
Proof
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?
