Ken
It would be something of a big deal; see below.
> How about the mythical "simple proof"?
If someone were to fine a simple proof, that would
be nice, and it would be a Good Thing profesionally
for the finder.
> Would the whole mathematical edifice come crashing down
In the event of a counterexample, there would be a lot
of scrambling to see just where Wiles' proof went wrong.
And it would refute the more important general result --
not put it doubt, but actually refute it.
> or would it just be
> a small chip on one of the statues in the courtyard?
A refutation of Taniyama/Shimura/Weil would cast doubt
on the general thrust of research in algebraic geometry
over the last three decades.
What with the amount of attention that's been given to
Wiles' proof by highly competent people over the past
decade. though -- the original proof has been corrected
and simplified, and the main result has been improved on
-- I find it hard to imagine it's fatally flawed.
A simple proof of FLT would get included in the next
edition of a lot of books about number theory.
> From the outset, I acknowlege this is a bit of a troll, but as an amateur,
> sitting in the bleachers, watching all this talk of FLT, I was wondering
> how big an impact a counter example have?
Tremendous.
> How about the mythical "simple
> proof"? Would the whole mathematical edifice come crashing down or would
> it just be a small chip on one of the statues in the courtyard?
Neither.
--
Robin Chapman, www.maths.ex.ac.uk/~rjc/rjc.html
"His mind has been corrupted by colours, sounds and shapes."
The League of Gentlemen
> How about the mythical "simple proof"?
> Would the whole mathematical edifice come crashing down or would
> it just be a small chip on one of the statues in the courtyard?
A counterexample would be quite an emergency, but a simple proof would not.
The importance of a simple proof would depend on the nature of that proof.
Doubtless everyone would want to see it at once, wondering whether some
Great New Idea had hatched, and what other applications that idea might
have. If the ideas involved are not so great, we will wonder -- at least we
should wonder -- why the proof was missed for so long.
But the Great New Idea is itself something of a mythical beast, n'est-ce
pas? Quite a bit could be written about it's "real and imaginary parts". Are
the truly valuable ideas as closely tied to specific problems or conjectures
as we like to think? I would say no, but the GNI is often mentioned in
connection with every major unsolved problem.
Larry
.. or rather its (it's ?-) "residue and carry"..
http://home.iae.nl/users/benschop/carry.htm (on revival of the 'carry')
(since Gauss' residues 1801)
> Are the truly valuable ideas as closely tied to specific problems or
> conjectures as we like to think? I would say no, but the GNI is
> often mentioned in connection with every major unsolved problem.
> - Larry
>> But the Great New Idea is itself something of a mythical
>> beast, n'est-ce pas? Quite a bit could be written about it's
>> "real and imaginary parts".
>
> .. or rather its (it's ?-) "residue and carry"..
> http://home.iae.nl/users/benschop/carry.htm (on revival of the 'carry')
> (since Gauss' residues 1801)
No, there's nothing great nor new about this idea :-(
If you're looking for somewhere to make your mark, try something that
hasn't been done and accepted yet.
Norm
> Ken Tozier <kent...@attbi.com> wrote in message
[snip]...
> If you're looking for somewhere to make your mark, try something that
> hasn't been done and accepted yet.
No mark-making intended, I was just trying to get some insight on the true
importance of FLT by finding out what would break if it was ever found to be
false.
>
> Norm
>
Ken
Right so.
The more amazing is it to see how the carry 'c' in: n = c.m + r
( natural modulus m>1, non-neg. residue r<m )
plays such little/no role, especially in additive nr-th problems,
like FLT and Goldbach. After attention is concentrated only on the
beautiful 'closure' properties of residues (Gauss), the necessary
extra step of including the carry (after proper choice of modulus m)
is skipped/forgotten, saying: Hensel blocks a direct proof of FLT...
[ of course: ignoring the carry will bring no global result;-)
E.g. use the powers of p+1 to represent the units mod p^k (odd
prime p, any precision k>1) uniquely as: u == g^i (p+1)^j mod p^k,
for some given primitive root g, non-neg pair (i,j): i<p, j<p^{k-1}.
'Trivial' in math does not imply 'unimportant', especially in applied
math. Or rather: 'trivial' in pure math may be 'most essential' but
not seen as such (yet) in applied math. Re Shannon noticing (1938)
Boolean algebra is an excellent model for digital (binary) circuits:
trivial by math standards, but most essential for engineering.
Well, a counter-ex to FLT (say case 1: x,y,z coprime to prime expon p)
would be most surprising & disastrous, since it would mean that exponent
p in (x+y)^p could distribute over a sum, as in fact is does
- but only for residues (!) -
in *each* solution of: x^p + y^p == z^p mod p^2.
The whole of arithmetic as-we-know-it would collapse!
> Robin Chapman wrote:
>>
>> Nico Benschop wrote:
>>
>> >> But the Great New Idea is itself something of a mythical
>> >> beast, n'est-ce pas? Quite a bit could be written about it's
>> >> "real and imaginary parts".
>> >
>> > .. or rather its (it's ?-) "residue and carry"..
>> > http://home.iae.nl/users/benschop/carry.htm
>> > (on revival of the 'carry', since Gauss' residues 1801)
>>
>> No, there's nothing great nor new about this idea :-(
>> -- Robin Chapman
>
> Right so.
> The more amazing is it to see how the carry 'c' in: n = c.m + r
> ( natural modulus m>1, non-neg. residue r<m )
> plays such little/no role, especially in additive nr-th problems,
> like FLT and Goldbach. After attention is concentrated only on the
> beautiful 'closure' properties of residues (Gauss), the necessary
> extra step of including the carry (after proper choice of modulus m)
> is skipped/forgotten, saying: Hensel blocks a direct proof of FLT...
> [ of course: ignoring the carry will bring no global result;-)
Yawn.
So what, Mr Benschop, have *you* achieved with your "residue and carry"
method? You have produced a paper claiming to prove Fermat's Last Theorem,
yet not doing so, and you have produced a paper claiming to prove
Goldbach's Conjecture, yet not doing so. Have you achieved *anything*
of value by using your "residue and carry" method at all, Mr Benschop?
Well, for instance, what is wrong with:
http://home.iae.nl/users/benschop/fst-root.dvi
"Complementing Fermat's Small Theorem:
the divisors r | p \pm 1 have distinct r^{p-1} mod p^3."
(re Wieferich's 1909 result regarding 2^p mod p^2, but now mod p^3)
Especially thm 2.1, which depends on idempotent e^2 == e mod p-1
(odd prime p) where e^2 > p only for natural e > 1.
So in: e^2 = c(p-1)+e = cp+(e-c)p < p^2,
carry c=0 only for trivial idempotent e=1.
--> Notice the idempotents 'e' of semigroup Z_{p-1}(.) mod p-1
are interpreted as naturals < p , essential for the 'carry'
to be re-introduced into number theory, since Gauss ignored it
in his residue arithmetic (1801) for obtaining nice 'closure'
properties of Z (mod m).
Re[*]: still sleeping, mr. Chapman?
Well, for instance, what is wrong with:
http://home.iae.nl/users/benschop/fst-root.dvi
"Complementing Fermat's Small Theorem:
the divisors r | p \pm 1 have distinct r^{p-1} mod p^3."
(re Wieferich's 1909 result regarding 2^p mod p^2, but now mod p^3)
Especially thm 2.1, which depends on idempotent e^2 == e mod p-1
(odd prime p) where e^2 > p for all natural idempotents e > 1.
So in: e^2 = c(p-1)+e = cp+(e-c)p < p^2,
carry c=0 only for trivial idempotent e=1.
--> Notice the idempotents 'e' of semigroup Z_{p-1}(.) mod p-1
are interpreted as naturals < p , essential for the 'carry'
to be re-introduced into number theory, since Gauss ignored it
in his residue arithmetic (1801) for obtaining nice 'closure'
properties of Z (mod m).
Re[*]: still sleeping, mr. Chapman?
^ Well, a counter-ex to FLT (say case 1: x,y,z coprime to prime expon p)
^ would be most surprising & disastrous, since it would mean that exponent
^ p in (x+y)^p could distribute over a sum, as in fact is does
^ - but only for residues (!) -
^ in *each* solution of: x^p + y^p == z^p mod p^2.
^ The whole of arithmetic as-we-know-it would collapse!
Hmm. If a genuine counter-example to FLT would cause the whole of
arithmetic as we know it to collapse, the genuine counter-example to
FLT (being nothing more than arithmetic) would collapse with it,
wouldn't it? And thus it would cease to be a counter-example,
whereupon arithmetic would uncollapse, and the counter-example would
be restored, and ...
I think I need to go for a little lie down.
Andy
--
sparge at globalnet point co point uk
Look after the sins of write-commission,
and the sins of read-omission will take care of themselves.
> Robin Chapman wrote:
>>
>>
>> Yawn. ...[*]
>>
>> So what, Mr Benschop, have *you* achieved with your "residue and
>> carry" method? You have produced a paper claiming to prove Fermat's
>> Last Theorem, yet not doing so, and you have produced a paper
>> claiming to prove Goldbach's Conjecture, yet not doing so.
>> Have you achieved *anything* of value by using your
>> "residue and carry" method at all, Mr Benschop? -- Robin Chapman
>
> Well, for instance, what is wrong with:
> http://home.iae.nl/users/benschop/fst-root.dvi
For a start you go from
e^2 = cp + (c-e) to
e^{2p} = (c-e)^p (mod p^3).
This is bollocks.
Incidentally, Mr Benschop, where have you submitted this tripe?
> ^ Well, a counter-ex to FLT (say case 1: x,y,z coprime to prime
> ^ expon p) would be most surprising & disastrous, since it would
> ^ mean that exponent p in (x+y)^p could distribute over a sum,
> ^ as in fact is does
> ^ - but only for residues (!) -
> ^ in *each* solution of: x^p + y^p == z^p mod p^2.
> ^ The whole of arithmetic as-we-know-it would collapse!
>
> Hmm. If a genuine counter-example to FLT would cause the whole of
> arithmetic as we know it to collapse, the genuine counter-example
> to FLT (being nothing more than arithmetic) would collapse with it,
> wouldn't it?
Indeed, we would be in _big_ trouble;-(
You can safely bet that FLT holds (if you can't follow Wiles' proof)
I have always felt that FLT is "at-the-core" of arithmetic, and not
some esoteric property. By Jove: (+) and (^) are essentially linked
there, with: the sum of any two p-th powers (p>2) is not a p-th power:
p-th powers are a very 'powerful' generating set of the integers
under (+) - http://home.iae.nl/users/benschop/anti-cl.htm -- NB
> And thus it would cease to be a counter-example,
> whereupon arithmetic would uncollapse, and the counter-example
> would be restored, and ...
> I think I need to go for a little lie down. -- Andy
Wiles's proof has been sufficiently scrutinized and can be assumed to
be correct. Note that while FLT is stated in elementary Peano
Arithmetic (PA), Wiles's proof is carried out using concepts that go
beyond PA, such as real numbers, complex numbers, sets of such numbers
(which may contain uncountably many numbers), etc. So Velleman
speculates that Wiles's proof can only be formalized in some theory
stronger than PA, but not quite as strong as Zermelo-Fraenkel set
theory (ZF). It follows that this theory, and hence ZF, would be
rendered inconsistent if the existence of a counter-example to FLT can
be demonstrated. This is certainly a big deal.
For an interesting challenge to classical logic and in particular, the
existence of infinite sets and all kinds of infinitary reasoning, such
as, the Cantor diagonalization argument, see my preprint
PITT-PHIL-SCI00000635 ("Quantum superposition justified in a new
non-Aristotelian finitary logic"), available from the link below:
http://philsci-archive.pitt.edu/documents/disk0/00/00/06/35/index.html
Sincerely,
R. Srinivasan srad...@in.ibm.com
OK, but e^{2p} == p.cp.(e-c)^{p-1} + (e-c)^p (mod p^3),
== cp^2 + (e-c)^p [mod p^3] with 0 <= c < p-1,
and natural e < p.
This still yields e^p mod p^3 *only* if carry c=0 (hence e=1),
sufficient for the results in section 3. But thanks anyway;-)
-- NB
> Robin Chapman wrote:
>>
>> Nico Benschop wrote:
>>
>> >
>> > Well, for instance, what is wrong with:
>> > http://home.iae.nl/users/benschop/fst-root.dvi
>>
>> For a start you go from
>> e^2 = cp + (e-c) to
> ^^^^^
>> e^{2p} = (e-c)^p (mod p^3). This is bollocks.
>
> OK, but e^{2p} == p.cp.(e-c)^{p-1} + (e-c)^p (mod p^3),
> == cp^2 + (e-c)^p [mod p^3] with 0 <= c < p-1,
> and natural e < p.
> This still yields e^p mod p^3 *only* if carry c=0 (hence e=1),
> sufficient for the results in section 3. But thanks anyway;-)
So the next non-sequitur is where you assert that
e^{2p} = e^p (mod p^3).
So where does this come from?
And now that you have made a correction to your MS are you going
to withdraw the previous one from submission?
From c=0 and FST (Fermat Small Thm): x^p == x mod p.
(e=c is excluded since that would imply e=0: non-unit)
> And now that you have made a correction to your MS are you going
> to withdraw the previous one from submission?
-- NB
Another nonsequitur.
Let's review what you are (feebly) attempting. You have
a number e with 1 < e < p-1 and with e^2 = e (mod p-1).
You are attempting to obtain a contradiction to e^3 = e (mod p^3).
We have e^2 = c(p-1) + e for some integer c, and indeed 0 < c < p.
Thus e^2 = c - e + pc and so e^{2p} = (c-e)^p + cp^2 (mod p^3).
You then say
"So e^{2p} = e^p = e (mod p^3) iff c = 0 (thus e = 1)."
But where does the congruence e^{2p} = e^p (mod p^3) come from?
You are attempting to get a contradiction from e^p = e (mod p^3)!
[It is trivial to get a contradiction from e^{2p} = e^p (mod p^3)
anyway.]
>> And now that you have made a correction to your MS are you going
>> to withdraw the previous one from submission?
>
>No need, see:
>http://groups.google.com/groups?hl=en&lr=&ie=UTF-8&safe=off&threadm=3DAE94C3.B47D8AEC%40chello.nl&rnum=2&prev=/groups%3Fq%3Dbenschop%2BHaugland%2Bgroup:sci.math.*%26hl%3Den%26lr%3D%26ie%3DUTF-8%26safe%3Doff%26scoring%3Dd%26selm%3D3DAE94C3.B47D8AEC%2540chello.nl%26rnum%3D2
I suppose there is no need. I mean there is no need for you
to adhere to any standards of moral and honest behaviour is
there? If you know that there is an error is a mathematical
paper it is dishonest and immoral to submit it to a journal.
Your cited webpage is only a sci.math posting from you,
which I quote:
> And if there is/are essential error(s) would naturally result
> in non- acceptance: the proper thing to do would be to point
> them out, no?
This is no justification for your immoral behaviour. It's
false too, and you have provided your own counterexample:
your paper in "Computers and Mathematics with Applications".
That was published despite there being essential errors.
That journal has an unusual submission policy, asking authors
to provide a list of six possible referees. Who were
your six nominees, Mr Benschop?
Robin Chapman