Verify my proof of the irrationality of $\sqrt{6}$. Here's my proof for the irrationality of $\sqrt{6}$:
Proof:
Let $\sqrt{6}$ be a rational number. Then
$\frac{a}{b}$ = $\sqrt{6}$ , where $a$ and $b$ are coprime.
Therefore,

$a^2 = 6 b^2$.
Here $a^2$ is divisible by both $2$ and $3$. Which means, $a$ too is divisible by both $2$ and $3$ (fundamental theorem of arithmetic).
Thus, $6b^2=(3\cdot2\cdot k)^2$ ; where $k$ is an integer.
$6b^2=36k^2$ or
$b^2=6k^2$.
Here $b^2$ is divisible by both $2$ and $3$. Which means, $b$ too is divisible by both $2$ and $3$. Therefore, $a$ and $b$ have common factors, which is a contradiction. Therefore $\sqrt{6}$ is not rational.

Is the above proof correct? Is there an easier way to prove this?

That works fine, but you don't need to do both $2$ and $3$. If:
$$a^2=6b^2$$
Then $a$ must be even, $a=2a_0$, and then: $4a_0^2=6b^2$ or $$2a_0^2=3b^2.$$
So $3b^2$ must be even, so by the fundamental theorem of arithmetic, $b$ must be even, and you are done.
