Let $r\in\mathbb{Q}[\sqrt2]$. Show $\phi(r) = r$ if and only if $r\in\mathbb{Q}$ 
Let $\mathbb{Q}[\sqrt2]=\{a+b\sqrt2 \mid a,b\in\mathbb{Q}\}$ and define $\phi:R \rightarrow R$ by $\phi(a+b\sqrt2)=a-b\sqrt2$. Show $\phi(r) = r$ if and only if $r\in\mathbb{Q}$.

My approach is that if $r\in\mathbb{Q}$, then the $b\sqrt2$ part must be zero since $\sqrt2$ is irrational, then I have $\phi(a)=a$. If $\phi(r)=r$, then r must be rational since $\sqrt2$ is irrational. I think this is just a very rough thought, can someone help shape it? Many thanks!

One direction looks fine, but needs a bit more justification. (You may find a contrapositive approach easier.) For the other, set $$a-b\sqrt2=:\phi(a+b\sqrt2)=a+b\sqrt2$$ and conclude from there.
