Prove that

Solution:

Let $5 \sqrt{2}$ is a rational number.

$\therefore 5 \sqrt{2}=\frac{p}{q}$, where $p$ and $q$ are some integers and $\operatorname{HCF}(p, q)=1$     ......(1)

$\Rightarrow 5 \sqrt{2} q=p$

$\Rightarrow(5 \sqrt{2} q)^{2}=p^{2}$

$\Rightarrow 2\left(25 q^{2}\right)=p^{2}$

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

$\Rightarrow p$ is divisible by 2 .....(2)

Let p = 2m, where m is some integer.

$\therefore 5 \sqrt{2} q=2 m$

$\Rightarrow(5 \sqrt{2} q)^{2}=(2 m)^{2}$

$\Rightarrow 2\left(25 q^{2}\right)=4 m^{2}$

$\Rightarrow 25 q^{2}=2 m^{2}$

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

$\Rightarrow q$ 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, $5 \sqrt{2}$ is irrational.