# Selina Solutions Class 9 Mathematics Solutions for Exercise 1(B) in Chapter 1 - Chapter 1- Rational and Irrational Numbers

Question 6 Exercise 1(B)

Prove that each of the following numbers is irrational:

(i) √𝟑 + √𝟐

(ii) 𝟑 − √𝟐

(iii) √𝟓 − 𝟐

(i) √𝟑 + √𝟐

Let √𝟑 + √𝟐 be a rational number.

⇒√𝟑 + √𝟐 = x

Squaring on both the sides, we get

\begin{array}{l} (\sqrt{3}+\sqrt{2})^{2}=x^{2} \\ \Rightarrow 3+2+2 \times \sqrt{3} \times \sqrt{2}=x^{2} \\ \Rightarrow x^{2}-5=2 \sqrt{6} \\ \Rightarrow \sqrt{6}=\frac{x^{2}-5}{2} \end{array}

Here, x is a rational number.

⇒ x² is a rational number.

⇒ x² - 5 is a rational number.

\frac{x^{2}-5}{2}is a rational number.

But is an irrational number.

is an irrational number.\sqrt{6}

x2- 5 is an irrational number.

⇒ x2 is an irrational number.

⇒ x is an irrational number.

Rational and Irrational Numbers

But we have assume that x is a rational number.

∴ we arrive at a contradiction.

So, our assumption that √𝟑 + √𝟐 is a rational number is wrong.

∴ √𝟑 + √𝟐 is an irrational number.

(ii) 𝟑 − √𝟐

Let 𝟑 − √𝟐 be a rational number.

⇒ 𝟑 − √𝟐 = x

Squaring on both the sides, we get

\begin{array}{l} (3-\sqrt{2})^{2}=x^{2} \\ \Rightarrow 9+2-2 \times 3 \times \sqrt{2}=x^{2} \\ \Rightarrow 11-x^{2}=6 \sqrt{2} \\ \Rightarrow \sqrt{2}=\frac{11-x^{2}}{6} \end{array}

Here, x is a rational number.

⇒ x2 is a rational number.

⇒ 11 - x2 is a rational number.

\frac{11-x^{2}}{6} is also a rational number.

\Rightarrow \sqrt{2}=\frac{11-x^{2}}{6} is a rational number.

But \sqrt{2} is an irrational number.

\Rightarrow \sqrt{2}=\frac{11-x^{2}}{6} is an irrational number.

⇒ 11 - x2 is an irrational number.

⇒ x2 is an irrational number.

⇒ x is an irrational number.

But we have assume that x is a rational number.

∴ we arrive at a contradiction.

So, our assumption that 𝟑 − √𝟐 is a rational number is wrong.

∴ 𝟑 − √𝟐 is an irrational number.

(iii) √𝟓 − 𝟐

Let √𝟓 − 𝟐 be a rational number.

⇒√𝟓 − 𝟐 = x

Squaring on both the sides, we get

$(\sqrt{5}-2)^{2}=x^{2}$ $\Rightarrow 5+4-2 \times 2 \times \sqrt{5}=x^{2}$ $\Rightarrow 9-x^{2}=4 \sqrt{5}$ $\Rightarrow \sqrt{5}=\frac{9-x^{2}}{4}$

Here, x is a rational number.

⇒ x2 is a rational number.

⇒ 9 - x2 is a rational number.

⇒ is also a rational number.

is a rational number.

But is an irrational number.

is an irrational number.

⇒ 9 - x2 is an irrational number.

⇒ x2 is an irrational number.

⇒ x is an irrational number.

But we have assume that x is a rational number.

∴ we arrive at a contradiction.

So, our assumption that √𝟓 − 𝟐 is a rational number is wrong.

∴ √𝟓 − 𝟐 is an irrational number.

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