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Math Help - N-th Root Irrationality Proof

  1. #1
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    N-th Root Irrationality Proof

    Hey all, any help with the following proof would be appreciated:

    The real number n-th root of 2 is irrational.


    A similar proof, which says that the square root of 2 is irrational, is proved in the following way by contradiction:

    Assume sqroot(2) = m/n for some m,n in Z (integers)
    Since it is rational, you can assume m and n have no common factors.
    2 = m^2 / n^2 implies m / n = 2n / m
    This means n divides m, but that means sqroot(2) is an integer, which is a contradiction.

    At our disposal, we have the fact that:
    -The real numbers sqroot(2) is irrational
    -If r in the Naturals is not a perfect square, then sqroot(r) is irrational
    -Let m and n be nonzero integers. Then (m/n)*sqroot(2) is irrational

    Thanks for the help!
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  2. #2
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by jstarks44444 View Post
    Hey all, any help with the following proof would be appreciated:

    The real number n-th root of 2 is irrational.


    A similar proof, which says that the square root of 2 is irrational, is proved in the following way by contradiction:

    Assume sqroot(2) = m/n for some m,n in Z (integers)
    Since it is rational, you can assume m and n have no common factors.
    2 = m^2 / n^2 implies m / n = 2n / m
    This means n divides m, but that means sqroot(2) is an integer, which is a contradiction.

    At our disposal, we have the fact that:
    -The real numbers sqroot(2) is irrational
    -If r in the Naturals is not a perfect square, then sqroot(r) is irrational
    -Let m and n be nonzero integers. Then (m/n)*sqroot(2) is irrational

    Thanks for the help!
    The following result is very useful:
    .

    Try to prove the above and then:

    assume

    from the above which is obviously wrong.
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  3. #3
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    Here is an overkill! You have already proven that √2 is irrational.
    Now suppose there exist p,q such that 2^{\frac{1}{m}} = \frac{p}{q}, for m \ge 3 then:

    2^{\frac{1}{m}} = \frac{p}{q} \Rightarrow 2 = \frac{p^m}{q^m} \Rightarrow 2q^m = p^m \Rightarrow q^m+q^m = p^m, ~~ m \ge 3

    which contradicts Fermat's Last Theorem, and hence the conclusion.
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  4. #4
    MHF Contributor Drexel28's Avatar
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    The following statement, which is a generalization, is true " a^{\frac{1}{m}} where a,m\in\mathbb{N} is rational if and only if a^{\frac{1}{m}} is an integer[/tex]" this can be seen as a corollary of the fact that the only rational algebraic integers are integers themselves. The proof is simple and basically follows from the rational root theorem. So, 2^{\frac{1}{n}} is never rational since it's never an integer since 1<2^{\frac{1}{n}}<2 for n\geqslant 2.
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  5. #5
    Senior Member abhishekkgp's Avatar
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    Quote Originally Posted by TheCoffeeMachine View Post
    Here is an overkill! You have already proven that √2 is irrational.
    Now suppose there exist p,q such that 2^{\frac{1}{m}} = \frac{p}{q}, for m \ge 3 then:

    2^{\frac{1}{m}} = \frac{p}{q} \Rightarrow 2 = \frac{p^m}{q^m} \Rightarrow 2q^m = p^m \Rightarrow q^m+q^m = p^m, ~~ m \ge 3

    which contradicts Fermat's Last Theorem, and hence the conclusion.
    i had never ever thought of it that way!!!
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  6. #6
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by abhishekkgp View Post
    i had never ever thought of it that way!!!
    Yes, because that was the first corollary of FLT that I thought of too.
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