
Re: FLT Discussion: Simplifying
Posted:
Jan 16, 2001 6:16 PM


In article <940akc$d9p$1@nntp.Stanford.EDU>, Michael Hochster <michael@rgmiller.Stanford.EDU> wrote: > > : Given x^2 + y^2 = 0, x and y nonzero integers, show that no solution > : exists. > > : Proof by contradiction: > > : (x+sqrt(1)y)(xsqrt(1)y) = x^2 + y^2 = 0, so > > : x = sqrt(1)y *or* x = sqrt(1)y. > > Still waiting for an explanation of this step.
Here's a case where I've left out what I think are obvious steps, and this person disagrees. Some may think it unnecessary for me to add them, others may not.
Here are the missing steps:
Starting from (x+sqrt(1)y)(xsqrt(1)y) = x^2 + y^2 = 0,
(x+sqrt(1)y)(xsqrt(1)= 0, so
x + sqrt(1)y = 0 or x sqrt(1)y = 0, so
x = sqrt(1)y or x = sqrt(1)y.
Some, for reasons I'd like them to explain, have complained that I don't know that x + sqrt(1)y = 0 or x sqrt(1)y = 0, if
(x+sqrt(1)y)(xsqrt(1)= 0.
(Sort of like if AB = 0, A or B = 0. These people are saying that must be proven, and that it is a "gap" in my proof that I don't do so.)
If so, I'd like them to say that is their position here and we can see if we can't work that one out.
> > : There doesn't exist an integers that multiplies times itself to give a > : negative number, and an integer can't be the product of an integer and > : a non integer, so there's a contradiction. > > An integer *can* be the product of an integer and a noninteger. > 2 * 0.5 = 1.
Yup, you're right.
And that folks outlines how people can *help* make a proof clearer.
Notice that the issue is clear and the resolution is as well.
Basically, I know that sqrt(1) isn't rational, and that an integer can't be the product of an integer and a non rational.
So, now I present a correction, and we see where things go from there.
Purists among you may note that I started out in integers as my ring, and that what I was doing was sticking to my ring.
Rationals are part of a field, and for some fraction to come in and invalidate my conclusion, some weird things have to happen.
But, with that said, I've decided to just go ahead and mention rationals, and see where the discussion goes.
> > So, your proof of this simple fact still needs work. >
Ok, let's say you're right, and it did need work, and you may think it still does. I don't have a problem with that.
What I want to emphasize is that there is a process that can lead to resolution and it is clear that there are those of you willing to engage in it based on the fact that you made those comments here.
So, a reasonable person may now ask, why hasn't that process played out this way with my claims of a simple proof of Fermat's Last Theorem?
James Harris
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