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Let $\frac{\sqrt{2}}{3}$ is a rational number.

$\therefore \frac{\sqrt{2}}{3}=\frac{p}{q}$, where $p$ and $q$ are some integers and $\operatorname{HCF}(p, q)=1$   .......(1)

$\Rightarrow \sqrt{2} q=3 p$

$\Rightarrow(\sqrt{2} q)^{2}=(3 p)^{2}$

$\Rightarrow 2 q^{2}=9 p^{2}$

⇒ p2 is divisible by 2
⇒ p is divisible by 2   ....(2)

Let p = 2m, where m is some integer.

$\therefore \sqrt{2} q=3 p$

$\Rightarrow \sqrt{2} q=3(2 m)$

$\Rightarrow(\sqrt{2} q)^{2}=(3(2 m))^{2}$

$\Rightarrow 2 q^{2}=4\left(9 p^{2}\right)$

$\Rightarrow q^{2}=2\left(9 p^{2}\right)$

⇒ q2 is divisible by 2
⇒ q is divisible by 2   ....(3)

From (2) and (3), 2 is a common factor of both p and q, which contradicts (1).
Hence, our assumption is wrong.

Thus, $\frac{\sqrt{2}}{3}$ is irrational.