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,

HCF 135,225 45

.

(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

members in 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.

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 2

15 15

1600

2 10

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

you say about the

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.