Lesson: Real Numbers
Exercise 1.1 (5)
Question: 1
Use Euclid’s division algorithm to find the HCF of:
(i) 135 and 225
(ii) 196 and 38220
(iii)
867 and 255
Solution:
(i) 135 and 225
Since,
225 135
.
Starting with the larger number 225.
Apply Euclid’s division algorithm,
225 135 1 90
Remainder
90 0
, so apply Euclid’s division
algorithm to
135
and
90
,
135 90 1 45
Since remainder
45 0
,
so apply Euclid’s division
algorithm to
90
and
45
,
90 45 2 0
Now, the remainder is zero and the divisor at the
end is 45.
Hence, the HCF of 135 and 225 is 45.
That is,
.
(ii) 196 and 38220
Since,
38220 196
.
Starting with the larger number 38220.
Apply Euclid’s division algorithm,
38220 196 195 0
Since, the remainder is zero and the divisor at the
end is
195
,
Hence, the HCF of 38220 and 196 is 195.
That is,
HCF 38220,196 195
.
(iii) 867 and 255
Since,
867 225
.
Starting with the larger number 867.
Applying Euclid’s division algorithm,
867 255 3 102
Remainder
102 0
, so apply Euclid’s division
algorithm to 255 and 102,
255 102 2 51
Since remainder
51 0
,
so apply Euclid’s division
algorithm to 102 and 51,
102 51 2 0
Now, the remainder is zero and the divisor at the
end is 51.
Hence, the HCF of 867 and 255 is 51.
That is,
HCF 867,225 51
.
Question: 2
Show that any positive odd integer is of the form
6q 1
, or
6q 3
, or
6q 5
, where
q
is some integer.
Solution:
Let two integers be a and b where
b 6
.
Using Euclid’s division algorithm,
a bq 1 ......(1)
The value of
r
varies from
0 r 6
.
Substitute
b 6
in equation
1
.
a 6q r ......(2)
Substitute
r 0
in equation (2).
a 6q
,
6q
is divisible by 6. Therefore,
6q
is even.
Substitute
r 1
in equation (2).
a 6q 1
,
6q 1
is not divisible by 2. Therefore,
6q 1
is odd.
Substitute
r 2
in equation (2).
a 6q 2
,
6q 2
is divisible by 2. Therefore,
6q 2
is even.
Substitute
r 3
in equation (2).
a 6q 3
,
6q 3
is not divisible by 2. Therefore,
6q 3
is odd.
Substitute
r 4
in equation (2).
a 6q 4
,
6q 4
is divisible by 2. Therefore,
6q 4
is even.
Substitute
r 5
in equation (2).
a 6q 5
,
6q 5
is not divisible by 2. Therefore,
6q 5
is odd.
This shows that, numbers of the form
6q
,
6q 2
and
6q 4
are even, and numbers of the form
6q 1
,
6q 3
and
6q 5
are odd.
Question: 3
An army contingent of 616 members is to march
behind an army band of
32
The two groups are to march in the same number of
columns. What is the maximum number of columns
in which they can march.
Solution:
The maximum number of columns is the HCF of
616 and 32.
Since,
616 32
.
Starting with the larger number 616.
Applying Euclid’s division algorithm,
616 32 19 8
Remainder
8 0
, so apply Euclid’s division
algorithm to 32 and 8,
32 8 4 0
Now, the remainder is zero and the divisor at the
end is
8
.
Hence, the HCF of
616
and
32
is 8.
That is,
HCF 616,32 8
.
Thus, the maximum number of columns in which
they can march is eight.
Question: 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
.
Solution:
Let two integers be a and b, where
b 3
.
Using Euclid’s division algorithm,
a bq r ......(1)
The value of
r
varies from
0 r 3
.
Substitute
b 3
in equation
1
,
a 3q r ......(2)
For
r 0
,
a 3q
For
r 1
,
a 3q 1
For
r 2
,
a 3q 2
Rewrite these terms in the form of square.
2 2
a 3q
or
2 2
a 3q 1
or
2 2
a 3q 2
2 2
a 3q
or
2
9q 6q 1
or
2
9q 12q 4
2
3 3q
or
2
3 3q 2q 1
or
2
3 3q 4q 1 1
Substitute
2
1
k 3q
,
2
2
k 3q 2q
and
2
3
k 3q 4q 1
. Here,
1
k
,
2
k
and
3
k
are positive
integers,
2
2
a 3 3q
or
2
3 3q 2q 1
or
2
3 3q 4q 1 1
1
3k
or
2
3k 1
or
3
3k 1
This shows that the square of any positive integer is
either of form
3m
or
3m 1
.
Question: 5
Use Euclid’s division lemma to show that the cube
of any positive integer is either of the form
9m
or
9m 1
for some integer
9m 8
.
Solution:
Let two integers be
a
and
b
, where
b 3
.
Using Euclid’s division algorithm,
a bq r ......(1)
The value of
r
varies from
0 r 3
.
Substitute
b 3
in equation
1
,
a 3q r ......(2)
For
r 0
,
a 3q
For
r 1
,
a 3q 1
For
r 2
,
a 3q 2
Rewrite these terms in the form of square.
3 3
a 3q
or
3 3
a 3q 1
or
3 3
a 3q 2
3 3
a 3q
or
3 2
27q 27q 9q 1
or
3 2
27q 54q 36q 8
2
3 3q
or
2
3 3q 2q 1
or
2
3 3q 4q 1 1
Substitute
2
1
k 3q
,
2
2
k 3q 2q
and
2
3
k 3q 4q 1
. Here,
1
k
,
2
k
and
3
k
are positive
integers,
2
2
a 3 3q
or
2
3 3q 2q 1
or
2
3 3q 4q 1 1
1
3k
or
2
3k 1
or
3
3k 1
3 3
a 3q
or
3 2
27q 27q 9q 1
or
3 2
27q 54q 36q 8
Substitute
3
1
k 3q
,
3 2
2
k 3q 3q q
and
3 2
2
k 3q 6q 4q
. Here,
1
k
,
2
k
and
3
k
are random
positive integers,
3
3
a 9 3q
or
3 2
9 3q 3q q 1
or
3 2
9 3q 6q 4q 1
=
1
9k
or
2
9k 1
or
3
9k 8
This shows that the square of any positive integer is
either of form
9m
or
9m 1
or
9m 8
.
Exercise 1.2 (7)
Question: 1
Express each number as product of its prime
factors:
(i)
140
(ii)
156
(iii)
3825
(iv)
5005
(v)
7429
Solution:
(i) Given number:
140
To find the prime factors of
140
.
140 2 2 5 7
2
2 5 7
Thus, the prime factors of
140
is
2
2 5 7
.
(ii) Given number:
156
To find the prime factors of
156
.
156 2 2 3 13
2
2 3 13
Thus, the prime factors of
156
is
2
2 3 13
.
(iii) Given number:
3825
To find the prime factors of
3825
.
3825 3 3 5 5 17
 
2 2
3 5 17
Thus, the prime factors of
3825
is
 
2 2
3 5 17
.
(iv) Given number:
5005
To find the prime factors of
5005
.
5005 5 7 11 13
Thus, the prime factors of
5005
is
5 7 11 13
.
(v) Given number:
7429
To find the prime factors of
7429
.
7429 17 19 23
Thus, the prime factors of
7429
is
17 19 23
.
Question: 2
Find the LCM and HCF of the following pairs of
integers and verify that
LCM HCF product or twonumbers.
(i)
26
and
91
(ii)
510
and
92
(iii)
336
and
54
Solution:
(i) Given:
26
and
91
Take factors of
26
and
91
.
26 2 13
91 7 13
Thus,
(26,H F 91)C 13
and
LCM(26,91) 2 7 13 182
The product of the given two numbers is
2366
, but,
the product of
HCF
and
LCM
is also
2366
.
That is,
product of twonumbers 26 91 2366
LCM HCF= 2 7 13 13 2366
This shows that the
product of twonumbers LCM HCF
(ii) Given:
510
and
92
Take factors of
510
and
92
.
510 2 3 5 17
92 2 2 23
Thus,
HCF 510,92 2
and
LCM(510,92) 2 2 3 5 17 23 23460
The product of the given two numbers is 46920. But,
the product of
HCF
and
LCM
is also 46920.
That is,
product of twonumbers 519 92 46920
LCM HCF 23460 2 46920
This shows that the
product of twonumbers LCM HCF
(iii) Given:
336
and
54
Take factors of
336
and
54
336 2 2 2 2 3 7
4
2 3 7
54 2 3 3 3
3
2 (3)
Thus,
HCF 336,54 2 3 6
and
4 3
LCM 336,54 2 3 7
3024
The product of the given two numbers is
18144
.
But, the product of
HCF
and
LCM
is also
18144
.
That is,
product of twonumbers=336 54=18144
LCM HCF=3024 6
=18144
This shows that the
product of twonumbers LCM HCF
Question: 3
Find the LCM and HCF of the following integers by
applying prime factorization method.
(i)
12, 15 and 21
(ii)
17, 23 and 290
(iii)
8, 9 and 25
Solution:
(i) Given:
12,15
and
21
Take factors of
12,15
and
21
.
12 2 2 3
15 3 5
21 3 7
Thus,
HCF 12,15,21 3
and
LCM 12,15,21 2 2 3 5 7 420
(ii) Given:
17,23
and
29
Take factors of
17,23
and
29
.
17 1 17
23 1 23
29 1 29
Thus,
HCF 17,23,29 1
and
LCM 17,23,29 1 17 23 29 11339
(iii) Given:
8,9
and
25
Take factors of
8,9
and
25
.
8 1 2 2 2
9 1 3 3
25 1 5 5
Thus,
HCF 8,9,25 1
and
LCM 8,9,25 1 2 2 2 3 3 5 5 1800
Question: 4
Given that
HCF 306,657 9
, find
LCM 306,657
.
Solution:
Given:
HCF 306,657 9
But, the
product of twonumbers LCM HCF
of
the two numbers.
The product of
306
and
657
is
306 657
Thus,
LCM HCF=306 657
. But,
HCF 306,657 9
.
That is,
LCM HCF=306 657
LCM 9=306 657
306 657
LCM=
9
22338
Hence, the value of
LCM 306,657
is
22338
.
Question: 5
Check whether can end with the digit 0 for any
n
6
natural number
n
.
Solution:
Any number that has end digit as
0
,
is divisible by
10
and its factors
2
and
5
that is,
10 2 5
.
Prime factorization of
n
n
6 2 3
The prime factors of
6
are only
2
and
3
but not
5
.
Therefore, the exponent cannot have
0
as the
n
6
last digit for any natural number
n
.
Question: 6
Explain why
7 11 13 13
and
7 6 5 4 3 2 1 5
are composite numbers.
Solution:
Given term:
7 11 13 13
Simplify the term,
7 11 13 13 13 7 11 1 1
13 77 1
13 78
13 13 6
Thus,
6
and
13
are the factors of this expression.
Thus, these are composite numbers.
Now, simplify the term
7 6 5 4 3 2 1 5
.
7 6 5 4 3 2 1 5 5 7 6 1 4 3 2 1 1
5 1008 1
5 1009
Thus,
5
and
1009
are the factors of this expression.
Thus, these are composite numbers.
Question: 7
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?
Solution:
Both Sonia and Ravi have started at the same time
and are going in the same direction. They will meet
again after completing one complete round of the
circular path.
Therefore, the LCM of both the values of time taken
by Sonia and Ravi will be the time at which they
meet again.
Take factors of
18
and
12
18 2 3 3
12 2 2 3
Thus,
LCM 18,12 2 2 3 3 36
This shows that Sonia and Ravi will meet after 36
minutes at the starting point.
Exercise 1.3 (3)
Question: 1
Prove that
5
is irrational.
Solution:
Let
5
be a rational number.
Then,
a
5
b
where
a
and
b
are two co-prime
numbers and
b 0
.
a
5
b
b 5 a
Taking square on both sides,
2 2
5b a ......(1)
This shows that
2
a
is divisible by
5
. Therefore,
a
will also be divisible by
5
.
That is,
a 5c
Substitute in equation
1
.
a 5c
2 2
5b a
2
2
5b 5c
2 2
5b 25c
2
2
b
c
5
2 2
b 5c
This shows that
2
b
is divisible by
5
. Therefore,
b
will also be divisible by
5
.
This means that both
a
and
b
have a common
factor
5
. This contradicts the assumption that
a
and
b
are co-prime numbers.
Since
5
cannot be expressed in the form of
p
q
,
5
is an irrational number.
Question: 2
Prove that
3 2 5
is irrational.
Solution:
Let
3 2 5
be a rational number.
Then,
a
3 2 5
b
, where
a
and
b
are two integers
and
b 0
a
3 2 5
b
a
2 5 3
b
1 a
5 3
2 b
This shows that
1 a
3
2 b
will be rational because
a
and
b
are integers. Therefore,
5
will also be
rational.
This contradicts to the proven fact that
5
is
irrational.
This proves that the assumption,
3 2 5
is rational,
is not true.
Hence,
3 2 5
is an irrational number.
Question: 3
Prove that the following are irrationals:
(i)
1
2
(ii)
7 5
(iii)
6 2
Solution:
(i) Let
1
2
be a rational number.
Then,
1 a
b
2
where
a
and
b
are two integers and
b 0
.
1 a
b
2
b
2
a
This shows that
b
a
will be rational because
a
and
b
are integers.
Thus,
2
will also be rational. This contradicts to
the proven fact that
2
is irrational.
This proves that the assumption,
1
2
is rational, is
not true.
Hence,
1
2
is an irrational number.
(ii) Let
7 5
be a rational number.
Then,
a
7 5
b
where
a
and
b
are two integers and
b 0
.
a
7 5
b
a
5
7b
This shows that will be rational because
a
and
a
7b
b
are integers.
Thus,
5
will also be rational. This contradicts to
the proven fact that
5
is irrational.
This proves that the assumption,
7 5
is rational, is
not true.
Hence,
7 5
is an irrational number.
(iii) Let
6 2
be a rational number.
Then,
a
6 2
b
where
a
and
b
are two integers and
b 0
a
6 2
b
a
2 6
b
This shows that
a
6
b
will be rational because
a
and
b
are integers.
Thus,
2
will also be rational. This contradicts to
the proven fact that
2
is irrational.
This proves that the assumption,
6 2
is rational,
is not true.
Hence,
6 2
is an irrational number.
Exercise 1.4 (3)
Question: 1
Without actually performing the long division, state
whether the following rational numbers will have a
terminating decimal expansion or a
non-terminating repeating decimal expansion.
(i)
13
3125
(ii)
17
8
(iii)
64
455
(iv)
15
1600
(v)
29
343
(vi)
3 2
23
2 5
(vii)
2 7 5
129
2 5 7
(viii)
6
15
(ix)
35
50
(x)
77
210
Solution:
(i)
13
3125
Factorise the denominator,
5
3125 5 5 5 5 5 5
Since the denominator is in the power of
5
that is
m
5
.
Its decimal expansion will be terminating.
Hence, the rational number
13
3125
is terminating.
(ii)
17
8
Factorizing the denominator.
3
8 2 2 2 2
Since the denominator is in the power of
2
that is
m
2
.
Its decimal expansion will be terminating.
Hence, the rational number
17
8
is terminating.
(iii)
64
455
Factorizing the denominator.
455 5 7 13
Since the denominator is neither in the power of
5
or
2
that is
m n
5 2
.
Its decimal expansion will be non-terminating
repeating.
Hence, the rational number
64
455
is non-terminating
repeating.
(iv)
15
1600
Factorizing the denominator.
1600 2 2 2 2 2 2 5 5
6 2
2 5
Since the denominator is in the power of both
5
and
2
that is
m n
5 2
.
Its decimal expansion will be terminating.
Hence, the rational number
15
1600
is terminating.
(v)
29
343
Factorizing the denominator.
343 7 7 7
3
7
Since the denominator is neither in the power of
5
or
2
that is
m n
5 2
.
Its decimal expansion will be non-terminating
repeating.
Hence, the rational number
29
343
is non-terminating
repeating.
(vi)
3 2
23
2 5
Clearly, the denominator is in the power of both
5
and
2
that is
m n
5 2
.
Its decimal expansion will be terminating.
Hence, the rational number
3 2
23
2 5
is
terminating.
(vii)
2 7 5
129
2 5 7
The denominator has a power of
7
. Entirely, it is
neither in the power of
5
or
2
that is
m n
5 2
.
Its decimal expansion will be non-terminating
repeating.
Hence, the rational number
2 7 5
129
2 5 7
is
non-terminating repeating.
(viii)
6
15
Simplifying the rational number.
1
6 2 3 2
15 3 5
5
Since, the denominator is in the power of
5
that is
m
5
Its decimal expansion will be terminating.
Hence, the rational number
6
15
is terminating.
(ix)
35
50
Simplifying the rational number.
35 5 7 7
50 5 10 10
Factorising the denominator.
10 2 5
Clearly, the denominator is in the power of both
2
and
5
that is
m n
5 2
.
Its decimal expansion will be terminating.
Hence, the rational number
35
50
is terminating.
(x)
77
210
Simplifying the rational number.
77 7 11 11
210 7 30 30
Factorising the denominator.
30 2 3 5
The denominator has a power of
3
. Entirely, it is
neither in the power of
5
or
2
that is
m n
5 2
.
Its decimal expansion will be non-terminating
repeating.
Hence, the rational number
77
210
is non-terminating
repeating.
Question: 2
Write down the decimal expansions of those
rational numbers in Question 1 above which have
terminating decimal expansions.
Solution:
(i) Given expression:
13
3125
Expand the rational number
13
3125
to get
terminating decimal expansions.
5
13 13
3125
5
5
5 5
13 2
5 2
5
416
10
0.00416
(ii) Given expression:
17
8
Expand the rational number
17
8
to get terminating
decimal expansions.
3
17 17
8
2
3
3 3
17 5
2 5
3
3
17 5
10
3
2125
10
2.125
(iii) Given expression:
64
455
The rational number
64
455
does not have a
terminating decimal expansion.
(iv) Given expression:
15
1600
Expand the rational number
15
1600
to get the
terminating decimal expansions.
4
4 2 4
15 5
2 10 5
6
9375
10
0.009375
(v) Given expression:
29
343
The rational number
29
343
does not have a
terminating decimal expansion.
(vi) Given expression:
3 2
23
2 5
Expand the rational number
3 2
23
2 5
to get the
terminating decimal expansions.
3 2
3 2 3 2 3 2
23 5 2
23
2 5 2 5 5 2
5
11500
10
0.115
(vii) Given expression:
2 7 5
129
2 5 7
The rational number
2 7 5
129
2 5 7
does not have
a terminating decimal expansion.
(viii) Given expression:
6
15
Expand the rational number
6
15
to get terminating
decimal expansions.
6 2
15 5
2 2
5 2
4
10
0.4
(ix) Given expression:
35
50
Expand the rational number
35
50
to get terminating
decimal expansions.
35 7
50 10
0.7
(x) Given expression:
77
210
The rational number
77
210
does not have a
terminating decimal expansion.
Question: 3
The following real numbers have decimal
expansions as given below. In each case, decide
whether they are rational or not. If they are
rational, and of the form
p
,
q
prime factors of
q
?
(i)
43.123456789
(ii)
0.120120012000120000
(iii)
43.123456789
Solution:
(i) Given number:
43.123456789
The number
43.123456789
has a terminating form
of decimal expansion. Therefore, it is a rational
number of the form
p
q
, where
q
is of the form
.
m n
2 5
(ii) Given number:
0.120120012000120000...
The number
0.120120012000120000...
has neither
terminating nor recurring form of decimal
expansion. Therefore, it is an irrational number.
(iii) Given number:
43.123456789
The number
43.123456789
has a non-terminating
but a recurring form of decimal expansion.
Therefore, it is a rational number of the form
p
q
,
where
q
is not of the form
. That is, the
m n
2 5
prime factorization of
q
will include factors except 2
or 5.