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

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) = a2 – b2
∴ (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 = a2 + b2 + 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) = a2 – b2]
= 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
$$\frac{1}{\sqrt{7}-2}=\frac{1}{\sqrt{7}-2} \times \frac{\sqrt{7}+2}{\sqrt{7}+2}$$
= $$\frac{\sqrt{7}+2}{7-4}$$
= $$\frac{\sqrt{7}+2}{3}$$

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