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Math Help - proving √n is irration (n=square free + integer)

  1. #1
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    proving √n is irration (n=square free + integer)

    Hi all!
    I have to show that √n is irrational, where n is "any square free positive integer". Ive never worked with "Square free" numbers, but I understand the term (no divisor can be a square).

    So I assume we proceed by contradiction, so
    √p=a/b ->p=(a/b)^2, now do we do the same "trick" where we find a common factor between a and b, or is there a different way to proceed because we are dealing with square free numbers instead of prime.
    Thanks!
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    Re: proving √n is irration (n=square free + integer)

    Quote Originally Posted by farmeruser1 View Post
    show that √n is irrational, where n is "any square free positive integer". Ive never worked with "Square free" numbers, but I understand the term (no divisor can be a square).
    Suppose that \sqrt{n}\notin\mathbb{Z}^+ but it is rational, then \exists p\in\mathbb{Z}^+[p\sqrt{n}\in\mathbb{Z}^+], WHY?

    By the well order of the positive integers, that is a least positive integer with at at property. Call it k.

    The greatest integer in \sqrt{n} is \left\lfloor {\sqrt n } \right\rfloor and we know that 0 < \sqrt n  - \left\lfloor {\sqrt n } \right\rfloor  < 1.

    Can you explain why \left( {\sqrt n  - \left\lfloor {\sqrt n } \right\rfloor } \right)k\in\mathbb{Z}^+~?

    Can you explain why \left( {\sqrt n  - \left\lfloor {\sqrt n } \right\rfloor } \right)k<k~?

    Can you explain why \left( {\sqrt n  - \left\lfloor {\sqrt n } \right\rfloor } \right)k\sqrt n\in\mathbb{Z}^+~?

    So we have found a smaller positive integer with the property of k that is a contradiction.
    Thanks from topsquark and farmeruser1
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    Re: proving √n is irration (n=square free + integer)

    Pardon my naivety, but I have never worked with Integer square roots before, Let me give this a crack.
    To answer your first question, if √n is rational, then √n=q/p, which is p√n=q, q is a positive integer, as wanted.
    Now, the next one, if we expand we get k√n-k[√n]. k√n is an integer that we just proved above, and k[√n] is a positive smaller integer by definition.
    So we have (√n-[√n])k is a positive integer.
    The next part by definition, <0(√n-[√n])<1 so this multiplied by some k is <k.
    Now the last part, lets expand, kn-k√n[√n], which by using the last 2 parts we know is a positive integer.
    Now because k√n<k, then it cannot be rational, is this correct?
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    Re: proving √n is irration (n=square free + integer)

    Quote Originally Posted by farmeruser1 View Post
    Pardon my naivety, but I have never worked with Integer square roots before, Let me give this a crack.
    To answer your first question, if √n is rational, then √n=q/p, which is p√n=q, q is a positive integer, as wanted.
    Now, the next one, if we expand we get k√n-k[√n]. k√n is an integer that we just proved above, and k[√n] is a positive smaller integer by definition.
    So we have (√n-[√n])k is a positive integer.
    The next part by definition, <0(√n-[√n])<1 so this multiplied by some k is <k.
    Now the last part, lets expand, kn-k√n[√n], which by using the last 2 parts we know is a positive integer.
    Now because k√n<k, then it cannot be rational, is this correct?
    That is correct. Well done.
    If fact, this proves a stronger result than asked for.
    For example, 18=2\cdot 3^2 is not square-free. But we have shown that \sqrt{18} is not rational.
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