Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP 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 -------------------------------------------------------------------------------- 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 --------------------------------------------------------------------------------