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Thread: Fermat's Theorem, Revisited

  1. #11
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    Default Re: Fermat's Theorem, Revisited

    Quote Originally Posted by BuleriaChk View Post
    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.....
    The right hand side does not simplify to that.

    ETA: The right hand side simplifies to be the same as the left hand side. If it didn't, the formulas wouldn't work. Shoulda been your first clue that you were making a mistake somewhere.
    Last edited by grapes; 12-20-2016 at 06:15 PM. Reason: ETA

  2. #12
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    Default Re: Fermat's Theorem, Revisited

    Quote Originally Posted by grapes View Post
    Here ya go. I marked it up in red, just as I did the first time:
    That's because a Pythagorean integer right triangle is and there is no other expression involving two independent integers that can be used to give Fermat's expression.

    Otherwise This forms a parallelogram of which half is a triangle along the vector, which forms a triangle of integers which is not a right triangle.

    Now evaluate .

    Or , for that matter.


    Study vectors....

    Village idiot....
    Last edited by BuleriaChk; 12-20-2016 at 06:26 PM.
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  3. #13
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    Default Re: Fermat's Theorem, Revisited

    Quote Originally Posted by BuleriaChk View Post
    That's because a Pythagorean integer right triangle is and there is no other expression involving two independent integers that can be used to give Fermat's expression.

    Study vectors....

    Village idiot....
    You said "If the triangle is not a right triangle." Not a right triangle. If it's not a right triangle, then

    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.

  4. #14
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    Default Re: Fermat's Theorem, Revisited

    Quote Originally Posted by grapes View Post
    Here ya go. I marked it up in red, just as I did the first time:
    That is correct, and obvious typo in context. Corrected without attribution.

    Doesn't invalidate my proof, not one iota.

    But thank you and fuck you....
    Last edited by BuleriaChk; 12-20-2016 at 06:34 PM.
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  5. #15
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    Default Re: Fermat's Theorem, Revisited

    Quote Originally Posted by grapes View Post
    The right hand side does not simplify to that.

    ETA: The right hand side simplifies to be the same as the left hand side. If it didn't, the formulas wouldn't work. Shoulda been your first clue that you were making a mistake somewhere.
    uh,

    what about the on the l.h.s.?

    The reason it vanishes is because it is a dot product. Euclid's formula doesn't generate triplets, it only shows that the formula (which is actually

    when the null vector is taken into account by eliminating the interaction element on the L.h.s.) is only consistent when you already know the triple.

    Village idiot....
    Last edited by BuleriaChk; 12-20-2016 at 06:39 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.

  6. #16
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    Default Re: Fermat's Theorem, Revisited

    Quote Originally Posted by BuleriaChk View Post
    uh,

    what about the on the l.h.s.?

    Village idiot....
    You seem to have misplaced a left parenthesis

  7. #17
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    Default Re: Fermat's Theorem, Revisited

    Quote Originally Posted by BuleriaChk View Post
    That is correct, and obvious typo in context. Corrected without attribution.

    Doesn't invalidate my proof, not one iota.

    But thank you and fuck you....
    You "corrected" it by taking out the "not"? Too funny. Now it's even worse.
    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 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.

  8. #18
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    Default Re: Fermat's Theorem, Revisited

    Quote Originally Posted by grapes View Post
    You seem to have misplaced a left parenthesis
    Not that I can see. Which one?

    (the + sigh was for village idiots. Perhaps that confused you. Sorry 'bout that)

    If that's all you got, I'm outta here until I write more about subjects I am interested in rather than explaining basic algebra and vectors to village idiots.....
    Last edited by BuleriaChk; 12-20-2016 at 06:48 PM.
    _______________________________________
    "Flamenco Chuck" Keyser
    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.

  9. #19
    Moderator grapes's Avatar
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    Default Re: Fermat's Theorem, Revisited

    Quote Originally Posted by BuleriaChk View Post
    uh,

    what about the on the l.h.s.?

    The reason it vanishes is because it is a dot product. Euclid's formula doesn't generate triplets, it only shows that the formula (which is actually

    when the null vector is taken into account by eliminating the interaction element on the L.h.s.) is only consistent when you already know the triple.

    Village idiot....
    Of course Euclid's formula generates Pythagorean triplets, that's the whole point of it. Just try it. Let m=2, n=1, see what you get.

  10. #20
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    Default Re: Fermat's Theorem, Revisited

    Quote Originally Posted by BuleriaChk View Post
    Not that I can see. Which one?
    You can't see it because you've blinded yourself. Look again.

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