If $a$ and $b$ are rational, then $a + b{\sqrt{2}} \ne {\sqrt{3}} $ 
If $a,b\in\mathbb{Q}$  then demonstrate:$$a + b{\sqrt{2}} \ne  {\sqrt{3}} $$

I raised and squared the equation but it didn't work.

I think you mean to show that there aren't $a$ and $b$ in $\Bbb{Q}$ for which equality holds. Squaring both sides leads to
$$a^2 + 2b^2 + 2ab\sqrt{2} = 3$$
If $ab \ne 0$, we rearrange to find
$$\sqrt{2} = \frac{3 - a^2 - 2b^2}{2ab} \in \Bbb{Q}$$
contradicting that $\sqrt{2}$ is not rational. If $ab = 0$, then $a = 0$ or $b = 0$, and these cases are easier to handle.
