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CHAPTER
1
NUMBER SYSTEMS
POINTS TO REMEMBER
1. Definition of a rational number. A number ‘r’ is called a rational number, if
p
it can be written in the form , where p and q are integers and q ≠ 0.
q
Note. We visit that q ≠ 0 because division by zero is not allowed.
2. Equivalent rational numbers. Rational numbers do not have a unique
p
representation in the form , where p and q are integers and q ≠ 0. For example,
q
3 4 5
1 2
= ...... , and so on, these are called equivalent rational
= = = =
6 8 10
2 4
numbers.
p
3. Standard form of a rational number. A rational number r = , q ≠ 0 is said
q
to be in its standard form if p and q are co-prime.
Note: Two integers are said to be co-prime when they have no common
factors other than 1.
4. In general, there lie infinitely many rational numbers between any two given
rational numbers.
5. Definition of an irrational number. A number ‘s’ is called an irrational number,
if it cannot be written in the form
p
, where p and q are integers and q ≠ 0.
q
6. Collection of real numbers. All rational numbers and all irrational numbers
taken together form the collection of real numbers. It is denoted by R obviously,
a real number is either rational or irrational.
7. Corresponding to every real number, there exists a unique point on the number
line. Also, corresponding to every point on the number line, there exists a
unique real number. This is why we call the number line, ‘The real number
line’.
8. Decimal expansion of a rational number. The decimal expansion of a
rational number is either terminating or non-terminating recurring. Conversely,
a number whose decimal expansion is terminating or non-terminating recurring
is rational.
12
GOLDEN SAMPLE PAPER (MATHEMATICS)–IX
9. Decimal expansion of an irrational number. The decimal expansion of an
irrational number is non-terminating non-recurring. Conversely, a number
whose decimal expansion is non-terminating non-recurring is irrational.
10. Few facts
(i) The sum or difference of a rational number and an irrational number is
irrational.
(ii) The product or quotient of a non-zero rational number with an irrational
number is irrational.
(iii) If we add, subtract, multiply or divide two irrationals, the result may be
rational or irrational.
11. Radical sign. Let a > 0 be a rational number and n be a positive integer. Then
n
a = b means bn = a and b > 0. Here, the symbol ‘ ’ is called the radical
sign. In particular, if a is a real number, then
a = b means b2 = a and b > 0.
12. Identities relating to square roots. Let a and b be positive real numbers,
then,
(i)
ab = a
b
(ii)
a
=
b
a
b
(iii) ( a + b ) ( a − b ) = a – b
(iv) ( a + b ) ( a − b ) = a2 – b
(v) ( a + b ) ( c + d ) =
ac + ad + bc + bd
(vi) ( a + b ) 2 = a + 2 ab + b
13. Laws of exponents
(i) am . an = am + n
(iii)
am
= am – n , m > n
an
(ii) (am)n = amn
(iv) am b m = (ab)m
Here, a, n and m are natural numbers. a is called the base and m and n are
called the exponents.
14. Definition. Let a > 0 be a real number. Let m and n be integers such that m and
n have no common factors other than 1, and n > 0. Then,
m
an =
d ai
n
m
= n am
15. Rationalisation. To rationalise the denominator of
by
a −b
, where a and b are integers.
a −b
1
, we multiply this
a +b
13
NUMBER SYSTEMS
NCERT TEXTBOOK EXERCISES WITH SOLUTIONS
EXERCISE 1.1 (Page 5 )
Example 1. Is zero a rational number ? Can you write it in the form
p
, where
q
p and q are integers and q ≠ 0 ?
Sol. Yes ! zero is a rational number. We can write zero in the form
p
, where p
q
and q are integers and q ≠ 0 as follows :
0
0
0
0=
=
=
etc.,
1
2
3
denominator q can also be taken as negative integer.
Example 2. Find six rational numbers between 3 and 4.
Sol.
⇒
⇒
⇒
3+4
2
7
3+
2
2
25
3+
8
2
97
3+
32
2
=
7
2
13
=
4
⇒
=
49
16
⇒
=
193
.
64
Thus, six rational numbers between 3 and 4 are
13
4 = 25
8
2
49
3+
16 = 97
32
2
3+
7 13 25 49 97
,
,
,
,
and
8 16 32
2 4
193
.
64
Aliter. We write 3 and 4 as rational numbers with denominator 6 + 1 (= 7), i.e.,
and
3=
21
3
3× 7
=
=
7
1
1× 7
4=
28
4
4×7
=
=
.
7
1
1× 7
Thus, six rational numbers between 3 and 4 are
22 23 24 25 26
,
,
,
,
and
7
7
7
7
7
27
.
7
Note. This is known as the method of finding rational numbers in one step.
Example 3. Find five rational numbers between
Sol.
3
30
=
,
5
50
3
4
and .
5
5
14
GOLDEN SAMPLE PAPER (MATHEMATICS)–IX
4
40
=
.
5
50
Therefore, five rational numbers between
4
3
31 32 33 34
and
are
,
,
,
,
5
4
50 50 50 50
35
.
50
Example 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.
Sol. (i) True, since the collection of whole numbers contains all natural numbers.
(ii) False, for example – 2 is not a whole number.
(iii) False, for example
1
is a rational number but not a whole number.
2
EXERCISE 1.2 (Page 8)
Example 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.
Sol. True, since collection of real numbers is made up of rational and irrational
numbers.
(ii) False, because no negative number can be the square root of any natural
number.
(iii) False, for example 2 is real but not irrational.
Example 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.
Sol. No. For example,
Example 3. Show how
4 = 2 is a rational number.
5 can be represented on the number line.
Sol. Representation of 5 on the number line
Consider a unit square OABC and transfer it onto the number line making sure
that the vertex O coincides with zero.
12 + 12 = 2
Construct BD of unit length perpendicular to OB.
Then
OB =
( 2)2 + 12 = 3
Construct DE of unit length perpendicular to OD.
Then
OD =
15
NUMBER SYSTEMS
( 3)2 + 12 = 4 = 2
Construct EF of unit length perpendicular to OE.
Then
OE =
Then
OF = 22 + 12 = 5
Using a compass, with centre O and radius OF, draw an arc which intersects
the number line in the point R. Then R corresponds to 5 .
E
1
1
4
D
3
= C 1 B
2
2 1
1 A
–1
O
1
5
F
–4
–3
–2
Representation of
2 R
3
4
5
Example 4. Classroom activity (Constructing the ‘square root spiral’) :
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 P1P2 perpendicular to OP1 of unit length [see figure]. Now draw a line segment
P2P3 perpendicular to OP2. Then draw a line segment P3P4 perpendicular to OP3.
Continuing in this manner, we can get the line segment Pn – 1Pn by drawing a line
segment of unit length perpendicular to OPn – 1. In this manner, we will have created
the points : P1, P2, P3, ......, Pn, ......, and joined them to create a beautiful spiral depicting
2,
3,
4 , ... .
P3
P2
1
Pn
3
2
P1
1
O
Constructing square root spiral.
EXERCISE 1.3 (Page 14)
Example 1. Write the following in decimal form and say what kind of decimal
expansion each has :
(i)
36
100
(iii) 4
(v)
1
8
2
11
1
11
3
(iv)
13
(ii)
(vi)
329
.
400
16
GOLDEN SAMPLE PAPER (MATHEMATICS)–IX
36
= 0.36
100
The decimal expansion is terminating.
Sol. (i)
(ii)
11) 1.000000 ( 0.090909......
99
100
99
100
99
1
1
= 0.090909...... = 0.09
11
The decimal expansion is non-terminating repeating.
∴
(iii)
4
33
1
4×8+1
32 + 1
=
=
=
8
8
8
8
8 ) 33.000 ( 4.125
32
10
8
20
16
40
40
×
1
= 4.125
8
The decimal expansion is terminating.
∴ 4
(iv)
13 ) 3.00000000000 ( 0.230769230769......
26
40
39
100
91
90
78
17
NUMBER SYSTEMS
120
117
30
26
40
39
100
91
90
78
120
117
3
3
= 0.230769230769 ...... = 0.230769 .
13
The decimal expansion is non-terminating repeating.
(v)
11 ) 2.0000 ( 0.1818......
11
∴
90
88
20
11
90
88
2
2
= 0.1818...... = 0.18 .
11
The decimal expansion is non-terminating repeating.
(vi)
400 ) 329.0000 ( 0.8225
3200
∴
900
800
1000
800
2000
2000
×
18
GOLDEN SAMPLE PAPER (MATHEMATICS)–IX
329
= 0.8225
400
The decimal expansion is terminating.
∴
Example 2. You know that
expansions of
1
= 0.142857 . Can you predict what the decimal
7
2 3 4 5 6
, , , ,
are, without actually doing the long division ? If so,
7 7 7 7 7
how ?
[Hint: Study the remainders while finding the value of
Sol. Yes ! We can predict the decimal expansions of
1
carefully.]
7
2 3 4 5 6
, , , , , without
7 7 7 7 7
actually doing the long division as follows :
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
Example 3. Express the following in the form
p
, where p and q are integers
q
and q ≠ 0.
(i) 0.6
(ii) 0.47
(iii) 0.001 .
Sol. (i) Let
x = 0.6 = 0.6666......
Multiplying both sides by 10 (since one digit is repeating), we get
⇒
⇒
10x = 6.666......
10x = 6 + 0.6666......
10x – x = 6
⇒
x=
⇒
2
3
2
0.6 =
3
Thus,
x=
6
9
⇒ 10x = 6 + x
⇒ 9x = 6
19
NUMBER SYSTEMS
Here
p=2
q = 3 (≠ 0)
(ii) Let
x = 0.47 = 0.47777......
Multiplying both sides by 10 (since one digit is repeating), we get
⇒
⇒
⇒
Thus,
Here
10x = 4.7777......
10x = 4.3 + 0.47777......
⇒
10x – x = 4.3
x=
⇒
10x = 4.3 + x
9x = 4.3
4.3
43
=
9
90
43
90
p = 43
4.7 =
q = 90 (≠ 0).
(iii) Let
x = 0.001 = 0.001001001......
Multiplying both sides by 1000 (since three digits are repeating), we get
⇒
⇒
⇒
Thus,
Here
1000x = 1.001001......
1000x = 1 + 0.001001001......
⇒
1000x – x = 1
x=
⇒ 1000x = 1 + x
999x = 1
1
999
1
999
p=1
0.001 =
q = 999 (≠ 0).
p
. Are you surprised by your
q
answer ? With your teacher and classmates discuss why the answer makes sense.
Sol. Let
x = 0.99999......
Example 4. Express 0.99999...... in the form
Multiplying both sides by 10 (since one digit is repeating), we get
10x = 9.9999......
⇒
⇒
⇒
10x = 9 + 0.99999......
10x – x = 9
x=
9
=1
9
Thus, 0.99999...... = 1 =
Here
p=1
q = 1.
1
1
⇒ 10x = 9 + x
⇒ 9x = 9
20
GOLDEN SAMPLE PAPER (MATHEMATICS)–IX
Since 0.99999...... goes on for ever, so there is no gap between 1 and 0.99999......
end hence they are equal.
Example 5. What can the maximum number of digits be in the repeating block
1
? Perform the division to check your answer.
of digits in the decimal expansion of
17
Sol. The maximum number of digits in the repeating block of digits in the
1
decimal expansion of
can be 16.
17
17 ) 1.000000000000000000000000000000
85
150
136
( 0.05882352941176470588235294117647......
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
100
85
21
NUMBER SYSTEMS
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
1
= 0.0588235294117647
17
By Long Division, the number of digits in the repeating block of digits in the
Thus,
1
= 16.
17
∴ The answer is verified.
decimal expansion of
p
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 ?
Example 6. Look at several examples of rational numbers in the form
22
GOLDEN SAMPLE PAPER (MATHEMATICS)–IX
Sol. The property that q must satisfy in order that the rational numbers in the
p
(q ≠ 0), where p and q are integers with no common factors other than 1, have
q
terminating decimal representation (expansions) is that the prime factorisation of q
has only powers of 2 or powers of 5 or both, i.e., q must be of the form 2m × 5n ; m =
0, 1, 2, 3, ......, n = 0, 1, 2, 3, ...... .
Example 7. Write three numbers whose decimal expansions and nonterminating non-recurring.
Sol.
0.01001 0001 00001......,
0.20 2002 20003 200002......,
0.003000300003......,
Example 8. Find three different irrational numbers between the rational
form
9
5
and
.
11
7
Sol.
numbers
7 ) 5.000000 ( 0.714285........
49
10
7
30
28
20
14
60
56
40
35
5
Thus,
5
= 0.714285...... = 0.714285
7
11 ) 9.0000 ( 0.8181......
88
20
11
90
88
20
11
9
23
NUMBER SYSTEMS
Thus,
9
= 0.8181...... = 0.81
11
Three different irrational numbers between the rational numbers
can be taken as
5
9
and
7
11
0.75 075007500075000075......
0.7670767000767......,
0.808008000800008......,
Example 9. Classify the following numbers as rational or irrational :
(i)
23
(iii) 0.3796
(v) 1.101001000100001......
Sol. (i)
4
87
949
9585
95908
959163
9591661
95916625
959166302
9591663043
(ii)
225
(iv) 7.478478......
4.795831523
23.00 00 00 00 00 00 00 00 00
16
700
609
9100
8541
55900
47925
797500
767264
3023600
2877489
14611100
9591661
501943900
479583125
2236077500
1918332604
31774489600
28774989129
2999500471
Thus,
23 = 4.795831523......
24
GOLDEN SAMPLE PAPER (MATHEMATICS)–IX
... The decimal expansion is non-terminating non-recurring.
∴
(ii)
23 is an irrational number.
15
1
225
1
25
125
125
×
225 = 15 =
∴
225 is a rational number.
Here
and
15
1
...
p = 15
q = 1(≠ 0).
.
.
(iii) . The decimal expansion is terminating.
∴ 0.3796 is a rational number.
(iv) 7.478478...... = 7.478
... The decimal expansion is non-terminating recurring.
∴ 7.478478...... is a rational number.
(v) 1.101001000100001......
... The decimal expansion is non-terminating non-recurring.
∴ 1.101001000100001...... is an irrational number.
EXERCISE 1.4 (Page 18 )
Example 1. Visualise 3.765 on the number line, using successive magnification.
Sol.
Example 2. Visualize 4.26 on the number line, up to 4 decimal places.
Sol.
4.26 = 4.262626......
25
NUMBER SYSTEMS
EXERCISE 1.5 (Page 24 )
Example 1. Classify the following numbers as rational or irrational:
(i) 2 –
(iii)
5
2 7
(ii) (3 +
(iv)
23 ) –
23
1
2
7 7
(v) 2π.
Sol. (i) ... 2 is a rational number and
∴ 2–
5 is an irrational number.
5 is an irrational number.
... The difference of a rational number and
an irrational number is irrational.
(ii) (3 + 23 ) – 23 = 3 + 23 –
which is a rational number.
23 = 3
2
7
7 7
which is a rational number.
(iv) ... 1(≠ 0) is a rational number and
(iii)
∴
2 7
1
2
=
2 (≠ 0) is an irrational number.
is an irrational number.
... The quotient of a non-zero rational number
with an irrational number is irrational.
.
.
(v) . 2 is a rational number and π is an irrational number.
∴ 2π is an irrational number.
... The product of a non-zero rational number
with an irrational number is irrational.
26
GOLDEN SAMPLE PAPER (MATHEMATICS)–IX
Example 2. Simplify each of the following expressions:
3 )(2 +
(i) (3 +
(iii) ( 5 +
Sol. (i) (3 +
2)
(ii) (3 +
2 )2
3 )(3 –
(iv) ( 5 –
3 )(2 +
2 ) = 3(2 +
2)+
3)
2 )( 5 +
3 (2 +
2 ).
2)
| Left Distributive law of multiplication over addition
= (3)(2) + 3 2 + ( 3 )(2) + ( 3 )( 2 )
| ...
= 6 + 3 2 + 2 3 + (3)(2)
=6+3 2 +2 3 +
(ii) (3 +
3 )(3 –
(iii) ( 5 +
a
b =
ab
a
b =
ab
6.
3 ) = (3)2 – ( 3 )2
= 9 – 3 = 6.
2 )2 = ( 5 )2 + 2 5
2 + ( 2 )2
| ...
= 5 + 2 10 + 2
= 7 + 2 10 .
(iv) ( 5 – 2 )( 5 + 2 ) = ( 5 )2 – ( 2 )2 = 5 – 2 = 3.
Example 3. Recall, π is defined as the ratio of the circumference (say c) of a
c
circle to its diameter (say d). That is, π = . This seems to contradict the fact that π is
d
irrational. How will you resolve this contradiction?
c
22
Sol. Actually
=
which is an approximate value of π.
d
7
Example 4. Represent
Sol.
9.3 on the number line.
9.3
10
9
Mark the distance 9.3 from a fixed point A on a given line to obtain a point B
such that AB = 9.3 units. From B mark a distance of 1 unit and mark the new point as
C. Find the mid-point of AC and mark that point as O. Draw a semi-circle with centre
O and radius OC. Draw a line perpendicular to AC passing through B and intersecting
the semi-circle at D. Then BD =
9.3 .
D
9.3
O
A
9.3
1
B
C
Example 5. Rationalise the denominators of the following:
(i)
1
7
(ii)
1
7− 6
27
NUMBER SYSTEMS
(iii)
1
5+ 2
1
Sol. (i)
=
7
1
7− 6
1
=
7 −2
7
×
.
| Multiplying and dividing by
7
7
=
(ii)
1
(iv)
7
.
7
1
7− 6
7+ 6
×
7+ 6
| Multiplying and dividing by
=
(iii)
1
5+ 2
7+ 6
=
7−6
1
=
5+ 2
7 +
×
=
1
7−2
=
5− 2
=
5−2
1
7−2
×
7+2
=
7−4
7+2
7+2
7+2
.
3
(ii) 321/5
(64)1/2 = (82)1/2
= 82 × 1/2 = 81 = 8.
321/5 = (25)1/5
= 25 × 1/5 = 21 = 2.
(iii)
1251/3 = (53)1/3
= 53 × 1/3 = 51 = 5.
(iii)
2
5− 2
.
3
(ii)
Example 2. Find:
(i) 93/2
(iii) 163/4
Sol. (i)
(ii)
5 –
5− 2
EXERCISE 1.6 (Page 26 )
Example 1. Find:
(i) 641/2
(iii) 1251/3.
Sol. (i)
6
5− 2
| Multiplying and dividing by
=
7 +
6.
| Multiplying and dividing by
(iv)
7
(ii) 322/5
(iv) 125–1/3.
93/2 = (91/2)3 = 33 = 27.
322/5 = (25)2/5 = 25 × 2/5 = 22 = 4.
163/4 = (24)3/4 = 24 × 3/4 = 23 = 8.
7 +2
28
GOLDEN SAMPLE PAPER (MATHEMATICS)–IX
125–1/3 = (53)–1/3
(iv)
= 53 × (–1/3) = 5–1 =
1
.
5
Example 3. Simplify:
(i)
22/3
(iii)
111/2
(iii)
(iv)
7
(iv) 71/2 . 81/2.
111/4
Sol. (i)
(ii)
.
 1 
(ii)  3 
3 
21/5
22/3 . 21/5 = 22/3 + 1/5 = 2
10 + 3
15
= 213/15.
7
1
17
 1
=
= 21 = 3–21.
 3
3 7
3
(3 )
3 
111/ 2
= 111/2 – 1/4 = 111/4.
111/ 4
71/2 . 81/2 = (7 . 8)1/2 = 561/2.