For all $x \in A$ we have $3x^2$ is rational number $\geq 0$ So $x=\sqrt{\frac{p}{q}}$ for some $p,q \in \mathbb{N}$ Let $a=\sqrt{\frac{p}{q}}$ and $b=\sqrt{\frac{r}{s}}$ Because $a,a,b \in A$ we have $a^2+2ab$ is rational number So $ab$ is rational number And $\frac{a}{b}=\frac{ab}{b^2}$ which is rational number because $ab,b^2$ is rational number.

ThE-dArK-lOrD wrote: P.S. Problem should state that $b\neq 0$ It says "positive reals".