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Topic: Proposal re James Harris
Replies: 16   Last Post: Dec 31, 2000 6:26 PM

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 Steven E. Landsburg Posts: 154 Registered: 12/8/04
Re: Proposal re James Harris
Posted: Dec 29, 2000 8:20 PM

Regarding my assertion that James's FLT proof and his sum-of-squares
proof are wrong in essentially the same way, David Ullrich writes:

> Actually there's a big difference: The proof that x^2 + y^2 <> 0
> is perfectly correct. Suppose that x and y are reals not both
> zero. If x^2 + y^2 = 0 then it _does_ follow that
>
> (x+iy)(x-iy) = 0

But *why* does it follow? It follows because this equation takes place
in a particular ring (the Gaussian integers or the complex numbers,
as you prefer) which we can prove has a certain property (namely
the property of being an integral domain).

And how do we know these rings are integral domains? We know it,
ultimately, because we know that -1 is not the square of any integer.
That's perilously close to the whole strength of what James is
trying to prove and hence perilously close to being circular.
Therefore I question the relevance of Prof. Ullrich's statement:

> Seriously: Suppose hypothetically you knew all about the
> real and complex numbers as _fields_, but you were somehow
> unaware that the reals could be ordered, so you didn't
> know the result about sums of squares of reals, because
> the word "positive" didn't exist. If someone showed you
> exactly what James wrote you would accept it as a proof
> that the sum of the squares of two non-zero reals cannot
> vanish.

The point is that it is not *possible* to "know about the complex
numbers as a field" without first knowing that -1 is not the square
of an integer. Without that knowledge, you cannot know that the
complex numbers form a field.

So I continue to maintain that:

a) James's sum-of-squares proof is wrong
b) It is wrong for exactly the same reason that his FLT proof is
wrong
and c) If James could be made to understand the reasons for a), then
he would understand most of what everybody has been trying to
explain to him.

Steven E. Landsburg
www.landsburg.com
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