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Thread: Proof of Fermat's Theorem by Modulus

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    Default Proof of Fermat's Theorem by Modulus

    Fermat's Theorem and Modulus

    See also

    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.
    -----------------------------------------------------------------------------
    Last edited by BuleriaChk; 12-26-2016 at 04:11 PM.
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    The Relativistic Unit Circle 03/28/2017 07:40 AM PST
    Proof of Fermat's Last Theorem Updates 03/19/2017 8:23 PM PST
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    Default Re: Proof of Fermat's Theorem by Modulus

    Quote Originally Posted by BuleriaChk View Post
    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.
    -----------------------------------------------------------------------------
    See the link at the top of your linked page, it describes how modulus works.
    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
    Last edited by grapes; 12-22-2016 at 05:13 PM. Reason: Add link

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    Default Re: Proof of Fermat's Theorem by Modulus

    Quote Originally Posted by grapes View Post
    See the link at the top of your linked page, it describes how modulus works.
    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)
    Last edited by BuleriaChk; 12-23-2016 at 02:28 AM.
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    The Relativistic Unit Circle 03/28/2017 07:40 AM PST
    Proof of Fermat's Last Theorem Updates 03/19/2017 8:23 PM PST
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    Default Re: Proof of Fermat's Theorem by Modulus

    Quote Originally Posted by BuleriaChk View Post
    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)
    Last edited by grapes; 12-23-2016 at 03:50 AM.

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    Default Re: Proof of Fermat's Theorem by Modulus

    Quote Originally Posted by grapes View Post
    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)....
    Last edited by BuleriaChk; 12-23-2016 at 11:04 AM.
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    The Relativistic Unit Circle 03/28/2017 07:40 AM PST
    Proof of Fermat's Last Theorem Updates 03/19/2017 8:23 PM PST
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    Default 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.
    Last edited by BuleriaChk; 12-23-2016 at 11:21 AM.
    _______________________________________
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    The Relativistic Unit Circle 03/28/2017 07:40 AM PST
    Proof of Fermat's Last Theorem Updates 03/19/2017 8:23 PM PST
    Ignore List -The Peanut Gallery.

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    Default Re: Proof of Fermat's Theorem by Modulus

    Quote Originally Posted by BuleriaChk View Post
    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)....

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    Default 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 ..)
    _______________________________________
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    The Relativistic Unit Circle 03/28/2017 07:40 AM PST
    Proof of Fermat's Last Theorem Updates 03/19/2017 8:23 PM PST
    Ignore List -The Peanut Gallery.

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    Default Re: Proof of Fermat's Theorem by Modulus

    Quote Originally Posted by grapes View Post
    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".
    Last edited by BuleriaChk; 12-23-2016 at 12:04 PM.
    _______________________________________
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    The Relativistic Unit Circle 03/28/2017 07:40 AM PST
    Proof of Fermat's Last Theorem Updates 03/19/2017 8:23 PM PST
    Ignore List -The Peanut Gallery.

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    Default Re: Proof of Fermat's Theorem by Modulus

    Quote Originally Posted by BuleriaChk View Post
    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 ..)

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