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Thread: Fermat's Theorem, Revisited

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    Default Fermat's Theorem, Revisited

    Fermat's Theorem, revisited

    Quote Originally Posted by BuleriaChk View Post
    I said:
    --------------------------
    Again, the (corrected) expansion is , which can only be valid if the interaction term on the l.h.s. , the conclusion reached before the correction. I.e., m=0 or n = 0 means that none of Euclid's equations are relevant.
    ---------------------------------------
    Suppose

    are on the same number line . Then the equation is inconsistent unless mn = 0 on the same number line.
    The equation is never inconsistent--it's an algebraic identity

    See below
    Actually, (they are scalars in two dimensions).

    it is that ), so and in the expansion, but the interaction term vanishes because of the dot product, and because of that, Euclid's formulae are consistent for Pythagorean triples.

    That is, m and n must be independent integers (m,n) for Euclid's formulae to be consistent.

    So Euclid's Formulae are actually constructed from:




    , ,

    and so forth....
    The left hand side of the equation is easily seen to be algebraically equivalent to the right hand side, including vectors:


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    Default Re: Fermat's Theorem, Revisited

    Since the village idiots have forced this thread in a new forum (supposedly unrelated to physics), I will just sum up by reiterating the fact that Fermat's Theorem refers to independent integers (a,b) which must be related as vectors in Cartesian sets with independent metrics, and so are represented by ), where the unit vectors and represent independent unit metrics in each dimension.

    This is represented in particular instances by the dot product of integers a and b by the null vector , which corresponds to the fact that corresponds to in Euclid's formulae in order for the expansion to be consistent for Pythagorean triples, a characterization which affirms that the triples form a right triangle, where the unit vectors are orthogonal, and so there is no projection between them.

    Thus, for Pythagorean triples, the integers are not on the same number line, but each in a dimension characterized by independent orthogonal number lines. As individual elements are concerned, this represents the concept that (e.g.) , but is a vector relation where



    so that



    and

    .

    but in a single dimension.

    (For the restriction to positive integers one uses both the left-hand and right-hand rule in the positive or negative quadrants of the relativistic unit circle (equal and opposite "spin", or ccw and cw Lorentz rotations) to represent the interaction term.

    That is, represent the same integer on each side of the equality in a single dimension (i.e., the integer represented by the l.h.s is identical to that represented by the r.h.s. in a single dimension). (as are all arithmetic operations operations in a single dimension)

    For two dimensions, then, three elements form a triangle characterized by a vector relationship.

    A Pythagorean triple forms a right triangle in which all the elements are integers, so that . If a right triangle is not Pythagorean, then at least one of its elements cannot be an integer, and therefore is not the subject of Fermat's theorem.

    If the triangle is a right triangle, then

    (If , then the l.h.s. and r.h.s represent the same integer c = d in one dimension; i.e., either a = 0 or b = 0).

    Then if a and b are independent, and do not form a Pythagorean triangle, , where 2ab is an interaction term, which can only be eliminated if a and b are orthogonal (as in the Pythagorean right triangle).

    Then by the Binomial Theorem (well proven since the 1600's), where rem(a,b,n) > 0 by construction.

    Therefore , which proves Fermat's theorem.

    (Any term that "destroys" rem(a,b,n) in the Binomial Theorem must be complex, where (e.g.) (which is equivalent to taking the dot product between a and b for orthogonal vectors and ) for n = 2 to retrieve the Pythagorean triple for n = 2 and all higher powers n > 2.) Note that a and b commute as scalar sums and products as scalar coefficients of the vectors in each dimension.

    QED

    (The relation to physics is via the relativistic unit circle and natural logarithms, where the relativistic unit circle can be considered as an integer generator in one dimension for , where integer generation rests on continuity instead of counting .... , so the real numbers are generated by Lorentz rotations in two dimensions rather than simply "justified" by Dedekind cuts. These are all mathematical concepts (since no physical values need be ascribed to v, c, t, or t'), since c and v are not interpreted physically but in the numerical relation in the vector space (ct,vt') where c,v,t, and t' are (continuous) real numbers in two dimensions, where the areas are related by either irrational (e.g. geometric polygons inscribed in the circle) or transcendental numbers (concentric circles where the radius (hypotenuse) "connects" the dependent and independent variables).

    Variations in are equivalent to variations in the metric tensor in the General Theory (which applies differential geometry to tensor analysis), and corresponds to "spin" in Quantum Field Theory, where positive and negative spin are introduced through the Pauli and Dirac equations (again, pure mathematics, and is also characterized by quaternions with parity (i.e. , corresponding to "left" and "right" hand rule - that is "mirror" reflections).

    This all may seem strange to village idiots, but others with a modicum of formal training will easily grasp the concepts involved; if not, pm me and I'll respond privately.
    Last edited by BuleriaChk; 12-20-2016 at 06:28 PM.
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    Default Re: Fermat's Theorem, Revisited

    Quote Originally Posted by BuleriaChk View Post
    Since the village idiots have forced this thread in a new forum (supposedly unrelated to physics), I will just sum up by reiterating the fact that Fermat's Theorem refers to independent integers (a,b) which must be related as vectors in Cartesian sets with independent metrics, and so are represented by ), where the unit vectors and represent independent unit metrics in each dimension.

    This is represented in particular instances by the dot product of integers a and b by the null vector , which corresponds to the fact that corresponds to in Euclid's formulae in order for the expansion to be consistent for Pythagorean triples, a characterization which affirms that the triples form a right triangle, where the unit vectors are orthogonal, and so there is no projection between them.

    Thus, for Pythagorean triples, the integers are not on the same number line, but each in a dimension characterized by independent orthogonal number lines. As individual elements are concerned, this represents the concept that (e.g.) , but is a vector relation where



    so that



    and

    .

    but in a single dimension.

    (For the restriction to positive integers one uses both the left-hand and right-hand rule in the positive or negative quadrants of the relativistic unit circle (equal and opposite "spin", or ccw and cw Lorentz rotations) to represent the interaction term.

    That is, represent the same integer on each side of the equality in a single dimension (i.e., the integer represented by the l.h.s is identical to that represented by the r.h.s. in a single dimension). (as are all arithmetic operations operations in a single dimension)

    For two dimensions, then, three elements form a triangle characterized by a vector relationship.

    A Pythagorean triple forms a right triangle in which all the elements are integers, so that . If a right triangle is not Pythagorean, then at least one of its elements cannot be an integer, and therefore is not the subject of Fermat's theorem.

    If the triangle is not a right triangle, then
    Besides the usual abuse and misuse of math symbols and terminology and basic trigonometry in the rest of your post, this is your fundamental mistake. You keep insisting on it. can never equal in a non-degenerate triangle, it is always less. That is the triangle inequality. No matter what you think your justification is, you cannot assert this. It is false.
    (If , then the l.h.s. and r.h.s represent the same integer c = d in one dimension; i.e., either a = 0 or b = 0).

    Then if a and b are independent, and do not form a Pythagorean triangle, , where 2ab is an interaction term, which can only be eliminated if a and b are orthogonal (as in the Pythagorean right triangle).

    Then by the Binomial Theorem (well proven since the 1600's), where rem(a,b,n) > 0 by construction.

    Therefore , which proves Fermat's theorem.

    (Any term that "destroys" rem(a,b,n) in the Binomial Theorem must be complex, where (e.g.) (which is equivalent to taking the dot product between a and b for orthogonal vectors and ) for n = 2 to retrieve the Pythagorean triple for n = 2 and all higher powers n > 2.) Note that a and b commute as scalar sums and products as scalar coefficients of the vectors in each dimension.

    QED

    (The relation to physics is via the relativistic unit circle and natural logarithms, where the relativistic unit circle can be considered as an integer generator in one dimension for , where integer generation rests on continuity instead of counting .... , so the real numbers are generated by Lorentz rotations in two dimensions rather than simply "justified" by Dedekind cuts. These are all mathematical concepts (since no physical values need be ascribed to v, c, t, or t'), since c and v are not interpreted physically but in the numerical relation in the vector space (ct,vt') where c,v,t, and t' are (continuous) real numbers in two dimensions, where the areas are related by either irrational (e.g. geometric polygons inscribed in the circle) or transcendental numbers (concentric circles where the radius (hypotenuse) "connects" the dependent and independent variables).

    Variations in are equivalent to variations in the metric tensor in the General Theory (which applies differential geometry to tensor analysis), and corresponds to "spin" in Quantum Field Theory, where positive and negative spin are introduced through the Pauli and Dirac equations (again, pure mathematics, and is also characterized by quaternions with parity (i.e. , corresponding to "left" and "right" hand rule - that is "mirror" reflections).

    This all may seem strange to village idiots, but others with a modicum of formal training will easily grasp the concepts involved; if not, pm me and I'll respond privately.

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    Default Re: Fermat's Theorem, Revisited

    Quote Originally Posted by grapes View Post
    Besides the usual abuse and misuse of math symbols and terminology and basic trigonometry in the rest of your post, this is your fundamental mistake. You keep insisting on it. can never equal in a non-degenerate triangle, it is always less. That is the triangle inequality. No matter what you think your justification is, you cannot assert this. It is false.

    in a (3,4,5) (Pythagorean triple) right triangle. What do you mean by "non-degenerate"? WTF kind of triangle (or triple, even on a straight line) are you talking about that I haven't covered?

    Pythagorean Triple right triangle, a straight line, non-Pythagorean triangle composed of integers, a triangle that is not a right triangle, a triangle that is not composed of integers?

    Which one?

    ?

    Are you saying I haven't mentioned that equation ever? That it isn't fundamental to my proof?

    Are you out of your mind? Or just lying ???


    But the point that rem(a,b,2) > 0 proves "Fermat's theorem" from the Binomial theorem for the case n=2 for a triangle that is non-Pythagorean (but still composed of integers). Which may be what you are saying above, and what I have said from the very beginning because of the interaction term 2ab. Read my post again, I cover all the possibilities...

    (and, of course, Fermat's theorem actually states n>2 to avoid idiot objections like your blatant lie about what I actually said).

    Sheesh. Village Idiot... and a "moderator", yet.

    (You're finally beginning to understand my proof. Be a man and come to terms with it. Because you're finally barely beginning to understand what I said from the very beginning)
    Last edited by BuleriaChk; 12-20-2016 at 04:09 PM.
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    Default Re: Fermat's Theorem, Revisited

    Quote Originally Posted by BuleriaChk View Post
    Quote Originally Posted by grapes
    Besides the usual abuse and misuse of math symbols and terminology and basic trigonometry in the rest of your post, this is your fundamental mistake. You keep insisting on it. can never equal in a non-degenerate triangle, it is always less. That is the triangle inequality. No matter what you think your justification is, you cannot assert this. It is false.
    in a (3,4,5) (Pythagorean triple) right triangle. What do you mean by "non-degenerate"?
    Mathwords: Degenerate
    WTF kind of triangle (or triple, even on a straight line) are you talking about that I haven't covered?

    Pythagorean Triple right triangle, a straight line, non-Pythagorean triangle composed of integers, a triangle that is not a right triangle, a triangle that is not composed of integers?

    Which one?
    All triangles.
    ?

    Are you saying I haven't mentioned that equation ever? That it isn't fundamental to my proof?
    Read my post. It is fundamental to your proof.

    That's why your proof is fallacious.
    Are you out of your mind? Or just lying ???


    But the point that rem(a,b,2) > 0 proves "Fermat's theorem" from the Binomial theorem for the case n=2 for a triangle that is non-Pythagorean (but still composed of integers). Which may be what you are saying above, and what I have said from the very beginning because of the interaction term 2ab. Read my post again, I cover all the possibilities...

    (and, of course, Fermat's theorem actually states n>2 to avoid idiot objections like your blatant lie about what I actually said).

    Sheesh. Village Idiot...

    (You're finally beginning to understand my proof. Be a man and come to terms with it. Because you're finally barely beginning to understand what I said from the very beginning)

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    Default Re: Fermat's Theorem, Revisited

    Quote Originally Posted by grapes View Post
    Mathwords: Degenerate

    All triangles.

    Read my post. It is fundamental to your proof.

    That's why your proof is fallacious.
    So a triangle that isn't a straight line is non-degenerate. Hey, ok, I agree with that. So we (finally) can omit straight lines from the discussion. So we can at last admit there are two dimensions necessary for a non-degenerate triangle. Sheesh!


    And your post is a lie about what I actually said ("moderator", my ass).

    (In addition, you are finally confirming my proof that I claimed from the very beginning; that Fermat's Theorem is true almost by inspection from the Binomial Theorem)


    That is, that (Fermat's Theorem for n > 2, a,b,c,n positive integers) because rem(a,b,n) > 0 (always) by the Binomial Theorem

    (and true for n=2 by the Binomial Theorem, excluding Pythagorean triple right triangles - eg (3,4,5) right triangles)

    because rem(a,b,2) = 2ab > 0, a positive integer).

    (That is, a and b are independent integer variables in each dimension, with c the dependent variable in the resultant dimension, and the vector analysis is confirmed by my analysis of Euclid's formulae, which simply says that a pair of integers forms a Pythagorean triple right triangle with a third if their dot product is the null vector).

    For village idiots - a is independent from b if changing a does not change b , represented by the pair (a,b).
    Last edited by BuleriaChk; 12-20-2016 at 05:00 PM.
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    The Relativistic Unit Circle 03/28/2017 07:40 AM PST
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    Default Re: Fermat's Theorem, Revisited

    Quote Originally Posted by BuleriaChk View Post
    So a triangle that isn't a straight line is non-degenerate. Hey, ok, I agree with that. So we (finally) can omit straight lines from the discussion. So we can at last admit there are two dimensions necessary for a non-degenerate triangle. Sheesh!


    And your post is a lie about what I actually said.
    I quoted you
    (In addition, you are finally confirming my proof that I claimed from the very beginning; that Fermat's Theorem is true almost by inspection from the Binomial Theorem)
    You think the proof is easy, "almost by inspection," and you call us village idiots?
    That is, that (Fermat's Theorem for n > 2, a,b,c,n positive integers) because rem(a,b,n) > 0 (always) by the Binomial Theorem

    (and true for n=2 by the Binomial Theorem, excluding Pythagorean triple right triangles - eg (3,4,5) right triangles)

    because rem(a,b,2) = 2ab > 0, a positive integer).

    (That is, a and b are independent integer variables in each dimension, with c the dependent variable in the resultant dimension, and the vector analysis is confirmed by my analysis of Euclid's formulae).

    For village idiots - a is independent from b if changing a does not change b , represented by the pair (a,b).

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    Default Re: Fermat's Theorem, Revisited

    Quote Originally Posted by grapes View Post
    I quoted you

    You think the proof is easy, "almost by inspection," and you call us village idiots?
    You quoted me, but you didn't understand my post at all or deliberately misread it in a desperate effort to prove me wrong by lying about it. (Instead of braying "nonsense" like a fathead donkey because you don't understand it).

    (there was an error in the line about the cross product, but I corrected it. It has nothing to do with your bullshit assessment.)

    The proof isn't easy for you because other people say it isn't, and you believe them, since you don't understand it yourself.

    If you did understand it, you would have realized that I made that point there.
    And ever since I began the discussion long, long ago. The fact is that you're just now realizing that it may be correct.

    My post #2 in this thread, that is, not your mistaken or lying characterization of it.

    You are a village idiot, but at least you may have gotten to the point where you can start learning about Descartes and vectors now.

    Like Neverfly, you are a fraud as a moderator, not to mention as a mathematician...


    But quote the section you disagree with or didn't understand, and I'll address it yet once again....
    (simply quoting the whole thing without further context is just a bullshit expression of your ignorance in its development.)
    Last edited by BuleriaChk; 12-20-2016 at 05:57 PM.
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    Default Re: Fermat's Theorem, Revisited

    Quote Originally Posted by grapes View Post
    Fermat's Theorem, revisited


    The equation is never inconsistent--it's an algebraic identity

    See below

    The left hand side of the equation is easily seen to be algebraically equivalent to the right hand side, including vectors:

    WTF?

    ??????

    Are you serious? (Talk about howlers)

    Village idiot... and a "moderator" at that.....
    Last edited by BuleriaChk; 12-20-2016 at 06:06 PM.
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    The Relativistic Unit Circle 03/28/2017 07:40 AM PST
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  10. #10
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    Default Re: Fermat's Theorem, Revisited

    Quote Originally Posted by BuleriaChk View Post
    You quoted me, but you didn't understand my post at all or deliberately misread it in a desperate effort to prove me wrong by lying about it. (Instead of braying "nonsense" like a fathead donkey because you don't understand it).

    (there was an error in the line about the cross product, but I corrected it. It has nothing to do with your bullshit assessment.)

    The proof isn't easy for you because other people say it isn't, and you believe them, since you don't understand it yourself.

    If you did understand it, you would have realized that I made that point there.
    And ever since I began the discussion long, long ago. The fact is that you're just now realizing that it may be correct.

    My post #2 in this thread, that is, not your mistaken or lying characterization of it.

    You are a village idiot, but at least you may have gotten to the point where you can start learning about Descartes and vectors now.

    Like Neverfly, you are a fraud as a moderator, not to mention as a mathematician...


    But quote the section you disagree with or didn't understand, and I'll address it yet once again....
    (simply quoting the whole thing without further context is just a bullshit expression of your ignorance in its development.)
    Here ya go. I marked it up in red, just as I did the first time:
    Quote Originally Posted by grapes View Post
    Quote Originally Posted by BuleriaChk View Post
    If the triangle is not a right triangle, then
    Besides the usual abuse and misuse of math symbols and terminology and basic trigonometry in the rest of your post, this is your fundamental mistake. You keep insisting on it. can never equal in a non-degenerate triangle, it is always less. That is the triangle inequality. No matter what you think your justification is, you cannot assert this. It is false.

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