NCERT Solutions
Class 9 - Mathematics - Chapter 1: NUMBER SYSTEMS
EXERCISE 1.4

Open NCERT PDFOpen NCERT Answers PDF

Question. 1

Classify the following numbers as rational or irrational:

(i) \(2 - \sqrt{5}\)
(ii) \((3 + \sqrt{23}) - \sqrt{23}\)
(iii) \(\dfrac{2\sqrt{7}}{7\sqrt{7}}\)
(iv) \(\dfrac{1}{\sqrt{2}}\)
(v) \(2\pi\)

Answer:

(i) Irrational

(ii) Rational

(iii) Rational

(iv) Irrational

(v) Irrational

Question. 2

Simplify each of the following expressions:

(i) \((3 + \sqrt{3})(2 + \sqrt{2})\)
(ii) \((3 + \sqrt{3})(3 - \sqrt{3})\)
(iii) \((\sqrt{5} + \sqrt{2})^2\)
(iv) \((\sqrt{5} - \sqrt{2})(\sqrt{5} + \sqrt{2})\)

Answer:

(i) \(6 + 3\sqrt{2} + 2\sqrt{3} + \sqrt{6}\)

(ii) 6

(iii) \(7 + 2\sqrt{10}\)

(iv) 3

Question. 3

Recall that \(\pi\) is defined as the ratio of the circumference (say \(c\)) of a circle to its diameter (say \(d\)). That is, \( \pi = \dfrac{c}{d} \). This seems to contradict the fact that \(\pi\) is irrational. How will you resolve this contradiction?

Answer:

There is no contradiction. Measurement using a scale only gives approximate rational values, so you may not realise that either \(c\) or \(d\) is irrational.

Question. 4

Represent \(\sqrt{9.3}\) on the number line.

Answer:

Refer Fig. 1.17.

Question. 5

Rationalise the denominators of the following:

(i) \(\dfrac{1}{\sqrt{7}}\)
(ii) \(\dfrac{1}{\sqrt{7} - \sqrt{6}}\)
(iii) \(\dfrac{1}{\sqrt{5} + \sqrt{2}}\)
(iv) \(\dfrac{1}{\sqrt{7} - 2}\)

Answer:

(i) \(\dfrac{\sqrt{7}}{7}\)

(ii) \(\sqrt{7} + \sqrt{6}\)

(iii) \(\dfrac{\sqrt{5} - \sqrt{2}}{3}\)

(iv) \(\dfrac{\sqrt{7} + 2}{3}\)