(i) $p: 125$ is divisible by 5 and 7 .

Let $q: 125$ is divisible by 5 .

$r: 125$ is divisible by 7 .

$q$ is true, $r$ is false.

$\Rightarrow q \wedge r$ is false.

[since, $p \wedge q$ has the truth value $F$ (false) whenever either $p$ or $q$ or both have the truth value F.]

Hence, $p$ is not valid.

(ii) $p: 131$ is a multiple of 3 or 11 .

Let $q: 131$ is multiple of 3 .

$r: 131$ is a multiple of 11 .

$p$ is true, $r$ is false.

$\Rightarrow \quad p \vee r$ is true. [since, $p \vee q$ has the truth value $T$ (true) whenever either $p$ or $q$ or both have the truth value T]

Hence, $q$ is valid.

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.