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Topic: Cantor's diagonal argument.
Replies: 24   Last Post: Oct 12, 2001 5:16 PM

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Giles Redgrave

Posts: 3
Registered: 12/13/04
Re: Cantor's diagonal argument.
Posted: Oct 4, 2001 6:39 AM
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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.







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