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Re: FLT Discussion: Simplifying
Posted:
Jan 16, 2001 6:04 PM
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In article <G78GGv.7ID@cwi.nl>,
"Dik T. Winter" <Dik.Winter@cwi.nl> wrote:
> In article <9400ps$b5g$1@nnrp1.deja.com> jstevh@my-deja.com writes:
> > Given x^2 + y^2 = 0, x and y nonzero integers, show that no
solution
> > exists.
> >
> > Proof by contradiction:
> >
> > (x+sqrt(-1)y)(x-sqrt(-1)y) = x^2 + y^2 = 0, so
> >
> > x = sqrt(-1)y *or* x = -sqrt(-1)y.
>
> James, you have to *prove* that last step. You cannot rely on the
standard
> proof because it uses: x^2 + y^2 <> 0 whenever x != 0 or y != 0.
I think some people may be surprised that you are essentially calling
this a gap.
My comment to them is that it's the same thing being done with the
proof of Fermat's Last Theorem.
As for your other statement, it doesn't make any sense to me.
What's going on is simple though. I have that x^2 + y^2 = 0, and I
know that x^2 + y^2 = (x+sqrt(-1)y)(x-sqrt(-1)y), so
(x+sqrt(-1)y)(x-sqrt(-1)y) = 0, so
x+sqrt(-1)y = 0 or x-sqrt(-1)y=0,
and that goes to the question of how long a proof has to be.
As you can see, some people will force you to outline simple steps that
others would find unnecessary.
That bit of arbitrariness is used by some people to claim that a proof
isn't.
>
> > First off, it's worth noting that I'm treating the sqrt(-1) as an
> > *operation*. I think some of you live under the false notion that
> > something like sqrt(2) is the actual number.
>
> I must be dense, but what the heck do you mean? If it is not a number
> you have to define how to do multiplication of operations with
numbers.
>
The sqrt(2) is a representation of the number that multiplies times
itself to give 2 (notice how circular that is).
To give you the perspective I'm trying to outline, consider that I can
call 2, 1+1 because 2 is the number that I get when I add 1 and 1
together.
We don't go around writing 1+1 because we can just write '2'.
However, we do go around writing sqrt(2) as if it were the number
instead of a definition for the number based on an operator, i.e. sqrt
(), for simple reasons.
My point is that no human being has ever seen or ever will see the
number sqrt(2) in the same way that she or he can see 2.
> So, James, how do I multiply a number with an operation, and what do I
> get?
Multiplication is an operation.
>
> > [(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}.
>
> Yup, except that this is necessarily not valid when z is a multiple
of 5.
>
> > It might not suprise you that now I go ahead and say that
> > [(v^5 + 1) z^2 -(5v^3 + sqrt(5v^6 - 20v))xy/2] = 0(mod (x+y+vz)/h)
> > *or*
> > [(v^5 + 1) z^2 +(5v^3 + sqrt(5v^6 - 20v))xy/2] = 0(mod (x+y+vz)/h.
>
> Like above, you have to *prove* this step. Off-hand I can say already
> that the step is unproven, because whenever x^5 + y^5 != z^5 this is
> clearly false. (The initial congruence above is not valid when that
> is the case.)
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?
Some of you have quietly sat back as if it only matters that some have
claimed that I haven't proven whether or not they're true.
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.
Then, an enterprising individual might suddenly wonder if the
statements are provable or not for what I'd think would be obvious
reasons.
However, you have seen people arguing for months that I haven't proven
those statements, and the insinuation is that the statements are false.
Why?
Again, I remind you that we're talking about Fermat's Last Theorem, the
most famous math problem in history.
If those statements are not false, are provably true, and I have not
proven that then someone else who did could get in the history books
(or at least math textbooks).
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.
>
> > The equivalent step with easy polynomials would be
> > [x^2 + y^2][x^2 - y^2] = 0(mod (x+y)), so
> > x^2 + y^2 = 0 (mod (x+y))
> > *or*
> > x^2 - y^2 = 0(mod (x+y))
> > (which is actually a bit misleading but I won't go into why at this
> > point to try and lessen the confusion level).
>
> Nope, it is utterly misleading because you do not have a condition in
the
> first congruence.
I'll stop here because you've just said something that seems indicates
a gross disregard for the truth to me.
I've made strong statements like that before and some may wonder if I'm
just mouthing off, or can I support it.
I think it should suffice to remind you all that x^5 + y^5 = z^5, which
means that z can be substituted out in the statements I've given, so
that you *do* have expressions that aren't based on a conditional.
Now, I've mentioned that to Winter and others many times before, so I
think it fair of me to believe that his statement is suspicious, and to
say so, since the public record is clear.
James Harris
Sent via Deja.com
http://www.deja.com/
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