Prove that

Solution:

$\frac{2}{\sqrt{7}}=\frac{2}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}}=\frac{2}{7} \sqrt{7}$

Let $\frac{2}{7} \sqrt{7}$ is a rational number.

$\therefore \frac{2}{7} \sqrt{7}=\frac{p}{q}$, where $p$ and $q$ are some integers and $\operatorname{HCF}(p, q)=1$

$\Rightarrow 2 \sqrt{7} q=7 p$

$\Rightarrow(2 \sqrt{7} q)^{2}=(7 p)^{2}$

$\Rightarrow 7\left(4 q^{2}\right)=49 p^{2}$

$\Rightarrow 4 q^{2}=7 p^{2}$

$\Rightarrow q^{2}$ is divisible by 7

$\Rightarrow q$ is divisible by 7   .........(2)

Let q = 7m, where m is some integer.

$\therefore 2 \sqrt{7} q=7 p$

$\Rightarrow[2 \sqrt{7}(7 m)]^{2}=(7 p)^{2}$

$\Rightarrow 343\left(4 m^{2}\right)=49 p^{2}$

$\Rightarrow 7\left(4 m^{2}\right)=p^{2}$

$\Rightarrow p^{2}$ is divisible by 7

$\Rightarrow p$ is divisible by 7   .......(3)

From (2) and (3), 7 is a common factor of both p and q, which contradicts (1).
Hence, our assumption is wrong.

Thus, $\frac{2}{\sqrt{7}}$ is irrational.