Prove that $\sqrt{2} + \sqrt{n}$ is irrational $\forall \ n \in \mathbb{N}$.

I have tried though contradiction but can't seem to come up with an answer

I did this where for $\frac ab$, $a \in \mathbb{N}, b \in \mathbb{Z}$ and in reduce form, and $0 \not \in \mathbb{N}$

$$\sqrt 2 + \sqrt n = \frac ab$$
$$\frac {2+n}{\sqrt 2 - \sqrt n} = \frac ab$$
$$\frac {(2+n)(\sqrt 2 + \sqrt n)}{2 + n} = \frac ab$$
$$\frac {(2+n)(a)}{(2-n)(b)}=\frac ab$$
$$\frac {2+n}{2-n}=1$$
$$2+n=2-n$$
$$2+2n=2$$
$$n=0$$
$$n \in \mathbb{N} \therefore n \not =0$$
$$\therefore \sqrt{2} + \sqrt{n} \not \in \mathbb{Q}$$

However this seems clearly flawed as even if $n = 0$ it still would be irrational, and that it subsitiuting $\sqrt 2 + \sqrt n$ for $\frac ab$ seems pretty incorrect.

More just a desperate attempt, pretty confident its not even close.