
Re: FLT Discussion: Simplifying
Posted:
Jan 18, 2001 6:53 PM


In article <943am4$59r$1@nnrp1.deja.com>, hale@mailhost.tcs.tulane.edu wrote: > In article <3a6506aa.270916173@news.newsguy.com>, > randyp@visionplace.com (Randy Poe) wrote: > > On Wed, 17 Jan 2001 01:57:50 GMT, hale@mailhost.tcs.tulane.edu wrote: > > > > >For example, I might be asked to justify the following step > > >(I will assume that I am working in the field of complex numbers, > > >which you have refused to do for some reason): > > > > The reason is that if he admits that saying "x and y are integers" is > > insufficient to discuss the behavior of (x+sqrt(1)y) here, he'd have > > to admit it's insufficient in the FLT proof. > > Yes.
Not exactly. I like simplicity. My feeling (and you can believe I'm wrong, or maybe prove it I guess) is that you're asking for unnecessary layering.
The reasoning I'm giving is relatively simple.
x,y and z are integers, so I'm working with integers.
I'm not forced from this assumption *until* I reach a contradiction.
But that's the whole point.
You say, I'm forced to act like I'm outside of integers at the start, but what if there were an integer solution to FLT?
Then wouldn't your objection fall away?
And you know why, because then the square root in my proof would give an integer for v with those x, y and z.
It'd have to.
Now Hale, you've been a serious and fairly reasonable person about all of this and I appreciate that. I want to emphasize to you what I just said. Isn't it true that if integer solutions existed that the square root that's caused all of this discussion would produce an integer with those solutions?
I've already emphasized that everything work with x,y and z in finite rings of integers. Why there's this push to claim that the transition to the infinite rings of integers makes it all invalid bemuses me.
> > But, nothing will be lost for James Harris if he would admit that > he is working in the field of complex numbers, and a lot would be > gained since he could use all the theorems proved for complex > numbers.
Unnecessary complexity.
> > However, this would only serve to make his statements to be > meaningful and allow him to define things like "mod" and > "fractional". He would still need to specify a subring of > the complex numbers, since the complex numbers contain "too > many" numbers for what he wants to do. > > James Harris has admitted that he is working in at least two > distinct rings: ring of integers and ring of polynomials. > His statements in the proof also imply that he is working > in the ring of complex numbers and the ring of symbolic > expressions (like sqrt(x^2+y^2)). > > My first impressions were that he was working in just a > subring of the complex numbers. When he claimed that > he was also working in the ring of polynomials, I thought > that would not be possible since he is using the equation > x^5+y^5 = z^5, which is not true in a polynomial ring. > But, he nicely got out of that problem by eliminating > the z and claiming that he is also working in the ring > of symbolic expressions. This clarification has helped > a lot. But, now he is going to the other extreme of > rejecting these known mathematical rings and he is > trying to create his own rings from scratch, which > will demand even more explanations and proofs than > what was originally required. >
To me, that's making it all way more complicated than necessary.
I can get a lot done by assuming certain things exist, like the factorization of x^5 + y^5  z^5. I can start off one of its factors as x +...+y+...z. Now, that's not a polynomial, but the question with respect to FLT is not that. The question with respect to FLT is whether or not for nonzero integer x, y and z is that an integer.
In that format, we can't tell.
That's why FLT is hard.
All I did is figure out how to use that factorization in a context where it gets wrapped into a radical.
Think about this.
sqrt(x^2 + y^2) = x+...+y, there are an infinite number of terms in there, and don't ask me what they look like. If I have x and y both equal 1, then all of that infinity is used to get 1.414...
But if x = 3 and y =4, that infinity of terms adds up to 5.
We casually deal with the first case and happily use sqrt(2) all over the place without caring about the higher abstraction, and act as if the operator with its object is the actual number, which is like using 1+1 for 2, except that it is more convenient than the alternative.
It just so happened that I had to care about that higher level of abstraction to prove FLT.
James Harris
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