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Seeking a Proof of Incommensurability

  1. Sep 9, 2008 #1
    Does anyone know of a theorem in number theory or other branch of mathematics that would prove the following: Plot the graph of the function y(x) = x ^ 2 in the first quadrant of a Cartesian x-y coordinate system. Drop a vertical line segment parallel to the y-axis, denoted as the end vertical, such that one end point of the end vertical is y(c) = c ^ 2, and the other end point is x = c. Drop another vertical line segment, denoted as the first vertical, where one end point is y(a) = a ^ 2, and the other end point is x = a. Drop another line segment, denoted as the second vertical, where one end point is y(b) = b ^ 2, and the other end point is x = b. Note, 0 < a < b < c. Define the area under the graph of the function bounded below by the x-axis and bounded to the right be the first vertical as area A. Define the area under the graph of the function bounded below by the x- axis, bounded to the left by the second vertical and bounded to the right by the end vertical as area B. Let area A = area B. Prove that the line segment on the x-axis from 0 to a is always incommensurable with the line segment from b to c, where points b and c are integral numbers.
     
    Last edited: Sep 9, 2008
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  3. Sep 9, 2008 #2

    HallsofIvy

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    I thought I understood what you were doing until I got here. Do you mean bounded to the right by the line x= a?

    Sounds to me like area A is
    [tex]\int_0^a x^2 dx= \frac{a^3}{3}[/tex]
    and that B is
    [tex]\int_b^c x^2 dx= \frac{c^3-b^3}{3}[/tex]
    and that those are equal. Now, you are asserting that (c-b)/a is not rational.
    Well, saying that those area are equal means that [itex](c^3- b^3)/a^3= 1[/itex]. We can factor [itex]c^3- b^3= (c- b)(c^2+ bc+ b^2)[/itex] so [itex](c^3- b^2)/a^2= [(c-b)/a][c^2+ bc+ c^2]= 1[/itex]. Saying that (c-b)/a is not rational is equivalent to saying that [itex]c^2+ bc+ c^2[/itex] is not rational. Since you could choose c and b arbitrarily, and then find a to fit, I don't believe that is true.
     
  4. Sep 9, 2008 #3
    Yes, I mean bounded to the right by the line x=a. I think you may have made some typo's on your equations. Check your powers and terms. You assert that c ^2 + bc + c ^ 2, (I think you mean c ^ 2 + bc + b ^ 2) is not rational. Whatever, either one will always be rational because I have given the condition that b and c are integers. So c ^ 2 + bc + c ^ 2 or c ^2 + bc + b ^ 2 must always be rational.
     
  5. Sep 10, 2008 #4

    Gib Z

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    I see nothing wrong with any of Hall's equations?

    Regardless, the given information implies a^3 + b^3 = c^3, and since its given a, b and c are integers, Fermat's Last Theorem shows that's quite wrong. I don't think its appropriate to use such a power theorem though lol. Although I know there were much simpler proofs for that case, n=3, so either use that fact or...

    [tex]a^3 = c^3 - b^3 = (c-b)(c^2 +bc+b^2)[/tex]

    What you want to prove is that [tex]\frac{c-b}{a}[/tex] is irrational, and arranging the previous equation gives:

    [tex]\frac{c-b}{a} = \frac{a^2}{b^2 + bc+c^2}[/tex], and since a, b and c are all integers, this implies that your theorem is in fact false- Just as Hall's has said.

    Of course, we can't actually find any counter-examples, as the original assertion that the areas are equal yet a, b and c are integers is incorrect.
     
  6. Sep 10, 2008 #5

    One problem. I did not assert explicitly or implicitly that "a" was an integer in the problem. I only explicitly stated "b" and "c" are integers.
     
  7. Sep 10, 2008 #6

    CRGreathouse

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    If b and c are integers and a = M/N is rational in your problem, then
    [tex]a^3+b^3=c^3[/tex]
    [tex](M/N)^3+b^3=c^3[/tex]
    [tex]M^3+b^3N^3=c^3N^3[/tex]
    [tex]M^3+(bN)^3=(cN)^3[/tex]
    a contradiction by the Fermat-Wiles theorem. So a is irrational, which may be what you wanted to prove (or at least a part of it). I couldn't understand the first post, sorry...
     
  8. Sep 10, 2008 #7
    Yes, you got it! But, what I am seeking now, is another proof, not one that cites Fermat-Wiles, that proves incommensurability for this problem. If such a proof existed, then we would have another round about proof of FLT.
     
  9. Sep 10, 2008 #8

    CRGreathouse

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  10. Sep 10, 2008 #9
    Well, actually I am seeking a general proof of the problem. The problem I stated comes to the same result for "a" being an irrational number, for all integral powers of a,b, and c, and for powers greater than 2, where b and c are integers. It's just another way to state FLT. I was hoping that maybe in the branch of topology or geometry or some other branch of mathematics there is a theorem that relates equal areas to each other and how they can be "morphed", one into another with certain restrictions, so that they don't lose their equality. For example, if you imagined area A, a 3-sided geometric object being transformed into area B, a 4-sided geometric object, are there theorems that state how this transformation can or cannot occur without losing their area equality? Is there a theorem that states as a necessary condition that at least one of the sides of the objects, a straight line segment, must be an irrational length for an equal transformation to occur? It appears that this is the case for how these objects are formed by the given conditions. If such a theorem exists, as I said, it would be a back-door way to prove flt.
     
  11. Sep 11, 2008 #10

    CRGreathouse

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    Realizing how many people tried to prove FLT and failed, I'd rate your chances as slim unless you can use some powerful result that was not available to the many FTL proof searchers. (Obviously, T-S would work...)
     
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