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Math Help - Square root Irrational

  1. #1
    Senior Member I-Think's Avatar
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    Square root Irrational

    Quick proof veracity test for this question

    Let d\in{N} that is not the square of another natural number. Prove that \sqrt{d} is not a rational number

    My proof
    Assume \sqrt{d}\in{Q}, hence \sqrt{d}=\frac{a}{b}, where a,b\in{N} and gcd(a,b)=1

    So d=\frac{a^2}{b^2}, so db^2=a^2. Note d|a^2, so d|a
    Hence  a=dn
    and a^2=d^2n^2

    Therefore
    d=\frac{d^{2}n^{2}}{b^2}, divide by b

    1=\frac{dn^2}{b^2}

    So b^2=n^{2}d and d=\frac{b^2}{n^2}
    So \sqrt{d}=\frac{b}{n}
    Thus \sqrt{d}=\frac{a}{b}=\frac{b}{n}
    Hence
    \frac{a}{b}=\frac{b}{n}\rightarrow{b^2=an}\rightar  row{\frac{b^2}{a}=n}
    Hence a|b
    But this gives a contradiction as gcd(a,b)=1
    Hence our initial assumption is wrong and \sqrt{d} is irrational
    QED
    Is this proof 100% correct?
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  2. #2
    Pim
    Pim is offline
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    To me it seems correct. Except for one thing: you say "divide by b" , but you're actually dividing by d. All the steps are correct though.
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