Download Class IX Math NCERT Solutions For Real Numbers Exercise–1.1

Class IX Math
NCERT Solutions For Real Numbers
Exercise–1.1
1.
Is zero a rational number? Can you write it in the form
p
, where p and q are integers and q  0 ?
q
Ans. Yes, zero is a rational number. We can write it in the form
That is zero 
such as
and
p
.
q
0
,
[any number]
0
0
0
 0,  0,
0
2
9
7000
0
 0.
 7 
2.
Find six rational numbers between 3 and 4.
Ans. We know that there are an infinite number of rational numbers between two rational numbers.
Therefore, six rational numbers between 3 and 4 can be:
Let x = 3 and y = 4, also n = 6
 y  x   4  3 1
d


 n  1  6  1 7
Since, the six rational numbers between x and y are:
 x  d  ,  x  2d ,  x  3d ,  x  4d ,  x  5d and  x  6d  .
 The six rational numbers between 3 and 4 are:
1 
2 
3 
4 
5

 3  ,  3  ,  3  ,  3  ,  3  

7 
7 
7 
7 
7
6

and  3  

7
22 23 24 25 26
27
,
,
,
,
i. e.
and
7 7 7 7 7
7
3.
Find five rational numbers between
3
4
and .
5
5
3
4
Ans. Let x  , y  and n  5
5
5
 4 3   1

(y  x)  5 5   5  1 1 1
d


=  
5 6 30
 5  1
(n  1)
6
 The Five rational numbers between ‘x’ and ‘y’ are:
 x  d  ;  x  2d;  x  3d;  x  4d
Page 1
and  x  5d .
3 1  3 2  3 3  3 4 
 
,  
,  
,  

 5 30   5 30   5 30   5 30 
3 5 
and  

 5 30 
3 1  3 1  3 1  3 2 
or  
,   ,    ,   
 5 30   5 15   5 10   5 15 
3 1
and   
5 6
 18  1   9  1   6  1   9  2 
or 
; 
; 
; 

 30   15   10   15 
 18  5 
and 

 30 
19 10 7 11
23
; ; ;
or
and
30 15 10 15
30
23
19 2 7 11
; ; ;
or
and
30
30 3 10 15
Another method
3 3  10 30

 
5 5  10 50
4 4  10 40


5 5  10 50
 Five rational numbers between
3
4
31 32 33 34
35
,
,
,
and are
and
5
5
50 50 50 50
50
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.
Ans. (i) True statement
[  The collection of all natural numbers
and 0 is called whole numbers]
(ii) False statement
[  Integers such as 1,  2 are non-whole numbers]
(iii) False statement
1
[  Rational number
is not a whole number]
2
Page 2
Exercise–1.2
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.
Ans. (i) True statement, because all rational numbers and all irrational numbers form the group
(collection) of real numbers.
(ii) False statement, because no negative number can be the square root of any natural
number.
(iii) False statement, because rational numbers are also a part of real numbers.
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:
Ans. No, if we take a positive integer, say 4 its square root is 2, which is a rational number.
REMEMBER
According to the Pythagoras theorem, in a right-angled triangle, the square of the hypotenuse is
equal to the sum of the squares of the other two sides.
In the figure:
OB2  OA 2  AB2
 OB2  12  2  OB  2
B
?
1 unit
O
1 unit
A
Base (OA) = 1 unit
Height (AB) = 1 unit
3.
Show how 5 can be represented on the number line.
Ans. Let us take the horizontal line XOX’ as the x-axis. Mark O as its origin such that it represents 0.
Cut off OA = 1 unit, AB = 1 unit.
 OB = 2 units
Draw a perpendicular BC  OX.
Cut off BC = 1 unit.
C
5
1 unit
90º
x’
O
A
B
x
2 unit
Since OBC is a right triangle.
OB2  OC2  OC2
Page 3
22  12  OC 2
4  1  OC2
OC2  5
 OC  5
With O as centre and OC as radius, draw an arc intersecting OX at D.
Since OC = OD
C
5
90º
x’
O
A
B
x
2 unit
5
 OD represents
5 on XOX’.
Exercise–1.3
1.
Write the following in decidmal form and say what kind of decimal expansion each has:
(i)
36
100
(ii)
1
11
1
8
(iv)
3
13
2
11
(vi)
329
400
(iii) 4
(v)
Ans.
(i) We have
36
 0.36
100
 The decimal expansion of
(ii) Dividing 1 by 11, we have:
Page 4
36
is terminating.
100
0.090909
11 1.00000
0
10
00
100
 99
10
 00
100
 99
10
 00
1 00
99
1

1
 0.090909 0.09
11
Thus, the decimal expansion is
“non-terminating repeating”.
Note:
The bar above the digits indicates the block of digits that repeats. Here, the repeating block is 09.
(iii) To wrote 4
p
1
in form,
q
8
1
1 32  1 33
 4 

8
8
8
8
Now, diving 33 by 8, we have:
we have 4
Page 5
4.125
8 33.000
32
10
8
20
 16
40
 40
0
Remainder = 0, means the process of division terminates.
33
1
 4  4.125
8
8
Thus, the decimal expansion is terminating.

(iv) Dividing 3 by 13, we have
13
0.23076923 ...
0000000
26
0

0
 00
1 0
 91
90
 78
120
 117
30
 26
40
 39
1
Page 6
Here, the repeating block of digits is 230769.
3
 0.23076923  0.230769
13
3
Thus, the decimal expansion of
is
13
“non-terminating repeating”.
(v) Dividing 2 by 11, we have

0.1818
11 2.0000
11
90
 88
20
 11
90
 80
2
Here, the repeating block of digits is 18.

2
 0.1818  0.0.18
11
Thus, the decimal expansion of
2
is
11
“non-terminating repeating”.
(iv) Dividing 329 by 400, we have
400
0.8225
0000
3200
900

0
 800
2000
 2000
0
Remainder = 0, means the process of division terminates.

329
 0.8225
400
Page 7
Thus, the decimal expansion of
2.
329
is terminating.
400
1
2 3 4 5 6
 0.142857. Can you predict what the decimal expansions of , , , , are,
7
7 7 7 7 7
without actually doing the long division? If so, how?
You know that
Ans.
We are given that
1
 0.142857.
7
2
1
 2   2   0.142857   0.285714
7
7
3
1
 3   3   0.142857   0.428571
7
7
4
1
 4   4   0.142857   0.571428
7
7
5
1
 5   5   0.142857   0.714285
7
7
6
1
 6   6   0.142857   0.857142
7
7
Thus, without actually doing the long division we can predict the decimanl expansions of the
above given rational numbers.
3.
Express the following in the form
p
,
q
where p and q are integers and q  0.
(i) Let x = 0.6
(ii) 0.47
(iii) 0.001
Ans.
(i) Let x = 0.6 = 0.6666 
Since, there is one repeating digit.
 We multiply both sides by 10,
10x = (0.666...) × 10
or 10x = 6.6666...
 10x  x  6.6666   0.6666 
or 9x = 6
or x =
6 2

9 3
Thus, 0.6 
Page 8
2
3
(ii) Let x = 0.47  0.4777
 10x = 10  (0.4777  ) or 10x = 4.777
and 100x = 47.777
Subtracting (1) from (2), we have
100x  10x   47.777. . .   4.777. . . 90x  43
or x 
43
90
Thus, 0.47 
43
90
(iii) Let x  0.001  0.001001. . .
Here, we have three repeating digits after the decimal point, therefore we multiply by 1000.
 1000x  1000   0.001
 1000  0.001001. . .
or Subtracting (1) from (2), we have
1000x  x  1.001. . .   0.001. . .
or 999x = 1
 x
1
999
Thus, 0.001 
4.
1
999
p
. Are you surprised by your answer? With your teacher and
q
classmates discuss why the answer makes sense.
Let x = 0.99999 . . . in the form
Ans.
Let x = 0.99999 . . .
Multiply both sides by 10, we have
[  There is only one repeating digit.]
10  x  10   0.99999 . . .
or 9x  9
9
or x   1
9
Thus, 0.9999 . . .  1
As 0.9999 . . . goes on forever, there is no gap between 1 and 0.9999 . . .
Hence both are equal.
5.
What can the maximum number of digits be in the repeating block of digits in the decimal expansion
of
1
? Perform the division to check your answer..
17
Ans.
Page 9
Since, the number of entries in the repeating block of digits is less than the divisor. In
1
, the
17
divisor is 17.
 The maximum number of digits in the repeating block is 16. To perform the long division, we
have
0.0588235294117647...
400
0000000000000000
 85
150

0
 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
The remainder 1 is the same digit from which we started the division.
Page 10

1
 0.0588235294117647
17
Thus, there are 16 digits in the repeating block in the decimal expansion of
1
. Hence, our answer
17
is verified.
6.
p
 q  0 , where p and q are integers
q
with no common factors other than 1 and having terminating decimal representations (expansions).
Can you guess what property q must satisfy?
Look at several examples of rational numbers in the form
Ans.
Let us look at decimal expansion of the following terminating rational numbers:
3 3  5 15


 1.5
2 2  5 10
1 1 2
2


 0.2
5 5  2 10
1
[Denominator  2  2 ]
[Denominator  5  51 ]
7 7  125 875


 0.875
8 8  125 1000
[Denominator  8  23 ]
8
88
64


 0.064
125 125  8 1000
[Denominator  125  53 ]
13 13  5
65


 0.65
20 20  5 100
[Denominator  20  22  51 ]
17 17  625 10625


 1.0625
16 16  625 10,000
[Denominator  16  24 ]
We observe that the prime factorisation of q (i.e. denominator) has only powers of 2 or powers of
5 or powers of both.
Note:
If the denominator of a rational number
(in its standard form) has prime factors either 2 or 5 or both, then and only then it can be represented
as a terminating decimal.
7.
Write three numbers whose decimal expansions are non-terminating
Page 11
non-recurring.
Ans.
2  1.414213562 . . .
3  1.732050808 . . .
5  2.236067978 . . .
8.
Find three different irrational numbers between the rational numbers
5
9
.
and
7
11
Ans.
To express decimal expansion of
5
9
, we have:
and
7
11
9
5
 0.8181. . .  0.81
 0.714285 and
11
7
0.714285...
5
 7 5.0
7
49
10
9
 11
11
7
30
 28
0.8181...
9.0
 88
20
 11
20
 14
60
 56
40
 35
90
 88
20
 11
9
5
As there are an infinite number of irrational numbers between 0.714285 and 0.81, any three of
them can be:
(i) 0.750750075000750. . .
(ii) 0.767076700767000767. . .
(iii) 0.78080078008000780. . .
9.
Classify the following numbers as rational or irrational:
Page 12
(i)
(ii)
23
225
(iii) 0.3796
(iv) 7.478478
(v) 1.101001000100001. . .
Ans.
(i)  23 is not a perfect square.

23 is an irrational number..
2
(ii)  225  15  15  15
 225 is a perfect square.
Thus, 225 is a rational number..
(iii)  0.3796 is a terminating decimal,
 It is a rational number..
(iv) 7.478478. . .  7.478
Since, 7.478 is a non-terminating and recurring (repeating) decimal.
 It is a rational number..
(v) Since, 1.101001000100001… is a non-terminating and non-repeating decimal number.
 It is an irrational number..
Exercise–1.4
1.
Visualise 3.765 on the number line, using successive magnification.
Ans.
3.765 lies between 3 and 4.
Let us divide the interval (3, 4) into 10 equal parts.
3.1 3.2 3.3 3.4 3.5 3.6 3.7 3.8
3
4
Fig. (i)
3.71 3.72 3.73 3.74
3
3.75 3.76 3.77 3.78 3.79
3.8
Fig. (ii)
3.761
3
3.9
3.762
3.763
3.764
3.765
3.767
3.766 3.768
3.769
3.77
Fig. (iii)
Since, 3.765 lies between 3.7 and 3.8 We again magnify the interval [3.7, 3.8] by dividing it further
into 10 parts and concentrate the distance between 3.76 and 3.77.
The number 3.765 lies between 3.76 and 3.77. Therefore we further magnify the interval [3.76,
3.77] into 10 equal parts.
Now, the point corresponding to 3.765 is clearly located, as shown in Fig. (iii) above.
2.
Visualise 4.26 on the number line, up to 4 decimal places.
Page 13
Ans.
Note: We can magnify an interval endlessly using successive magnification.
To visualize 4.26 or 4.2626. . . on the number line up to 4 decimal places, we use the following
steps.
I. The number 4.2626. . . lies between 4 and 5. Divide the interval [4, 5] in 10 smaller parts:
4.1 4.2 4.3 4.4 4.5 4.6 4.7 4.8 4.9
4
5
Fig.(i)
II. Obviously, the number 4.2626. . . lies between 4.2 and 4.3. We magnify the interval [4.2,
4.3]:
4.28 4.29
4.21 4.22 4.23 4.24 4.25 4.26 4.27
4.2
4.3
Fig.(ii)
III. Next, we magnify the interval [4.26, 4.27]:
4.261
4.263
4.262
4.26
4.265
4.264
4.267
4.269
4.27
4.266
4.268
Fig. (iii)
IV. Finally magnify the interval [4.262, 4263]:
4.2621
4.262
4.2623 4.2625 4.2627 4.2629
4.2622
4.2624 4.2626
4.2628
4.264
Fig.(iv)
In Fig. (iv), we can easily observe the number 4.2626. . . or 4.26.
Exercise–1.5
Page 14
1.
Classify the following numbers as rational or irrational:
(i) 2  5
(iii)
3 
(ii)
2 7
(iv)
7 7
23   23
1
2
(v) 2
Ans.
(i) 2  5
Since it is a difference of a rational and irrational number,
 2  5 is an irrational number..
(ii)
3 
23   23
We have:
3 
23   23  3  23  23  3,
which is a rational number.
(iii)
2 7
7 7
Since,
2 7

7 7
2 7
7 7

2 7
7 7

2
, which is a rational number..
7
is a rational number..
1
(iv)
2
 The quotient of rational and irrational is an irrational number..
1

2
is an irrational number..
(v) 2
 2  2    Product of a rational and an irrational (which is an irrational number)
 2 is an irrational number..
2.
Simplify each of the following expressions:
(i)
3 
3   2  2  (ii)
(iii)

5  2
(iv)

5  2  5  2 
3 
3  3  3 
2
Ans.
Page 15
(i)
3  3   2  2   2 3 
  2  3  2 3   3 2 
3   2 3  3 
2  3
 62 3 3 2  6
Thus,
(ii)
3 
3  2  2   6  2 3  3 2  6
3 
3  3  3    3    3 
2
2
[   a  b  a  b   a2  b2 ]
 32  3  9  3  6
 3 
(iii)

3  3  3   6.
2
2
2
5  2    5    2   2 5   2 
2
[  a  b   a 2  b 2  2ab]
 5  2  2 10
 7  2 10

(iv)


2
5  2   7  2 10
2
5  2  5  2    5    2 
2
[  a  b  a  b   a2  b2 ]

3.

5  2  5  2   3
Recall, is defined as the ratio of the circumference (say c) of a circle to its diameter (say d). That
c
. This seems to contradict the fact that  is irrational. How will you resolve this
d
contradiction?
is,  
Ans.
When we measure the length of a line with a scale or with any other device, we only get an
approximate rational value, i.e. c and d both are irrational.
c
is irrational and hence  is irrational.
d
Thus, there is no contradiction in saying that  is irrational.

4.
Rationalise the denominators of the following:
(i)
Page 16
1
7
(ii)
1
7 6
1
(iii)
(iv)
5 2
1
7 2
Ans.
1
(i)
7
1 7

7 7
1
(ii)


7  6
2
1

 
x y ]
[ RF of

x  y is

 
x y ]
5  2  5  2 
5  2

5   2

5  2
3
7  2
x y 
1 5  2 

2
1

  7  6
7 6

[RF of
7  6
76
7  6
  7  6
1
2
1



Thus,
(iv)

7  6
5 2

7  6  7  6 
2
1



Thus,
(iii)
7
7
1  7  6 

7 6




5  2



5  2
52
5 2
3
1   7  2


7  2 7  2


7  2
 7 2   2 2
[ R.F. OF  7  2  IS  7  2 ]


7  2

74
Thus,
7 2
3
1

7  2


7  2
3
Page 17
Exercise–1.6
1.
Find:
(i) 641/2
(iii) 1251/3
(ii) 321/5
Ans.
(i)  64 = 8 x 8 = 82
 (64)1/2 = (82)1/2 = 82  1/2
[  (am)n = am  n]
81
=8

Thus, 641/2 = 8
(ii)  32 = 2× 2× 2× 2× 2× = 25
1/5
 (32)1/5   25 
 251/5
[ (am )n  amn ]
 21  2
1/5
Thus,  32   2
(iii)  125 = 5 × 5 × 5 = 53
1/3
 125 
2.
1/3
=  53 
= 53×1/3  51
Thus, 1251/3  5
Find:
(i) 93/2
(ii) 322/5
3/4
(iii) 16
(iv) 125-1/3
Ans.
(i)  9 = 3 × 3 = 32
 9
3/2
 32 3/2 = 32×3/2 = 33 = 27
Thus, 93/2  27
(ii)  32  2  2  2  2  2  25
  32 
2/5
  25 
2/5
 22  4
Thus, 322/5  4
(iii)  16  2  2  2  2  24
 16 
3/ 4
  24 
Thus, 163/4  8
Page 18
3/ 4
 2 43/4  23  8
(iv)  125  5  5  5  53
 125 
 5 1 
3.
1/3
  53 
1/3
1
5
Simplify:
7
(i)
22/3

 1
(ii)  3 
3 
21/5
1
(iii)
1 12
(iv) 71/2  81/2
1
1 14
Ans.
(i)  am  an  amn
 22/3  21/5  22/3  1/5
 2 1 10  3 13 
 3  5  15  15 
 213/15
Thus, 22/3 . 21/5 = 213/15
(ii) 
1
1
n
m
a
 am and  a m   a mn 
mn
a
7
7
1
 1
  3    3 3   337  321  21
3
3 
7
1
 1
Thus,  3   21
3
3 
1
(iii) 
112
1
114
 111/2  111/4  111/21/ 4
 am

mn
 n  a 
 a

= 111/4
1
112
 111/4
Thus,
1
11
4
Page 19
(iv)  am  bm = (ab)m
 71/2  81/2 = (7 × 8)1/2 = (56)1/2
Thus, 71/2  81/2 = (56)1/2
Page 20