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Thread: Proof

  1. #1
    Junior Member
    Dec 2005


    I have to prove that numbers $\displaystyle {\sqrt2}$ and $\displaystyle {\sqrt3}$ can't be solutions of quadratic equation if coefficients are rational.

    My solution is based on using Viete's formulas.

    $\displaystyle {\sqrt2}+{\sqrt3} = -\frac{b}{a}$
    $\displaystyle {\sqrt2}{\sqrt3} = \frac{c}{a} $

    From $\displaystyle {\sqrt2}{\sqrt3} = \frac{c}{a} $ we get that
    $\displaystyle {a} = \frac{c}{{\sqrt2}{\sqrt3}} $.

    So $\displaystyle {a} $ is then irrational number.

    Knowing that, then $\displaystyle {b} $ is irrational number because
    $\displaystyle -{b} = {a}({\sqrt2}+{\sqrt3}) =\frac{{c}({\sqrt2}+{\sqrt3})}{{\sqrt2}{\sqrt3}} $

    c is then easy to prove that its also irrational.

    Is my proof correct?
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  2. #2
    Global Moderator

    Nov 2005
    New York City
    Correct, Good job
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