Is $\sqrt{2}$ a rationnal number ?

Squaring the equation gives $2=\frac{a^2}{b^2}$ and taking the $2$-adic valuation leads to:
$$1=v_2(2)=v_2\left(\left(\frac{a}{b}\right)^2\right)=2\,v_2\left(\frac{a}{b}\right)$$But $v_2\left(\frac{a}{b}\right)$ is an integer and $1$ is not an even integer so this is absurd.