Re: Cantor's diagonal proof wrong?

From: Rasmus Villemoes (burner+usenet_at_imf.au.dk)
Date: 11/14/04 

Date: Sun, 14 Nov 2004 18:44:07 +0100

curt@kcwc.com (Curt Welch) writes:

> José_Carlos_Santos <jcsantos@fc.up.pt> wrote:
>
>> Could you please tell me then which integer corresponds to the real
>> number 1/9 (or 0.11111111111111111111... if you prefer)?
>
> ....1111111
>
> Why is it ok to write 0.111... but not ...11111 ?

Because 0.111... has a well defined interpretation in terms of the
limit of the convergent series

sum_{i=1 to infinity} 1·10^{-i}

whereas the ...11111 is just nonsense.

> My point is that 1/9 is in fact an algorithm for generating a real
> value. It is not the real value itself.

Yes, it _is_ the real number "one ninth".

> It's just a name we use to talk about the real value which is 0.1111
> repeating forever. And I can just as easily define the integer of 1
> repeating forver. The only reason we do not do that is a matter of
> convention. It's not (so I claim) in violation of what integers are.

Then your notion of an integer is in violation with the generally
accepted notion. If you would post a system of axioms (similar to the
Peano axioms) for your "integers" in which ...11111 has a well defined
interpretion as an "integer", then we can continue the discussion.

> If you start with 0, and continue to apply the +1 function to it,
>and ignore all the values you come up with which does not have all
>1's, you find you have the exact same type of defintion that gives
>you 1/9 when you generate a string of one's running to the right,
>instead of running to the left.

Yes, there is an infinite family of integers consisting of "only 1's",
namely 1 (one), 11 (eleven), 111 (one hundred and eleven) etc., just
as there is an infinite family of reals between 0 and 1 whose decimal
representation starts with a number of 1's and has 0's elsewhere
(namely .1, .11, .111 etc.). The _difference_ is that the latter has a
real number (namely 1/9) as a limit, whereas the former has no limit
at all. (Well, it diverges in such a nice way that one sometimes says
it converges to infinity, but infinity is not an integer nor a real).

> Just change the implied 0, to an implied 1, and there you have it.
> The integer pair for the real 1/9.

No. What you have is an infinite string of 1's which do NOT represent
an integer. 

-- 
Rasmus Villemoes
<http://home.imf.au.dk/burner/>

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