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Revisiting an old problem with startling new mathematical results

  1. What's wrong with the math in this paper?

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  2. Do we actually use equality testing, if only partially, when solving equations?

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    0.0%
  3. "definitional loops," are they fixed with polynomial equations and their solutions?

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    0.0%
  4. The math is simple. Where is the fallacy?

    0 vote(s)
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Multiple votes are allowed.
  1. Jun 24, 2012 #1
    I've got the attached paper... It's interesting, because it deal with an old complex problem is a seemingly simple form, and just using high school algebra. Check it out!

    Here is the attackment:
     

    Attached Files:

  2. jcsd
  3. Jun 24, 2012 #2

    micromass

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    You seem to like to set f=d. Why can you do that? Certainly f is not equal to d in general.
     
  4. Jun 24, 2012 #3


    You seem to believe that very trivial, old-known results are new (like that "never before known axiom or postulate", as you called it)...

    I'm not even sure what you think you've proved: is is true that you believe that you've proved Fermat's Last Theorem in that paper of yours? Because if you do then, with all due respect, you're dead wrong.

    I think you need to begin taking some basic mathematics courses, at least to know how to write mathematics and/or how to use logic, and then perhaps go back to your paper and, if you still think it is worthwhile, re-write it all through.

    DonAntonio
     
  5. Jun 24, 2012 #4
    DonAntonio... you took too many words to reply to this. And yes, I proved the Theorem... as simple as that. Just follow the arguments and don't get excited about nothing. This is not religion or politics. Exactly where is my arguments wrong?
     
  6. Jun 24, 2012 #5

    mathwonk

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    why should we do your work for you? if you have a proof of fermat, submit it to a journal and the editor will try to find a referee willing to read it.
     
  7. Jun 24, 2012 #6

    micromass

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    I showed you were it is wrong. If you're not going to reply to our comments, then this will be locked.
     
  8. Jun 24, 2012 #7

    pwsnafu

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    Statements such as
    when you clearly wrote ##b = (b^2 – d^2) / 2d## the line before are not valid in mathematics. If you write an equality it is an equality. You are not allowed to retrospectively change it. As Don said, you clearly do not understand mathematical logic or the requirements of writing a paper.
     
  9. Jun 24, 2012 #8


    Just what I thought...but you know what? Sometimes things are SO wrong that it isn't easy to point out where "exactly" they are wong. I can tell you where you begin to go astray from logic, and mathematics, which is pretty close to the beginning, but I don't think that will do any good to anyone.

    As already said, it is obvious you're not a mathematician as it is obvious you don abide neither by its logical rules nor by the standard ways to communicate mathematical ideas, so again: I think the best for you is to begin studying some basic maths and THEN, after you already know some, go back to your paper and check whether it is worthwhile to keep it as it is, to improve it, to change it...or to toss it away.

    DonAntonio
     
  10. Jun 24, 2012 #9
    I had this before:

    b = (a^2 - d^2) / 2d is an equality. Once I changed it to b = (b^2 - d^2) /2d I turned it into an inequality means that b ≠ (b^2 - d^2) /2d, even though the quadratic equation
    b^2 - 2db - d^2 = 0 has a solution that is irrational... and here I'm talking about rational solutions.
     
  11. Jun 24, 2012 #10

    micromass

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    Why can you set f=d??
     
  12. Jun 24, 2012 #11
    Another thing... try the formulas. If they work, I'm doing correct math. If they don't, I'm all yours to lapidate. But don't get stuck in minucia. I'm no Newton, but if he had followed the suggestions and objections of Bishop Berkeley, we've probably have no Calculus today, if in addition, no one had understood Mr. Leibnitz own nomenclature.
     
  13. Jun 24, 2012 #12
    micromass... because f can actually be equal to d. Try this 3^2 + 4^2 = 5^2, and a = b -1 = 4 -1 =3, while c = b + 1 = 4 +1 = 5, which means tha f = d =1.
     
  14. Jun 24, 2012 #13

    pwsnafu

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    But micromass's point is that you have not proved the case ##f\neq d##.
     
  15. Jun 24, 2012 #14

    micromass

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    OK, this proves you have no idea how math works. If you want your mistake, there it is. f does not have to equal d.

    Thread locked. Do not post this again.
     
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