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Path: utzoo!watmath!clyde!burl!hou3c!hocda!houxm!ihnp4!inuxc!pur-ee!uiucdcs!uiuccsb!grunwald
From: grunwald@uiuccsb.UUCP
Newsgroups: net.math
Subject: Re: Pedantic Question - (nf)
Message-ID: <5806@uiucdcs.UUCP>
Date: Wed, 22-Feb-84 22:31:54 EST
Article-I.D.: uiucdcs.5806
Posted: Wed Feb 22 22:31:54 1984
Date-Received: Fri, 24-Feb-84 01:32:09 EST
Lines: 43

#R:uiuccsb:9700020:uiuccsb:9700022:000:1384
uiuccsb!grunwald    Feb 22 12:33:00 1984

Here's the best answer I've recieved so far. A couple of other people sent mail
which had somewhat circular arguments. Thanks to all who responded to this

Dirk Grunwald
ihnp4 ! uiucdcs ! grunwald

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	From: ihnp4!allegra!sdcrdcf!trwrb!pearlman
	Date: Tuesday, 21 Feb 1984 18:10-PST
	To: sdcrdcf!allegra!ihnp4!inuxc!pur-ee!uiucdcs!uiuccsb!grunwald
	Subject: Cantor proof (net.math article)

	Hi --

	You asked why a version of Cantor's diagonal proof could not be
	used to show that the natural numbers weren't uncountably infinite.
	If M is the number "left out" of the ordering, you can prove that
	M is not a natural number by contradiction:

	Let n be any natural number.  Then n < 2**n, by induction:

	For n = 1, it's trivial:
		1 < 2 = 2**1

	If it's true for n, then it's true for n+1:
		2**(n+1) = 2(2**n)
			 = 2**n + 2**n
		n < 2**n, and 1 < n < 2**n, so
		     n+1 < 2**(n+1)

	Let M be the number "left out" of your ordering.  You've admitted
	that M has an infinite number of leading ones, so M must be greater
	than 2**n for any natural n.  Thus, if M is a natural number, then
	M > 2**M, which is a contradiction.

	"Trust me -- I was a math major"

			-- Laura
			...{sdcrdcf, ucbvax, decvax}!trwrb!pearlman
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