Consider the set $Q=\{p+q \sqrt2 : p,q \in\Bbb Q\}$. Prove that if $a\in Q\setminus\{0\}$ then $1/a\in Q$

Question

Given (For all $a,b\in Q$, $a+b\in Q$ and $ab\in Q$)

This was a two part question.

Part a) is to prove that $Q$ is closed under addition and multiplication.

Part b) is prove that if $a\in Q$ and $a\ne0$, then $\frac1a\in Q$.

I proved part a but I'm stuck on part b... I know that I have to let $a=c+d\sqrt2$ and some how try and move things around to try and make it "look" like $p+q\sqrt2$ but I can't seem to get it.

Answer 1

$\frac{1}{a+b\sqrt{2}}=\frac{a-b\sqrt{2}}{(a+b\sqrt{2})(a-b\sqrt{2})}=\frac{a-b\sqrt{2}}{a^2-2b^2}=\frac{a}{a^2-2b^2}-\frac{b}{a^2-2b^2}\sqrt{2}$