Re: Cantor's diagonal proof wrong?

From: Brian Chandler (imaginatorium_at_despammed.com)
Date: 11/22/04 

Date: 22 Nov 2004 10:14:07 -0800

"Shmuel (Seymour J.) Metz" <spamtrap@library.lspace.org.invalid> wrote in message news:<41a145fa$7$fuzhry+tra$mr2ice@news.patriot.net>...
> In <f2c35871.0411180147.6252ca24@posting.google.com>, on 11/18/2004
> at 01:47 AM, imaginatorium@despammed.com (Brian Chandler) said:
>
> >Well, since the OP clearly intended 8 to mean eight, and 7 to mean
> >seven, then the sum is naturally defined using the standard carry.
>
> The standard carry is only defined for decimal expansions of integers.
> It is not defined for infinite sequences of integers. You can provide
> a definition, but if you do it will break some expected properties of
> the integers.

I'm thinking about the 10-adics (I believe; thanks to Robin Chapman
for pointing me in the right direction). Think of a decimal register
extending indefinitely far to the left, and representing zero by
...000, positive integers by (eg) ...000163 and negative by (eg)
...9997 (for -3).

It seems obvious that any integer is represented within a finite
number of digits of this register, and since a standard carry
mechanism for any finite number of digits works for integers up to
(about one less than) that number of digits, a notional infinite
register will work for (proper) integers in exactly the normal way. I
therefore cannot see how any definition that extends the normal
behaviour will 'break' the integers.

Anyway, keep the mechanism of the register exactly the same, but allow
arbitrary non-integer values, including any sequence to the left.
Surely, with a bit of hand-waving (any finite segment of the adding
machine 'works'; so in the limit so does the whole thing ^_^), we can
see that the whole set of values at least satisfies the ring axioms,
but doesn't look at all like the integers, in that there are zero
divisors, and some funny rationals: like ...3334 = 2/3.

And the limit of the sequence 5, 5^2, 5^4, 5^8, ... is a funny value
that ends in ...625, (call it phyve), and phyve^2 = phyve. So
phyve*(1-phyve) = 0.
My calculator broke on 625^4, and I haven't got round to writing a
page using the php bignumber stuff... but it looks as though squaring
- going from 5^n to 5^(2n) gets you at least two more constant decimal
digits.)

> >Am I right, though, in saying that it is clear that these things at
> >least form a ring?
>
> Not unless you define + and * in such a way that the ring properties
> hold. In fact, it will take some work to even make them form a group.

Here's the definition of +. Consider two sequences P and Q, number the
digits leftwards from 0 (so in ...625 p2=6)
To find digit n of P+Q, take the final n digits of P and Q, add them
in an n-digit register, ignoring the carry, and take the nth digit of
the result.
Obviously one could formalise this in fancy typesetting, but basically
what more do I need to do?

Brian Chandler

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