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Thread: Square root Irrational

  1. #1
    Senior Member I-Think's Avatar
    Apr 2009

    Square root Irrational

    Quick proof veracity test for this question

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

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

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

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

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

    So $\displaystyle b^2=n^{2}d$ and $\displaystyle d=\frac{b^2}{n^2}$
    So $\displaystyle \sqrt{d}=\frac{b}{n}$
    Thus $\displaystyle \sqrt{d}=\frac{a}{b}=\frac{b}{n}$
    $\displaystyle \frac{a}{b}=\frac{b}{n}\rightarrow{b^2=an}\rightar row{\frac{b^2}{a}=n}$
    Hence $\displaystyle a|b$
    But this gives a contradiction as $\displaystyle gcd(a,b)=1$
    Hence our initial assumption is wrong and $\displaystyle \sqrt{d}$ is irrational
    Is this proof 100% correct?
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  2. #2
    Pim is offline
    Dec 2008
    The Netherlands
    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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