
Re: Cantor's diagonal argument.
Posted:
Oct 4, 2001 6:39 AM


Randy Poe <rpoe@nospamatl.lmco.com> wrote in message news:<3BBB4A4D.7025D757@nospamatl.lmco.com>... > briggs@encompasserve.org wrote: > > Classic issue. The resolution is to realize that this number is > > not a natural number. All natural numbers have finitely many > > decimal digits. This number has infinitely many. > > But it's important to note, as it was in this thread, that > this property (finite number of digits) should not be > taken as an axiom. Instead it follows by induction from > the construction of the natural numbers: > > 1 has a single digit (in all bases). > Let a be any natural number with a finite > number of digits, call it n_a. Then (a+1) has > at most (n_a+1) digits, which is finite.
But can't you use the same argument to show that the number of natural numbers is finite.
Call the set of integers from 1 to n A_n.
The number of members of A_1 is finite. If the number of members of A_n is finite then so is the number of members of A_{n+1} (A_n + 1). Therefore there are a finite number of natural numbers.
I can't see how the infinity of natural numbers can be represented by a finite number of digits. In decimal the number of integers representable by n digits is 10^n. How can 10^n be infinite when n is finite.
Or to put it onother way, doesn't log n exceeds any finite value as n > infinity.
I must not understand what is meant by infinite and finite, is there a clear mathematical definition of these terms. Help, I'm still confused!
Giles.

