# Thread: Proof of Fermat's Theorem using vectors

1. ## Proof of Fermat's Theorem using vectors

Proof by Vector

There is a more general .pdf at "The Relativistic Unit Circle" for those interested.

The proof that follows references all scalar operations to the null vector. The null vector signifies equal and opposite interactions, which is why the light vector is characterized as the null vector in STR by virtue of simultaneity (e.g., "equal and opposite" signals). For Pauli, this means "equal and opposite" spin, and for Dirac it means "equal and opposite" mass.

It amounts to referencing positive scalar interactions between two numbers, particles (signals,photon masses, whatever) as equal and opposite vectors to the origin of the Cartesian Coordinate system (e.g., the center of the relativistic unit circle, by setting

and

so that , and if there is a product , it can't be an integer.

The only integers for the relativistic unit circle occur when either , or vice versa (i.e., as unit vectors), which is analagous to the condition in the Binomial theorem that rem(a,b,n) = 0 only if a=0 or b= 0 (or both, in which case c=0).

That is, the fact that in the Binomial Theorem, the interaction terms do not vanish for n=2 (excluding Pythagorean triples).

In addition, invoking the relativistic unit circle means the proof is valid for all numbers, not just integers.

(the fact that c must be irrational at the end of the proof is a consequence of invoking Cartesian, rather than radial coordinates in the proof)

Since the relativistic unit circle covers all possible cases in which the invariant (final result) is that of the above, it shows that the interaction term in the Binomial theorem cannot be integers for independent variables in two dimensions.

In any case, it is consistent with my other lines of thinking, I'm pretty sure, so if it is wrong, then the others are probably wrong as well. But if it is right...

a and b orthogonal

If and are orthogonal , then there is no interaction term, since , for so the term .

We can assign this operation as the scalar coefficient of a vector , so that That is, , where the arithmetic operations have taken place on a single number line.

We can just as easily assign the coefficient to the null vector, so that

(a vector translation to the origin (0,0)

----- Dot Product () -----

If and are not orthogonal, then the term ,

Then where we have assigned the scalar product as the coefficient of the null vector.

----- Cross Product () -----

-------------------
From the cross product:

From the dot product:

----------------------------

(suggestion - set a = b = 1, and note the relation to quantized spin or angular momentum and the area of the unit circle)

Therefore, c cannot be an integer.

(in all the above cases, the scalar operatioins have been assigned to the null vector, so are related to the origin for both dot and cross products).

(I'm still thinking about the consequences of this one; for the relativistic unit circle, set , and for the contradiction...)

2. ## Re: Proof of Fermat's Theorem using vectors

Originally Posted by BuleriaChk
The proof that follows references all scalar operations to the null vector.

I'm still trying to figure out if it makes sense, but I think it amounts to referencing scalar operations to the origin of the Cartesian Coordinate system (e.g., the center of the relativistic unit circle, by setting

and

so that , and if there is a product , it can't be an integer.

The only integers for the relativistic unit circle occur when either , or vice versa (i.e., as unit vectors), which is analagous to the condition in the Binomial theorem that rem(a,b,n) = 0 only if a=0 or b= 0 (or both, in which case c=0).

That is, the fact that in the Binomial Theorem, the interaction terms do not vanish for n=2 (excluding Pythagorean triples).

In addition, invoking the relativistic unit circle means the proof is valid for all numbers, not just integers.

(the fact that c must be irrational at the end of the proof is a consequence of invoking Cartesian, rather than radial coordinates in the proof)

Since the relativistic unit circle covers all possible cases in which the invariant (final result) is that of the above, it shows that the interaction term in the Binomial theorem cannot be integers for independent variables in two dimensions.

In any case, it is consistent with my other lines of thinking, I'm pretty sure, so if it is wrong, then the others are probably wrong as well. But if it is right...

a and b orthogonal

If and are orthogonal , then there is no interaction term, since , for so the term .

In order for the vector c to equal the vector a plus the vector b, the coefficients have to satisfy the triangle inequality. For instance, use the values a=3 and b=4 in those equations. If the vectors are orthogonal, then their sum is the hypotenuse of a right triangle. The math works out, but c equals 5, and never does equal a+b. That is, 5 does not equal 3+4, but instead .

When you assume c=a+b in your proofs (so you can invoke the binomial theorem), you start from an erroneous assumption. That guarantees failure.
We can assign this operation as the scalar coefficient of a vector , so that That is, , where the arithmetic operations have taken place on a single number line.

We can just as easily assign the coefficient to the null vector, so that

(a vector translation to the origin (0,0)

----- Dot Product () -----

If and are not orthogonal, then the term ,

Then where we have assigned the scalar product as the coefficient of the null vector.

----- Cross Product () -----

-------------------
From the cross product:

From the dot product:

----------------------------

Therefore, c cannot be an integer.

(in all the above cases, the scalar operatioins have been assigned to the null vector, so are related to the origin for both dot and cross products).

(I'm still thinking about the consequences of this one; for the relativistic unit circle, set , and for the contradiction...)

3. ## Re: Proof of Fermat's Theorem using vectors

Originally Posted by grapes
In order for the vector c to equal the vector a plus the vector b, the coefficients have to satisfy the triangle inequality. For instance, use the values a=3 and b=4 in those equations. If the vectors are orthogonal, then their sum is the hypotenuse of a right triangle. The math works out, but c equals 5, and never does equal a+b. That is, 5 does not equal 3+4, but instead .

When you assume c=a+b in your proofs (so you can invoke the binomial theorem), you start from an erroneous assumption. That guarantees failure.
I start from not where indicates that all operations in the coefficient take place on a single number line (which is precise statement of your wrong, repeat WRONG claim about what I actually say).

The Binomial theorem implicitly assumes the independence of a and b in the Binomial Expansion.
This proof shows the foundation of that assumption.

You continue to ignore what I assert about this, including ignoring the proof above where it comes into importance in the cross product (which I have referred to in all my threads on the subject).

Either that or you baldfacedly lie that I am not using vectors in my analysis.
Stop making claims that are simply not true.

Read the title of the post again
Look under the section entitled "cross product"

Try to understand the proof (I don't hold out much hope on this one...)

Unethical Village idiot!

4. ## Re: Proof of Fermat's Theorem using vectors

Fermat's Last Theorem

Originally Posted by BuleriaChk
Consider x, y, z, k and n positive integers (n , n > 2).

Then xn, yn, zn, and xn-k yk are also positive integers.

Then x + y = z is a positive integer, and (x + y)n = zn is a positive integer as well.

By the Binomial expansion,

(x + y)n = zn = xn + yn + f(x, y, n, k),

where f(x, y, n, k) represents the remaining terms in the Binomial expansion; each term is a function of powers n,k of positive integers (x,y)

multiplied by the binomial coefficient

(from Pascal's triangle) for that term - each such term is a nonzero positive integer.

f(x, y, n, k) is the sum of these terms, which are also non zero positive integers, so f(x, y, n, k) is a positive integer as well.

For Fermat's theorem to be false, zn = xn + yn , so that f(x, y, n, k) would have to be equal to zero, which is a contradiction, since f(x, y, n, k) 0 by construction/observation.

Fermat's theorem is therefore proved. (Q.E.D.)

I would like my Abel prize in small unmarked currency bills, in paper sacks.....

Note: I spent weeks on this project as an undergraduate at UCSB in the 60's ...

-------------------------------------------------------------

Easy version for Grapes:

Let x, y, z, n, be positive integers, n > 2

For some z, let (x + y) = z

Then (x + y)n = zn

Then zn =(x + y)n = xn + yn + f(positive integers) ( from Binomial Theorem if not by inspection)

zn = xn + yn iff f(positive integers) = 0

f(positive integers) 0 ( Binomial Theorem, or inspection)

Therefore, zn xn + yn for n > 2 (Fermat's Theorem)

QED

5. ## Re: Proof of Fermat's Theorem using vectors

The very characterization of positive integers in the Binomial Theorem assumes their independence in the coordinate plane (a,b). Thus the Binomial Theorem (and Fermat) did not feel it necessary to specify that condition. Neither did I.

(although I have specified it over, and over, and over, and over again.... (sigh)

e.g. (assuming positive integers)

on the number line represented by
on the number line represented by

replace

with
with

Do your best to imagine how the Binomial Theorem works.....
(assuming all positive coefficients ot the cross products in the Binomial Theorem)

--------------------------
The null vector (via the cross product) has to do with:

equal and opposite interaction forces in physics.
or
equal and opposite interactions between integers a and b in the Binomial Theorem.
or
equal and opposite interactions between real numbers in the Binomial Theorem
or
equal and opposite interactions between fractions in the Binomial Theorem
or
equal and opposite interactions between complex numbers in the Binomial Theorem
or
equal and opposite interactions between irrational numbers in the Binomial Theorem
or
equal and opposite interactions between transcendental numbers in the Binomial Theorem
or
equal and opposite charge interactions in electrodynamics
or
equal and opposite effect between matter and anti-matter in the Dirac equatiion.
or ....
equal and opposite spin in the Pauli equation
or ....
equal and opposite attraction in gravitating masses. (The "speed" of light is the null vector in STR by virtue of "simultaneity").
or ....
or ....
or......

In particular, it has to do with (interaction) energy always being positive definite in physics, which only changed with Dirac and anti-matter in applying Relativity to Quantum Mechanics. That is why Rem(a,b,n) is always positive in the Binomial Theorem, since it represents and interaction energy in terms of positive interations between the integers a and b. (Which is why I started my analysis in the Physics section).

(And why the Binomial Theorem starts at n=2) In fact, even Pythagrean triples are positive definite; there just is no interaction term.)

(Newton, where mass is assumed positive before Dirac. BTw, what happened to the other half? If you don't know, talk to your local high school physics teacher).

6. ## Re: Proof of Fermat's Theorem using vectors

Again you imbecile, why are you using tensor product for cross product? You don't even understand basic notation.

7. ## Re: Proof of Fermat's Theorem using vectors

{}^{ }As far as Grapes' contention that there is something wrong with c = a + b, I would point out that its extension to the Binomial expansion is agnostic as far as the ordeing is concerned.

That is c=(a+b), a=(c+b), b = (a + c) all work in the expansion, since rem(a,b,n), rem(c,b,n) and rem(a,c,n) all are greater than zero - that is, if you pick any three positive integers, it doesn't matter which ones go where in that equation. The ordering is agnostic.

So statements like "c <(a+b)" are just as irrelevant a<(c+b) or b <(a+c). That is why these statements can't be true for all integers. c<(a+b) does not include the set c>(a+b) so the inequalities are irrelevant.

My proof works because

IFF works just as well as
IFF and
IFF

for ANY three positive integers a,b, c no matter what the order in the expression.
And if that is true, the rest follows because rem(x,y,n) > 0 ALWAYS.

I don't have to express the proof in vectors; but the relativistic unit circle is very interesting to me.

The only issue is that a and b (or whiever elements appear in the sum) represent different integers in the expression without any other variables or constants.

(So, for example, in the first case a=b says c = 2a = d, so since they are on the same number line. That is why I would use the notation (a,b) or (c,b) or (a,c) respectively in the examples above, since they indicate different number lines (cartesian coordinates).

That said, I have been inconsistent technically in distinguishing between vector spaces and coordinate systems, but I think the distinction and usage are generally clear from the context. There are imporatnt difference between coordinate systems and vectors.

But if one doesn't know what a coordinate system is to begin with, the question is moot....

8. ## Re: Proof of Fermat's Theorem using vectors

Originally Posted by emperorzelos
Again you imbecile, why are you using tensor product for cross product? You don't even understand basic notation.
What symbol would you use? It is the symbol for outer product, which is just a generalization of vector product for to vector transformation. Anyway, I have made my usage clear, use whatever symbol turns you on....

The context should be clear to everyone but village idiots.

Village idiot..

9. ## Re: Proof of Fermat's Theorem using vectors

Originally Posted by BuleriaChk
{}^{ }As far as Grapes' contention that there is something wrong with c = a + b, I would point out that its extension to the Binomial expansion is agnostic as far as the ordeing is concerned.

That is c=(a+b), a=(c+b), b = (a + c) all work in the expansion, since rem(a,b,n), rem(c,b,n) and rem(a,c,n) all are greater than zero - that is, if you pick any three positive integers, it doesn't matter which ones go where in that equation. The ordering is agnostic.
If you have three distinct integers, there will always be one that is the largest, and one will be the smallest. So, letting c represent the largest is not unusual.
So statements like "c <(a+b)" are just as irrelevant a<(c+b) or b <(a+c). That is why these statements can't be true for all integers.
Of course they can't be true for all integers. If c=12, and a=2 and b=5, then they cannot represent the three sides of a triangle, because they do not satisfy the triangle inequality. 2<5+12, 5<2+12, but 12 is not less than 5+2

If c represents the largest integer, then a<c+b and b<c+a, obviously. But there are three possibilities for the relationship between c and a+b:

c > a+b, or c=a+b, or c<a+b

Only one of those satisfies the Triangle Inequality.

So,

assume we do have a solution to the equality, and equals

Then

If we add 1 to the right hand side, but not to the left, then there will be an inequality:

and that's true if we add any positive value:

The right hand side is just your binomial:

And taking nth roots of both sides gives us:

In other words, if there is a counter example to Fermat's Last Theorem, it must satisfy the Triangle Inequality. The largest of the three integers c must be less than the sum of the other two, a+b. Of course, a<c+b and b<c+a as well.

c<(a+b) does not include the set c>(a+b) so the inequalities are irrelevant.

My proof works because

IFF works just as well as
IFF and
IFF

for ANY three positive integers a,b, c no matter what the order in the expression.
And if that is true, the rest follows because rem(x,y,n) > 0 ALWAYS.

I don't have to express the proof in vectors; but the relativistic unit circle is very interesting to me.

The only issue is that a and b represent different integers in the expression without any other variables or constants.
(So, for example, in the first case a=b says c = 2a =d, so since they are on the same number line.
Yes, they must be distinct, and nonzero, integers. is true, but it is not a counter example to Fermat's Last Theorem.

When you start off your proof by assuming that c=a+b, that is equivalent to assuming already that (a,b,c) is not a counter example to the Theorem.

As difficult as it may be (and it was exceedingly difficult), you have to start with c<a+b

(a,b,c) has to satisfy the Triangle Inequality, to have any chance of being a counterexample to Fermat's Last Theorem.

10. ## Re: Proof of Fermat's Theorem using vectors

Originally Posted by BuleriaChk
What symbol would you use? It is the symbol for outer product, which is just a generalization of vector product for to vector transformation. Anyway, I have made my usage clear, use whatever symbol turns you on....

The context should be clear to everyone but village idiots.

Village idiot..
the symbol for CROSS product is the CROSS .

you pleb. The idiot here is you and forever you.

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