# Thread: Proof of Fermat's Theorem by Modulus

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

Fermat's Theorem and Modulus

The Relativistic Unit Circle

"Modulus" refers to one dimension because it is simply a way of characterizing integer division by a counting mechanism - if the result isn't Mod 0, (e.g. 4/3) one has a fraction, and therefore the result is not an integer, so the counting procedure is not successful (that is, the integer generation fails, the fraction is not in the set of integers on the number line in a single dimension.

For example, in my (Fermat's) proof for a,b,c,n positive integers, n> 2:

?

?

only if and ?

However, and

Therefore

The fact that in the Binomial Theorem expansion is not necessary, but true, since the fraction is not in the set of integers, so is not in a vector space consisting of integers.
----------------------------------------------------------------------------
That said, the application of the concept of Modulo is moot for my (Fermat's) proof of his theorem since the Binomial Expansion is valid for all positive numbers; integers, rationals, irrationals, transcendentals, and complex.
-----------------------------------------------------------------------------

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

Originally Posted by BuleriaChk
Fermat's Theorem and Modulus

"Modulus" refers to one dimension because it is simply a way of characterizing integer division by a counting mechanism - if the result isn't Mod 0, (e.g. 4/3) one has a fraction, and therefore the result is not an integer, so the counting procedure is not successful (that is, the integer generation fails, the fraction is not in the set of integers on the number line in a single dimension.

For example, in my (Fermat's) proof for a,b,c,n positive integers, n> 2:

?

?

only if and ?

However, and

Therefore

The fact that in the Binomial Theorem expansion is not necessary, but true, since the fraction is not in the set of integers, so is not in a vector space consisting of integers.
----------------------------------------------------------------------------
That said, the application of the concept of Modulo is moot for my (Fermat's) proof of his theorem since the Binomial Expansion is valid for all positive numbers; integers, rationals, irrationals, transcendentals, and complex.
-----------------------------------------------------------------------------
https://en.m.wikipedia.org/wiki/Modulo_operation

You specify a modulus, m, and compare two numbers by their remainders when divided by m.

Say the modulus is 5, then 7 = 2 mod 5, because 7 = 5n+2 for some integer n

But then you can't have a modulus of zero, because division by zero is not defined.

12 = 7 mod 5

7 does not equal 2 mod 3, but it does equal 1 mod 3

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

Originally Posted by grapes
https://en.m.wikipedia.org/wiki/Modulo_operation

You specify a modulus, m, and compare two numbers by their remainders when divided by m.

Say the modulus is 5, then 7 = 2 mod 5, because 7 = 5n+2 for some integer n

But then you can't have a modulus of zero, because division by zero is not defined.

12 = 7 mod 5

7 does not equal 2 mod 3, but it does equal 1 mod 3
Thank you.

The concept of Modulus is irrrelevant to my proof of Fermat's theorem, however, since the Binomial Theorem holds as well for positive real numbers, so the question of modulus is moot. As a matter of fact it holds because the numbers it refers to are conceptuallized as single valued (i.e., invariant under arithmetic operations, and that the BT is a relation between independent number lines)

(I wrote the post because I had mistakenly thought the Naked Emporer was referring to "Modulus"; actually he was referring to "module" as a term in abstract algebra, which is equally irrelevant; not only to Fermat's proof, but to a whole lot of other concepts in mathematics - e.g., independent variables

That doesn't mean that number theorists and abstract algebraicists are bad; it just means that one likes some people a whole lot better if they don't know them very well....

Maybe some day I'll go back and look at it again (I'll re-analyze the lines containing modulo reference, because they may be incorrect as written, but there is a correct version, since I proved that could not be an integer from the relativistic unit circle...

(BTW, Grapes, did you ever think that someone else might have a relevant point to make in my thread about Fermat's Theorem, besides your wrong and/or irrelevant comments (except for minor typo molehills that you tried to make into mountains? And that I might want to respond to someone who had an intelligent contribution to make in the interest of the free flow of ideas other than braying "nonsense" like a donkey?)

Or was it just to stop me from posting in my own thread?

(Moderator, my ass)

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

Originally Posted by BuleriaChk
Thank you.
You're welcome
The concept of Modulus is irrrelevant to my proof of Fermat's theorem, however, since the Binomial Theorem holds as well for positive real numbers, so the question of modulus is moot. As a matter of fact it holds because the numbers it refers to are conceptuallized as single valued (i.e., invariant under arithmetic operations, and that the BT is a relation between independent number lines)

(I wrote the post because I had mistakenly thought the Naked Emporer was referring to "Modulus"; actually he was referring to "module" as a term in abstract algebra, which is equally irrelevant; not only to Fermat's proof, but to a whole lot of other concepts in mathematics - e.g., independent variables

That doesn't mean that number theorists and abstract algebraicists are bad; it just means that one likes some people a whole lot better if they don't know them very well....

Maybe some day I'll go back and look at it again (I'll re-analyze the lines containing modulo reference, because they may be incorrect as written, but there is a correct version, since I proved that could not be an integer from the relativistic unit circle...

(BTW, Grapes, did you ever think that someone else might have a relevant point to make in my thread about Fermat's Theorem, besides your wrong and/or irrelevant comments (except for minor typo molehills that you tried to make into mountains? And that I might want to respond to someone who had an intelligent contribution to make in the interest of the free flow of ideas other than braying "nonsense" like a donkey?)

Or was it just to stop me from posting in my own thread?

(Moderator, my ass)
You can still post in this forum, but you don't own threads. The threads are public.

The mistake in your proof (using the binomial theorem), is that you first assume c=a+b

However, any counter example (c,a,b) to Fermat's Last Theorem has to satisfy the Triangle Inequality, c<a+b

The Triangle Inequality is true even for Pythagorean triples, e.g. (5,3,4), where 5<3+4 (and, 3<5+4 and 4<5+3 as well)

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

Originally Posted by grapes
You're welcome

You can still post in this forum, but you don't own threads. The threads are public.

The mistake in your proof (using the binomial theorem), is that you first assume c=a+b

However, any counter example (c,a,b) to Fermat's Last Theorem has to satisfy the Triangle Inequality, c<a+b

The Triangle Inequality is true even for Pythagorean triples, e.g. (5,3,4), where 5<3+4 (and, 3<5+4 and 4<5+3 as well)
So whtat? All you've specified is an expression that can't represent Fermat's equation in the first place, and so is irrelevant to my proof.
In fact, specifying inequalities like that is irrelevant to a WHOLE LOT of proofs, come to think of it...

And I can't post in my own thread (or in yours, not that I'd ever want to except to point out stupid mistakes relating to my posts) in this forum because you closed it, you stupid shit..

(I can't believe this guy is a moderator)....

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

The reason Fermat's expression can be related to the Binomial Theorem is becauset the integers a and b are independent. That is, with the exception of Pythagorean triples, is not an integer for as generated from the expression c = a + b, where where rem(a,b,n) > 0 for positive integers a,b,c,n so the Binomial Theorem applies to Fermat's Proof.

That means that with the exception of Pythagorean triples, the coefficients each vector are a ring without division, which is why fractions consisting only of individual number lines as coefficients of vectors are independent.

E.g. For fractions such as z = y/x, x and y are generally independent (which it is why it is the slope of a straight line in two dimensions), but not fractions such as z = f(x,y)/g(x,y). So they are also independent for , which is why they can form the bases of linear systems consisting of polynomials with constant coefficients.

(One subject of tensor analysis is using polynomials with constant coeffiicients as the as coefficients for yet another vector space (e.g., the metric tensor) in GTR.

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

Originally Posted by BuleriaChk
So whtat? All you've specified is an expression that can't represent Fermat's equation in the first place, and so is irrelevant to my proof.
Not true. That's why you think it is easy to prove Fermat's Last Theorem

It is the equation c=a+b that is irrelevant, since it violates the Triangle Inequality. You yourself have shown that c<a+b

Start there, and try to prove Fermat's Last Theorem. That's what everybody else has done. It's not easy
In fact, specifying inequalities like that is irrelevant to a WHOLE LOT of proofs, come to think of it...

And I can't post in my own thread (or in yours, not that I'd ever want to except to point out stupid mistakes relating to my posts) in this forum because you closed it, you stupid shit..

(I can't believe this guy is a moderator)....

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

is not an integer for .

Think real hard, I mean really, really hard about why.

(I can't believe the idea that Fermat's theorem hasn't been proved long ago has been a practical joke in math departments for several centuries ..)

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

Originally Posted by grapes
Not true. That's why you think it is easy to prove Fermat's Last Theorem

Start there, and try to prove Fermat's Last Theorem. That's what everybody else has done. It's not easy
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". Are you sure about "everybody else"? Have you done a poll of humanity? Or even a real literature search? (I admit that I haven't; I'm waiting for input from someone who actually has... i.e., other than urban legend. Or at least someone intelligent.)

(I cann't believe that the idea the Fermat's Theorem hasn't been proven for several centuries isn't a huge practical joke in math departments. OTH, I had had a suspicion about the Binomial Theorem for decades, I just couldn't put my finger on it until a several of years ago).

(That said, the cases c < a+b and c > a + b are covered in my proof from the relativistic unit circle, where c = 1, and and ) That satisfies the triangle equality, which is true for all indpendent variables, integers or not.

Which is true for all numbers (x,y) related by the scaling factors t, and t' so that (xt,yt') are orthogonal (independent). All invariants are independent, in the sense that changing a or b in the pair (a,b) does not change a or b. I.e., in the pair (3,x), changing x does not change "3".

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

Originally Posted by BuleriaChk
is not an integer for .
Yes it is. It is equal to 1. Did you mean something else?
Think real hard, I mean really, really hard about why.

(I can't believe the idea that Fermat's theorem hasn't been proved long ago has been a practical joke in math departments for several centuries ..)

Page 1 of 4 123 ... Last

#### Posting Permissions

• You may not post new threads
• You may not post replies
• You may not post attachments
• You may not edit your posts
•