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Thread: Proof

  1. #1
    Junior Member
    Dec 2005


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

    My solution is based on using Viete's formulas.

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

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

    So {a} is then irrational number.

    Knowing that, then {b} is irrational number because
    -{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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