Lol. Welcome to the club!
Fermat's was a note in the margin of a book claiming he had devised a simple proof that a^n+b^n=/=c^n for n>2.
Here's a simple proof.
when a=b, a^n+b^n<c^n for n>2. As a approaches 0, b approaches c. Therefore as a approaches 0, a^n+b^n approaches c^n for n>2.
End of story.
Lies have the stench of death and defeat.
The Binomial Theorem states it exactly (as does my proof) for positive integers, n >2:
, where (so only if or
(so or , respectively.
But I'm happy to see that Grapes seems to understands your version, even though it doesn't involve integers (). (There is a derivative relation between the Binomial Theorem and GTR as I've just posted in the other thread, where the Jacobian of the metric tensor is characterized as consisting of integers (v'/c' invariant as an "acceleration" )
The (general) relativistic relation is where a is identified with c in the relation v/c =b/c, so if a goes to zero v=b goes to zero as well and there is no integer to begin with (c=0) with all elements identified with differentials rather than integers.
(i.e, there is no metric tensor characterizing stress-energy)
(there is no curvature because there is nothing in the Universe... not even a parking lot..)
Proof of Fermat's Theorem
If Fermat's expression were true (), , , and would be a Pythagorean Triple. They are not a Pythagorean Triple by the Binomial Theorem, where rem(a,b,n) > 0.
Therefore, , for a,b,c,n positive integers, n>2
Apologies in advance if I don't format this right, I have very little experience with latex
I just wanted to point out that your proof doesn't really work.
You're just saying that:
And then setting (a+b)^n equal to a^n+b^n,
so of course a or b has to be zero.
All you're saying is (a+b)^n=a^n+b^n IFF a or b=0
What you want to prove is that c^n = a^n+b^b IFF a or b=0
Those are different things
Last edited by Icarus496; 02-26-2017 at 04:49 PM.
Right, I understand the binomial expansion, but you can only apply the binomial theorem if you have something of the form (a+b)^n. If you set that equal to c^n, like you have to do for your proof, that reduces to c=a+b, which is not what you want to prove.
If you read the Wiki page you cited (under "Theorem statement"), it says , not for any c.
Added Link for user:
Cited Wikipedia Page
Last edited by Neverfly; 02-27-2017 at 03:22 AM.