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Thread: Proof of Fermat's Theorem by Modulus

  1. #31
    Moderator grapes's Avatar
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    Default Re: Proof of Fermat's Theorem by Modulus

    Quote Originally Posted by BuleriaChk View Post
    Time will tell. My .pdf is on my website.... and you're on record for everyone to see...
    Hmmm, let's see, less than half an hour since you posted that.

    Here is your claim:
    Quote Originally Posted by grapes View Post
    So, this is "true for all positive numbers, not just integers:"

    , n > 2 (Fermat's Theorem, my proof)

    It appears that your proof does show that.
    So, let

    Yep, you're wrong

  2. #32
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    Default Re: Proof of Fermat's Theorem by Modulus

    Quote Originally Posted by grapes View Post
    Hmmm, let's see, less than half an hour since you posted that.

    Here is your claim:


    So, let

    Yep, you're wrong
    WTF? This is nonsense. Grapes is a Village Idiot.

    , a,b,c positive integers
    , x,y,z positive real numbers.
    Last edited by BuleriaChk; 02-27-2017 at 12:46 AM.
    _______________________________________
    "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.

  3. #33
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    Default Re: Proof of Fermat's Theorem by Modulus

    Quote Originally Posted by BuleriaChk View Post
    I do prove the equality false in Fermat's last theorem via the Binomial Expansion for n >2.
    (also for n=2, if one excludes Pythagorean triples; but not necessary for Fermat's proof.)
    ------------------------
    iff

    The point is that a and b must be independent (a,b); i.e., not on the same number line, but on orthogonal number lines. If they are on the same number line, then c =(a+b) = d, so ; that is, the identity of a and b is not preserved in the Binomial expansion because the lhs and rhs of c=(a +b) refer to the same single integer.

    By the Binomial Expansion, true for all positive numbers, not just integers:

    That is, (once again)

    since rem(a,b,n) > 0

    , n > 2 (Fermat's Theorem, my proof)

    QED

    (Try really, really hard to understand Fermat's Theorem AND the Binomial Expansion)

    rem(a,b,n) is simply everything that is not or in the Binomial Expansion.
    ------------------
    c < (a+b) merely says so

    which is a true expression, but is a metric expression, not a proof of Fermat's Theorem, so I don't have to prove it, trivial as it is. It is irrelevant to the proof. (In particular, it is not true for all integers a, b, c, in particular for c > (a + b) - c cannot satisfy both equations.

    c=(a+b) sets the foundation for the proof (by contradiction)

    My proof is true for all integers (and numbers) in Fermat's equation. And I don't really need all the analysis in terms of vectors, the Binomial Theorem is just fine as it is. (I had been trying to justify the Binomial Theorem in terms of vectors by including Pythagorean triples for the case n = 2, but Fermat specifies n>2, so that analysis is also irrelevant to the actual proof).
    You can't do that, binomial require equality, and FLT requires INequality, that makes them incompatible. The initial assumption are in direct contradiction.

  4. #34
    Moderator grapes's Avatar
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    Default Re: Proof of Fermat's Theorem by Modulus

    Quote Originally Posted by BuleriaChk View Post
    Quote Originally Posted by grapes View Post
    Hmmm, let's see, less than half an hour since you posted that.

    Here is your claim:


    So, let

    Yep, you're wrong
    WTF? This is nonsense. Grapes is a Village Idiot.

    , a,b,c positive integers
    , x,y,z positive real numbers.
    You claimed that cn didn't equal an+bn, for all positive numbers

    I gave you an example that showed you were wrong

    You are confusing the two things. Yes, if c=a+b, then cn cannot equal an+bn, that's obvious. But cn could equal an+bn for some a and b--and that's why your "proof" doesn't work.

  5. #35
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    Default Re: Proof of Fermat's Theorem by Modulus

    Quote Originally Posted by grapes View Post
    You claimed that cn didn't equal an+bn, for all positive numbers

    I gave you an example that showed you were wrong

    You are confusing the two things. Yes, if c=a+b, then cn cannot equal an+bn, that's obvious. But cn could equal an+bn for some a and b--and that's why your "proof" doesn't work.
    c=a+b means the symbols on both sides of the equality refer to a single integer; the r.h.s. is not two separate integers but just another name for c.
    _______________________________________
    "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.

  6. #36
    Moderator grapes's Avatar
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    Default Re: Proof of Fermat's Theorem by Modulus

    Quote Originally Posted by BuleriaChk View Post
    c=a+b means the symbols on both sides of the equality refer to a single integer; the r.h.s. is not two separate integers but just another name for c.
    The Right Hand Side would be a single integer, but that single integer is represented as being the sum of two other integers, b and c

    This is a two month old thread, the same subject is treated in anothe thread:

    Fermat's Theorem, Relativity, Quantum Field Theory

    This thread is closed.
    Last edited by grapes; 02-27-2017 at 10:39 AM.

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