These NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Ex 1.5 Questions and Answers are prepared by our highly skilled subject experts.

## NCERT Solutions for Class 9 Maths Chapter 1 Number Systems Exercise 1.5

Question 1.

Classify the following numbers as rational or irrational:

(i) 2 – √5

(ii) (3 + √23) – √23

(iii) \(\frac{2 \sqrt{7}}{7 \sqrt{7}}\)

(iv) \(\frac{1}{\sqrt{2}}\)

(v) 2π

Solution:

(i) We have, 2 – √5

or, 2 – 2.236067977…….

or, -0.236067977……..

which is non terminating non recurring.

So, it is an irrational number.

(ii) We have, (3 + √23) – √23

or, 3 + √23 – √23

or, 3

which is a rational number.

(iii) We have, \(\frac{2 \sqrt{7}}{7 \sqrt{7}}\)

or \(\frac{2}{7}\)

which is a rational number.

(iv) We have, \(\frac{1}{\sqrt{2}}\)

or, \(\frac{1}{1.414213562 \ldots}\)

Therefore the quotient of this number is irrational.

(v) We have,

2π = 2 × 3.1415926535……. = 6.2831853070……..

which is non-terminating non-recurring.

Therefore, it is an irrational number.

Question 2.

Simplify each of the following expressions:

(i) (3 + √3) (2 + √2)

(ii) (3+ √3) (3 – √3)

(iii) (√5 + √2)^{2}

(iv) (√5 – √2) (√5 + √2)

Solution:

(i) We have, (3 + √3) (2 + √2)

or 6 + 3√2 + 2√3 + √6

(ii) We have, (3 + √3) (3 – √3)

We know that (a + b) (a – b) = a^{2} – b^{2}

∴ (3 + √3) (3 – √3) = (3)^{2} – (√3)^{2}

= 9 – 3

= 6

So, (3 + √3) (3 – √3) = 6

(iii) We have, (√5 + √2)^{2}

We know that (a + b)^{2} = a^{2} + b^{2} + 2ab

∴ (√5 + √2)^{2} = (√5)^{2} + (√2)^{2} + 2(√5)(√2) = 5 + 2 + 2√10

or, (√5 + √2)^{2} = 7 + 2√10

(iv) We have,

(√5 – √2) (√5 + √2) = (√5)^{2} – (√2)^{2}

[∴ (a + b) (a – b) = a^{2} – b^{2}]

= 5 – 2

= 3

∴ (√5 – √2) (√5 + √2) = 3

Question 3.

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

Solution:

Circumference = 2πR (Irrational number) of the circle (c)

R → Radius of the circle

Diameter of circle (D) = 2R (Rational number)

\(\frac{C}{D}=\frac{2 \pi R}{2 R}=\pi=\frac{\text { Irrational number }}{\text { Rational number }}\) = Irrational number

As we know irrational number divided by a rational number results in an irrational number.

Question 4.

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

Solution:

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

We mark a point B on the number line so that AB = 9.3 units.

Again mark a point C so that BC = 1 unit.

Now take the midpoint of AC and mark that point as O.

Draw a semicircle with centre O and radius OC.

Draw the line perpendicular to AC passing through B and intersecting the semicircle at D. Then, BD = \(\sqrt{9.3}\)

Again taking BD as a radius and draw an arc which intersecting the number line at E.

Then E represents \(\sqrt{9.3}\) on number line.

Take b at zero on number line, then point E represent \(\sqrt{9.3}\).

Question 5.

Rationalise the denominators of the following:

(i) \(\frac{1}{\sqrt{7}}\)

(ii) \(\frac{1}{\sqrt{7}-\sqrt{6}}\)

(iii) \(\frac{1}{\sqrt{5}+\sqrt{2}}\)

(iv) \(\frac{1}{\sqrt{7}-2}\)

Solution:

(i) We have, \(\frac{1}{\sqrt{7}}\)

Multiply √7 both numerator and denominator.

\(\frac{1}{\sqrt{7}} \times \frac{\sqrt{7}}{\sqrt{7}}=\frac{\sqrt{7}}{7}\)

(ii) We have, \(\frac{1}{\sqrt{7}-\sqrt{6}}\)

Multiply √7 + √6 both numerator and denominator

\(\frac{1}{\sqrt{7}-\sqrt{6}} \times \frac{\sqrt{7}+\sqrt{6}}{\sqrt{7}+\sqrt{6}}=\frac{\sqrt{7}+\sqrt{6}}{7-6}\)

= \(\frac{\sqrt{7}+\sqrt{6}}{1}\)

= √7 + √6

(iii) We have, \(\frac{1}{\sqrt{5}+\sqrt{2}}\)

Multiply √5 – √2 both numerator and denominator

\(\frac{1}{\sqrt{5}+\sqrt{2}} \times \frac{\sqrt{5}-\sqrt{2}}{\sqrt{5}-\sqrt{2}}=\frac{\sqrt{5}-\sqrt{2}}{5-2}=\frac{\sqrt{5}-\sqrt{2}}{3}\)

(iv) We have

\(\frac{1}{\sqrt{7}-2}=\frac{1}{\sqrt{7}-2} \times \frac{\sqrt{7}+2}{\sqrt{7}+2}\)

= \(\frac{\sqrt{7}+2}{7-4}\)

= \(\frac{\sqrt{7}+2}{3}\)