
Re: FLT Discussion: Simplifying
Posted:
Jan 18, 2001 7:16 PM


In article <943094$sq2$1@nnrp1.deja.com>, hale@mailhost.tcs.tulane.edu wrote: > In article <942k1a$i6f$1@nnrp1.deja.com>, > jstevh@mydeja.com wrote: > > I think I should mention that there's also a question of trueness. > > > > Granted, there's the issue of whether or not a given person can prove > > this or that statement, but there's also the question of truth. > > > > Are those statements true, or not? > > Since you are doing a proof by contradiction and you are assuming > at the start that you have integers x, y, and z that are solutions > of the Fermat equation, then even false statements will be true > (under your assumptions).
Well, it turns out that once that is shown to be true, we have the proof.
Of course, the reason is simple. There *are* solutions to x^p + y^p = z^p, with p an odd prime.
Every expression I give in my proof must be true for those solutions.
And they are.
Remember, I start with x+y+vz = x+y+vz, which is true.
I then use elementary operations and the conditional statement x^p + y^p = z^p; therefore, the statements I give must be true unless I make an error.
None of you have been able to show an error.
There wouldn't even be discussion of a gap if (x+y+vz)/h were x+y+vz instead. It's that (x+y)^{1/p} (for the simplest case) that's giving problems.
Some have tried to find fault with my method using the n=2 case, which *does* use x+y+vz. I think those are interesting, so I'll go over them quickly.
x+y+vz = 0(mod (x+y+vz)), jumping ahead
x^2 + 2xy + y^2 = v^2z^2, and jumping to the end,
(v^2  1)z^2  2xy = 0(mod (x+y+vz)), with x^2 + y^2 = z^2.
Some have argued, why not factor what's on the left to get
(sqrt(v^2  1)z  sqrt(2xy))(sqrt(v^2  1)z + sqrt(2xy)) = 0,
so (sqrt(v^2  1)z  sqrt(2xy)) = 0(mod (x+y+vz)?
Is there a really simple answer that's rigorous? I guess so, but the answer I give is you look at the exponents.
To me it's a simple matter of asking what expression can multiply times (x+y+vz) to get (sqrt(v^2  1)z  sqrt(2xy))?
I'll let you folks mull that one over.
> > We could make a distinction between "derivable" and "true". > Thus, we could say that all of your statements in your proof of > FLT might be derivable, some of which are true, and others false. >
Again, the statements are true if x^p + y^p = z^p. Note, as written that is saying that that statement is true. That is, that we have some x, y and z that fit.
One thing I know as an absolute fact is that you cannot prove otherwise.
As far as the math is concerned, as long as you take away that assertion that x, y and z are nonzero integer, everything is fine, correct, and perfect.
> This gets back to what several people have requested that you do. > If a statement is true independent of the Fermat counterexample, > then you should separate out such statements as lemmas or > propositions. This would make analysis of your proof easier.
Hey, I tried something like that with my p=5 proof. I started by producing the partial factorization that was true independent of the rest of the FLT business. People got upset with me on the ring issue. I said I was giving a factorization and that many rings would fit, and they wouldn't go for that.
Hey, that proof is still there in that format if you don't believe me at
http://www.mindspring.com/~jstev/FLTp5.htm.
I've debated starting it out with something like x,y and z being in a commutative ring, but I'm wary of being chased into complexity by those of you who love to try and do that whenever I attempt to get a bit more technical.
> > > But, hey, if they're true (ignoring the question of whether or not > > I've proven them for the moment) then a simple proof of Fermat's > > Last Theorem quickly follows. > > This does not follow at all. There are many true statements in > mathematics that do not have simple proofs.
My point is that if those statements are true, and you assume that then the proof is handily had.
So, the question of whether or not they are true is of interest for that reason.
I mentioned that because some of you have seen fit to act like casting doubt on whether or not I proved them to be true is sufficient.
The analogy that comes to mind is Ribet's result that proved FLT if TS were true.
Hey, that *is* a good analogy.
Let me repeat, objectors have claimed there is a gap in my proof of those statements, but the statements are either true or false. If true, FLT follows simply for prime p>3.
So, that question is the same type that faced Wiles when he set out to prove TS to prove FLT.
> > > However, you have seen people arguing for months that I haven't > > proven those statements, and the insinuation is that the > > statements are false. > > There is no insinuation that your statements are false. In fact, > David Libert has already proved they are true by invoking Wiles > theorem.
Consider this statement given what I said above.
> > > I doubt that many of you would believe that such simple math isn't > > provably true or not; therefore, I submit that it is reasonable to > > conclude that these people are claiming the statements are false. > > It might be reasonable to you, but we are not claiming the statements > are false. We are claiming that you have not derived them. >
Which would be like when Ribet didn't have proof of TS.
And let me emphasize that I'm not saying there is a gap because there isn't it. I still say you guys have pushed the ring issue out of context, since x,y and z are assumed to be integers, and that you're not being reasonable about how symbolic expressions must behave.
James Harris
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