How to prove $a+b\sqrt{2}=0$ iff $a=b=0$ I want to prove for $a,b \in \mathbb{Q}$
\begin{align}
a+b\sqrt{2}=0 \quad \Leftrightarrow \quad a=b=0
\end{align}
For $\Leftarrow$ is trivial, now I am in trouble of showing $\Rightarrow$ 
Simply I can say $\sqrt{2}$ is irrational and go proceed, but I want some formal mathematical proof. 

Assume:($a\neq b\neq 0$)
$$a+b\sqrt{2}=0$$
$$(a+b\sqrt{2})(\frac{a-b\sqrt{2}}{a-b\sqrt{2}})=0$$
$$\frac{a^2-2b^2}{a-b\sqrt{2}}=0$$
$$\left\{\begin{matrix}
a^2-2b^2=0 &  & \\ 
 a-b\sqrt{2}\neq 0&  & 
\end{matrix}\right.$$
$$\left\{\begin{matrix}
\frac{a}{b}=\sqrt{2} &  & \\ 
 \frac{a}{b}\neq \sqrt{2}&  & 
\end{matrix}\right.$$
Contradiction
So:($a=b=0$) for $a+b\sqrt{2}=0$
