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Topic: JSH: FLT Proof is still a test.
Replies: 47   Last Post: Mar 8, 2002 9:50 PM

 Messages: [ Previous | Next ]
 Nico Benschop Posts: 1,708 Registered: 12/6/04
Re: JSH: FLT Proof is still a test.
Posted: Mar 4, 2002 6:26 AM

James Harris wrote:
>
> Nico Benschop <n.benschop@chello.nl> wrote in message news:<3C822261.D11813F9@chello.nl>...

> > James Harris wrote:
> > >
> > > Nico Benschop <n.benschop@chello.nl> wrote in message news:<3C80D27D.96150E03@chello.nl>...

> > > > James Harris wrote:
> > > > >
> > > > > [...]
> > > > > However, the truth is there's no such thing. Instead there's a
> > > > > worldwide collaboration of academic and working mathematicians
> > > > > who produce mathematical works and proof each other. So you have
> > > > > a group of people who check for each other's mistakes and they
> > > > > tell the rest of the world when their work is correct, and when
> > > > > it's important.
> > > > > The problem with that is that only the work that comes through
> > > > > that pipeline is accepted. It's like some kind of big club

> > > >
> > > > ...and don't forget that the competition *in* that club is fierce,
> > > > which is a pretty sharp filter, James!

> > >
> > > Not as sharp as many people seem to think.
> > > Mathematicians don't gain points by proving each other wrong.
> > > They have a group that works more by consensus than by fierce
> > > attacks on each other's positions.

> >
> > I think you're wrong here, James. In maths, by its very nature, a
> > fault in the logic of a proof is deadly, and to be avoided at any
> > cost. Rather an extremely long but correct proof (Wiles' FLT) ..[*]
> > than a short one with lingering doubts.

>
> And how do you know it's correct? Seriously, why don't you answer that
> question, and tell me what makes you believe that Wiles' work is correct?

In fact, Wiles proved the Tanyiama-Shimura conjecture for special types
of
elliptic curves (EC) being modular. I have not seen that paper, I admit,
but I have no reason to doubt the correctness of that result - which has
by others been extended to a larger (less restrictive) class of EC's.

Furthermore, regarding one of its corollaries: the truth of the FLT
conjecture on the sum of to integer p-th powers (prime p>2) not being
a p-th power (Frey's impetus of 1985 to Wiles to start digging in):
I *do* have a lingering doubt, or rather a methodical question,
regarding the *residue_method* ('local') of proof that Wiles used,
to derive a result such as FLT which holds for *integers* ('global'),
at least as I understand it from a simple description by Simon Singh
in his book 'Enigma...'

I voiced this concern as a question here in this NG several times:
my problem is equivalent to the argument used for the impossibility
of deriving a global result (on integers, such as the FLT inequality)
from a local method, viz. using only residues.

Wiles uses a 'spectral' method of Dirichlet L-series based on residues
mod p (prime p) to count the number of solutions of an equation
(elliptic:
of degree 2 and 3 in two variables resp.) with induction on all primes
p.
The same is then done with modular forms, to establish 1-1
correspondence between EC's and MF's. All fine & dandy.

The problem is: HOW can in this case a (local) residue method yield a
(global) result on integers, such as FLT. In detail:
WHERE and HOW precisely does Wiles in the FLT corollary 'escape'
the local (residue) method in order to derive the FLT 'global'
result?

While in fact, THE argument against any direct proof of FLT by using
ONLY residues mod p^k cannot possibly yield a (global) result such as
the FLT inequality. Phrased by some mathematicians as: the Hensel lift
cannot be
broken ( see http://home.iae.nl/users/benschop/sgrp-flt.htm ), which
indeed is true, so an EXTRA property of solutions of x^p + y^p == z^p
mod p^k must hold, in order to derive the FLT integer p-th power
inequality:
in fact ALL such residue solutions have exponent p distributing over a
sum
(the 'EDS' poperty) or are equivalent (via a linear transformation that
preserves equivalence) to a solution with that property. So that is *my*
trick to break the Hensel lift; *what* is Wiles' trick to derive FLT
with
a residue method ??
- No answer yet, on this question of mine. But maybe later in my
pension,
with more hobby-time, I can dig into his paper to find out for myself;-)

It is similar to Galois' theory of the unsolvability in radicals of the
roots of *general* >4 degree polynomials (based on the alternating
groups
A_n being simple for n>4). This does *not* mean that any 5-degree
polynomial
cannot be resolved, because a polynomial which is a product of say 2-nd
and 1-st degree polynomials is trivially solvable, with the
corresponding
roots of quadratics and linears. Here: the solutions of the FLT mod p^k
equivalence all have (up to a linear transformation) a special property
(EDS) that *does* allow to derive the required result for integers.

> Like I said, mathematicians work more by consensus and suppose that
> human eyes can catch every fault in the logic. Once a proof from a
> colleague is in front of them, they hope it's right. I'd just as soon
> they viciously attacked it with everything they had ..[*]
> in an attempt to find any kind of fault, which they could throw back
> at the mathematician who came up with it to see if they could fix it.
> Then I might have more faith in the system as in my mind that's the
> only way to do it.

Re[*]:
Don't you worry, they *do*... human nature being (competitive) what it
is.
Moreover, if the claimed result is important for other parts of math,
they
better be *sure* it is correct, otherwise the whole building collapses,
with them in it!

>
> >That's why he (W) got so emotional
> > in the BBC interview, explaining the moment he could correct the flaw
> > in his proof of the T-S conjecture: coming out (after 7 years in the
> > attic) with an incorrect proof is disastrous for a professional
> > mathematician!

>
> Yet Wiles did just that.
> Hey, I don't know if his work is correct or not. I'm just not going
> to go by someone else's say-so on it because I've seen how
> mathematicians can say all kinds of things (many of them quite crude).

^^^^^^^^^^^^^^^^^^^^^^^^
As can you, I'm convinced by now ;-)

> > Hence the extreme secrecy... The reason for this
> > attitude comes, moreover, from the 'single leg' they stand on (only
> > logic) - compared to physics where reality & measurements _also_
> > plays an important role, next to the theory (for which the axioms do
> > have a link to reality, while i math only self consistency counts).

>
> You got it backwards. In physics a theory both has to be logically
> consistent AND it has to fit the real world.

That's precisely what I mean: having an alterate context for extra
verification makes it *easier* to check results, doesn't it?

> Consider Ptolemy's spheres: beautiful and CORRECT mathematics that
> didn't completely fit the real world.

Nice example, but the 77 circles & epi-circles required provided quite
accurate orbit prediction. Copernik's shift to the Sun as center
reduced this to 43, for equivalent accuracy. And Kepler's ellipses
with the Sun in one of the two foci still simplified the model further.
Who knows, using a center of gravity might still further simplify &
improve the model...

It is just a matter of model_complexity vs. accuracy,
that makes one model more 'true' than another.
And in the FLT proof case: the simpler the better. Wiles' corollary
is a start, but not a very convincing one, at least regarding FLT...
(there *must* be a more direct way, than via the Northpole of EC = MF,
...and I'm convinced the EDS property of FLT mod p^k solutions does
it;-)

> Mathematicians have it EASIER than physicists and other scientists
> because of what you mention, not the other way around.

No, I don't think so: their balancing act (on one leg) is more
difficult,
and 'scary', than that of a physicist (on two legs).

> > > Scientists are better about that as a physicist or chemist or
> > > biologist knows that any new proposition is likely to face fierce
> > > scrutiny,

Quite the opposite:
the smaller the context, the more fiercely territory is protected...

> > > and where it's well-known that even well accepted positions can
> > > turn out to be wrong or incomplete. Which is all part of what makes
> > > science fun.
> > >

> > > > > where you have to play arcane rules or they simply dismiss
> > > > > your work. I like breaking such senseless rules.

> > > >
> > > > If your approach is so good, why not simply satisfy their rules,
> > > > just in order to publish and be done with it. ...

> > >
> > > Good question. My answer is that it's my choice.
> > > It's a matter of my free will and what I choose to do.
> > > Right now I choose not to play by silly rules just to get acceptance
> > > from a group of people who I think should care more about the truth
> > > than that people play their game.

> >
> > You mean: you devise your own 'logic' - preventing a priori agreement.

>
> No, I mean that mathematicians shouldn't force me to submit such a
> short, direct proof to a journal before they'll admit that it's
> correct. It seems to me to be a useless formality.
>

> > You should know such 'proof' does not hold normal water by definition.
> > And don't keep talking about 'truth': even in math it is long known
> > that only self consistency counts, the axiom's are "true" by consent
> > only, as are the logic derivation rules. It is a 'game' purely by
> > consent alone, albeit by very strict rules (that are agreed upon).

>
> You're babbling nonsense. I doubt you'll even find many mathematicians
> who'll agree with you that axioms are true by consent only.
> Next you'll be saying that 2+2 = 4 only because human beings believe it,
> or saying that 2+2 = 1 mod 3, or anything else like that).

The symbols 2, +, = and 4 have semantics ONLY agreed upon by
convention.
(as BTW do all words and strings in this exchange;-)

> > There are even different types of math, of 'sub clubs' that do not
> > agree with one particular rule (like not excluding the middle in a
> > binary true/false system). [...]

>
> > > For instance, you have a^3 == 1 mod p^k, and (a+1)^p == a^p + 1^p
> > > mod p^k, which comes from (z/y)^p == (x/y)^p + 1^p (NB: in residues)
> > > so your result is (z/y-x/y)^p == 1 mod p^k, ..[*]
> > > which is (z-x)^p = y^p mod p^k. [...]

>
> Sorry dude, I notice that you follow this with a whole lot of
> verbiage, none of which changes the fact that all of your results
> can be checked, and shown to be incorrect by using z/y and x/y.

Well, for FLT case_1 : only units (such as y or y^p) have an inverse,
by scaling (divide by -z^p) yielding normal form a^p + b^p == -1 mod
p^k,
with a == -x/z, b== -y/z (mod p^k). The base p code of -1 = \sum
(p-1)p^i
(i=0.. k-1), hence with k digits p-1, making extension of p-th power
residues to any precision easy for k>1 : just make (a^p)_i + (b^p)_i =
p-1
digit-wise, they remain p-th power resides (by the Hensel lift lemma;-)

> > [*]: No, you equivalence (z/y-x/y)^p == (z/y)^p - (x/y)^p mod p^k,
> > in other symbols (;-): (a-b)^p == a^p - b^p mod p^k, which in general
> > does not hold because the exponent p does not distribute over a sum.
> > But in the special case of a^p == a, b^p == b, (a-b)^p == a-b mod p^k
> > it can hold indeed, thus a solution of a^p - b^p == c^p mod p^k,

>
> <deleted>
>
> That last is what should clue you in.
> You see, x^p + y^p = z^p means that x^p + y^p = z^p mod p^k.

Indeed, and for any precision (base p) of k >0.
For inst: k=1, using FST (Small Thm): x^p == x, y^p == y, z^p == z,
..[&]
then x+y == z mod p (viz: each last digit must add).
All I do is generalize this to any k>1, by noting [&] also hold for
any solution 'in core' A_k mod p^k (k>1), which are of type n^{p^[k-1]}
forming a p-1 cycle with of course n^p == n mod p^k, similar to FST mod
p.

> You see, the Fermat expression is THE king expression for all of your
> congruence results, so z/y and x/y can be used at any point in your
> work instead of your variable, and when that's done, it's clear that
> you can't have proven Fermat's Last Theorem.
> In fact, you apparently are going in circles. {...] -- James Harris

Sorry, James, but you apparently don't understand the principles...

-- NB - http://home.iae.nl/users/benschop