Lesson: Real Numbers

EXERCISE 1.1 (10)(Multiple Choice Questions

and Answers)

Choose the correct answers from the given four

options in the following questions:

Question: 1

For some integer , every even integer is of the form

m

a.

m

b.

m 1

c.

2m

d.

2m 1

Solution

(c)

An integer is a number, without fractional

component.

Thus,

m 2, 1,0,1,2,...

2m 4, 2,0,2,4,...

It is clear that, every integer of the form is even.

2m

Hence, for some integer , every even integer is of

m

the form .

2m

Question: 2

For some integer , every odd integer is of the form:

q

a.

q

b.

q 1

c.

2q

d.

2q 1

Solution

(d)

An integer is a number, without fractional

component.

Thus,

q 2, 1,0,1,2,...

2q 1 3, 1,1,3,5,...

It is clear, that every integer of the form is

2q 1

odd.

Hence, for some integer , every odd integer is of the

q

form .

2q 1

Question: 3

is divisible by , if is:

2

n 1

8

n

a. an integer

b. a natural number

c. an odd integer

d. an even integer

Solution

(c)

Let

2

p n 1 ......(1)

There can be two possibilities for .

n

Case 1: When is an even integer.

n

Let is an integer such that,

q

n 2q

Put in equation (1)

n 2q

2

p 2q 1

2

p 4q 1 ......(2)

Put in equation (2)

q 1

2

p 4 1 1

p 4 1

p 3

is not divisible by .

p 3

8

Now, put in equation (2)

q 0

2

p 4 0 1

p 1

is not divisible by .

p 1

8

Case 2: When is an odd integer.

n

Let is an integer such that,

q

n 2q 1

Put in equation (1)

n 2q 1

2

p 2q 1 1

2

p 4q 1 4q 1

2

p 4q 4q

p 4q q 1 ......(3)

Put in equation (3)

q 1

p 4 1 1 1

p 0

Put in equation (3)

q 0

p 4 0 0 1

p 0

Put in equation (3)

q 1

p 4 1 1 1

p 4 2

p 8

It is clear that, and are both divisible by .

p 0

8

8

Hence, is divisible by when is an odd

2

n 1

8

n

integer.

Question: 4

If the HCF of and is expressible in the form

65

117

, then the value of is:

65m 117

m

a. 4

b. 2

c. 1

d. 3

Solution

(b)

By Euclid’s division algorithm,

where,

a bq r,

0 r b

Here, is the dividend, is the divisor, is the

a

b

q

quotient, and is the remainder.

r

By the Euclid’s division lemma for 117 and 65.

117 65 1 52

Remainder .

52 0

Now, using division lemma for 65 and 52.

65 52 1 13

Remainder .

13 0

Now, using division lemma for 52 and 13.

52 13 4 0

Remainder and divisor at the end is .

0

13

Thus, the HCF of 117 and 65 is .

13

Now, according to the question,

65m 117 13

65m 13 117

65m 130

130

m

65

m 2

Hence, the value of is .

m

2

Question: 5

The largest number which divides 70 and 125, leaving

remainders 5 and 8 respectively, is:

a. 13

b. 65

c. 875

d. 1750

Solution

(a)

The required number leaves remainders 5 and 8 on

dividing the numbers 70 and 125.

Thus, the required number is the HCF of and

70 5

.

125 8

Now, find the HCF of 65 and 117.

By Euclid’s division algorithm,

where,

a bq r,

0 r b

Here, is the dividend, is the divisor, is the

a

b

q

quotient, and is the remainder.

r

By Euclid’s division lemma for 117 and 65.

117 65 1 52

Remainder .

52 0

Now, using division lemma for 65 and 52.

65 52 1 13

Remainder .

13 0

Now, using division lemma for 52 and 13.

52 13 4 0

Remainder and the divisor at the end is 13.

0

Thus, the HCF of 117 and 65 is 13.

Hence, the largest number which divides 70 and 125,

leaving remainders 5 and 8 respectively is 13.

Question: 6

If two positive integers and are written as

a

b

and , where are prime numbers,

3 2

a x y

3

b xy

x,y

then is

HCF a,b

a.

xy

b.

2

xy

c.

3 3

x y

d.

2 2

x y

Solution

(b)

Given: and

3 2

a x y

3

b xy

To find the HCF of and .

a

b

Factories .

3 2

a x y

a x x x y y

Factories .

3

b xy

b x y y y

Product of least powers of common factors and

a

b

is .

x y y

Thus, the HCF of and is .

a

b

x y y

Hence, .

2

HCF a,b xy

Question: 7

If two positive integers and can be expressed as

p

q

and , where being prime numbers,

2

p ab

3

q a b

a,b

then is equal to

LCM p,q

a.

ab

b.

2 2

a b

c.

3 2

a b

d.

3 3

a b

Solution

(c)

Given: and

2

p ab

3

q a b

Now, find the LCM of and .

p

q

Factories .

2

p ab

p a b b

Factories .

3

q a b

q a a a b

The product of the highest powers of common prime

factors of and is .

p

q

a b b a a

Thus, LCM of and is .

p

q

a b b a a

Hence, .

3 2

LCM p,q a b

Question: 8

The product of a non-zero rational and an irrational

number is

a. always irrational

b. always rational

c. rational or irrational

d. one

Solution

(a)

Consider rational number and irrational number

3

2

.

8

Calculate the product of and .

3

2

8

3 3

8 2 2

2 2

3 2

is an irrational number.

3 2

Hence, the product of a non-zero rational and an

irrational number is always irrational.

Question: 9

The least number that is divisible by all the numbers

from 1 to 10 (both inclusive)

a. 10

b. 1000

c. 504

d. 2520

Solution

(d)

The least number divisible by all the numbers from 1

to 10 is the LCM of all the numbers from 1 to 10.

Calculate .

LCM 1,2,3,4,5,6,7,8,9,10

Factories the numbers from 1 to 10,

1 1

2 1 2

3 1 3

4 1 2 2

5 1 5

6 1 2 3

7 1 7

8 1 2 2 2

9 1 3 3

10 1 2 5

The product of the highest powers of common prime

factors is .

1 2 2 2 3 3 5 7

Thus,

LCM 1,2,3,4,5,6,7,8,9,10 1 2 2 2 3 3 5 7

2520

Hence, the least number divisible by all the numbers

from 1 to 10 is .

2520

Question: 10

The decimal expansion of rational number will

14587

1250

terminate after

a. one decimal place

b. two decimal places

c. three decimal places

d. four decimal places

Solution

(d)

Let the rational number .

14587

a

1250

Factories the denominator .

1250

14587

a

2 5 5 5 5

Now, multiply numerator and denominator by

3

2

3

3

14587 2

a

2 5 5 5 5 2

14587 8

a

10 1000

116696

a

10000

a 11.6696

Hence, the decimal expansion of rational number

will terminate after four decimal places.

14587

1250

EXERCISE 1.2 (10)

Question: 1

Write whether every positive integer can be of the

form , where is an integer. Justify your

4q 2

q

answer.

Solution

No, every positive integer cannot be of the form

.

4q 2

So, by Euclid’s division lemma,

where,

b aq r

0 r a

Since,

(dividend divisor quotient remainder)

Here, is any positive integer and ,

b

a 4

b 4q r

Where, i.e.,

0 r 4

r 0,1,2,3

So, this must be of the form , , ,

4q

4q 1

4q 2

4q 3

.

Question: 2

The product of two consecutive positive integers is

divisible by ‘2’. Is this statement true or false? Give

reason.

Solution

Yes, the statement is true.

Let the two consecutive integers be and .

n

n 1

So, one number out of these two must be divisible by

2.

Thus, product of the numbers is also divisible by 2.

Let us consider some examples for this,

is divisible by ,

3 4

2

is divisible by ,

11 12

2

is divisible by and so on.

33 34

2

Question: 3

‘The product of three consecutive positive integers is

divisible by ’. Is this statement true or false? Justify

6

your answer.

Solution

Yes, the statement is true.

Let the three consecutive integers be , and

n

n 1

.

n 2

So, one number out of these three must be divisible

by and another one must be divisible by .

2

3

Hence, the product of numbers is divisible by .

6

Let us consider some examples for this,

is divisible by 6,

3 4 5

is divisible by 6,

10 11 12

is divisible by and so on.

82 83 84

6

Question: 4

Write whether the square of any positive integer can

be of the form , where is a natural number.

3m 2

m

Justify your answer.

Solution

No, this statement is not true.

By Euclid’s division lemma, where,

b aq r

0 r a

.

Here, is any positive integer and ,

b

a 3

b 3q r

for .

0 r 2

So, any positive integer is of the form , or

3k

3k 1

.

3k 2

Now, [where, ]

2

2

3k 9k 3m

2

m 3k

and

2

2

3k 1 9k 6k 1

2

3 3k 2k 1

[where, ]

3m 1

2

m 3k 2k

Also,

2

2

3k 2 9k 12k 4

2

2 2

a b a 2ab b

2

9k 12k 3 1

2

3 3k 4k 1 1

[where, ]

3m 1

2

m 3k 4k 1

Which is of the form and .

3m

3m 1

Hence, square of any positive number cannot be of

the form .

3m 2

Question: 5

A positive integer is of the form , being a

3q 1

q

natural number. Can you write its square in any

form other than , i.e., or for some

3m 1

3m

3m 2

integer ? Justify your answer.

m

Solution

No, the statement cannot be true.

By Euclid’s division lemma, where,

b aq r

0 r a

.

Here, is any positive integer and ,

b

a 3

b 3q r

for .

0 r 2

So, this must be of the form , or .

3q

3q 1

3q 2

Now, [where, ]

2 2

3q 9q 3m

2

m 3q

And,

2

2

3q 1 9q 6q 1

2

3 3q 2q 1

[where, ]

3m 1

2

m 3q 2q

Also,

2

2

3q 2 9q 12q 4

2

9q 12q 3 1

2

3 3q 4q 1 1

3m 1

[where, ]

2

m 3q 4q 1

Hence, square of a positive integer is of the form

is always in the form for some integer

3q 1

3m 1

.

m

Question: 6

The numbers 525 and 3000 are both divisible only by

3, 5, 15, 25, and 75. What is HCF ? Justify

525,3000

your answer.

Solution

To calculate the HCF of and 3000.

525

By Euclid’s division lemma,

where,

a bq r,

0 r b

Here, is the dividend, is the divisor, is the

a

b

q

quotient, and is the remainder.

r

Since, .

dividend divisor quotient remainder

Apply, Euclid’s division lemma for 3000 and 525.

3000 525 5 375

Remainder .

375 0

Now, apply Euclid’s division lemma for 525 and 375.

525 375 1 150

Remainder .

150 0

Now, apply Euclid’s division lemma for 375 and 150.

375 150 2 75

Remainder .

75 0

So, again apply Euclid’s division lemma for 150 and

75.

150 75 2 0

Remainder .

0

The numbers 3, 5, 15, 25 and 75 divide the numbers

525 and 3000. It means, these terms are common in

both 525 and 3000.

So, the highest common factor among these is 75.

Question: 7

Explain why is a composite number.

3 5 7 7

Solution

Let us consider, .

a 3 5 7 7

a 3 5 7 7

7 3 5 1

7 16

Since, has more than two factors (as 1, 7, 16 and

a

a

are factors).

So, it is a composite number.

Question: 8

Can two numbers have 18 as their HCF and 380 as

their LCM? Give reasons.

Solution

No, it is not possible.

This is because, HCF is always a factor of LCM but

here, 18 is not a factor of 380.

Question: 9

Without actually performing the long division, find if

will have terminating or non-terminating

987

10500

(repeating) decimal expansion. Give reasons for your

answer.

Solution

Yes, it is a terminating decimal expansion.

If the simplified denominator has factor in the form

of .

m n

2 5

So, this is a terminating decimal.

Now, let us simplify the given expression,

987 47

10500 500

3 2

47 2

5 2 2

3 3

94

5 2

94

1000

0.094

Hence, is a terminating decimal.

0.094

Question: 10

A rational number in its decimal expansion is

. What can you say about the prime factors

327.7081

of , when this number is expressed in the form ?

q

p

q

Give reasons.

Solution

The given number, is a terminating decimal

327.7081

number.

So, it represents a rational number and its

denominator must have the form .

m n

2 5

Thus,

3277081

327.7081

10000

p

q

Since, .

q 10000

Now, find the factors of .

q

4

q 10

2 2 2 2 5 5 5 5

4 4

2 5

Hence, the prime factorization of contains only the

q

factors 2 and 5.

EXERCISE 1.3 (14)

Question: 1

Show that the square of any positive integer is either

of the form or for some integer .

4q

4q 1

q

Solution

Let us consider, as an arbitrary positive integer.

a

Then, by Euclid’s division algorithm, for the positive

integers and , there is an existence of non–

a

4

negative integers and , such that,

m

r

, where

a 4m r

0 r 4

2

2

a 4m r

2 2

16m r 8mr ......(1)

Where,

0 r 4

Now, let us consider the following cases:

Case :

1

When , then put the value of in equation (1),

r 0

r

2 2

a 16m

2

4 4m

4q

Where, is an integer.

2

q 4m

Case :

2

When , then put the value of in equation (1),

r 1

r

2 2

a 16m 1 8m

2

4 4m 2m 1

4q 1

Where, is an integer.

2

q 4 4m 2m

Case :

3

When , then put the value of in equation (1),

r 2

r

2 2

2a 16m 4 16m

2

4 4m 4m 1

4q

Where, is an integer.

2

q 4m 4m 1

Case :

4

When , then put the value of in equation (1),

r 3

r

2 2

a 16m 24m 8 1

2

4 4m 6m 2 1

4q 1

Where, is an integer.

2

q 4m 6m 2

Hence, the square of any positive integer is either of

the form or for some integer .

4q

4q 1

q

Question: 2

Show that cube of any positive integer is of the form

, or , for some integer .

4m

4m 1

4m 3

m

Solution

Let us consider, as an arbitrary positive integer.

a

Then, by Euclid’s division algorithm, for the positive

integers and , there is an existence of non-

a

4

negative integers and such that, where

q

r

a 4q r

.

0 r 4

3

3

a 4q r

3 3 2 2

64q r 12qr 48q r

3 2 2 3

64q 48q r 12qr r ......(1)

Where,

0 r 4

Now, let us consider the following cases:

Case 1:

When , put the value of in equation (1)

r 0

r

3 3

a 64q

3

4 16q

4m

Where is an integer.

3

m 16q

Case 2:

When , put the value of in equation (1).

r 1

r

3 3 2

a 64q 48q 12q 1

3 2

4 16q 12q 3q 1

4m 1

Where is an integer.

3 2

m 16q 12q 3q

Case 3:

When , put the value of in equation (1),

r 3

r

3 3 2

a 64q 144q 108q 27

3 2

64q 144q 108q 24 3

3 2

4 16q 36q 27q 6 3

4m 3

3 3 2

a 64q 144q 108q 27

3 2

64q 144q 108q 24 3

3 2

4 16q 36q 27q 6 3

4m 3

Where is an integer.

3 2

m 16q 36q 27q 6

Hence, the cube of any positive integer is of the form

, , or for some integer .

4m

4m 1

4m 3

m

Question: 3

Show that the square of any positive integer cannot

be of the form or for any integer .

5q 2

5q 3

q

Solution

Let us consider as an arbitrary positive integer.

a

Then, by Euclid’s divisions algorithm, for the positive

integers and , there is an existence of non-

a

5

negative integers and such that,

m

r

a 5m r

where

0 r 5

Take square on both sides,

2

2

a 5m r

2 2

25m r 10mr

2 2

5 5m 2mr r ......(1)

Now, let us consider the following cases:

Case :

1

When , put the value of in equation (1).

r 0

r

2 2

a 5 5m

5q

Where, is an integer.

2

q 5m

Case :

2

When , put the value of in equation (1).

r 1

r

2 2

a 5 5m 2m 1

5q 1

Where, is an integer.

2

q 5m 2m

Case :

3

When , put the value of in equation (1).

r 2

r

2 2

a 5 5m 4m 4

5q 4

Where, is an integer.

2

q 5m 4m

Case :

4

When , put the value of in equation (1).

r 3

r

2 2

a 5 5m 6m 9

2

5 5m 6m 5 4

2

5 5m 6m 1 4 5q 4

Where, is an integer.

2

q 5m 6m 1

Case :

5

When , put the value of in equation (1).

r 4

r

2 2

a 5 5m 8m 16

2

5 5m 8m 15 1

2

5 5m 8m 3 1

5q 1

Where, is an integer.

2

q 5m 8m 3

Hence, the square of any positive integer cannot be

of the form or for any integer .

5q 2

5q 3

q

Question: 4

Show that the square of any positive integer cannot

be of the form or for any integer .

6m 2

6m 5

m

Solution

Let us consider as an arbitrary positive integer,

a

Then, by Euclid’s division algorithm, for the positive

integers and , there is an existence of non-

a

6

negative integers and such that,

q

r

, where

a 6q r

0 r 6

2

2

a 6q r

2 2 2

a 36q r 12qr

2 2 2

a 6 6q 2qr r ......(1)

Where,

0 r 6

Now, let us consider the following cases,

Case 1:

When , put the value of in equation (1),

r 0

r

2

2

a 6 6q

6m

Where, is an integer.

2

m 6q

Case 2:

When , put the value of in equation (1),

r 1

r

2 2

2

a 6 6q 2q 1

a 6m 1

Where, is an integer.

2

m 6q 2q

Case 3:

When , put the value of in equation (1).

r 2

r

2 2

a 6 6q 4q 4

2

a 6m 4

Where, is an integer.

2

m 6q 4q

Case 4:

When , put the value of in equation (1),

r 3

r

2 2

a 6 6q 6q 9

2

6 6q 6q 6 3

2

6 6q 6q 1 3

6m 3

Where, is an integer.

2

m 6q 6q 1

Case 5:

When , put the value of in equation (1),

r 4

r

2 2

a 6 6q 8q 16

2

6 6q 8q 12 4

2

6 6q 8q 2 4

6m 4

Where, is an integer.

2

m 6q 8q 2

Case 6:

When , put the value of in equation (1),

r 5

r

2 2

a 6 6q 10q 25

2

6 6q 10q 24 1

2

6 6q 10q 4 1

6m 1

Where, is an integer.

2

m 6q 10q 1

Hence, the square of any positive integer cannot be

of the form or for any integer .

6m 2

6m 5

m

Question: 5

Show that the square of any odd integer is of the

form , for some integer .

4m 1

m

Solution

By Euclid’s division algorithm,

a bq r,0 r 4 ......(1)

Put the value of in equation (1).

b

, where i.e

a 4q r

0 r 4

r 0,1,2,3

Now, let us put the different values of in the

r

equation .

a 4q r

Case 1:

If ,

r 0

Then, , which is divisible by .

a 4q

2

Thus, is even.

4q

Case 2:

If ,

r 1

Then,

a 4q 1 ......(2)

Which is not divisible by

2

Case 3:

If ,

r 2

, which is divisible by

a 4q 2

2 2q 1

2

Thus, is even

2 2q 1

Case :

4

If ,

r 3

Then,

a 4q 3 ......(3)

Which is not divisible by

2

So, for any positive integer , and are

q

4q 1

4q 3

odd integers.

Take square on both sides of the equation (2)

2

2

a 4q 1

2

16q 1 8q

2

4 4q 2q 1

2

2 2

a b a b 2ab

Which is of the form , where is

4m 1

2

m 4q 2q

an integer.

Now, take square on both sides of the equation (3)

2

2

a 4q 3

2

16q 9 24q

2

4 4q 6q 2 1

2

2 2

a b a b 2ab

Which is of the form , where

4m 1

2

m 4q 6q 2

is an integer.

Hence, for some integer , the square of any odd

m

integer is of the form .

4m 1

Question: 6

If is an odd integer, then show that is

n

2

n 1

divisible by 8.

Solution

Let us consider,

2

a n 1 ......(1)

It is given that, is an odd integer.

n

n 1,3,5,...

Now, put the value of in equation (1).

n

2

a 1 1

1 1

0

Which is divisible by 8.

Now, put the value of in equation (1)

n

2

a 3 1

9 1

8

Which is divisible by .

8

Now, put the value of in equation (1).

n

2

a 5 1

25 1

24

Which is divisible by .

8

Hence, is divisible by , when is an odd

2

n 1

8

n

integer.

Question: 7

Prove that, if and are both odd positive integers,

x

y

then is even but not divisible by .

2 2

x y

4

Solution

Let us consider, and where is

x 2m 1

y 2m 3

m

an odd positive integer.

Then, put the values of and in the expression

x

y

.

2 2

x y

2 2

2 2

x y 2m 1 2m 3

2 2

4m 1 4m 4m 9 12m

2

8m 16m 10

2

2 4m 8m 5

Which is even.

Now, to find out, if the given expression is divisible

by .

4

, which is not

2 2

2 4m 8m 5 4 2m 2m 2 2

divisible by .

4

Hence, is even for every odd positive integer,

2 2

x y

but not divisible by .

4

Question: 8

Use Euclid’s division algorithm to find HCF of ,

441

and .

567

693

Solution

Let us consider, , and

a 693

b 567

c 441

Now, by Euclid’s division algorithm,

a bq r,0 r b ......(1)

dividend divisor quotient remainder

Using Euclid’s division lemma for and .

693

567

693 567 1 126

r 0

Remainder .

126 0

Now, using Euclid’s division lemma for and

567

126

567 126 4 63

r 0

Remainder .

63 0

Now, using Euclid’s division lemma for and .

126

63

126 63 2 0

Here,r 0

Remainder .

0

HCF of and .

693

567

63

Now, let us take and and put it in the

c 441

d 63

Euclid’s division algorithm,

c dq r 0 r d ......(2)

441 63 7 0

Here,r 0

Thus, HCF of (693, 567 and 441) = 63.

Question: 9

Using Euclid’s division algorithm, find the largest

number that divides 1251, 9377 and 15628 leaving

remainders 1, 2 and 3 respectively.

Solution

Since, 1, 2 and 3 are the remainders of 1251, 9377 and

15628, respectively.

Now, subtract these remainders from the given

numbers.

We get the following numbers,

,

1251 1 1250

and

9377 2 9375

,

15628 3 15625

which are divisible by the required number.

Now, the required number is equal to the HCF of

1250, 9375 and 15625.

So, by Euclid’s division algorithm,

a bq r,0 r b ......(1)

dividend divisor quotient remainder

Now, to find the largest number, use Euclid’s division

lemma for and .

15625

9375

15625 9375 1 6250

r 0

Remainder .

6250 0

Now, use Euclid’s division lemma for and .

9375

6250

9375 6250 1 3125

r 0

Remainder .

3125 0

Again, use Euclid’s division lemma for and

6250

3125

.

6250 3125 2 0

Here,r 0

Now, the remainder is .

0

HCF of and .

15625

9375

3125

Now, let us take and and put it in

a 1250

b 3125

the Euclid’s division algorithm,

a bq r,0 r b ......(2)

Use Euclid’s division lemma for and .

3125

1250

3125 1250 2 625

r 0

Remainder .

625 0

Now, use Euclid’s division lemma for and .

1250

625

1250 625 2 0

Here,r 0

Remainder .

0

HCF of (1250, 9375 and 15625) = 625.

Hence, 625 is the largest number which divides 1251,

9377 and 15628 leaving remainders 1, 2 and 3

respectively.

Question: 10

Prove that is irrational.

3 5

Solution

Let us suppose that the given expression is

3 5

rational.

Now, let , where is rational.

a 3 5

a

So, .

3 a 5

Take square on both sides,

2 2

3 a 5

2

3 a 5 2a 5

2

2a 5 a 2

2

a 2

5

2a

Which is a contradiction.

This is because on the right-hand side is a rational

number, while on the left-hand side is , which is

5

an irrational number.

Hence, is irrational.

3 5

Question: 11

Show that cannot end with the digit or for

n

12

0

5

any natural number .

n

Solution

If there is any number which ends with the digits

0

or , it is always divisible by .

5

5

So, if the number, ends with the digit zero, it

n

12

must be divisible by .

5

This is possible, only if there is a prime number , in

5

the prime factorization of .

n

12

Now, factories the number .

12

12 2 2 3

2

2 3

Thus, .

n

n 2 2n n

12 2 3 2 3

Since, the factorisation of the number does not

n

12

contain .

5

Hence, there is no value of (Natural number) for

n

which ends with digit zero of five.

n

12

Question: 12

On a morning walk, three persons step off together

and their steps measure 40 cm, 42 cm, and 45 cm

respectively. What is the minimum distance each

should walk, so that each cover the same distance in

complete steps?

Solution

The measurement of steps of three persons 40 cm, 42

cm, and 45 cm.

Now, to find the required minimum distance,

first find the LCM of 40 cm, 42 cm, and 45 cm .

40 2 2 2 5

and

42 2 3 7

45 3 3 5

40,42and 45 2 3 5 2 2L M of 3C 7

30 12 7

210 12

2520

Thus, the minimum distance each should walk is

. Also, each can cover the same distance in

2520cm

complete steps.

Question: 13

Write the denominator of rational number in

257

5000

the form , where are non-negative

m n

2 5

m,n

integers. Hence, write its decimal expansion, without

actual division.

Solution

Denominator of the given rational number is

257

5000

.

5000

Now,

5000 2 2 2 5 5 5 5

3 4

2 5

which is of the type , where and

m n

2 5

m 3

n 4

are non-negative integers.

So, simplify the rational number .

257

5000

3 4

257 257 2

5000 2 5 2

4 4

514

2 5

4

514

10

514

10000

0.0514

Thus, is the required decimal expansion of the

0.0514

rational number and it is also a terminating

257

5000

decimal number.

Question: 14

Prove that is irrational, where and are

p q

p

q

primes.

Solution

Let us consider, to be rational.

p q

Again, let , where is rational.

p q a

a

So, .

q a p

Take square on both sides,

2

q a p 2a p

2

2 2

a b a b 2ab

Therefore, ,

2

a p q

p

2a

which is a contradiction, as on the right-hand side is

a rational number, while on the left-hand side, is

p

an irrational number, since and are prime

p

q

numbers.

Hence, is irrational.

p q

EXERCISE 1.4 (5)

Question: 1

Show that the cube of positive integer of the form

, where is an integer and is also

6q r

q

r 0,1,2,3,4,5

of the form .

6m r

Solution

Let us consider, as an arbitrary positive integer.

a

Then, by Euclid’s division algorithm, for positive

integers and , there is an existence of non-

a

6

negative integers and such that,

q

r

, where

a 6q r

0 r 6 ......(1)

Take cube on both sides of the equation (1),

3

3

a 6q r

3 3

216q r 18qr 6q r

3 2 2 3

216q 108q r 18qr r ......(2)

Where,

0 r 6

Case 1:

When ,put the value of in equation (2),

r 0

r

3 3

a 216q

3

6 36q

6m

Where, is an integer.

3

m 36q

Case 2:

When , put the value of in equation (2),

r 1

r

3 3 2

a 216q 108q 18q 1

3 3 2

a 6 36q 18q 3q 1

3

a 6m 1

Where is an integer.

3 2

m 36q 18q 3q

Case 3:

When , put the value of in equation (2),

r 2

r

3 3 2

a 216q 216q 72q 8

3 3 2

a 216q 216q 72q 6 2

3 3 2

a 6 36q 36q 12q 1 2

3

a 6m 2

Where is an integer.

3 2

m 36q 36q 12q 1

Case 4:

When , put the value of in equation (2),

r 3

r

3 3 2

a 216q 324q 162q 27

3 3 2

a 216q 324q 162q 24 3

3 3 2

a 6 36q 54q 27q 4 3

3

a 6m 3

Where is an integer.

3 2

m 36q 54q 27q 4

Case 5:

When , put the value of in equation (2),

r 4

r

3 2 2

a 216q 432q 288q 64

3 3 2

a 6 36q 72q 48q 60 4

3 3 2

a 6 36q 72q 48q 10 4

3

a 6m 4

Where is an integer.

3 2

m 36q 72q 48q 10

Case 6:

When ,put the value of in equation (2),

r 5

r

3 3 2

a 216q 540q 450q 125

3 3 2

a 216q 540q 450q 120 5

3 3 2

a 6 36q 90q 75q 20 5

3

a 6m 5

Where is an integer.

3 2

m 36q 90q 75q 20

Hence, the cube of a positive integer of the form

where, is an integer and is also

6q r

q

r 0,1,2,3,4,5

of the forms , , , , , and

6m

6m 1

6m 2

6m 3

6m 4

i.e .

6m 5

6m r

Question: 2

Prove that one and only one out of , and

n

n 2

is divisible by , where is any positive

n 4

3

n

integer.

Solution

Let us consider , and .

n

n 2

n 4

a,b,c n,n 2,n 4

Where is any positive integer i.e.

n

n 1,2,3...

At ;

n 1

a,b,c 1,1 2,1 4 1,3,5

At ;

n 2

a,b,c 2,2 2,2 4 2,4,6

At ;

n 3

a,b,c 3,3 2,3 4 3,5,7

At ;

n 4

a,b,c 4,4 2,4 4 4,6,8

At ;

n 5

a,b,c 5,5 2,5 4 5,7,9

At ;

n 6

a,b,c 6,6 2,6 4 6,8,10

At ;

n 7

a,b,c 7,7 2,7 4 7,9,11

At ;

n 8

a,b,c 8,8 2,8 4 8,10,12

Thus, in each set only one number is a

a,b,c

multiple of .

3

Hence, only one out of , and is

n

n 2

n 4

divisible by , where, is any positive integer.

3

n

Question: 3

Prove that one of any three consecutive positive

integers must be divisible by .

3

Solution

Any three consecutive positive integers must be of

the form , and , where is any

n

n 1

n 2

n

natural number, i.e.,

n 1,2,3...

Let us consider , and .

a n

b n 1

c n 2

, where

a,b,c n,n 1,n 2

n 1,2,3...

At ;

n 1

a,b,c 1,1 1,1 2 1,2,3

At ;

n 2

a,b,c 2,2 1,2 2 2,3,4

At ;

n 3

a,b,c 3,3 1,3 2 3,4,5

At ;

n 4

a,b,c 4,4 1,4 2 4,5,6

At ;

n 5

a,b,c 5,5 1,5 2 5,6,7

At ;

n 6

a,b,c 6,6 1,6 2 6,7,8

At ;

n 7

a,b,c 7,7 1,7 2 7,8,9

At ;

n 8

a,b,c 8,8 1,8 2 8,9,10

Thus, in each set only one number is a

a,b,c

multiple of .

3

Hence, one out of any three consecutive positive

integers must be divisible by .

3

Question: 4

For any positive integer , prove that is

n

3

n n

divisible by .

6

Solution

Let us consider ..

3

a n n

2

a n n 1

a n n 1 n 1

2 2

a b a b a b

a n 1 .n n 1

Observe that this is the product of three consecutive

positive integers.

The product of three consecutive positive integers is

divisible by and . So, it must be divisible by .

2

3

6

Hence, is always divisible by , where is any

3

n n

6

n

positive integer.

Question: 5

Show that one and only one out of , , ,

n

n 4

n 8

, and is divisible by , where is any

n 12

n 16

5

n

positive integer.

Solution

Given numbers are , , , , and ,

n

n 4

n 8

n 12

n 16

where is any positive integer.

n

Let us consider, , where

n 5q r

0 r 5

Now, , , , , for any

n 5q

5q 1

5q 2

5q 3

5q 4

natural number [by Euclid’s division algorithm].

q

Let us consider the following cases,

Case 1:

When

n 5q

is divisible by .

n 5q

5

is not divisible by .

n 4 5q 4

5

n 8 5q 8

5q 5 3

is not divisible by .

5 q 1 3

5

n 12 5q 12

5q 10 2

is not divisible by .

5 q 2 2

5

n 16 5q 16

5q 15 1

is not divisible by .

5 q 3 1

5

Case 2:

When

n 5q 1

is not divisible by .

n 5q 1

5

n 4 5q 1 4

5q 5

is divisible by .

5 q 1

5

n 8 5q 1 8

5q 5 4

is not divisible by .

5 q 1 4

5

n 12 5q 1 12

5q 10 3

is not divisible by .

5 q 2 3

5

n 16 5q 1 16

5q 15 2

is not divisible by .

5 q 3 2

5

Case 3:

When

n 5q 2

is not divisible by .

n 5q 2

5

n 4 5q 2 4

5q 5 1

is not divisible by .

5 q 1 1

5

n 8 5q 2 8

5q 10

is divisible by .

5 q 2

5

n 12 5q 2 12

5q 10 4

is not divisible by .

5 q 2 4

5

n 16 5q 2 16

5q 15 3

is not divisible by .

5 q 3 3

5

Case :

4

When

n 5q 3

is not divisible by .

n 5q 3

5

n 4 5q 3 4

5q 5 2

is not divisible by .

5 q 1 2

5

n 8 5q 3 8

5q 10 1

is not divisible by .

5 q 2 1

5

n 12 5q 3 12

5q 15

is divisible by .

5 q 3

5

n 16 5q 3 16

5q 15 4

is not divisible by .

5 q 3 4

5

Case 5:

When

n 5q 4

is not divisible by .

n 5q 4

5

n 4 5q 4 4

5q 5 3

is not divisible by .

5 q 1 3

5

n 8 5q 4 8

5q 10 2

is not divisible by .

5 q 2 2

5

n 12 5q 4 12

5q 15 1

is not divisible by .

5 q 3 1

5

n 16 5q 4 16

5q 20

is divisible by .

5 q 4

5

Hence, in each case, one and only one out of , ,

n

n 4

, , and is divisible by , where is

n 8

n 12

n 16

5

n

any positive integer.