Is this a correct proof of $r+\sqrt{2}\in\Bbb{Q}$?

Question

My proof goes as follows:

Suppose that $r+\sqrt{2}$ is rational, then $r+\sqrt{2}=p/q\leftrightarrow2=(\dfrac{p-rq}{q})^2$. Hence $\dfrac{p-rq}{q} | 2$ thus $\dfrac{p-rq}{q}=2k$ where $k\in\Bbb{N}$. Therefore we have $2=(2k)^2=4k^2\leftrightarrow (k=\sqrt{1/2}\lor k=-\sqrt{1/2})$. Due to the contradiction $k\in\Bbb{N}\land k\notin\Bbb{N}$, we conclude that $r+\sqrt{2}$ is irrational.

Answer 1

You can actually use this property:

If $r+x\in\mathbb{Q}$ for $r\in\mathbb{Q}$ then also $x\in \mathbb{Q}$.

From this it is easy to see that for your theorem it is sufficient to prove that $\sqrt{2}$ is not rational, which is a well-known fact, that you can find here, for example Irrational number $\sqrt2$ proof.

Answer 2

Suppose that $r+\sqrt{2}$ is rational, then $r+\sqrt{2}=p/q\leftrightarrow2=(\dfrac{p-rq}{q})^2$. Hence $\dfrac{p-rq}{q} | 2$...

No, because you don't know that $\dfrac{p-rq}{q}$ is even an integer. It's just a rational number.