 Lesson: Rational Number
Exercise 1.1
Question 1
Using appropriate properties, find:
(i)
2 3 5 3 1
3 5 2 5 6
(ii)
2 3 1 3 1 2
5 7 6 2 14 5



(i) 2 (ii)
11
28
Solution:
(i)
2 3 5 3 1
3 5 2 5 6
=
2 3 3 1 5 3 2 1 5
3 5 5 6 2 5 3 6 2








[Using a (b
c) = ab - ac]
3 4 1 5 3 5 5 1 5 1 5 4
2
5 6 2 5 6 2 2 2 2 2




(ii)
2 3 1 3 1 2
5 7 6 2 14 5



=
2 3 1 2 1 3
5 7 14 5 6 2









=
2 3 1 1 3
5 7 14 6 2





[Using a (b + c) = ab + ac]
1 1 4 7 11
7 4 28 28 Question 2
Write the additive inverse of each of the following:
(i)
2
8
(ii)
5
9
(iii)
6
5
(iv)
2
9
(v)
19
6
(i)
2
8
(ii)
5
9
(iii)
6
5
(iv)
2
9
(v)
19
6
Solution:
2
8
is
2
8
as
2 2 2 2 0
0
8 8 8 8
5
9
is
5
9
as
5 5 5 5 0
0
9 9 9 9





6
5
is
6
5
as
6 6 6 6 6 6 0
0
5 5 5 5 5
5




2
9
is
2
9
as
2 2 2 2 2 2 0
0
9 9 9 9 9
9








19
6
is
19
6 as
19 19 19 19 19 19 0
0
6 6 6 6 6
6








Question 3
Verify that ( x) = x for : (i) x =
11
15
(ii) x =
13
17
i.e, ( x) = x. is verified
Solution:
(i) We have,
x =
11
15
The additive inverse of x =
11
15
is x =
11
15
as
11
15
+
11
15




= 0
The same equality
11
15
+
11
15




= 0 shows that the additive inverse of
11
15
is
11
15
.
This
11
15




=
11
15
, i.e, ( x ) = x. is verified.
(ii) We have,
x =
13
17
The additive inverse of x =
13
17
is x =
13
17
as
13
17
+
13
17
= 0.
13
17
=
(
13
17
)
The same equality
13
17
+
13
17
= 0 shows that the additive inverse of
13
17 is
13
17
. This
13 13
( ( ))
17 17

i.e, ( x) = x. is verified.
Question 4
Find the multiplicative inverse of the following
(i) 13 (ii)
13
19
(iii)
1
5
(iv)
53
87

(v) 1 ×
2
5
(vi) 1
(i)
1
13
(ii)
19
13
(iii) 5 (iv)
56
15
(v)
5
2
(vi) 1
Solution:
(i) The multiplicative inverse of 13 is
1
13
as
1
13 1
13
(ii) The multiplicative inverse of
13
19
is
19
13
as
13 19
1
19 13
(iii) The multiplicative inverse of
1
5
is 5 as
1
51
5

.
(iv) The multiplicative inverse of
5 3 8 7
is
87
53



=
56
15
as
5 3 8 7
1
87
53










(v) The multiplicative inverse of 1 ×
2
5
is 1 ×
5
2
=
5
2
as
(vi) The multiplicative inverse of 1 is 1 as
1 1 1
. 2
1 1
2
5
5
1







Question 5
Name the property under multiplication used in each of the following:
(i)
4
5
× 1 = 1 ×
4
5
=
4
5
(ii)
13 2 2 13
17 7 7 17
(iii)
19 29
1
29
19

(i) Existence of Multiplicative Identity (1 is the multiplicative identity)
(ii) Commutatively
(iii) Existence of Multiplicative Inverse
Solution:
(i)
4
5
× 1 = 1 ×
4
5
=
4
5
as 1is the multiplicative identity.
(ii)
13 2 2 13
17 7 7 17
as multiplication is commutative.
(iii)
19 29
1
29
19

as
29
19
is the multiplicative inverse of
19
29
.
Question 6
Multiply
6
13
by the reciprocal of
7
16
.
96
91 Solution:
Reciprocal of
7 16
is
16
7
.
Now,
6 16 96
13 91
7




Question 7
What property allows you to compute?
1 4 1 4
6 as 6
3 3 3 3
Associativity of Multiplication
Solution:
For any three rational numbers a, b and c, a × (b × c) = (a × b) × c.
The multiplication is associative for rational numbers.
The associativity property allows us to compute
1 4 1 4
6 as 6 ,
3 3 3 3







1 4 1 24 8
× 6 ×
3 3 3 3 3



and
1 4 6 4 8
6
3 3 3 3 3



Here,
1 4 1 4
66
3 3 3 3
Question 8
Is
8
9
the multiplicative inverse of 1
1
8
? Why or why not?
No, because the product is not 1
Solution:
No, because the product of
8
9
and 1
1
8
is not 1 Question 9
Is 0.3 the multiplicative inverse of
1
3
3
? Why or Why not?
Yes, because
10 3 10
0.3 1
3 10 3
Solution:
Yes, because
1 10
3
33
and
10 3 10
0.3 1
3 10 3
Question 10
Write:
(i) The rational number that does not have a reciprocal.
(ii) The rational numbers that are equal to their reciprocals.
(iii) The rational number that is equal to its negative.
(i) 0 (ii) 1 and
1 (iii) 0
Solution:
(i) Zero has no reciprocal.
(ii) 1 is equal to its reciprocal as
1 1 1
.
1 is also equal to its reciprocal as 1 × ( 1) = 1
(iii) Zero is equal to its negative as
0 0 0
.
Question 11
Fill in the blanks:
(i) Zero has ____________ reciprocals.
(ii) The numbers ____________ and ____________ are their own reciprocals.
(iii) The reciprocal of 5 is ____________ (iv) Reciprocal of
1
x
, where x
0 is ____________
(v) The product of two rational numbers is always a ____________
(vi) The reciprocal of a positive rational number is ____________
(i) No (ii) 1, 1 (iii)
1
5
(iv) x
(v) Rational number
(vi) Positive
Solution:
(i) No (ii) 1, 1 (iii)
1
5
(iv) x (v) rational number
(vi) a positive rational number.
Exercise 1.2
Question 1
Represent these numbers on the number line (i)
7
4
(ii)
5
6
(i) (ii)
Solution:
(i) To represent
7
4
, we make 7 markings of distance
1
4
each on the right of 0.
Thus, the point A represents
7
4
.
(ii) To represent
5
6
, we make 6 markings of distance
1
6
each on the left of zero.
Thus, the point B represents
5
6
. Question 2
Represent
2 5 9
,,
11 11 11
on the number line.
Solution:
We draw a line l. We take a point O on it. From O, we mark 11 markings of distance
1
11
each on the left of O
(i.e., 0)
Thus, the points A, B and C represent
2 5 9
, and
11 11 11
respectively. Question 3
Write five rational numbers which are smaller than 2.
Some of these are
11
1, , 0, 1,
22
(There can be many more such rational numbers)
Solution:
Let’s take the two numbers 0 and 2. 0 is smaller than 2.
Now, 2 can be written as
20
10
and 0 as
0
10
.
Thus, we have
19 18 17 16 15 14 1
, , , , , .........
10 10 10 10 10 10 10
between 2 and 0.
We can take any five of these values.
Question 4
Find ten rational numbers between
2
5
and
1
2
.
7 6 5 8 9
, , , ........ ,
20 20 20 20 20
(There can be many more such rational numbers)
Solution:
First, we make the same denominator of the given rational numbers.
2 2 2 4 4 2 8
5 10 20
5 2 10 2

and
1 1 5 5 5 2 10
2 10 20
2 5 10 2

 Note: We make denominator 20 because when the denominator is 10, then we
can conveniently find out only 8 rational numbers.
Thus, we have
7 6 5 8 9
, , , ........ ,
20 20 20 20 20
We can take any ten of these values.
Question 5
Find five rational numbers between
(i)
24
and
35
(ii)
35
and
23
(iii)
11
and
42
(i)
41 7 11 23
, , ,
60 10 15 30
(ii)
8 7 6 5 4 8 9
, , , , ..... ,
6 6 6 6 6 6 6
(iii)
11 10 9 8 7
,,,,
24 24 24 24 24
(There can be many more such rational numbers)
Solution:
(i)
24
and
35
First Method:
We first convert
2
3
and
4
5
to rational numbers with the same denominator i.e.
2 2 5 10 10 3 30
3 15 45
3 5 15 3


and
4 4 3 12 12 3 36
5 15 45
5 3 15 3


Thus, we have,
35 34 33 32 31
, , , and
45 45 45 45 45
as five ration numbers between
24
and
35
Another Method:
We know that, if a and b are two rational numbers, then
ab
2
is a rational
number between a and b such that a <
ab
2
< b. We find the mean of the given rational numbers,
2 4 10 12 1 22 1 11
2
3 5 15 2 15 2 15








So,
2 11 4
3 15 5

We now find another rational number between
2 11
and
3 15
.
For this, we again find the mean of
2 11
and
3 15
, i.e.,
2 11 30 33 1 63 1 7 1 7
2
3 15 45 2 45 2 5 2 10








So,
2 7 11
3 10 15

or
2 7 11 4
3 10 15 5
Further, we find another rational number between
11
15
and
4
5
For this, we again find the mean of
11
15
and
4
5
i.e,
11 4 11 12 1 23 1 23
2
15 5 15 2 15 2 30








.
So,
11 23 4
15 30 5

or
2 7 11 23 4
3 10 15 30 5
Now, we find another rational number between
2
3
and
7
10
For this, we again find the mean of
2
3
and
7
10
,
i.e,
2 7 20 21 1 41 1 41
2
3 10 30 2 30 2 60








So,
2 41 7
3 60 10

or
2 41 7 11 23 4
3 60 10 15 30 5
Also, we find another rational number between
23
30
and
4
5 For this, we again find the mean of
23
30
and
4
5
, i.e.
23 4 1
30 5 2





23 24 1 47 1 47
30 2 30 2 60




So,
23 47 4
30 60 5

or
2 41 7 11 23 47 4
3 60 10 15 30 60 5
Thus, the five rational numbers between
2
3
and
4
5
are
41 7 11 23
, , ,
60 10 15 30
and
47
60
(ii)
35
and
23
We first convert
35
and
23
to rational numbers with the same denominator
i.e.,
3 3 3 9
26
23

and
5 5 2 10
36
23

Thus, we have
8 7 6 5 4 8 9
, , , , ..... ,
6 6 6 6 6 6 6
We can take any five of these values as rational numbers between
35
and
23
.
(iii)
1
4
and
1
2
We first convert
1
4
and
1
2
to rational numbers with the same denominator
i.e.,
1 1 6 6
4 24
46

and
1 1 12 12
2 24
2 12

.
Thus, we have,
11 10 9 8 7
, , , and
24 24 24 24 24
as rational numbers between
1
4
and
1
2
. Question 6
Write five rational numbers greater than 2.
-1. 0. 1, 2, 3 (There can be many more such rational numbers)
Solution:
Since, every integer is a rational number, therefore, we can pick up any five integers
greater than -2.
Thus five rational numbers greater than -2 are -1, 0, 1, 2, 3.
Question 7
Find ten rational numbers between
3
5
and
3
4
.
59 58 57 56 55 54 53 52 51 50 49
, , , , , , , , , ,
80 80 80 80 80 80 80 80 80 80 80
(There can be many more such rational numbers)
Solution:
First we make the denominator of the given rational numbers equal
3
5
=
3 4 12 24 48
20 40 80
54
and
3
4
=
3 5 15 30 60
20 40 80
45

Thus, we have,
59 58 57 56 55 54 53 52 51 50 49
, , , , , , , , , ,
80 80 80 80 80 80 80 80 80 80 80
as rational numbers between
3
5
and
3
4
.
We can take any ten of these values.