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From: csc@watmath.UUCP (Computer Sci Club)
Newsgroups: net.math
Subject: Re: Pedantic Question - (nf)
Message-ID: <6996@watmath.UUCP>
Date: Thu, 23-Feb-84 14:34:38 EST
Article-I.D.: watmath.6996
Posted: Thu Feb 23 14:34:38 1984
Date-Received: Fri, 24-Feb-84 00:45:14 EST
References: <5710@uiucdcs.UUCP>
Organization: U of Waterloo, Ontario
Lines: 34

   The replies to the net discussing the question of Cantors diagonalization
proof apllied to the whole numbers have all been adressed to the question
"Can a whole number have an infinite number of non zero digits", and have
shown that the answer is no.  This invalidates the "proof" as applied to
the whole numbers, as we do not know whether or not the sequence of
zero's and ones created by the diagonalization process has a finite number
of ones. (The statement was made that M, the sequence created, would have
an infinite number of leading ones.  This is only true if the usual ordering
of the whole numbers is used, but the sequence might start
        
                 100000000 ...
                 010000000 ...
                 000000000 ...
                 110000000 ...
                 100010000 ...
                   .
                   .
                   .

in which case M would start 00110) I here provide a proof that the number of
1's in the sequence M is infinite.
   Let the ordering of the whole numbers be represented by a function f.
(i.e. f(0) is the first number in the sequence, f(1) the second etc.).
As all the whole numbers must appear f must be a bijection, i.e. if a is
a whole number then there exists b with f(b)=a.  Now let us assume
the sequence M has a finite number of ones, then there exists N (N>3), such
that for all n>N the nth element of M is zero.  Then (n>N) f(n) must
have a one in the nth position, ie f(n) => 2**N.  Hence all numbers
c < 2**N must equal f(d) d < N.  Now there are 2**N whole numbers
less than 2**(N) (don't forget zero).  These cannot all fit into
N positions as N < 2**N.  Therefore M must contain an infinite number 
of ones.
                           William Hughes