# Thread: Proof of Fermat's Theorem by Modulus

1. ## Re: Proof of Fermat's Theorem by Modulus

Originally Posted by BuleriaChk
It is easy, c < a+b is irrelevant (since it can't generate the Fermat equation to begin with), and the rest of your post is an abject admission of incompetence and a retreat to a blind faith in an urban legend that the theorem is "unproveable".
Apparently it has been proven.

Just not by you.
(I cann't believe that the idea the Fermat's Theorem hasn't been proven for several centuries is a huge practical joke in math departments.)

2. ## Re: Proof of Fermat's Theorem by Modulus

Originally Posted by grapes
Yes it is. It is equal to 1. Did you mean something else?
So that a + b = c?

That is, 1 = 1? some counter-proof... Sheesh!

so that a = .5 and b = .5 means that a + b = 1 (note, please, that a and b are not integers...)

Now calculate for a = 1, b = 1...

and then, maybe try a = m*1, b = n*1 ... in Euler's formulae ...

again, the point is that for a and b independent variables (a,b)

for (n = 2), vanishes in the case of Pythagorean triples, where is the interaction vector .

If a and b are dependent, the a = mb so c = mb +mb , so .

Where did a go?

If they are independent, then so , and the interaction term does not vanish.

, where is the positive unit vector for both cw and ccw Lorentz rotations in the positive quadrant of the relativistic unit circle (+1,+1) as well as the negative quadrant (-1,-1).

That is,

Furthermore, for the interaction profuct ( <1; if , then = 0, and vice versa)

3. ## Re: Proof of Fermat's Theorem by Modulus

Originally Posted by BuleriaChk
So that a + b = c?
You're the one that said
Originally Posted by BuleriaChk
is not an integer for .
That is, 1 = 1? some counter-proof... Sheesh!
??
Why can't c equal a+b? That seems like that's what you specified by that last equation, where c divided by a+b equals 1. That's the same as c=a+b
so that a = .5 and b = .5 means that a + b = 1 (note, please, that a and b are not integers...)

Now calculate for a = 1, b = 1...

and then, maybe try a = m*1, b = n*1 ... in Euler's formulae ...

again, the point is that for a and b independent variables (a,b)

for (n = 2) vanishes in the case of Pythagorean triples, where is the interaction vector.

If a and b are dependent, the a = mb so c = mb +mb , so , so

4. ## Re: Proof of Fermat's Theorem by Modulus

Originally Posted by grapes
You're the one that said

??
Why can't c equal a+b? That seems like that's what you specified by that last equation, where c divided by a+b equals 1. That's the same as c=a+b
c = a+b in that expression in context means that the l.h.s. and r.h.s. refer to the same number. Call it d, so that c = a+b = d. (e.g. a = .5, b=.5, (a+b) = c = d = 1

so

The point is that (a,b) are independent vectors which have different addition rules than arithmetic. That's why the Binomial theorem is true for ALL a and b independent variables over two orhtogonal number lines (each of which satisfy arithmetic operations as vector coefficients in each dimension, but the coefficients in each number line do not interact arithmetically with the coefficients in the other; the vectors interact. That is why a and b retain their identities in the Binomial Theorem.

(All the so-called paradoxes in physics and math vanish with a little thought...)

5. ## Re: Proof of Fermat's Theorem by Modulus

Originally Posted by BuleriaChk
c = a+b in that expression in context means that the l.h.s. and r.h.s. refer to the same number. Call it d, so that c = a+b = d. (e.g. a = .5, b=.5, (a+b) = c = d = 1

so

The point is that (a,b) are independent vectors which have different addition rules than arithmetic. That's why the Binomial theorem is true for ALL a and b independent variables over two orhtogonal number lines (each of which satisfy arithmetic operations as vector coefficients in each dimension, but the coefficients in each number line do not interact arithmetically with the coefficients in the other; the vectors interact. That is why a and b retain their identities in the Binomial Theorem.

(All the so-called paradoxes in physics and math vanish with a little thought...)
Which paradox vanishes? Specifics?

6. ## Re: Proof of Fermat's Theorem by Modulus

The Bob and Mary paradox in STR (space-time). The Barber Paradox in Logic. The EPR paradox in quantum mechanics (causality). Schroedinger's cat in quantum mechanics (causality), etc., etc.....

e.g., The Barber Paradox: "A barber shaves all those and only those who don't shave themseleves. Does the barber shave himself?" (Bertrand Russell)

7. ## Re: Proof of Fermat's Theorem by Modulus

Originally Posted by BuleriaChk
The Bob and Mary paradox in STR. The Barber Paradox. The EPR paradox in quantum mechanics....
etc., etc., etc.....
The barber paradox? That's a physics and math paradox?

Never mind. What you're saying is, those paradoxes aren't really paradoxes, right? You just have to frame them in the right terms, and there really is no conflict, within the theory?

Of course, general relativity vs quantum mechanics is not a paradox either, just an unresolved conflict. Is that what you meant?

8. ## Re: Proof of Fermat's Theorem by Modulus

Originally Posted by grapes
The barber paradox? That's a physics and math paradox?

Never mind. What you're saying is, those paradoxes aren't really paradoxes, right? You just have to frame them in the right terms, and there really is no conflict, within the theory?

Of course, general relativity vs quantum mechanics is not a paradox either, just an unresolved conflict. Is that what you meant?
I would say those statements are valid. The "proof" isn't. But the statement that paradoxes are resolvable by more accurate theory is valid.

9. ## Re: Proof of Fermat's Theorem by Modulus

Originally Posted by Neverfly
I would say those statements are valid. The "proof" isn't. But the statement that paradoxes are resolvable by more accurate theory is valid.
You say the "proof" isn't valid, but have no fricken idea why it is not. It is a polite way of braying "nonsense" like a polite donkey, but nevertheless a donkey just the same. And the other statement is without content unless you actually know how to resolve the paradoxes. I may write some posts on them after I'm finished with Euler's idenity and its relation to the relativistic unit circle.

But I doubt that you (collectively) understand the paradoxes at all, much less have tried to resolve them yourselves. Just trivial platitiudes without content "The paradoxes are resolvable by more accurate theory", without understanding the paradoxes or anything about a relevant theory whatever.

It is an opinion without any substance whatever, like Grapes, based on urban legend rather than a technical analysis of my (and, I strongly suspect Fermat's) any sort. In fact, you don't seem to have any idea of the fundamental prnciple on which my proof is based. Like Grapes, you confuse arithmetic operations with vector operations, and I doubt you understand the latter at all.

The real question you should be asking is "why does one need vectors"?. The answer is that they represent a characterization of independent variables in the definition of a function of two variables - e.g. z=f(x,y)=y/x.

e.g. A(x,y)= (y-b)/x for the slope of a straight line. The simplest possible example in analytic geometry, and the foundation of calculus, based on, uh, Descartes and the, uh, Cartesian coordinate system. The Naked Emperor wants to deny the mathematical existence of slopes by clinging desperately to his fundamentalist religion..... believing he has found god, and is thus superior to every one else, when in truth it is doubtful that he has a technical degree from anywhere, just like you and Grapes.

Maybe you could read a couple of Wiki articles on them before you respond again.

fricken village idiots....

(like the last comment from Grapes, where c < a+b can be used to create any number relation; e.g. 1 < (1 + 1) and so and therefore 2 < (3 + 5), and is in no way related to Fermat's equation), and so is irrelevant, whereas actually does mean something, provided you understand and and the relativistic unit circle.

Oh well, at least you were more polite than usual. But still a donkey...

And if you have nothing to say besides braying and fan-boy remarks, then as a moderator you should shut the fuck up, instead of spamming threads where technical issues are being discussed, and cheapening the content of the discussion any more than it already is.

10. ## Re: Proof of Fermat's Theorem by Modulus

The conflict between Quantum Mechanics and General Relativity is the same as the conflict between Special and General Relativity.

It is resolved by the fact that energy is not conserved in General Relativity if the Stress-Energy and Metric tensors are not diagonal (unless spin is introduced), but if they are diagonal, then Special Relativity and Quantum Mechanics apply where energy is conserved. (Of course, the tensors have to be finite dimensional to even make sense.. another problem with continuous functions in GTR)

Those may sound like buzz words (the conceptual ideas upon which all physics and math are founded are diffiicult with centuries of work by brilliant people behind them), but if I get an interesting question about them, I'll try to explain... (without much hope...), starting with Newton, Descartes, and Maxwell... (well, ok, before that, Gauss, Coulomb, Ampere, and Faraday). Well, ok, before them, there was ...

Why not start with the slope of a straight line? Go check out the Math is Fun website if you have questions about it... Then when you finally understand that, we can move on to circles....

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