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Square root of a Mersenne Number is irrational

  1. Apr 10, 2014 #1
    1. The problem statement, all variables and given/known data

    A user on math.se wanted to prove that any Mersenne number m = 2^n - 1 has an irrational square root for n > 1. So, it can be proved rather easily that any non-perfect square has an irrational root, and that all of the numbers to be considered are not perfect squares. However, the user requested a solution that did not use this idea, and that just focused on the Mersenne numbers themselves. I'm not sure why one would do this besides the challenge.

    A few other posters proved this by using binary. I don't know anything about that. This discouraged me, as I submitted a simple proof by contradiction using just basic algebra. Can anyone tell me if my proof actually works?


    2. Relevant equations



    3. The attempt at a solution

    Assume by way of contradiction that there exist integers a, b, such that gcd(a,b) = 1 and (a/b) = sqrt(2^n - 1).

    Then
    [itex]\frac{a^{2}}{b^{2}} = 2^{n} - 1[/itex]
    [itex]\frac{a^{2}}{b^{2}} + 1 = 2^{n}[/itex]

    Since 2^n is even, a^2/b^2 is odd.
    Since the quotient a^2/b^2 is odd, either both a^2 and b^2 are even or both a^2 and b^2 are odd.

    Case #1; a^2 and b^2 are both even.

    Then, a is even and b is even.
    Contradiction, gcd(a,b) = 1.

    Case #2; a^2 and b^2 are both odd.

    Then, a is odd and b is odd.
    Let a = 2x + 1 and b = 2y + 1

    [itex]\frac{a^{2}}{b^{2}} + 1 = 2^{n}[/itex]

    [itex]a^{2} + b^{2} = 2^{n}b^{2}[/itex]

    [itex](2x + 1)^{2} + (2y + 1)^{2} = 2^{n}b^{2}[/itex]

    [itex]4x^{2} + 4x + 4y^{2} + 4y + 2 = 2^{n}b^{2}[/itex]

    [itex]2(2x^{2} + 2x + 2y^{2} + 2y + 1) = 2^{n}b^{2}[/itex]

    [itex]2x^{2} + 2x + 2y^{2} + 2y + 1 = 2^{n-1}b^{2}[/itex]

    [itex]2(x^{2} + x + y^{2} + y) + 1 = 2^{n-1}b^{2}[/itex]

    Contradiction; 2^(n-1) is even since n > 1, and b^2 is odd. Therefore since an odd times an even is always even, the RHS is even. The LHS is odd by definition.

    The shows that a contradiction occurs in all cases, so there is no such a, b.
     
  2. jcsd
  3. Apr 11, 2014 #2

    jbunniii

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    Your proof looks valid to me.
     
  4. Apr 11, 2014 #3
    Thanks for the response. It was pretty fun to write, I'm glad it's ok.
     
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