
Re: FLT Discussion: Simplifying
Posted:
Jan 16, 2001 11:04 AM


jstevh@mydeja.com wrote: > [snip] > 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. > > 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. > > First off, it's worth noting that I'm treating the sqrt(1) as an > *operation*.
*Operation*???  do you mean something like a lobotomy?
How can you multiply an *operantion* by an integer?  there you go again making up inane terminology (*operation*) with no definition or discussion of its properties  using it in ways that are not defined for your new (and useless!) terminology. You must be the consummate bovine fecal artist on this planet.
Certainly, you do not understand what is wrong with your proof, even though it contains some (*possibly*) correct formulae, but no explanation or reasoning to put them in context and to justify them.
If you cannot get this proof straight or understand just what part of mathematics is necessary to prove this simple claim, just think how much worse your attempts to prove FLT look to the rest of the world.
Evidently, you do not understand that, in order to prove that the complex numbers are a field (or integral domain), you need *first* to know that, for any real numbers x, y not both zero, x^2 + y^2 != 0  and thus you *cannot* use the complex numbers to prove x^2 + y^2 != 0 if at least one of the real numbers x, y is nonzero.
As I pointed out a long time ago  but you seem to block all my emails and postings  the proof that x^2 + y^2 > 0 (and hence != 0) relies on the *order* properties of the reals  namely, that the square of any nonzero real is positive and the sum of two nonnegative reals is positive if at least one of them is positive.
I think some of you live under the false notion that > something like sqrt(2) is the actual number. Nope. It's an operation > that tells you something about that number, like that it multiplies > times itself to give 2. We also use it handily as a shorthand > representation for the actual number, which is also sometimes > represented by 1.414... >
sqrt( ) is an "operation"  i.e. a function  at least to most people. It associates to x one (or more than one, depending upon the context) square root of x, i.e. an element t such that t^2 = x.
> No human being has ever seen or ever will see the actual number that we > can conveniently represent by sqrt(2). >
I see sqrt(2) almost every time I look at an isoceles right triangle  don't you?
> We're just limited that way. >
You may be limited that way  others are not.
> So, I use sqrt(1) in a proof that's dealing with integers, and I don't > care that there's such a thing as i = sqrt(1). All I care about is > that I've found this thing that isn't an integer. >
If you are using it in a proof that deals with integers, then, at the very least, it [ sqrt(1) ] had doggone well better lie in a ring that contains the integers, such as the complex numbers or the Gaussian integers  *not* in the integers modulo 5 (or 6)  "integers modulo m" (m > 0) do *not* contain the integers!
> Ok, jumping to my proof of FLT for p=5. I prove (and no one disputes > this) that >
Jumping from one false argument to another...
> [(v^5 + 1) z^2 (5v^3 + sqrt(5v^6  20v))xy/2] [(v^5 + 1) z^2 (5v^3  > sqrt(5v^6  20v))xy/2] = 0(mod (x+y+vz)/h), when x^5 + y^5 = z^5, and > for the simplest case h = (x+y)^{1/5}. > [snip]
Chuck Giffen

