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CHAPTER1
POINTS TO REMEMBER
1. Definition of a rational number.A number r is called a rational number, if
it can be written in the formp
q, wherepand qare integers and q0.
Note.We visit that q0 because division by zero is not allowed.
2. Equivalent rational numbers. Rational numbers do not have a unique
representation in the formp
q, wherepand qare integers and q0. For example,
1
2
2
4= =
3
6
4
8
5
10= = = ......, and so on, these are called equivalent rational
numbers.
3. Standard form of a rational number.A rational number r= pq
, q0 is said
to be in its standard form ifpand qare 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 formp
q, wherepand qare integers and q0.
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.
NUMBER SYSTEMS
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12 GOLDEN SAMPLE PAPER (MATHEMATICS)IX
9. Decimalexpansion of an irrational number.The decimal expansion of an
irrational number is non-terminating non-recurring. Conversely, a numberwhose 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 nbe a positive integer. Then
an = bmeans bn
= aand b> 0. Here, the symbol is called the radicalsign. In particular, if ais a real number, then a = bmeans b2= aand 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 (ii) (am)n= amn
(iii)a
a
m
n = am n, m> n (iv) am bm= (ab)m
Here, a, nand mare natural numbers. a is called the base and mand n are
called the exponents.
14. Definition.Let a> 0 be a real number. Let mand nbe integers such that mand
nhave no common factors other than 1, and n> 0. Then,
a a amn n
m mn= =d i
15. Rationalisation.To rationalise the denominator of1
a b+, we multiply this
bya b
a b
, where aand bare integers.
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NUMBER SYSTEMS 13
NCERT TEXTBOOK EXERCISES WITH SOLUTIONS
EXERCISE 1.1 (Page 5)
Example 1.Is zero a rational number ? Can you write it in the formp
q, where
p and q are integers and q 0 ?
Sol.Yes ! zero is a rational number. We can write zero in the formp
q, wherep
and qare integers and q0 as follows :
0 =0
1=
0
2=
0
3etc.,
denominator qcan also be taken as negative integer.
Example 2.Find six rational numbers between 3 and 4.
Sol.+3 4
2=
7
2
+7
32
2=
13
4
+13
34
2=
25
8
+25
38
2=
49
16
+49
316
2=
97
32
+97
332
2= 193
64.
Thus, six rational numbers between 3 and 4 are7
2,
13
4,
25
8,
49
16,
97
32and
193
64.
Aliter.We write 3 and 4 as rational numbers with denominator 6 + 1 (= 7), i.e.,
3 =3
1=
3 7
1 7=
21
7
and 4 =4
1
=4 7
1 7
=28
7.
Thus, six rational numbers between 3 and 4 are22
7,
23
7,
24
7,
25
7,
26
7and
27
7.
Note.This is known as the method of finding rational numbers in one step.
Example 3.Find five rational numbers between3
5and
4
5.
Sol.3
5=
30
50,
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14 GOLDEN SAMPLE PAPER (MATHEMATICS)IX
4
5 =
40
50 .
Therefore, five rational numbers between3
4and
4
5are
31
50,
32
50,
33
50,
34
50,
35
50.
Example 4. State whether the following statements are true or false. Givereasons 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 example1
2is a rational number but not a whole number.
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, 4 = 2 is a rational number.
Example 3.Show how 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.
Then OB = 2 21 1+ = 2
Construct BD of unit length perpendicular to OB.
Then OD = 2 2( 2) 1+ = 3
Construct DE of unit length perpendicular to OD.
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NUMBER SYSTEMS 15
Then OE =
2 2
( 3) 1+ = 4 = 2Construct EF of unit length perpendicular to OE.
Then OF = 2 22 1+ = 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 .
3 2 1 O 1 2 3
1 1
1
5
3
21
1
4=2
C B
DE
F
R
A
4 4
Representation of 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 OP1of unit length. Draw a line
segment P1P
2perpendicular to OP
1of unit length [see figure]. Now draw a line segment
P2P
3 perpendicular to OP
2. Then draw a line segment P
3P
4 perpendicular to OP
3.
Continuing in this manner, we can get the line segment Pn 1
Pnby drawing a line
segment of unit length perpendicular to OPn 1
. In this manner, we will have created
the points : P1, P
2, P
3, ......, P
n,......, and joined them to create a beautiful spiral depicting
2,
3,
4, ... .
1
1
P3 P
2
P1
3
2
O
Pn
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(ii)
1
11
(iii) 41
8(iv)
3
13
(v)2
11(vi)
329
400.
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16 GOLDEN SAMPLE PAPER (MATHEMATICS)IX
Sol.(i)
36
100 = 0.36
The decimal expansion is terminating.
(ii) 11) 1.000000 ( 0.090909......
99
100
99
100
99
1
1
11= 0.090909...... = 0.09
The decimal expansion is non-terminating repeating.
(iii) 41
8=
4 8 1
8
+=
32 1
8
+=
33
8
8 ) 33.000 ( 4.125
32
10
8
2016
40
40
41
8= 4.125
The decimal expansion is terminating.
(iv) 13 ) 3.00000000000 ( 0.230769230769......
26
40
39
100
91
90
78
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NUMBER SYSTEMS 17
120
117
30
26
40
39
100
91
90
78
120
117
3
3
13= 0.230769230769 ...... = 0.230769 .
The decimal expansion is non-terminating repeating.
(v) 11 ) 2.0000 ( 0.1818......
11
90
88
20
11
90
88
2
2
11= 0.1818...... = 0.18 .
The decimal expansion is non-terminating repeating.
(vi) 400 ) 329.0000 ( 0.8225
3200
900
800
1000
800
2000
2000
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18 GOLDEN SAMPLE PAPER (MATHEMATICS)IX
329
400 = 0.8225
The decimal expansion is terminating.
Example 2.You know that1
7= 0.142857. Can you predict what the decimal
expansions of2
7,
3
7,
4
7,
5
7,
6
7are, without actually doing the long division ? If so,
how ?
[Hint:Study the remainders while finding the value of1
7carefully.]
Sol.Yes ! We can predict the decimal expansions of2
7
,3
7
,4
7
,5
7
,6
7
, without
actually doing the long division as follows :
2
7= 2
1
7= 2 0.142857 = 0.285714
3
7= 3
1
7= 3 0.142857 = 0.428571
4
7= 4
1
7= 4 0.142857 = 0.571428
5
7= 5
1
7= 5 0.142857 = 0.714285
67
= 6 17
= 6 0.142857 = 0.857142 .
Example 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 .
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= 6 +x
10xx= 6 9x= 6
x=6
9
x=2
3
Thus, 0.6 =2
3
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NUMBER SYSTEMS 19
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
10x= 4.7777......
10x= 4.3 + 0.47777...... 10x= 4.3 +x
10xx= 4.3 9x= 4.3
x=4.3
9=
43
90
Thus, 4.7 =43
90
Here p= 43
q= 90 (0).
(iii) Let x= 0.001 = 0.001001001......
Multiplying both sides by 1000 (since three digits are repeating), we get
1000x= 1.001001......
1000x= 1 + 0.001001001...... 1000x= 1 +x
1000xx= 1 999x= 1
x=1
999
Thus, 0.001 =1
999
Here p= 1
q= 999 (0).
Example 4.Express 0.99999...... in the formp
q. Are you surprised by your
answer ? With your teacher and classmates discuss why the answer makes sense.
Sol.Let x= 0.99999......
Multiplying both sides by 10 (since one digit is repeating), we get
10x= 9.9999......
10x= 9 + 0.99999...... 10x= 9 +x 10xx= 9 9x= 9
x=9
9= 1
Thus, 0.99999...... = 1 =1
1
Here p= 1
q= 1.
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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
of digits in the decimal expansion of1
17? Perform the division to check your answer.
Sol.The maximum number of digits in the repeating block of digits in the
decimal expansion of1
17can be 16.
17 ) 1.000000000000000000000000000000
85 ( 0.05882352941176470588235294117647......
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
100
85
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NUMBER SYSTEMS 21
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
17= 0.0588235294117647
By Long Division, the number of digits in the repeating block of digits in the
decimal expansion of1
17= 16.
The answer is verified.
Example 6.Look at several examples of rational numbers in the formp
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 ?
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22 GOLDEN SAMPLE PAPER (MATHEMATICS)IX
Sol.The property that qmust satisfy in order that the rational numbers in the
formp
q(q0), wherepand qare integers with no common factors other than 1, have
terminating decimal representation (expansions) is that the prime factorisation of q
has only powers of 2 or powers of 5 or both, i.e.,qmust 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 non-
terminating non-recurring.
Sol. 0.01001 0001 00001......,
0.20 2002 20003 200002......,
0.003000300003......,
Example 8. Find three different irrational numbers between the rational
numbers5
7and
9
11.
Sol. 7 ) 5.000000 ( 0.714285........
49
10
7
30
28
20
14
60
56
40
35
5
Thus,5
7= 0.714285...... = 0.714285
11 ) 9.0000 ( 0.8181......
8820
11
90
88
20
11
9
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NUMBER SYSTEMS 23
Thus,
9
11 = 0.8181...... = 0.81
Three different irrational numbers between the rational numbers5
7and
9
11can be taken as
0.75 075007500075000075......
0.7670767000767......,
0.808008000800008......,
Example 9.Classify the following numbers as rational or irrational :
(i) 23 (ii) 225
(iii) 0.3796 (iv) 7.478478......(v) 1.101001000100001......
Sol.(i) 4.795831523
4 23.00 00 00 00 00 00 00 00 00
16
87 700
609
949 9100
8541
9585 55900
47925
95908 797500
767264
959163 3023600
2877489
9591661 14611100
9591661
95916625 501943900
479583125
959166302 2236077500
1918332604
9591663043 31774489600
28774989129
2999500471
Thus, 23 = 4.795831523......
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24 GOLDEN SAMPLE PAPER (MATHEMATICS)IX
... The decimal expansion is non-terminating non-recurring.
23 is an irrational number.
(ii)
15
1 225
1
25 125
125
... 225 = 15 = 151
225 is a rational number.
Here p= 15
and 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 4.26on the number line, up to 4 decimal places.
Sol. 4.26 = 4.262626......
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EXERCISE 1.5 (Page 24)
Example 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) 2.
Sol.(i) ... 2 is a rational number and 5 is an irrational number.
2 5 is an irrational number.... The difference of a rational number and
an irrational number is irrational.
(ii) (3 + 23 ) 23 = 3 + 23 23 = 3
which is a rational number.
(iii)2 7
7 7=
2
7
which is a rational number.
(iv) ... 1(0) is a rational number and 2 (0) is an irrational number.
1
2 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 is an irrational number.
2is an irrational number.... The product of a non-zero rational number
with an irrational number is irrational.
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NUMBER SYSTEMS 27
(iii)
1
5 + 2 (iv)
1
7 2 .
Sol.(i)1
7=
1
7
7
7| Multiplying and dividing by 7
=7
7.
(ii)1
7 6=
1
7 6
7 6
7 6
+
+
| Multiplying and dividing by 7 + 6
= 6
7 6
7
+
= 7 + 6 .
(iii)1
5 2+=
1
5 2+
5 2
5 2
| Multiplying and dividing by 5 2
=5 2
5 2
=
5 2
3
.
(iv)1
7 2=
1
7 2
7 2
7 2
+
+
| Multiplying and dividing by 7 + 2
=7 2
7 4
+
=
7 2
3
+.
EXERCISE 1.6 (Page 26)
Example 1.Find:
(i) 641/2 (ii) 321/5
(iii) 1251/3.
Sol.(i) (64)1/2= (82)1/2
= 82 1/2= 81= 8.
(ii) 321/5= (25)1/5
= 25 1/5= 21= 2.
(iii) 1251/3= (53)1/3
= 53 1/3= 51= 5.
Example 2.Find:
(i) 93/2 (ii) 322/5
(iii) 163/4 (iv) 1251/3.
Sol.(i) 93/2= (91/2)3= 33= 27.
(ii) 322/5= (25)2/5= 25 2/5= 22= 4.
(iii) 163/4= (24)3/4= 24 3/4= 23= 8.
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28 GOLDEN SAMPLE PAPER (MATHEMATICS)IX
(iv) 1251/3= (53)1/3
= 53 (1/3)= 51=1
5.
Example 3.Simplify:
(i) 22/3. 21/5 (ii)
7
3
1
3
(iii)
1/2
1/4
11
11(iv) 71/2. 81/2.
Sol.(i) 22/3. 21/5= 22/3 + 1/5 =
10 3
152
+
= 213/15.
(ii)7
3
1
3
=7
3 7
1
(3 )= 21
1
3= 321.
(iii)1/ 2
1/ 4
11
11= 111/2 1/4= 111/4.
(iv) 71/2. 81/2= (7 . 8)1/2= 561/2.