Proof

**DenMac21** · January 6th 2006, 05:40 PM

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?

**ThePerfectHacker** · January 6th 2006, 05:56 PM

Correct, Good job

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