Let $\sim$ be a relation on $\mathbb R$ by $x \sim y \iff x-y \in\mathbb Q$. Show that $\sqrt{3} \notin \left[\sqrt{2}\right]$.
We will use the fact that $\sqrt{6}$ is irrational. (Small proof) Suppose $\sqrt 6 = \frac ab,$ is rational and $\gcd (a,b)=1$. Then we have $\sqrt 6 = \frac ab \implies 6 = \frac {a^2}{b^2} \implies 2(3b^2) = a^2 \implies a=2k, k\in\mathbb Z$. We have $2(3b^2) = 4k^2 \implies 3b^2 = 2k^2 \implies b$ is even, a contradiction.
Assume that $\sqrt{3} \sim \sqrt{2} \implies \sqrt{3} - \sqrt{2} \in \mathbb Q$. Let $\sqrt{3} - \sqrt{2} = \frac ab$ for $a,b\in\mathbb Z$. Squaring both sides we have $\frac{a^2}{b^2} = 3 - 2\sqrt{6} + 2$. By "Small proof" above, $\sqrt{6}$ is irrational. Thus, we have an irrational number on the right and a rational on the left. A contradiction. Thus $\sqrt 3 \notin \left[\sqrt{2}\right]$.
Is this a valid argument/proof? In addition, is it correct to say $\left[\sqrt{2}\right] = \left\{\sqrt 2\right\}$? (No, refer to comments below)