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Thread: Fermat's last, and mine too.

  1. #1
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    Default Fermat's last, and mine too.

    Fermat's was a note in the margin of a book claiming he had devised a simple proof that a^n+b^n=/=c^n for n>2.


    Here's a simple proof.
    when a=b, a^n+b^n<c^n for n>2. As a approaches 0, b approaches c. Therefore as a approaches 0, a^n+b^n approaches c^n for n>2.

    End of story.
    Lies have the stench of death and defeat.

  2. #2
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    Default Re: Fermat's last, and mine too.

    Lol. Welcome to the club!

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    Default Re: Fermat's last, and mine too.

    Quote Originally Posted by astrotech View Post
    Fermat's was a note in the margin of a book claiming he had devised a simple proof that a^n+b^n=/=c^n for n>2.


    Here's a simple proof.
    when a=b, a^n+b^n<c^n for n>2. As a approaches 0, b approaches c. Therefore as a approaches 0, a^n+b^n approaches c^n for n>2.

    End of story.
    The problem is in the term "approach" if a,b, and c are integers, which is a catchall phrase for infinitesimals (derivatives).

    The Binomial Theorem states it exactly (as does my proof) for positive integers, n >2:

    , where (so only if or

    (so or , respectively.

    But I'm happy to see that Grapes seems to understands your version, even though it doesn't involve integers (). (There is a derivative relation between the Binomial Theorem and GTR as I've just posted in the other thread, where the Jacobian of the metric tensor is characterized as consisting of integers (v'/c' invariant as an "acceleration" )

    The (general) relativistic relation is where a is identified with c in the relation v/c =b/c, so if a goes to zero v=b goes to zero as well and there is no integer to begin with (c=0) with all elements identified with differentials rather than integers.

    (i.e, there is no metric tensor characterizing stress-energy)
    (there is no curvature because there is nothing in the Universe... not even a parking lot..)
    Last edited by BuleriaChk; 02-21-2017 at 01:47 PM.
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    The Relativistic Unit Circle Updates 03/02/2017 12:07 PM PST
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    Default Re: Fermat's Theorem, Revistited

    Proof of Fermat's Theorem


    If Fermat's expression were true (), , , and would be a Pythagorean Triple. They are not a Pythagorean Triple by the Binomial Theorem, where rem(a,b,n) > 0.

    Therefore, , for a,b,c,n positive integers, n>2

    QED

    Triangle.png
    _______________________________________
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    The Relativistic Unit Circle Updates 03/02/2017 12:07 PM PST
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    Default Re: Fermat's last, and mine too.

    Hey, buleria,
    Apologies in advance if I don't format this right, I have very little experience with latex
    I just wanted to point out that your proof doesn't really work.
    You're just saying that:

    c=a+b
    c^n=(a+b)^n
    And then setting (a+b)^n equal to a^n+b^n,
    so of course a or b has to be zero.


    All you're saying is (a+b)^n=a^n+b^n IFF a or b=0
    What you want to prove is that c^n = a^n+b^b IFF a or b=0
    Those are different things
    Last edited by Icarus496; 02-26-2017 at 04:49 PM.
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    Default Re: Fermat's last, and mine too.

    Quote Originally Posted by Icarus496 View Post
    Hey, buleria,
    Apologies in advance if I don't format this right, I have very little experience with latex
    I just wanted to point out that your proof doesn't really work.
    You're just saying that:

    c=a+b
    c^n=(a+b)^n
    And then setting (a+b)^n equal to a^n+b^n,
    so of course a or b has to be zero.


    All you're saying is (a+b)^n=a^n+b^n IFF a or b=0
    What you want to prove is that c^n = a^n+b^b IFF a or b=0
    Those are different things
    The Binomial theorem expresses this, which is my proof, where iff or for Fermat's expression. Since rem(a.b.n) is not equal to zero (is a positive integer for n>2), Fermat's theorem is proved.

    Look up the Binomial Expansion.. in Wiki...

    Last edited by BuleriaChk; 02-26-2017 at 04:56 PM.
    _______________________________________
    "Flamenco Chuck" Keyser
    The Relativistic Unit Circle Updates 03/02/2017 12:07 PM 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: Fermat's last, and mine too.

    Right, I understand the binomial expansion, but you can only apply the binomial theorem if you have something of the form (a+b)^n. If you set that equal to c^n, like you have to do for your proof, that reduces to c=a+b, which is not what you want to prove.
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    Default Re: Fermat's last, and mine too.

    Quote Originally Posted by Icarus496 View Post
    Right, I understand the binomial expansion, but you can only apply the binomial theorem if you have something of the form (a+b)^n. If you set that equal to c^n, like you have to do for your proof, that reduces to c=a+b, which is not what you want to prove.


    It only reduces to c = a+b if a=0 or b= 0, in which case c = b or c=a, respectively.
    _______________________________________
    "Flamenco Chuck" Keyser
    The Relativistic Unit Circle Updates 03/02/2017 12:07 PM 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: Fermat's last, and mine too.

    Quote Originally Posted by BuleriaChk View Post


    It only reduces to c = a+b if a=0 or b= 0, in which case c = b or c=a, respectively.
    and all cases has nothing to do with FLT

  10. #10
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    Default Re: Fermat's last, and mine too.

    Quote Originally Posted by BuleriaChk View Post


    It only reduces to c = a+b if a=0 or b= 0, in which case c = b or c=a, respectively.
    Let c=2 and a=b=1.

    If you read the Wiki page you cited (under "Theorem statement"), it says , not for any c.

    Added Link for user:
    Cited Wikipedia Page
    Last edited by Neverfly; 02-27-2017 at 03:22 AM.

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