Chapter – 1

Real Number

EXERCISE 1.1
1. Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225 (ii) 196 and 38220 (iii) 867 and 255
Sol. (i) 135 and 225
Since 225 > 135, we apply the division lemma to 225 and 135 to obtain
225 = 135 �1 + 90
Since remainder 90 �0, we apply the division lemma to 135 and 90 to obtain
135 = 90 �1 + 45
We consider the new divisor 90 and new remainder 45, and apply the division lemma to obtain 90 = 2 �45 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 45,
Therefore, the HCF of 135 and 225 is 45.

(ii) 196 and 38220

Since 38220 > 196, we apply the division lemma to 38220 and 196 to obtain
38220 = 196 �195 + 0
Since the remainder is zero the process stops.
Since the divisor at this stage is 196,
Therefore, HCF of 196 and 38220 is 196.

(iii) 867 and 255

Since 867 > 255, we apply the division lemma to 867 and 255 to obtain
867 = 255�3 + 102
Since remainder 102 �0, we apply the division lemma to 255 and 102 to obtain
255 = 102 �2 + 51
We consider the new divisor 102 and new remainder 51, and apply the division lemma to obtain
102 = 51 �2 + 0
Since the remainder is zero, the process stops.
Since the divisor at this stage is 51,
Therefore, HCF of 867 and 255 is 51.

2. Show that any positive odd integer is of the form 6q+1 or 6q+3 or 6q+5 , where q is some integer.
Sol. Let a be any positive integer and b = 6. Then, by Euclid’s algorithm, a = 6q + r for some integer q �0, and
r = 0,1, 2,3, 4,5 because 0 �r < 6 .

Therefore, a = 6q or 6q + 1 or 6q + 2 or 6q + 3 or 6q + 4 or 6q + 5

Also,
6q + 1 = 2 �3q + 1 = 2k1 + 1, where k1 is a positive integer
6q + 3 = (6q + 2) + 1 = 2(3q + 1) + 1 = 2k2 + 1, where k2 is an integer
6q + 5 = (6q + 4) + 1 = 3(3q + 2) + 1 = 2k3 + 1, where k3 is an integer

Clearly, 6q + 1,6q + 3,6q + 5 are of the form 2k + 1, where k is an integer.

Therefore 6q + 1, 6q + 3,6q + 5 are not exactly divisible by 2.
Hence, these expressions of numbers are odd numbers.
And therefore, any odd integer can be expressed in the form 6q + 1,
 Or 6q + 3,
Or 6q + 5
3. An army contingent of 616 members is to march behind an army band of 32 members is a parade. The two
groups are to march in the same number of columns. What is the maximum number of columns in which they
can march?
Sol. The maximum number of columns would be the HCF of (616, 32)
We can find the HCF of 616 and 32 by using Euclid Division Algorithm.
Therefore,

616 = 19 �32 + 8
32 = 4 �8 + 0
So, HCF of 616 and 32 is 8

Hence, the maximum number of columns in which they can march is 8.

4. Use Euclid’s division lemma to show that the square of any positive integer is either of form 3m or 3m + 1 for
some integer m.
[Hint: Let x be any positive integer then it is of the form 3q, 3q + 1 or 3q + 2. How square each of these and
show that they can be rewritten in the form 3m or 3m + 1.]
Sol. Let a be any positive integer and b = 3.
Then a = 3q + r for some integer q �0
And r = 0, 1, 2 because 0 �r < 3
Therefore, a = 3q or 3q + 1 or 3q + 2
Or,
a 2 = (3q ) 2 or (3q + 1) 2 or (3q + 2) 2
a 2 = (9q 2 )or 9q 2 + 6q + 1 or 9q 2 + 12q + 4
= 3 �(3q 2 ) or 3(3q 2 + 2q) + 1 or 3(3q 2 + 4q + 1) + 1
= 3k1 or 3k2 + 1 or 3k3 + 1
k ,k k
Where 1 2 and 3 are some positive integers
Hence, it can be said that the square of any positive integer is either of the form 3m or 3m + 1.

5. Use Euclid’s division lemma to show that the cube of any positive integer is of the form 9m, 9m + 1 or 9m + 8.
Sol. Let a be any positive integer and b = 3
a = 3q + r , where q �0 and 0 �r < 3
\a = 3q or 3q +1 or 3q + 2
Therefore, every number can be represented as these three forms.
There are three cases.
Case 1: When a = 3q,
a 3 = (3q)3 = 27 q 3 = 9(3q 3 ) = 9m,

Where m is an integer such that m = 3q

3

Case 2: When a = 3q + 1,
a 3 = (3q + 1)3
a 3 = 27 q3 + 27 q 2 + 9q + 1
a 3 = 9(3q 3 + 3q 2 + q) + 1
 a 3 = 9m + 1
Where m is an integer such that m = (3q + 3q + q )
3 2

Case 3: When a = 3q + 2,
a 3 = (3q + 2)3
a 3 = 27q 3 + 54q 2 + 36q + 8
a 3 = 9(3q 3 + 6q 2 + 4q) + 8
a 3 = 9m + 8
Where m is an integer such that m = (3q + 6q + 4q )
3 2

Therefore, the cube of any positive integer is of the form 9m, 9m + 1, or 9m + 8.

EXERCISE 1.2
6. Express each number as product of its prime factors:
(i) 140 (ii) 156 (iii) 3825 (iv) 5005
(v) 7429
Sol. Express each number as a product of its prime factors:
(i) 140
140 = 2 �2 �5 �7
= 22 �5 �7

(ii) 156
156 = 2 �2 �13 �3
= 22 �3 �13

(iii) 3825
3825 = 3 �3 �5 �5 �17
= 32 �52 �7

(iv) 5005
5005 = 5 �7 �11�13

(v) 7429
7429 = 17 �19 �23
Hence,

7. Find the LCM and HCF of the following pairs of integers and verify that LCM �HCF = product of the two
numbers.
(i) 26 and 91 (ii) 510 and 92 (iii) 336 and 54
Sol. (i) 26 and 91
26 = 2 �13
91 = 7 �13
HCF = 13
LCM = 2 �7 �13 = 182
Product of the two numbers = 26 �91= 2366
HCF �LCM = 13 �182 = 2366
Hence, product of two numbers = HCF �LCM
 (ii) 510 and 92
510 = 2 �3 �5 �17
92 = 2 �2 �23
HCF = 2
LCM = 2 �2 �3 �5 �17 �23 = 23460
Product of the two numbers = 510 �92 = 46920
HCF �LCM = 2 �23460
= 46920
Hence, product of two numbers = HCF �LCM

(iii) 336 and 54

336 = 2 �2 �2 �2 �3 �7
336 = 24 �3 �7
54 = 2 �3 �3 �3
54 = 2 �33
HCF = 2 �3 = 6
LCM = 2 4 �33 �7 = 3024
Product of the two numbers = 336 �54 = 18144
HCF×LCM = 6 �3024 = 18144
Hence, product of two numbers = HCF �LCM

8. Find the LCM and HCF of the following integers by applying the prime factorisation method.
(i) 12, 15 and 21 (ii) 17, 23 and 29 (iii) 8, 9 and 25
Sol. (i) 12, 15 and 21
12 = 22 �3
15 = 3 �5
21 = 3 �7
HCF = 3
LCM = 22 ×3×5× 7 = 240

(ii) 17, 23 and 29

17=1 �17
23 = 1 �23
29 = 1 �29
HCF = 1
LCM = 17 �23 �29 = 11339

(iii) 8, 9 and 25
8 = 2 �2 �2
9 = 3 �3
25 = 5 �5
HCF = 1
LCM = 2 �2 �2 �3 �3 �5 �5 = 1800

9. Given that HCF (306, 657) = 9, find LCM (306, 657).

Sol. We know, HCF �LCM = Product of two numbers
i.e. 9 �LCM = 306 �657
 306�657
LCM = = 22338
9

n
10. Check whether 6 can end with the digit 0 for any natural number n.
n n
Sol. If the number 6 ends with the digit zero, then it is divisible by 5.Therefore the prime factorization of 6
n
contains the prime 5.This is not possible because the only prime in the factorization of 6 is 2 and 3 and the
uniqueness of the fundamental theorem of arithmetic guarantees that these are no other prime in the
n
factorization of 6
n
Hence, it is very clear that there is no value of n in natural number for which 6 ends with the digit zero.

11. Explain why 7 �11 �13 + 13 and 7 �6 �5 �4 �3 �2 �1 + 5 are composite numbers.

Sol. Numbers are of two types – prime and composite. Prime numbers can be divided by 1 and only itself, whereas
composite numbers have factors other than 1 and itself.
It can be observed that
7 �11 �13 + 13 = 13 �(7 �11 + 1) = 13 �(77 + 1)
= 13 �78
= 13 �13 �6
The given expression has 6 and 13 as it factors. Therefore, it is a composite number.
7 �6 �5 �4 �3 �2 �1 + 5 = 5 �(7 �6 �4 �3 �2 �1 + 1)
= 5 �(1008 + 1) = 5 �1009
1009 cannot be factorised further. Therefore, the given expression has 5 and 1009 as its factors. Hence, It is a
composite number.

12. There is a circular path around a sports field. Sonia takes 18 minutes to drive one round of the field, while Ravi
takes 12 minutes for the same. Suppose they both start at the same point and at the same time, and go in the
same direction. After how many minutes will they meet again at the starting point?
Sol. Since, Both Sonia and Ravi move in the same direction and at the same time, the method to find the time when
they will be meeting again at the starting point is LCM(18,12) is 2 �3 �3�2 �1 = 36
Therefore, Sonia and Ravi will meet again after 36 minutes.

EXERCISE 1.3
13. Prove that 5 is irrational.

Sol. Let 5 is a rational number.

a
5=
Therefore, we can find two integers a, b(b �0) such that b
Let a and b have a common factor other than 1. Then we can divide them by the common factor, and assume
that a and b are co-prime.
a = 5b
a 2 = 5b 2
2
Therefore, a is divisible by 5 and it can be said that a is divisible by 5.
Let a = 5k, where k is an integer
(5k )2 = 5b2
b 2 = 5k 2 . This means that b 2 is divisible by 5 and hence, b is divisible by 5.
This implies that a and b have 5 as a common factor.
And this is a contradiction to the fact that a and b are co-prime.
 p
Hence, 5 cannot be expressed as q or it can be said that 5 is irrational.

14. Prove that 3 + 2 5 is irrational.

Sol. Let 3 + 2 5 is rational.

Therefore, we can find two integers a, b(b �0) such that
a
3+ 2 5 =
b
a
2 5= -3
b
1 �a �
5 = � - 3�
2 �b �
1 �a �
� - 3�
Since a and b are integers, 2 �b �will also be rational and therefore, 5 is rational.

This contradicts the fact that 5 is irrational. Hence, our assumption that 3 + 2 5 is rational is false.

Therefore, 3 + 2 5 is irrational.

15. Prove that the following are irrationals:

1
(i) 2 (ii) 7 5 (iii) 6 + 2
1
Sol. (i) 2
1
Let 2 is rational.
Therefore, we can find two integers a, b (b �0)
1 a
=
2 b
b
2=
a
b
a is rational as a and b are integers.

Therefore, 2 is rational which contradicts to the fact that 2 is irrational.

1
Hence, our assumption is false and 2 is irrational.

(ii) 7 5

Let 7 5 is rational.
Therefore, we can find two integers a, b (b �0) such that
a
7 5=
b for some integers a and b
a
\ 5=
7b
 a
7b is rational as a and b are integers.

Therefore, 5 should be rational.

This contradicts the fact that 5 is irrational. Therefore, our assumption that 7 5 is rational is false. Hence,
7 5 is irrational.

(iii) 6 + 2
Let 6 + 2 be rational.
Therefore, we can find two integers, a, b(b �0)
a
6+ 2=
b
a
2= -6
b
a
-6
Since a and b are integers, b is also rational and hence, 2 should be rational. This contradicts the fact that
2 is irrational. Therefore, our assumption is false and hence, 6 + 2 out assumption is false and hence,
6 + 2 is irrational.

EXERCISE 1.4
16. Without actually performing the long division, state whether the following rational numbers will have a
terminating decimal expansion or a non-terminating repeating decimal
13 17 64 15
17. (i) 3125 (ii) 8 (iii) 455 (iv) 1600
29 23 129 6
3 2 2 7 5
(v) 343 (vi) 2 5 (vii) 2 5 7 (viii) 15
35 77
(ix) 50 (x) 210
13
Sol. (i) 3125

3125 = 55
m
The denominator is of the form 5 .

13
Hence, the decimal expansion of 3125 is terminating.

17
(ii) 8
8 = 23
m
The denominator is of the form 2 .
17
Hence, the decimal expansion of 8 is terminating.
64
(iii) 455
455 = 5 �7 �13
 m n
Since the denominator is not in the form 2 �5 , and it also contains 7 and 13 as its factors, its decimal
expansion will be non-terminating repeating.
15 3
=
(iv) 1600 320
320 = 26 �5
m n
The denominator is of the form 2 �5 .
15
Hence, the decimal expansion of 1600 is terminating.
29
(v) 343
343 = 73
29
m n
Since the denominator is not in the form 2 �5 , and it has 7 as its factor, the decimal expansion of 343 is
non-terminating repeating.
23
3 2
(vi) 2 5
m n
The denominator is of the form 2 �5 .
23
3 2
Hence, the decimal expansion of 2 �5 is terminating.
129
2 7 5
(vii) 2 5 7
m n
Since the denominator is not of the form 2 �5 , and it also has 7 as its factor, the decimal expansion of
129
2 �57 �75 is non-terminating repeating.
2

6
(viii) 15
6 2 �3 2
= =
15 3 �5 5
n
The denominator is of the form 5 .
6
Hence, the decimal expansion of 15 is terminating.
35
(ix) 50
35 7 �5 7
= =
50 10 �5 10
10 = 2 �5
m n
The denominator is of the form 2 �5 .
35
Hence, the decimal expansion of 50 is terminating.
77
(x) 210
77 11 �7 11
= =
210 30 �7 30
30 = 2 �3 �5
 77
m n
Since the denominator is not of the form 2 �5 , and it also has 3 as its factors, the decimal expansion of 210
is non-terminating repeating.
18. Write down the decimal expansions of those rational numbers in Question 1 above which have terminating
decimal expansions.
13
= 0.00416
Sol. (i) 3125

17
= 2.125
(ii) 8

15
= 0.009375
(iv) 1600

23 23
3 2
= = 0.115
(vi) 2 �5 200
 6 2 �3 2
= = = 0.4
(viii) 15 3 �5 5

35 7
= = 0.7
(ix) 50 10

19. The following real numbers have decimal expansions as given below. In each case, decide whether they are
p
rational or not. If they are rational, and of the form q , what can you say about the prime factors of q?
(i) 43.123456789
(ii) 0.120120012000120000…
(iii) 43.123456789
Sol. (i) 43.123456789
p
Since this number has a terminating decimal expansion, it is a rational number of the form q and q is of the
m n
form 2 �5
i.e., the prime factors of q will be either 2 or 5 or both.
(ii) 0.120120012000120000…
The decimal expansion is neither terminating nor recurring. Therefore, the given number is an irrational
number.
(iii) 43.123456789
p
Since the decimal expansion is non-terminating recurring the given number is a rational number of the form q
m n
and q is not of the form 2 �5