|
|
Re: FLT Discussion: Simplifying
Posted:
Jan 16, 2001 8:19 PM
|
|
In article <942k1a$i6f$1@nnrp1.deja.com> jstevh@my-deja.com writes:
> In article <G78GGv.7ID@cwi.nl>,
> "Dik T. Winter" <Dik.Winter@cwi.nl> wrote:
> > In article <9400ps$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.
I think not.
> My comment to them is that it's the same thing being done with the
> proof of Fermat's Last Theorem.
Yup, indeed. You do something you do not prove because it is "obvious",
but it is *not* obvious. You assume that whenever AB = 0 for your funny
things that either A = 0 or B = 0 or both. But you have to give a proof
of that because there are many funny things in mathematics where that
does *not* hold. And it has been *proven* for the complex numbers, but
that proof actually uses that x^2 + y^2 can not be 0 unless both x and
y are 0. Just the thing you want to prove. So because you rely on that
fact for the complex numbers you are reasoning in a circular fashion. It
is like (with either x or y nonzero or both):
x^2 + y^2 != 0 because whenever AB = 0 either A = 0 or B = 0 and the
latter is true because x^2 + y^2 != 0.
> > > 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).
Well, for the mathematicians it is just a number.
> 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.
Yes, so what?
> 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.
Not "as if it were a number". It *is* a number. You can also write it as:
{ x <= 0 or (x > 0 and x^2 < 2) | x > 0 and x^2 >= 2}
or whatever other formulation you fancy. They are just the same.
In the same way 15 is the definition of a number based on a radix and
stuff like that.
> 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.
I don't know. Ever looked at the diagonal of a square with unit sides?
> > So, James, how do I multiply a number with an operation, and what do I
> > get?
>
> Multiplication is an operation.
Yes, I knew that. But if you say that something is not a number but
something else, you have to define that operation of multiplying a
number with such a something else. Mathematics does not define it yet.
> > > 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 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.
Ok, I can do that; I substitute z = (x^5 + y^5)^(1/5). Now I have a
completely different kind of objects. And now I have something like:
f1(x,y,v) * f2(x,y,v) = 0 (mod f3(x,y,v)).
But with these funny objects you have to *prove* that either f1 or f2
is 0 mod f3; because that is no longer clear (and I think that David
Libert has actually proven that it is false). Note that on
http://www.cwi.nl/~dik/english/mathematics/jsh.html
I can do the substitution z = ((x^5 + y^5)/n)^(1/5) and obtain the
same fallacy.
However, you snipped a part:
> [(v^2-1)z + sqrt((v^2-1).2xy)][(v^2-1)z - sqrt((v^2-1).2xy)] =
> 0 (mod (x+y+vz)) when x^2 + y^2 = z^2.
Factually true.
> So
> (v^2-1)z + sqrt((v^2-1).2xy) = 0 (mod (x+y+vz))
> *or*
> (v^2-1)z - sqrt((v^2-1).2xy) = 0 (mod (x+y+vz))
by the same reasoning. Why is this one wrong (or the one on my web
page) and yours correct? Just point to a single step of the proof
on the web page and state explicitly why that step is wrong. Just
once, please, pretty please.
--
dik t. winter, cwi, kruislaan 413, 1098 sj amsterdam, nederland, +31205924131
home: bovenover 215, 1025 jn amsterdam, nederland; http://www.cwi.nl/~dik/
|
|
