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Re: FLT Discussion: Simplifying
Posted:
Jan 20, 2001 4:20 PM
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In article <3a683db5.481647214@news.newsguy.com>,
randyp@visionplace.com (Randy Poe) wrote:
> On Fri, 19 Jan 2001 11:24:12 GMT, "Dik T. Winter" <Dik.Winter@cwi.nl>
> wrote:
>
> > > I'm looking for something that pushes me outside of integers
because
> > > that's what I want to prove must happen!
> >
> >Tsk. Your initial use of sqrt(-1) pushes you out of the integers
already.
>
> You misunderstood this statement. He believes that the fact that
> sqrt(-1) pushes him out of the integers is the contradiction and that
> the proof can end at that point, just with the factorization. That
> writing down a statement not in integers, though his first equation
> was in the integers, is a contradiction.
>
Here's where there's the issue between what I've recently called
patterns and regular rings.
My understanding is that mathematicians have avoided this through
coupling.
That is, polynomials have counting number exponents *and* counting
number coefficients.
However, I have argued that this coupling is simply one way of doing
things and there's no mathematical or logical requirement for it.
My example has been
(x+sqrt(-1)y)(x-sqrt(-1)y) = x^2 + y^2,
which as a pattern is simply true for valid x's and y's.
Here validity simply means they work with the given operators.
That basically means you don't add apples and oranges.
What I've shown is a problem with the coupling so many of you have been
heatedly defending, since the statement is valid with x and y members
of an infinite number of rings, where the ring of integers is *one*
such, which happens to not have the result of sqrt(-1) as a member.
Note the distinction between calling sqrt(-1) a result, since we're
looking at a number with an operator, versus the casual error of
concluding that it is the number itself. After all, in once case that
result is i, for another it could be, oh, 2. This casual error is
common with other cases besides sqrt(-1).
The only remaining issue then has been an insistence that it must be
proven that (x+sqrt(-1)y)(x-sqrt(-1)y) = 0 means that x+sqrt(-1)y = 0
or x - sqrt(-1)y = 0, as some have insisted I must move to complex
numbers to get this result.
But, I've repeatedly brought up the fact that it's true for other
rings, so why if you guys are acting like using rings is so fundamental
and important, do you wish to make a result that's a hack depending on
what ring you're using?
Don't understand what I mean?
Let's say that I'm in some other ring besides integers. All the
results I've given are still true. That is,
(x+sqrt(-1)y)(x-sqrt(-1)y) = x^2 + y^2, and if x^2 + y^2 = 0, then
(x+sqrt(-1)y)(x-sqrt(-1)y) = 0.
You've all been insisting that I have to know what ring x and y are in
to prove whether or not this means that
(x+sqrt(-1)y) = 0 *or* (x-sqrt(-1)y) = 0.
When my simple question to you is, when would it not be?
The issue has been that sqrt(-1) is not in the ring of integers, when I
have a *proof by contradiction* that starts by saying that the other
objects x and y *are* nonzero integers and that x^2 + y^2 = 0.
So, if all of you are right, then it's not possible to have the general
result over a large number of rings that (x+sqrt(-1)y)(x-sqrt(-1)y) = 0
means that (x+sqrt(-1)y) = 0 *or* (x-sqrt(-1)y) = 0.
Then, basically all of you are arguing that it's impossible to prove
this result generally, and that you have to go ring, by ring.
Well folks, there are an infinite number of rings that x and y could
belong to for which the above would be valid.
You've all essentially been arguing that we can't know that result is
true for those rings, since we must check each ring individually.
Seems like an awful lot of effort and added complexity to me.
James Harris
Sent via Deja.com
http://www.deja.com/
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