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Number Systems1Exercise 1.1Exercise 1.1

Question 1. Is zero a rational number? Can you write it in the formp

q,

where p and q are integers and q ≠ 0?

Solution Yes, write0

1(where 0 and 1 are integers andq = 1which is not equal

to zero).

Question 2. Find six rational numbers between 3 and 4.

Solution There can be infinitely many rationals between 3 and 4, one way is

to take them 321

7= and 4

28

7= . (Q6 1 7+ = )

First�rational�number�between�3�and�4

q1 = (rational�number�between21

7and

28

7) =

+= =

21

7

28

72

49

72

7

2

∴ 21

7

7

2

28

7< <

Second�rational�number�between�3�and�4


7and

7

2) =

+=

21

7

7

22

91

28

∴ 21

7

91

28

7

2

28

7< < <

Third�rational�number�between�3�and�4


2and

28

7) =

+=

7

2

28

72

105

28

∴ 21

7

91

28

7

2

105

28

28

7< < < <

Similarly,21

7

175

56

91

28

7

2

203

56

105

28

217

56

28

7< < < < < < <

Hence, the six rational numbers175

56

91

28

7

2

203

56

105

28

217

56, , , , , are all lying

between�3�and�4.

Question 3. Find�five�rational�numbers�between3

5and

4

5.

Solution Let a = 3

5and b = 4

5

A�rational�number�between a and ba b= +

2

∴ A�rational�number�between3

5and

4

5

3

5

4

52

7

5 2

7

10=

+=

×=

∴ 3

5

7

10

4

5< <

Now,�a�rational�number�between3

5and

7

10

3

5

7

102

=+

= +×

=( )6 7

10 2

13

20

7

Mathematics-IX Number System

Question 2. Find six rational numbers between 3 and 4.

Solution There can be infinitely many rationals between 3 and 4, one way is

to take them 321

7= and 4

28

7= . (Q6 1 7+ = )

First�rational�number�between�3�and�4


7and

28

7) =

+= =

21

7

28

72

49

72

7

2

∴ 21

7

7

2

28

7< <

Second�rational�number�between�3�and�4


7and

7

2) =

+=

21

7

7

22

91

28

∴ 21

7

91

28

7

2

28

7< < <

Third�rational�number�between�3�and�4


2and

28

7) =

+=

7

2

28

72

105

28

∴ 21

7

91

28

7

2

105

28

28

7< < < <

Similarly,21

7

175

56

91

28

7

2

203

56

105

28

217

56

28

7< < < < < < <

Hence, the six rational numbers175

56

91

28

7

2

203

56

105

28

217

56, , , , , are all lying

between�3�and�4.

Question 3. Find�five�rational�numbers�between3

5and

4

5.

Solution Let a = 3

5and b = 4

5

A�rational�number�between a and ba b= +

2

∴ A�rational�number�between3

5and

4

5

3

5

4

52

7

5 2

7

10=

+=

×=

∴ 3

5

7

10

4

5< <

Now,�a�rational�number�between3

5and

7

10

3

5

7

102

=+

= +×

=( )6 7

10 2

13

20

7

∴ 3

5

13

20

7

10

4

5< < <

Similarly,25

40

27

40

15

20, , are�rational�numbers�between

3

5and

4

5.

Hence,�required�rational�numbers�are25

40

13

20

27

40

7

10

15

20, , , , .

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, because natural numbers are 1, 2, 3, 4,...., ∞ and whole

numbers are 0, 1, 2, 3, 4, 5,....,∞.

or

The�collection�of�whole�numbers�contain�all�the�natural�numbers.

(ii) False (Qnegative integers are not included in the list of whole numbers.)

(iii) False Q1

3

6

7

10

19, ,

are not whole numbers.

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 anatural�number.

(iii) Every�real�number�is�an�irrational�number.

Solution (i) True (QReal numbers = Rational numbers + Irrational numbers.)

(ii) False (Q no negative number can be the square root of any natural

number.)

(iii) False (Qrational numbers are also present in the set of real numbers.)

Question 2. Are the square roots of all positive integers irrational? Ifnot, give an example of the square root of a number that is a rationalnumber.

Solution No, the square roots of all positive integers are not irrational.

e g. ., 16 4=Here,�‘4’�is�a�rational�number.

Number Systems1Exercise 1.2

∴ 3

5

13

20

7

10

4

5< < <

Similarly,25

40

27

40

15

20, , are�rational�numbers�between

3

5and

4

5.

Hence,�required�rational�numbers�are25

40

13

20

27

40

7

10

15

20, , , , .

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, because natural numbers are 1, 2, 3, 4,...., ∞ and whole

numbers are 0, 1, 2, 3, 4, 5,....,∞.

or

The�collection�of�whole�numbers�contain�all�the�natural�numbers.

(ii) False (Qnegative integers are not included in the list of whole numbers.)

(iii) False Q1

3

6

7

10

19, ,

are not whole numbers.

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 anatural�number.

(iii) Every�real�number�is�an�irrational�number.

Solution (i) True (QReal numbers = Rational numbers + Irrational numbers.)

(ii) False (Q no negative number can be the square root of any natural

number.)

(iii) False (Qrational numbers are also present in the set of real numbers.)

Question 2. Are the square roots of all positive integers irrational? Ifnot, give an example of the square root of a number that is a rationalnumber.

Solution No, the square roots of all positive integers are not irrational.

e g. ., 16 4=Here,�‘4’�is�a�rational�number.

Question 3. Show how 5 can be represented on the number line.

Solution We know that, 5 4 1= +

= +2 12 2

Draw�of�right�angled�triangleOQP ,�such�that

OQ = 2 units

PQ = 1unit

and ∠ = °OQP 90

Now,�by�using�Pythagoras�theorem,�we�have

OP OQ PQ2 2 2= += +2 12 2

⇒ OP = +4 1

= 5

Now, takeO as centreOP = 5 as radius, draw an arc, which intersects the lineat�pointR.

Hence,�the�pointR represents 5.

Question 4. Classroom activity (constructing the ‘square root spiral’).

Solution Take a large sheet of paper and construct the ‘square root spiral’ in

the following fashion. Start with a point O and draw a line segment OP1 of unit

length. Draw a line segment P P1 2 perpendicular to OP1 of unit length

(see figure).

9

3210–1–2

P

Q

5

1

O R2

32

P1

P2P3

O

Constructing�square�root

spiral




= +2 12 2


OQ = 2 units

PQ = 1unit

and ∠ = °OQP 90


OP OQ PQ2 2 2= += +2 12 2

⇒ OP = +4 1

= 5







(see figure).

9

3210–1–2

P

Q

5

1

O R2

32

P1

P2P3

O


spiral



= +2 12 2


OQ = 2 units

PQ = 1unit

and ∠ = °OQP 90


OP OQ PQ2 2 2= += +2 12 2

⇒ OP = +4 1

= 5







(see figure).

9

3210–1–2

P

Q

5

1

O R2

32

P1

P2P3

O


spiral

Now, draw a line segment P P2 3 perpendicular toOP2. Then draw a line segment

P P3 4 perpendicular to OP3. Continuing in this manner, you can get the line

segment P Pn n− 1 by drawing a line segment of unit length perpendicular to

OPn − 1. In this manner, you will have created the points P P Pn2 3, , , , ,K K and

joined�them�to�create�a�beautiful�spiral�depicting 2 3 4, , , K .

Exercise 1.3

Question 1. Write the following in decimal form and say what kind ofdecimal expansion each has

(i)36

100(ii)

1

11(iii) 4

1

8(iv)

3

13(v)

2

11(vi)

329

400

Solution (i) Clearly,36

100can be written as 0.36

∴ 36

100036= . (Terminating decimal)

(ii) Dividing 1 by 11, we get

0.0909

11 ) 100

99

100

99

1

∴ 1

110090909 009= =. .... . (Non-terminating repeating)

(iii) We have, 41

8

4 8 1

8= × + = 33

8

Dividing�33�by�8,�we�get

4.125

8 33)

32

10

8

20

16

40

40

×

∴ 33

84125= . (Terminating)

10







Exercise 1.3


(i)36

100(ii)

1

11(iii) 4

1

8(iv)

3

13(v)

2

11(vi)

329

400



∴ 36



0.0909

11 ) 100

99

100

99

1

∴ 1


(iii) We have, 41

8

4 8 1

8= × + = 33

8


4.125

8 33)

32

10

8

20

16

40

40

×

∴ 33


10






Exercise 1.3


(i)36

100(ii)

1

11(iii) 4

1

8(iv)

3

13(v)

2

11(vi)

329

400



∴ 36



0.0909

11 ) 100

99

100

99

1

∴ 1


(iii) We have, 41

8

4 8 1

8= × + = 33

8


4.125

8 33)

32

10

8

20

16

40

40

×

∴ 33


10


(iv) We have, 3 13/

Dividing 3 by 13, we get

0.230769

13 30)

26

40

39

100

91

90

78

120

117

3

∴ 3

1330769= 0.2 (Non-terminating repeating)

(v) We have, 2 11/


0.1818

11 20)

11

90

88

20

11

90

88

2

∴ 2

110181818= . ... = 018. (Non-terminating repeating)

(vi) We have, 320 400/


0.8225

400 3290)

3200

900

800

1000

800

2000

2000

×

∴ 329 400 08225/ .= (Terminating)

11

(iv) We have, 3 13/


0.230769

13 30)

26

40

39

100

91

90

78

120

117

3

∴ 3

1330769= 0.2 (Non-terminating repeating)

(v) We have, 2 11/


0.1818

11 20)

11

90

88

20

11

90

88

2

∴ 2

110181818= . ... = 018. (Non-terminating repeating)

(vi) We have, 320 400/


0.8225

400 3290)

3200

900

800

1000

800

2000

2000

×

∴ 329 400 08225/ .= (Terminating)

11


Question 2. You know that1

70 142857= . . Can you predict what the

decimal expansions of2

7

3

7

4

7

5

7

6

7, , , , are , without actually doing the

long division? If so, how?

[Hint Study the remainders while finding the value of1

7carefully.]

Solution We have,1

70142857= .

∴ 2

72

1

7= ×

= ×2 0142857.

⇒ 2

70 285714= .

3

73

1

7= ×

= ×3 428570.1

⇒ 3

70 428571= .

4

74

1

7= ×

= ×4 428570.1

⇒ 4

771428= 0.5

5

75

1

7= ×

= ×5 0142857.

⇒ 5

70714285= .

6

76

1

7= ×

= ×6 428570.1

⇒ 6

757142= 0.8

Question 3. Express the following in the formp

q, where p and q are

integers and q ≠ 0.

(i) 0 6. (ii)�0.47 (iii) 0.001

12


Solution (i) Let x = =0. 0.66 66... ...(i)

Multiplying�Eq.�(i)�by�10,�we�get

10 6666x = . ... …(ii)

On�subtracting�Eq.�(ii)�from�Eq.�(i),�we�get

( ) ( ...) ( ...)10 66 0 66x x− = −6.6 .6

9 6x =x = 6 9/

⇒ x = 2 3/

(ii) Let x = =047 04777. . ... ...(iii)

Multiplying�Eq.�(iii)�by�10,�we�get

10 4777x = . ... …(iv)

Multiplying�Eq.�(iv)�by�10,�we�get

100 47777x = . …(v)

On�subtracting�Eq.�(v)�from�Eq.�(iv),�we�get

( ) ( . ...) ( . ...)100 10 47 777 4 777x x− = −

90 43x = ⇒ x = 43

90

(iii) Let x = =0.0 0.001 01001001... …(vi)

Multiplying�Eq.�(vi)�by�(1000),�we�get

1000 1001001001x = . ... …(vii)

On�subtracting�Eq.�(vii)�by�Eq.�(vi),�we�get

( ) ( . ....) ( . ....)1000 1001001001 0 001001001x x− = −999 1x =

⇒ x = 1

999

Question 4. Express 0.99999... in the formp

q. Are you surprised by

your answer? With your teacher and classmates discuss why the answermakes sense.

Solution Let x = 0 99999. ... …(i)

Multiplying�Eq.�(i)�by�10,�we�get

10 9 99999x = . ... …(ii)

On�subtracting�Eq.�(ii)�by�Eq.�(i),�we�get

( ) ( . ...) ( . ...)10 999999 099999x x− = −9 9x =

⇒ x = 9

9

x = 1

13


Question 5. What can the maximum number of digits be in the

repeating block of digits in the decimal expansion of1

17? Perform the

division to check your answer.

Solution The maximum number of digits in the repeating block of digits in

the decimal expansion of1

17is 17 1 16− =

we�have,

0.0588235294117647

17 100)

85

150

136

140

136

40

34

60

51

90

85

50

34

160

153

70

68

20

17

30

17

130

119

110

102

80

68

120

119

1

Thus,1

170 0588235294117647= . .... ,�a�block�of�16�digits�is�repeated.

14


Question 6. Look at several examples of rational numbers in the formp

qq( )≠ 0 . Where, p and q are integers with no common factors other that

1 and having terminating decimal representations (expansions). Can you

guess what property q must satisfy?

Solution Consider many rational numbers in the formp

qq( )≠ 0 , where p and

q are integers with no common factors other that 1 and having terminating

decimal representations.

Let�the�various�such�rational�numbers�be1

2

1

4

5

8

36

25

7

125

19

20

29

16, , , , , , etc.

In all cases, we think of the natural number which when multiplied by their

respective�denominators�gives�10�or�a�power�of�10.

1

2

1 5

2 5

5

1005= ×

×= = .

1

4

1 25

4 25

25

1000 25= ×

×= = .

5

8

5 125

8 125= ×

×

= =625

10000625.

36

25

36 4

25 4= ×

×= =144

1001.44

7

125

7 8

125 8= ×

×= =56

1000560.0

19

20

19 5

20 5

95

1005= ×

×= = 0.9

29

16

29 625

16 625= ×

×= 18125

10000= 18125.

From the above, we find that the decimal expansion of above numbers are

terminating. Along with we see that the denominator of above numbers are in the

form 2 5m n× , wherem and n are natural numbers. So, the decimal representation

of rational numbers can be represented as a terminating decimal.

Question 7. Write three numbers whose decimal expansions are

non-terminating non-recurring.

Solution 0.74074007400074000074...

0.6650665006650006650000...

0.70700700070000...

15

Question 8. Find three different irrational numbers between the

rational numbers5

7and

9

11.

Solution To find irrational numbers, firstly we shall divide 5 by 7 and 9 by 11,So,

0.714285...

7 50

49

10

7

30

28

20

14

60

56

40

35

5

Thus,5

7= =0.714285... 0.714285

11 900.8181...

88

20

11

90

88

20

11

9

Thus,9

110 81= =0.8181K .

∴ The�required�numbers�are

0.73073007300073000073...

0.7650765007650007650000...

0.80800800080000...

Question 9. Classify the following numbers as rational or irrational

(i) 23 (ii) 225 (iii)�0.3796 (iv)�7.478478...

(v)�1.101001000100001...

16


Question 8. Find three different irrational numbers between the

rational numbers5

7and

9

11.

Solution To find irrational numbers, firstly we shall divide 5 by 7 and 9 by 11,So,

0.714285...

7 50

49

10

7

30

28

20

14

60

56

40

35

5

Thus,5

7= =0.714285... 0.714285

11 900.8181...

88

20

11

90

88

20

11

9

Thus,9

110 81= =0.8181K .

∴ The�required�numbers�are

0.73073007300073000073...

0.7650765007650007650000...

0.80800800080000...

Question 9. Classify the following numbers as rational or irrational

(i) 23 (ii) 225 (iii)�0.3796 (iv)�7.478478...

(v)�1.101001000100001...

16

Solution

(i) 23 (irrationalQit is not a perfect square.)

(ii) 225 15= (rational) (whole number.)

(iii) 0.3796 = rational (terminating.)

(iv) 7.478478… = 7.478 = rational (non-terminating repeating.)

(v) 1.101001000100001… = irrational (non-terminating non-repeating.)

Exercise 1.4

Question 1. Visualise 3.765 on the number line, using successivemagnification.

Solution We know that, 3.765 lies between 3 and 4. So, let us divide the part

of the number line between 3 and 4 into 10 equal parts and look at the portion

between 3.7 and 3.8 through a magnifying glass. Now 3.765 lies between 3.7

and 3.8 [Fig. (i)]. Now, we imagine to divide this again into ten equal parts. The

first mark will represent 3.71, the next 3.72 and soon. To see this clearly,we

magnify this as shown in [Fig. (ii)].

Again 3.765 lies between 3.76 and 3.77 [Fig. (ii)]. So, let us focus on this portion

of the number line [Fig. (iii)] and imagine to divide it again into ten equal parts

[Fig. (iii)]. Here, we can visualise that 3.761 is the first mark and 3.765 is the 5th

mark in these subdivisions. We call this process of visualisation of

representation of numbers on the number line through a magnifying glass as the

process of successive magnification.

So, we get seen that it is possible by sufficient successive magnifications of

visualise the position (or representation) of a real number with a terminating

decimal expansion on the number line.

17

3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 4

3.71 3.72 3.73 3.74 3.75 3.76 3.77 3.78 3.79 3.83.7

3.76 3.761 3.762 3.763 3.764 3.765 3.766 3.767 3.768 3.769 3.77

(i)

(ii)

(iii)

M1

N1 P

N2

M2

N2

N1

3


Solution

(i) 23 (irrationalQit is not a perfect square.)

(ii) 225 15= (rational) (whole number.)

(iii) 0.3796 = rational (terminating.)

(iv) 7.478478… = 7.478 = rational (non-terminating repeating.)

(v) 1.101001000100001… = irrational (non-terminating non-repeating.)

Exercise 1.4

Question 1. Visualise 3.765 on the number line, using successivemagnification.

Solution We know that, 3.765 lies between 3 and 4. So, let us divide the part

of the number line between 3 and 4 into 10 equal parts and look at the portion

between 3.7 and 3.8 through a magnifying glass. Now 3.765 lies between 3.7

and 3.8 [Fig. (i)]. Now, we imagine to divide this again into ten equal parts. The

first mark will represent 3.71, the next 3.72 and soon. To see this clearly,we

magnify this as shown in [Fig. (ii)].

Again 3.765 lies between 3.76 and 3.77 [Fig. (ii)]. So, let us focus on this portion

of the number line [Fig. (iii)] and imagine to divide it again into ten equal parts

[Fig. (iii)]. Here, we can visualise that 3.761 is the first mark and 3.765 is the 5th

mark in these subdivisions. We call this process of visualisation of

representation of numbers on the number line through a magnifying glass as the

process of successive magnification.

So, we get seen that it is possible by sufficient successive magnifications of

visualise the position (or representation) of a real number with a terminating

decimal expansion on the number line.

17

3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8 3.9 4

3.71 3.72 3.73 3.74 3.75 3.76 3.77 3.78 3.79 3.83.7

3.76 3.761 3.762 3.763 3.764 3.765 3.766 3.767 3.768 3.769 3.77

(i)

(ii)

(iii)

M1

N1 P

N2

M2

N2

N1

3

Question 2. Visualise 4 26. on the number line, upto 4 decimal places.

Solution We adopt process by successive magnification and successivelydecrease the lengths of the portion of the number line in which 4.26 is located.Since 4.26 is located between 4 and 5 and is divided into 10 equal parts[Fig. (i)]. In further, we locate 4 26. between 4.2 and 4.3 [Fig. (ii)].

To get more accurate visualisation of the representation, we divide this portioninto 10 equal parts and use a magnifying glass to visualise that 4.26 liesbetween 4.26 and 4.27. To visualise 4.26 more clearly we divide again between4.26 and 4.27 into 10 equal parts and visualise the repsentation of 4.26 between4.262 and 4.263 [Fig. (iii)].

Now, for a much better visualisation between 4.262 and 4.263 is agin dividedinto 10 equal parts [Fig. (iv)]. Notice that 4.26 is located closer to 4.263 then to4.262 at 4.2627.

�

Exercise 1.5

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

(i) 2 5− (ii) ( )3 23 23+ − (iii)2 7

7 7

(iv)1

2(v) 2π

18

4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 5

4.2 4.21 4.22 4.23 4.24 4.25 4.26 4.27 4.28 4.29 4.3

4.25 4.261 4.262 4.263 4.264 4.265 4.266 4.267 4.268 4.269 4.27

4.262 4.2621 4.2622 4.2623 4.2624 4.2625 4.2626 4.2627 4.2628 4.2629 4.263

M1 M2 (i)

M2M1P1

P2P1 (iii)

(ii)

N1 N2(iv)P

P2

N2N1

P1






�

Exercise 1.5


(i) 2 5− (ii) ( )3 23 23+ − (iii)2 7

7 7

(iv)1

2(v) 2π

18

4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 5

4.2 4.21 4.22 4.23 4.24 4.25 4.26 4.27 4.28 4.29 4.3

4.25 4.261 4.262 4.263 4.264 4.265 4.266 4.267 4.268 4.269 4.27

4.262 4.2621 4.2622 4.2623 4.2624 4.2625 4.2626 4.2627 4.2628 4.2629 4.263

M1 M2 (i)

M2M1P1

P2P1 (iii)

(ii)

N1 N2(iv)P

P2

N2N1

P1





�

Exercise 1.5


(i) 2 5− (ii) ( )3 23 23+ − (iii)2 7

7 7

(iv)1

2(v) 2π

18

4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 5

4.2 4.21 4.22 4.23 4.24 4.25 4.26 4.27 4.28 4.29 4.3

4.25 4.261 4.262 4.263 4.264 4.265 4.266 4.267 4.268 4.269 4.27

4.262 4.2621 4.2622 4.2623 4.2624 4.2625 4.2626 4.2627 4.2628 4.2629 4.263

M1 M2 (i)

M2M1P1

P2P1 (iii)

(ii)

N1 N2(iv)P

P2

N2N1

P1






�

Exercise 1.5


(i) 2 5− (ii) ( )3 23 23+ − (iii)2 7

7 7

(iv)1

2(v) 2π

18

4 4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9 5

4.2 4.21 4.22 4.23 4.24 4.25 4.26 4.27 4.28 4.29 4.3

4.25 4.261 4.262 4.263 4.264 4.265 4.266 4.267 4.268 4.269 4.27

4.262 4.2621 4.2622 4.2623 4.2624 4.2625 4.2626 4.2627 4.2628 4.2629 4.263

M1 M2 (i)

M2M1P1

P2P1 (iii)

(ii)

N1 N2(iv)P

P2

N2N1

P1

Solution (i) IrrationalQ2 is a rational number and 5 is an irrational number.

∴2 5− is�an�irrational�number.

(QThe difference of a rational number and an irrational number is irrational)

(ii) 3 23 23 3+ − = (rational)

(iii)2 7

7 7

2

7= (rational)

(iv)1

2(irrational) Q 1 0≠ is a rational number and 2 0≠ is an irrational

number.

∴ 1

2is�an�irrational�number.

(QThe quotient of a non-zero rational number with an irrational number isirrational).

(v) 2 π (irrational)Q2 is a rational number and π is an irrational number.

∴2x is an irrational number. (QThe product of a non-zero rational numberwith an irrational number is an irrational)

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 32 2( ) [ ( ) ( ) ]Q a b a b a b+ − = −2 2

= −9 3 = 6

(iii) ( )5 2 2+ = + × +( ) ( )5 2 5 2 22 2 [ ( ) ]Q a b a ab b+ = + +2 2 22

= + +5 2 10 2

= +7 2 10

(iv) ( ) ( )5 2 5 2− + = −( ) ( )5 22 2 = −5 2 = 3 [Q( ) ( ) ]a b a b a b− + = −2 2

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 = c

d. This seems to contradict

the fact that π is irrational. How will you resolve this contradiction?

Solution Actuallyc

d= 22

7which is an approximate value of π.

Question 4. Represent 9 3. on the number line.

Solution Firstly we draw AB = 9.3 units. Now, from B, mark a distance of

1 unit. Let this point beC. LetO be the mid-point of AC. Now, draw a semi-circle

with centre O and radius OA. Let us draw a line perpendicular to AC passing

through pointB and intersecting the semi-circle at pointD .

19


Solution (i) IrrationalQ2 is a rational number and 5 is an irrational number.

∴2 5− is�an�irrational�number.

(QThe difference of a rational number and an irrational number is irrational)

(ii) 3 23 23 3+ − = (rational)

(iii)2 7

7 7

2

7= (rational)

(iv)1

2(irrational) Q 1 0≠ is a rational number and 2 0≠ is an irrational

number.

∴ 1

2is�an�irrational�number.

(QThe quotient of a non-zero rational number with an irrational number isirrational).

(v) 2 π (irrational)Q2 is a rational number and π is an irrational number.

∴2x is an irrational number. (QThe product of a non-zero rational numberwith an irrational number is an irrational)

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 32 2( ) [ ( ) ( ) ]Q a b a b a b+ − = −2 2

= −9 3 = 6

(iii) ( )5 2 2+ = + × +( ) ( )5 2 5 2 22 2 [ ( ) ]Q a b a ab b+ = + +2 2 22

= + +5 2 10 2

= +7 2 10

(iv) ( ) ( )5 2 5 2− + = −( ) ( )5 22 2 = −5 2 = 3 [Q( ) ( ) ]a b a b a b− + = −2 2

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 = c

d. This seems to contradict

the fact that π is irrational. How will you resolve this contradiction?

Solution Actuallyc

d= 22

7which is an approximate value of π.

Question 4. Represent 9 3. on the number line.

Solution Firstly we draw AB = 9.3 units. Now, from B, mark a distance of

1 unit. Let this point beC. LetO be the mid-point of AC. Now, draw a semi-circle

with centre O and radius OA. Let us draw a line perpendicular to AC passing

through pointB and intersecting the semi-circle at pointD .

19

∴The distanceBD = 93. .

Draw an arc with centre B and radius BD, which intersects the number line atpoint E , then the point E represents 9.3.

Question 5. Rationalise the denominator of the following

(i)1

7(ii)

1

7 6−(iii)

1

5 2+(iv)

1

7 2−

Solution (i)1

7

7

7× = 7

7(Multiplying and dividing by 7)

(ii)1

7 6

7 6

7 6−× +

+⇒ 7 6

7 62 2

+−( ) ( )

= +−

7 6

7 6= +7 6

(Multiplying and dividing by 7 6+ )

(iii)1

5 2

5 2

5 2+× −

−= −

−5 2

5 22 2( ) ( )(Multiplying and dividing by 5 2− )

= −−

5 2

5 2= −5 2

3

(iv)1

7 2

7 2

7 2−× +

+= +

−7 2

7 22 2( )(Multiplying and dividing by 7 2+ )

= +−

= +7 2

7 4

7 2

3

Exercise 1.6

Question 1. Find

(i) 64

1

2 (ii) 32

1

5 (iii) 125

1

3

Solution (i) 64

1

2 = ×( )8 8

1

2 =×

82

1

2 = 8 [ ( ) ]Q a am n mn=

(ii) 32

1

5 = × × × ×( )2 2 2 2 2

1

5 =×

25

1

5 = 2

(iii) 125

1

3 = × ×( )5 5 5

1

3 =×

53

1

3 = 5

20

9.3 9.3

ECBOA

1 unit9.3 units

D


∴The distanceBD = 93. .

Draw an arc with centre B and radius BD, which intersects the number line atpoint E , then the point E represents 9.3.

Question 5. Rationalise the denominator of the following

(i)1

7(ii)

1

7 6−(iii)

1

5 2+(iv)

1

7 2−

Solution (i)1

7

7

7× = 7

7(Multiplying and dividing by 7)

(ii)1

7 6

7 6

7 6−× +

+⇒ 7 6

7 62 2

+−( ) ( )

= +−

7 6

7 6= +7 6

(Multiplying and dividing by 7 6+ )

(iii)1

5 2

5 2

5 2+× −

−= −

−5 2

5 22 2( ) ( )(Multiplying and dividing by 5 2− )

= −−

5 2

5 2= −5 2

3

(iv)1

7 2

7 2

7 2−× +

+= +

−7 2

7 22 2( )(Multiplying and dividing by 7 2+ )

= +−

= +7 2

7 4

7 2

3

Exercise 1.6

Question 1. Find

(i) 64

1

2 (ii) 32

1

5 (iii) 125

1

3

Solution (i) 64

1

2 = ×( )8 8

1

2 =×

82

1

2 = 8 [ ( ) ]Q a am n mn=

(ii) 32

1

5 = × × × ×( )2 2 2 2 2

1

5 =×

25

1

5 = 2

(iii) 125

1

3 = × ×( )5 5 5

1

3 =×

53

1

3 = 5

20

9.3 9.3

ECBOA

1 unit9.3 units

D

Question 2. Find

(i) 9

3

2 (ii) 32

2

5 (iii)16

3

4 (iv) 125

1

3−

Solution (i) 9

3

2 = ( )32

3

2 = 33 = 27

(ii) 32

2

5 = ( )25

2

5 = 22 = 4

(iii) 16

3

4 = ( )24

3

4 = 23 = 8

(iv) 125

1

3−

=−

( )53

1

3 = −5 1 = 1

5

Question 3. Simplify

(i) 2 2

2

3

1

5⋅ (ii)1

33

7

(iii)11

11

1 2

1 4

/

/(iv) 7 8

1

2

1

2⋅

Solution (i) 2 2

2

3

1

5⋅ =+

2

2

3

1

5 (Qx x xa b a b⋅ = + )

=+

2

10 3

15 = 2

13

15

(ii)1

3

1

33

7

3 7

=×

= = −1

33

21

21 [Q( )x xa b ab= ]

(iii)11

11

1

2

1

4

= × =− −

11 11 11

1

2

1

4

1

2

1

4 =−

11

2 1

4Qx

xx

a

b

a b=

−

= 11

1

4

(iv) 7 8

1

2

1

2⋅ = ⋅( )7 8

1

2 = ( )56

1

2 [Qx y xya a a⋅ = ( ) ]

Selected NCERT Exemplar�Problems

Exercise 1.1

� Choose�the�correct�answer�from�the�given�four�options.

Question 1. Every rational number is

(a) a�natural�number (b)�an�integer

(c) a�real�number (d)�a�whole�number

Solution (c) Since, real numbers are the combination of rational and irrationalnumbers.

21

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