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

Answer:

(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]

2 6 1 1 3 2 5 1 3

5 14 6 2 5 14 6 2

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

Answer:

(i)

2

8

(ii)

5

9

(iii)

6

5

(iv)

2

9

(v)

19

6

Solution:

(i) The additive inverse of

2

8

is

2

8

as

2 2 2 2 0

0

8 8 8 8

(ii) The additive inverse of

5

9

is

5

9

as

5 5 5 5 0

0

9 9 9 9

(iii) The additive inverse of

6

5

is

6

5

as

6 6 6 6 6 6 0

0

5 5 5 5 5

5

(iv) The additive inverse of

2

9

is

2

9

as

2 2 2 2 2 2 0

0

9 9 9 9 9

9

(v) The additive inverse of

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

Answer:

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

Answer:

(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

Answer:

(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

.

Answer:

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

Answer:

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?

Answer:

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?

Answer:

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.

Answer:

(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 ____________

Answer:

(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

Answer:

(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.

Answer:

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.

Answer:

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

.

Answer:

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

Answer:

(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.

Answer:

-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

.

Answer:

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.