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Thread: Fermat's Theorem, revisited

  1. #21
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    Default Re: Fermat's Theorem, revisited

    Quote Originally Posted by grapes View Post
    Almost a year:

    Fermat's Last Theorem
    Quote Originally Posted by grapes View Post
    You said , which is the same as c = a+b, when you're dealting with positive integers

    Elementary algebra, it's what I've been pointing out for months, and you ignore it.
    see post #17 - And demonstrably wrong the whole time?
    (Jeez, just like John Gabriel - a "new Algebra"...
    Last edited by BuleriaChk; 08-11-2016 at 09:06 PM.
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  2. #22
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    Default Re: Fermat's Theorem, revisited

    Quote Originally Posted by BuleriaChk View Post
    see post #17 -
    OK
    Quote Originally Posted by BuleriaChk View Post
    Try some numbers; e.g. a = 3, b = 4, c = 7

    c2 = (a + b)2 = a2 + 2ab + b2 = 9 + 24 + 16 = 49

    ( (a+b) = c = 7)

    Hey, but it is equal to c2 = 72
    What are you trying to say with that not-equals sign?
    And demonstrably wrong the whole time?
    (Jeez, just like John Gabriel - a "new Algebra"...
    You're being foolish

  3. #23
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    Default Re: Fermat's Theorem, revisited

    Quote Originally Posted by grapes View Post
    What are you trying to say with that not-equals sign?
    Uh, lessee now...

    That (a + b)2 = c2 is not the same as a + b = c, maybe?

    (unless the 2's and parentheses are invisible....

    Keep working on it; I'm outta here except for revising my first post maybe...
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  4. #24
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    Default Re: Fermat's Theorem, revisited

    Quote Originally Posted by BuleriaChk View Post
    Uh, lessee now...

    That (a + b)2 = c2 is not the same as a + b = c, maybe?
    For positive integers, yes, it is the same thing.

    Keep working on it; I'm outta here except for revising my first post maybe...
    You said that before.

    The only way to improve your first post is to delete it. It's been saved.

  5. #25
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    Default Re: Fermat's Theorem, revisited

    Quote Originally Posted by BuleriaChk View Post
    Here is the real solution to Fermat's Theorem (see my ignore list)

    The Binomial Expansion is of the form (a + b)n = an + bn + rem(a,b) where rem (a,b) all powers of a and b and coefficients are positive integers. Fermat's Theorem is:

    an + bn cn for n > 2, a and b positive integers

    In order for Fermat's Theorem to be false (i.e., an + bn = cn) for some n > 2, a and b positive integers, rem(a,b) must be equal to zero, which it is not, by inspection (since rem(a,b) > 0, with a and b and the coefficients are positive integers for all powers of a and b and all n > 2.

    (Edit: I had a senior moment where I wrote "true" a short while ago) I have written it correctly numerous times in other threads. If this is what Grapes is referring to, I can only say "my bad"; sorry, Grapes if you couldn't figure that out from the context.) .. (There is a whole other thread where I say it correctly, but WAY over engineer the problem - I simply couldn't believe it was that easy. I'm starting to believe the whole thing is a huge joke, like believing Trump can be president (just ask Vince McMahon if it is just another WWE passion play). Well, ok, if it weren't so serious...

    ===============================
    For the case n = 2 (note: Fermat's Theorem is not relevant here, so this is actually a fool's errand w.r.t. to the hypothesis),



    we must have

    c2 - (2ab) = a2 + b2 is a positive integer

    Therefore for Fermat's theorem to be true for this case, we must have a = 0 or b = 0, so the equation is not satisfied.
    (there must be an (integer) metric for the integers as a ring without division, so 0 is the midpoint of all possible integer "distances" (lengths) defined in relation to 0 (a - a) = (b - b) = 0 for all a and b.)

    QED

    Try calling the Abel committee again... they may be finished their conversation with John....
    (humbly, I await your applause.....
    I can only say that this is not even wrong. There is nothing even remotely correct in your attempt of a proof. This is on the level of Gabriel kind of idiocy. You are so severely inept in reason, proof and rational thought that you have nothing but incoherency in your attempt here. You establishes no relations that you try to demonstrate, absolutely nothing. This isn't a proof even worthy of highschool students as many of them out do you.

    If you want to know how a proof is done, either read a book or read my posts where I make them, they are properly done. If you were one of my students comming with that when you are NOT younger than 18, I'd be laughing and asking if you are serious. This is nothing short of pointless rambling by a lunatic with delusions of grandure. Thinking with your infantile understanding of mathematics you could possibly get an Abels prize and solve one of the most difficult mathematical questions in less than one page is nothing short of hubris.
    Last edited by emperorzelos; 08-12-2016 at 01:15 AM.

  6. #26
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    Default Re: Fermat's Theorem, revisited

    Quote Originally Posted by BuleriaChk View Post
    I just realized why Fermat's Last Theorem is correct, using the Binomial Theorem:

    My Take on Fermat's Last Theorem

    (I suspect that he didn't have space to write the Binomial theorem on his page; the Binomial Theorem was already known in his time, he just put 2 and 2 together. He just didn't have space on his paper to quote it.

    My suspicion is that Wiles notwithstanding, it is a huge mathematical joke

    If I'm right the joke is on me, since I spent a ridiculous of time thinking about it, by trying to WAY over-engineer the problem (sigh) ... If so, however, there must be a vast mathematical conspiracy...

    (jump to post #12)

    (Ignore Grapes; he hasn't a clue (well, ok, except for my senior moment)....

    (Ignore emporerzelos; he has his head up his ass, where he is probably talking to Grapes. Not only that, but neither of them has a sense of humor. See My Ignore List in my signature)

    (Edit: the following approach in terms of complex numbers is unnecessary; to save time and agony, jump to post #12 (Actually the Binomial Expansion suffices by inspection)

    My take is that one uses complex numbers for the proof, but not for the theorem itself. The reason is that i2 can be thought of as the additive "destructor" of positive integers (e.g., a), since a2 + (ia)2 = a2 - a2 = 0 .

    Now consider the equation (a + b)2 = a2 + 2ab + b2. Then 2ab = a2 + b2 , so that the l.h.s. can be destroyed by multiplying each of the positive integers by i so that -2ab = 2(ia)(ib) = 2i2(ab) so that a2 + b2 +2ab - 2ab = 0 if the destructor is included, and the equation is valid (is a circle - since eliminating a and b is equivalent to eliminating the ab translation at the origin).

    But the Binomial Expansion for n > 2 will always have odd elements like a3 and b3 (eg like a3b2) which means that a corresponding destructor cannot be found, since (e.g.) (ia)3 = i(i2)a= -ia which are not only negative, but also complex, and therefore cannot be eliminated by subtracting the expansion, since at least one of the terms in the expansion will be imaginary for the terms containing both odd and even powers.

    QED

    (This has very interesting consequences for STR and QFT, which is why I got to thinking about it. There are many more consequences; I'm absolutely sure they have been published somewhere or are covered in a course I missed somewhere, but I haven't seen it, which is why I'm posting this.) I think it is why Einstein failed in his goal to provide a set of laws that were globally covariant, which is what I'm working (slowly, painfully) towards.. (lots of indices...

    That said, so far I have only been wrong once, and that was when I thought I made a mistake...
    I see now that you've "fixed" your first post by jumping out of it to post #12 (which is now quoted by empororzelos)

    My objections to your attempts have nothing to do with confusion over whether the theorem is true or false.

    You seem to be focusing on trying to shoehorn the theorem into the binomial form , where rem(a,b) represents the other terms of the binomial expansion)

    BUT

    We know that c cannot be (a+b), nor similarly can cn be (a+b)n

    That means your approach is irrelevant to the proof of Fermat's Last Theorem

  7. #27
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    Default Re: Fermat's Theorem, revisited

    Quote Originally Posted by grapes View Post
    I see now that you've "fixed" your first post by jumping out of it to post #12 (which is now quoted by empororzelos)

    My objections to your attempts have nothing to do with confusion over whether the theorem is true or false.

    You seem to be focusing on trying to shoehorn the theorem into the binomial form , where rem(a,b) represents the other terms of the binomial expansion)

    BUT

    We know that c cannot be (a+b), nor similarly can cn be (a+b)n

    That means your approach is irrelevant to the proof of Fermat's Last Theorem
    It is astounding to me that he thinks something that trivial can be the solution when clearly people have thought of that long ago and already dismissed it as a fruitless attempt

  8. #28
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    Default Re: Fermat's Theorem, revisited

    What you have proved, is that (a+b)n cannot equal an+bn, except when n=1, but

    you have not proven that (a+b-1)n cannot equal an+bn

    nor have you proven that (a+b-2)n cannot equal an+bn. In fact, we know it can, for n=2, but not for any other n

  9. #29
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    Default Re: Fermat's Theorem, revisited

    Quote Originally Posted by grapes View Post
    What you have proved, is that (a+b)n cannot equal an+bn, except when n=1, but

    you have not proven that (a+b-1)n cannot equal an+bn

    nor have you proven that (a+b-2)n cannot equal an+bn. In fact, we know it can, for n=2, but not for any other n
    I leave it to the reader to try to figure this one out....
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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.

  10. #30
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    Default Re: Fermat's Theorem, revisited

    Quote Originally Posted by emperorzelos View Post
    It is astounding to me that he thinks something that trivial can be the solution when clearly people have thought of that long ago and already dismissed it as a fruitless attempt
    Not by you, that's for sure. Get out of my thread or stop whining that you didn't figure it out and say something intelligent. And even then, get out of my thread.
    Last edited by BuleriaChk; 08-12-2016 at 01:11 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
    Ignore List -The Peanut Gallery.

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