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Math Help - Proof of square root being irrational

  1. #1
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    Proof of square root being irrational

    Let N be a Natural Number such that square root of N is not an integer. Prove that then square root of N is even irrational.

    *The problem gives you a hint stating that:
    Assume square root of N is in the irrational numbers. Then the Set X := {x is in the real numbers : x multiplied by square root N is in the Natural Numbers} is nonempty. Show that if x is in X and x' := (x multiplied by square root of N) minus (x[square root of N]) then x' is in X and x' < x. thus square root of 2, 3, 5,... are irrational.

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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by ml692787 View Post
    Let N be a Natural Number such that square root of N is not an integer. Prove that then square root of N is even irrational.

    *The problem gives you a hint stating that:
    Assume square root of N is in the irrational numbers. Then the Set X := {x is in the real numbers : x multiplied by square root N is in the Natural Numbers} is nonempty. Show that if x is in X and x' := (x multiplied by square root of N) minus (x[square root of N]) then x' is in X and x' < x. thus square root of 2, 3, 5,... are irrational.

    Thanks
    Someone posted this:

    If you know the rational root theorem we have:

    let y=sqrt(n), then:

    y^2-n=0

    and if this has rational roots they are amoung the factors (positive or
    negative) of n.

    So we have sqrt(n) is an integer, and [sqrt(n)]^2=n, that is n is a
    perfect square.
    So we have that if sqrt(n) is rational then sqrt(n) is an integer, hence
    if sqrt(n) is not an integer it is irrational


    RonL
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  3. #3
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    I'm pretty sure we have not covered the rational root theorem in class, and as much sense as that theorem makes, I don't think I can use that to solve the problem on a test if we have not covered it yet.
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