**Question:**

Check the validity of the statements given below by the method given against it.

Check the validity of the statements given below by the method given against it.

(i) *p*: The sum of an irrational number and a rational number is irrational (by contradiction method).

(ii) $q$ : If $n$ is a real number with $n>3$, then $n^{2}>9$ (by contradiction method).

**Solution:**

(i) The given statement is as follows.*p*: the sum of an irrational number and a rational number is irrational.

Let us assume that the given statement, *p*, is false. That is, we assume that the sum of an irrational number and a rational number is rational.

Therefore, $\sqrt{a}+\frac{b}{c}=\frac{d}{e}$, where $\sqrt{a}$ is irrational and $b, c, d$, e are integers.

$\Rightarrow \frac{d}{e}-\frac{b}{c}=\sqrt{a}$

But here, $\frac{d}{e}-\frac{b}{c}$ is a rational number and $\sqrt{a}$ is an irrational number.

But here, $\frac{d}{a}-\frac{b}{\rho}$ is a rational number and $\sqrt{a}$ is an irrational number.

This is a contradiction. Therefore, our assumption is wrong.

Therefore, the sum of an irrational number and a rational number is rational.

Thus, the given statement is true.

(ii) The given statement, *q*, is as follows.

If $n$ is a real number with $n>3$, then $n^{2}>9$.

Let us assume that $n$ is a real number with $n>3$, but $n^{2}>9$ is not true.

That is, $n^{2}<9$

Then, $n>3$ and $n$ is a real number.

Squaring both the sides, we obtain

$n^{2}>(3)^{2}$

$\Rightarrow n^{2}>9$, which is a contradiction, since we have assumed that $n^{2}<9$.

Thus, the given statement is true. That is, if $n$ is a real number with $n>3$, then $n^{2}>9$.

Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.