Prove by direct method that for any integer ' $n$ ', $n^3-n$ is always even.
Here, two cases arise
Case I When $n$ is even,
$$\begin{aligned} & \text { Let } \quad n=2 K, K \in N \\ & \Rightarrow \quad n^3-n=(2 K)^3-(2 K)=2 K\left(4 K^2-1\right) \\ & =2 \lambda \text {, where } \lambda=K\left(4 K^2-1\right) \end{aligned}$$
Thus, $\left(n^3-n\right)$ is even when $n$ is even.
Case II When $n$ is odd,
$$\begin{aligned} \text{Let}\quad n & =2 K+1, K \in N \\ \Rightarrow \quad & n^3-n & =(2 K+1)^3-(2 K+1) \\ & =(2 K+1)\left[(2 K+1)^2-1\right] \\ & =(2 K+1)\left[4 K^2+1+4 K-1\right] \\ & =(2 K+1)\left(4 K^2+4 K\right) \\ & =4 K(2 K+1)(K+1) \\ & =2 \propto, \text { when } \alpha=2 K(K+1)(2 K+1) \end{aligned}$$
Then, $n^3-n$ is even when $n$ is odd.
So, $n^3-n$ is always even.
Check validity of the following statement.
(i) $p: 125$ is divisible by 5 and 7.
(ii) $q: 131$ is a multiple of 3 or 11 .
(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.
Prove the following statement by contradiction method
$p$ : The sum of an irrational number and a rational number is irrational.
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.
Prove by direct method that for any real number $x, y$ if $x=y$, then $x^2=y^2$.
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.
Using contrapositive method prove that, if $n^2$ is an even integer, then $n$ is also an even integer.
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.