Relay-Version: version B 2.10 5/3/83; site utzoo.UUCP Posting-Version: version B 2.10.1 6/24/83; site watmath.UUCP Path: utzoo!watmath!csc 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