Let $p$ is false i.e., sum of an irrational and a rational number is rational.

Let $\sqrt{m}$ is irrational and $n$ is rational number.

$\Rightarrow\quad\sqrt{m}+n=r\quad \text{[rational]}$

$\Rightarrow\quad\sqrt{m}=r-n$

$\sqrt{\mathrm{m}}$ is irrational, where as $(r-n)$ is rational. This is contradiction.

Then, our supposition is wrong.

Hence, $p$ is true.

Let $p: x=y, \quad x, y \in R$

On squaring both sides,

$$\begin{gathered} x^2=y^2: q \quad\text{[say]}\\ p \Rightarrow q \end{gathered}$$

Hence, we have the result.

Let $p: n^2$ is an even integer.

$q: n$ is also an even integer.

Let $\sim p$ is true i.e., $n$ is not an even integer.

$\Rightarrow n^2$ is not an even integer. $\quad$ [since, square of an odd integer is odd]

$\Rightarrow \sim p$ is true.

Therefore, $\sim q$ is true $\Rightarrow \sim p$ is true.

Hence proved.