user templates|Irrational e]]&lt;br /&gt;
&amp;lt;/noinclude&amp;gt;&lt;/div&gt;</summary>
		<author><name>imported&gt;Voidxor</name></author>
		
	</entry>
</feed>