Re: FLTAA: Field structure p = 2, p>2


The Dougster schrieb:

hagman wrote:
The Dougster schrieb:
I mean, can you imaging replacing the 200 page proof of Wiles et al
with:

M2p </= Mxyz, ergo FLT

or something similar. Wow. That's like 20 characters, not 200 pages.

Yeah, that would be really amazing, esp. as you would have found the
proof
without ever reading an introductory book on the subject of group
theory.

I can reply to the rest later with thanks for your taking the time to
write it all out.

I just finished not only an introductory *book* on group theory (the
text by Fraliegh) but an introductory *course* in abstract algebra. And
earned an A in the course. Not that that implies I am facile with the
knowledge, or that I understand the topic, only that there is no need
to be snide or cruel. Unlike JSH, I see no conspiracies here and am
willing to take my knocks as I learn and work things out. You might say
I downloaded the stuff and am now installing it. It's in process.

In that case please accept my apologies.
But a few of the problems that surfaced in this thread should have
been dealt with fairly explicitly in a not too shallow introduction to
group theory.
My reaction was too easily triggered by the suggested possibility of
a short and simple proof of FLT requiring merely beginners knowledge in
group theory.
One tends to have in mind a preprinted reply form
"Dear ___,
thank you for the suggested proof of FLT that you submitted.
The first mistake is on page ___, line ___.
Yours,
___"
As a matter of fact, real instances of such forms were (still are?)
common
to handle the masses of submitted false FLT proofs. ;)


I am not saying I have this proof, only that there is room for the key
point of what could be a proof. I see there's a key point in there. I
haven't got the point itself, but the frame around it is strong:

If there are no solutions prime p > 2 to

( z/y, z/x, x/y) ^p = ( 1, 1, -1) (and I have never found one )
in Mx x My x Mz ~= Mxyz, then FLT follows.

If somebody could show me *one* solution, I could drop this whole three
or four year exploration and be happy to go work on my electric
bicycle.

How about x=2, y=3, z=5, p=5:
Since we have z=y mod x, z=x mod y, x=-y mod z,
it follows that
(z/y,z/x,x/y)^p = (1,1,-1)^5 = (1,1,-1) in Mx x My x Mz.

After all, this is not my field of expertise; I am a
machinist, not a mathematician. :)

On the other hand, one cannot learn without asking questions.


Beginning 15 January, I have Discrete Math with logic, proofs, and more
number theory, much of which comes from abstract algebra.

Beware, there are many pitfalls in these areas as well where a beginner
might think
"Why do all experts consider problem X difficult? It /almost/ follows
from ..."

.

Relevant Pages

  • Re: FLTAA: Field structure p = 2, p>2
    ... M2p </= Mxyz, ergo FLT ... I just finished not only an introductory *book* on group theory (the ...
    (sci.math)
  • Re: How do I do this problem without a calculator?
    ... But when you have to use FLT you firstly have to use the theorem [a = y ... member of ^x, so (because 11 is prime, this is important, it ... using squares, or if you want to use any theorem, use a calculator. ...
    (sci.math)
  • Re: How do I do this problem without a calculator?
    ... But when you have to use FLT you firstly have to use the theorem [a = y ... member of ^x, so (because 11 is prime, this is important, it ... using squares, or if you want to use any theorem, use a calculator. ...
    (sci.math)
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