NCERT

Solution
Class 9 Chapter 1
 Exercise 1.1
𝑝
Question 1: Is zero a rational number? Can you write it in the form𝑞 , where p and q
are integers and q ≠0?

0 0 0
Solution : Yes. Zero is a rational number as it can be represented as1 , 2 , 3 etc.
Question 2: Find six rational numbers between 3 and 4.
Solution : There are infinite rational numbers in between 3 and 4.
21 28
3 and 4 can be represented as and respectively.
7 7

Therefore, rational numbers between 3 and 4 are

22 23 24 25 26 27
, , , , ,
7 7 7 7 7 7
 3 4
Question 3: Find five rational numbers between5and5 .
3 4
Solution : There are infinite rational numbers between5and5.
3 4 18 24
and can be represented as and respectively.
5 5 30 30

Therefore, rational numbers between are

19 20 21 22 23
, , , ,
30 30 30 30 30
Question 4: State whether the following statements are true or false. Give reasons
for your answers.

(i) Every natural number is a whole number.

(ii) Every integer is a whole number.

(iii) Every rational number is a whole number.

Solution : (i) True; since the collection of whole numbers contains all natural
numbers.

(ii) False; as integers may be negative but whole numbers are positive. For example:
−5 is an integer but not a whole number.

(iii) False; as rational numbers may be fractional but whole numbers may not be.
2
For example: is a rational number but not a whole number.
7
 Exercise 1.2

Question 1: State whether the following statements are true or false. Justify your
answers.

(i) Every irrational number is a real number.

(ii) Every point on the number line is of the form √𝑚, where m is a natural number.

(iii) Every real number is an irrational number.

Solution : (i) True; since the collection of real numbers is made up of rational and
irrational numbers.

(ii) False; as negative numbers cannot be expressed as the square root of any other
number.

(iii) False; as real numbers include both rational and irrational numbers. Therefore,
every real number cannot be an irrational number.
Question 2: Are the square roots of all positive integers irrational? If not, give an
example of the square root of a number that is a rational number.

Solution : If numbers such as√4 = 2, √9 = 3 are considered,

Then here, 2 and 3 are rational numbers. Thus, the square roots of all positive
integers are not irrational.
Question 3: Show how√5 can be represented on the number line.
Solution : We know that,

√22 + 12 = √5

Mark a point ‘A’ representing 2 on the number line. Now, construct AB of unit length
perpendicular to OA. Then, taking O as center and OB as radius, draw an arc
intersecting number line at C.

C is representing .
 Exercise 1.3

Question 1: Write the following in decimal form and say what kind of decimal
expansion each has:

36 1 1
(i) (ii) (iii)4
100 11 8

3 2 329
(iv) (v) (vi)
13 11 400

36
Solution : (i)100 = 0.36

Terminating
1
(ii) = 0.09090909. . . . . . . . . = 0. 09
11

Non-terminating repeating
 1
(iii)4 = 4.125
8

Terminating
3
(iv) = 0.230769230769. . . . . . . . = 0. 230769
13

Non-terminating repeating
2
(v) = 0.18181818. . . . . . = 0. 18
11

Non-terminating repeating
329
(vi) = 0.8225
400

Terminating
 1
Question 2: You know that7 = 0. 142857 . Can you predict what the decimal
2 3 4 5 6
expansion of , , , , are, without actually doing the long division? If so, how?
7 7 7 7 7

1
[Hint: Study the remainders while finding the value of carefully.]
7

Solution : Yes. It can be done as follows.

2 1
= 2 × = 2 × 0. 142857 = 0. 285714
7 7
Same as
3
= 0. 428571
7
4
= 0. 571428
7
5
= 0. 714285
7
6
= 0. 857142
7
 𝑝
Question 3: Express the following in the form𝑞 , where p and q are integers and q ≠
0.

(i) 0. 6 (ii)0.47 (iii)0. 001

Solution : (i) 0. 6 = 0.6666. . . ..

Let x = 0.666…

10x = 6.666…

10x = 6 + x

9x = 6
2
x=
3
 4 0.7777
(ii) 0.47 = 0.47777. . . . . . . . = +
10 10

𝐿𝑒𝑡 𝑥 = 0.777. ..

10𝑥 = 7.777 …

10𝑥 = 7 + 𝑥
7
𝑥=
9
(iii) 0. 001 = 0.001001001. ..

𝐿𝑒𝑡 𝑥 = 0.001001 …

1000𝑥 = 1.001001 …

1000𝑥 = 1.001001 …

999𝑥 = 1
1
𝑥=
999
 𝑝
Question 4: Express 0.99999…in the form𝑞 . Are you surprised by your answer?
With your teacher and classmates, discuss why the answer makes sense.

Solution : 𝐿𝑒𝑡 𝑥 = 0.9999 …

10x = 9.9999…

10x = 9 + x

9x = 9

x=1
Question 5: What can the maximum number of digits be in the repeating block of
1
digits in the decimal expansion of ? Perform the division to check your answer.
17

Solution : It can be observed that,

1
= = 0. 05882352941
17

1
There are 16 digits in the repeating block of the decimal expansion of .
17
 𝑝
Question 6: Look at several examples of rational numbers in the form 𝑞 (q≠ 0),
where p and q are integers with no common factors other than 1 and having
terminating decimal representations (expansions). Can you guess what property q
must satisfy?

Solution : Terminating decimal expansion will occur when denominator q of

𝑝
rational number is either 2, 4, 5, 8, 10, and so on…
𝑞

9
= 2.25
4
11
= 1.375
8
27
= 5.4
5
It can be observed that terminating decimals may be obtained in the situation where
prime factorisation of the denominator of the given fractions has the power of 2
only or 5 only or both.
Question 7: Write three numbers whose decimal expansions are non-terminating
non-recurring.

Solution : 3 numbers whose decimal expansions are non-terminating non-recurring

are as follows.

0.505005000500005000005…

0.7207200720007200007200000…

0.080080008000080000080000008…
Question 8: Find three different irrational numbers between the rational numbers
5 9
and .
7 11

Solution :
5
= 0. 714285
7
9
= 0. 81
11
3 irrational numbers are as follows.

0.73073007300073000073…

0.75075007500075000075…

0.79079007900079000079…
Question 9: Classify the following numbers as rational or irrational:

(i) √23

(ii) √225

(iii) 0.3796

(iv) 7.478478

(v) 1.101001000100001…

Solution :

(i)√23 = 4.79583152331. . ..

As the decimal expansion of this number is non-terminating non-recurring,

therefore, it is an irrational number.

(ii)√225 = 15
It is a rational number as it can be represented in form.
(iii) 0.3796

As the decimal expansion of this number is terminating, therefore, it is a rational

number.

(iv) 7.478478 …=7. 478

As the decimal expansion of this number is non-terminating, therefore, it is a

rational number.

(v) 1.10100100010000 …

As the decimal expansion of this number is non-terminating and non-repeating,

therefore, it is an irrational number.
 Exercise 1.4

Question 1: Visualize 3.765 on the number line using successive magnification.

Solution : 3.765 can be visualized as in the following steps.
Question 2: Visualize on the number line,
up to 4. 26 decimal places.

Solution :
4. 26 = 4.2626…

4.2626 can be visualized as in the

following steps.
 Exercise 1.5

Question 1: Classify the following numbers as rational or irrational:

(i)2 − √5 (ii)(3 + √23) − √23 2√7

(iii)
7√7
1
(iv) (v)2𝜋
√2

Solution :

(i)2 − √5 = 2 − 2.2360679…

= − 0.2360679…

As the decimal expansion of this expression is non-terminating non-recurring,

therefore, it is an irrational number.

(ii)(3 + √23) − √23 = 3

𝑝
As it can be represented in form, therefore, it is a rational number.
𝑞
 2√7 2
(iii) = x
7√7 7
𝑝
As it can be represented in form, therefore, it is a rational number.
𝑞

1 √2
(iv) = = 0.70710678118. . . ..
√2 2

As the decimal expansion of this expression is non-terminating non-recurring,

therefore, it is an irrational number.

(v) 2π = 2(3.1415 …)

= 6.2830 …

As the decimal expansion of this expression 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)(3 + √3)(2 + √2) = 3(2 + √2) + √3(2 + √2)

6 + 3√2 + 2√3 + √6

(ii)(3 + √3)(3 − √3) = (3)2 + (√3)2

=9−3=6

(iii)(√5 + √2)2 = (√5)2 + (√2)2 + 2√10

= 5 + 2 + 2√10 = 7 + 2√10
(iv)(√5 − √2)(√5 + √2) = (√5)2 + (√2)2

=5−2=3
Question 3: Recall, π is defined as the ratio of the circumference (say c) of a circle to
𝑐
its diameter (say d). That is,𝜋 = . This seems to contradict the fact that π is
𝑑
irrational. How will you resolve this contradiction?

Solution : There is no contradiction. When we measure a length with scale or any

other instrument, we only obtain an approximate rational value. We never obtain an
exact value. For this reason, we may not realize that either c or d is irrational.
𝑐
Therefore, the fraction is irrational. Hence, π is irrational.
𝑑
Question 4: Represent√9.3 on the number line.

Solution : Mark a line segment OB = 9.3 on number line. Further, take BC of 1 unit.
Find the midpoint D of OC and draw a semi-circle on OC while taking D as its center.
Draw a perpendicular to line OC passing through point B. Let it intersect the semi-
circle at E. Taking B as center and BE as radius, draw an arc intersecting number line
at F. BF is√9.3.
Question 5: Rationalize the denominators of the following:
1
(i)
√7

1
(ii)
√7−√6

1
(iii)
√5+√2

1
(iv)
√7−2

Solution :
1 1×√7 √7
(i) = =
√7 √7×√7 7

1 1(√7+√6)
(ii) =
√7−√6 (√7−√6)(√7+√6

1(√7 + √6) 1(√7 + √6)

=
(7)2 − (6)2 7−6
= 1(√7 + √6)
 1 1(√5−√2)
(iii) =
√5+√2 (√5+√2)(√5−√2)

1(√5 − √2) 1(√5 − √2)

=
(√5)2 − (√2)2 5−2

√5 − 2
=
3
1 1(√7+2)
(iv) =
√7−2 (√7−2)(√7+2)

1(√7 + 2) (√7 + 2)
=
(√7)2 − (2)2 7−4

(√7 + 2)
3
 Exercise 1.6

Question 1: Find:
1
(i) 64 2

1
(ii) 32 5

1
(iii) 125 3

Solution :
1 1
(i) 64 = 2 (26 )2
1
6×
2 2 = 23 = 8
1 1
(ii) 32 = 5 (25 )5
 1
5×
2 5 = 21 = 2
1 1
(iii) 125 =
3 (53 )3
1
3×
5 3 = 51 = 5
Question 2: Find:
3
(i) 9 2

2
(ii) 32 5

3
(iii) 16 4

−1
(iv) 125 3

Solution :
3 3
(i) 9 = 2 (32 )2
3
2×
3 2 = 33 = 27
2 2
(ii) 32 = 5 (25 )5
2
5×
2 5 = 22 = 4
 3 3
(iii) 16 =
4 (24 )4
3
4×
2 4 = 23 = 8
−1 −1
(iv) 125 3 = (53 ) 3

3×
−1 1
5 3 = 5−1 =
5
Question 3: Simplify:
2 1
(i) 2 . 2
3 5

1 7
(ii) ( 3 )
3
1
(iii) 112
1114

1 1
(iv) 72 . 82

Solution :
2 1 2 1
+
(i) 2 . 2 = 2
3 5 3 5

10 + 3 10
2 = 215
15
1 7 1
(ii) ( 3 ) =
3 33×7
 1
21 = 3−21
3
1 1−1
(iii) 112 = 11 2 4
1114

4−2 2 1
11 8 = 118 = 114

11−4
1 1 1
(iv) 7 . 8
2 2 = (7 × 8)2

1
562