Contents

**Real Numbers**

## Question 1:

**Exercise 1A**

The numbers of the form , where p and q are integers and q ≠ 0 are known as rational numbers.

Ten examples of rational numbers are:

, , , , , , , , 1,

## Question 2:

- 5
- -3

(v) 1.3

(vi) -2.4

(vii)

## Question 3:

A rational number lying between and is

Therefore, we have < < < < Or we can say that, < < < < That is, < < < <

Therefore, three rational numbers between and are

, and

## Question 5:

Let and

Then, x < y because <

Or we can say that, That is, < .

We know that, 8 < 9 < 10 < 11 < 12 < 13 < 14 < 15.

Therefore, we have, < < < < < < < Thus, 5 rational numbers between, < are:

, , , and

## Question 6:

Let x = 3 and y = 4

Then, x < y, because 3 < 4 We can say that, < .

We know that, 21 < 22 < 23 < 24 < 25 < 26 < 27 < 28.

Therefore, we have, < < < < < < < Therefore, 6 rational numbers between 3 and 4 are:

, , , and

## Question 7:

Let x = 2.1 and y = 2.2

Then, x < y because 2.1 < 2.2 Or we can say that, < Or,

That is, we have, <

We know that, 2100 < 2105 < 2110 < 2115 < 2120 < 2125 < 2130 < 2135 < 2140 <

2145 < 2150 < 2155 < 2160 < 2165 < 2170 < 2175 < 2180 < 2185 < 2190 < 2195 <

2200

Therefore, we can have,

Therefore, 16 rational numbers between, 2.1 and 2.2 are:

So, 16 rational numbers between 2.1 and 2.2 are:

2.105, 2.11, 2.115, 2.12, 2.125, 2.13, 2.135, 2.14, 2.145, 2.15, 2.155, 2.16, 2.165, 2.17,

2.175, 2.18

# Exercise 1B

## Question 1:

(i)

If the prime factors of the denominator are 2 and/or 5 then the rational number is a terminating decimal.

Since, 80 has prime factors 2 and 5, is a terminating decimal.

(ii)

If the prime factors of the denominators of the fraction are other than 2 and 5, then the rational number is not a terminating decimal.

Since, 24 has prime factors 2 and 3 and 3 is different from 2 and 5, is not a terminating decimal.

(iii)

If the prime factors of the denominators of the fraction are other than 2 and 5, then the rational number is not a terminating decimal.

Since 12 has prime factors 2 and 3 and 3 is different from 2 and 5, is not a terminating decimal.

(iv)

If the prime factors of the denominators of the fraction are other than 2 and 5, then the rational number is not a terminating decimal.

Since 35 has prime factors 5 and 7, and 7 is different from 2 and 5, is not a terminating decimal.

(v)

If the prime factors of the denominator are 2 and/or 5 then the rational number is a terminating decimal.

Since 125 has prime factor 5 only is a terminating decimal.

## Question 2:

(i)

= 0.625

(ii)

= 0.5625

(iii)

= 0.28

(iv)

= 0.458

(v)

= 2.41

## Question 3:

- Let x =

i.e x = 0.333 …. (i)

⇒ 10x = 3.333 …. (ii) Subtracting (i) from (ii), we get 9x = 3

⇒ x = = Hence, 0. =

- Let x = 1.

i.e x = 1.333 …. (i)

⇒10x = 13.333 …. (ii)

Subtracting (i) from (ii) we get; 9x = 12

⇒ x = =

Hence, 1. =

- Let x = 0.

i.e x = 0.3434 …. (i)

⇒ 100x = 34.3434 …. (ii)

Subtracting (i) from (ii), we get 99x = 34

⇒ x =

Hence, 0. =

- Let x = 3.

i.e x = 3.1414 …. (i)

⇒ 100x = 314.1414 …. (ii)

Subtracting (i) from (ii), we get 99x = 311

⇒ x =

Hence, 3. =

- Let x = 0.

i.e. x = 0.324324 ….(i)

⇒ 1000x = 324.324324….(ii)

Subtracting (i) from (ii), we get 999x = 324

⇒ x = =

Hence, 0. =

- Let x = 0.

i.e. x = 0.177 …. (i)

⇒ 10x = 1.777 …. (ii)

and 100x = 17.777…. (iii) Subtracting (ii) from (iii), we get 90x = 16

⇒ x = =

Hence, 0. =

- Let x = 0.

i.e. x = 0.544 …. (i)

⇒ 10 x = 5.44 …. (ii) and 100x = 54.44 ….(iii)

Subtracting (ii) from (iii), we get 90x = 49

⇒ x =

Hence, 0. =

(vii) Let x = Let x = 0.1 i.e. x = 0.16363 …. (i)

⇒ 10x = 1.6363 …. (ii)

and 1000 x = 163.6363 …. (iii) Subtracting (ii) from (iii), we get

990x = 162

⇒ x = = Hence, 0.1 =

## Question 4:

- True. Since the collection of natural number is a sub collection of whole numbers, and every element of natural numbers is an element of whole numbers
- False. Since 0 is whole number but it is not a natural number.
- True. Every integer can be represented in a fraction form with denominator 1.
- False. Since division of whole numbers is not closed under division, the value of , p and q are integers and q ≠ 0, may not be a whole number.
- True. The prime factors of the denominator of the fraction form of terminating

decimal contains 2 and/or 5, which are integers and are not equal to zero.

- True. The prime factors of the denominator of the fraction form of repeating decimal contains integers, which are not equal to zero.
- True. 0 can considered as a fraction , which is a rational number.

## Question 1:

**Exercise 1C**

Irrational number: A number which cannot be expressed either as a terminating decimal or a repeating decimal is known as irrational number. Rather irrational numbers cannot

be expressed in the fraction form, , p and q are integers and q ≠ 0

For example, 0.101001000100001 is neither a terminating nor a repeating decimal and so is an irrational number.

Also, etc are examples of irrational numbers.

## Question 2:

(i)

We know that, if n is a perfect square, then is a rational number. Here, 4 is a perfect square and hence, = 2 is a rational number. So, is a rational number.

(ii)

We know that, if n is a perfect square, then is a rational number. Here, 196 is a perfect square and hence is a rational number. So, is rational.

(iii)

We know that, if n is a not a perfect square, then is an irrational number. Here, 21 is a not a perfect square number and hence, is an irrational number. So, is irrational.

(iv)

We know that, if n is a not a perfect square, then is an irrational number. Here, 43 is not a perfect square number and hence, is an irrational number. So, is irrational.

(v)

, is the sum of a rational number 3 and irrational number .

Theorem: The sum of a rational number and an irrational number is an irrational number.

So by the above theorem, the sum, , is an irrational number.

(vi)

= + (-2) is the sum of a rational number and an irrational number.

Theorem: The sum of a rational number and an irrational number is an irrational number.

So by the above theorem, the sum, + (-2) , is an irrational number.

So, is irrational.

(vii)

= × is the product of a rational number and an irrational number .

Theorem: The product of a non-zero rational number and an irrational number is an irrational number.

Thus, by the above theorem, × is an irrational number.

So, is an irrational number.

- 0.

Every rational number can be expressed either in the terminating form or in the non- terminating, recurring decimal form.

Therefore, 0. = 0.6666

## Question 3:

Let X’OX be a horizontal line, taken as the x-axis and let O be the origin. Let O represent 0.

Take OA = 1 unit and draw BA ⊥ OA such that AB = 1 unit, join OB. Then,

With O as centre and OB as radius, drawn an arc, meeting OX at P. Then, OP = OB = units

Thus the point P represents on the real line.

Now draw BC ⊥ OB such that BC = 1 units Join OC. Then,

With O as centre and OC as radius, draw an arc, meeting OX at Q. The, OQ = OC = units

Thus, the point Q represents on the real line. Now draw CD ⊥ OC such that CD = 1 units

Join OD. Then,

Now draw DE ⊥ OD such that DE = 1 units Join OE. Then,

With O as centre and OE as radius draw an arc, meeting OX at R. Then, OR = OE = units

Thus, the point R represents on the real line.

## Question 4:

Draw horizontal line X’OX taken as the x-axis Take O as the origin to represent 0.

Let OA = 2 units and let AB ⊥ OA such that AB = 1 units

Join OB. Then,

With O as centre and OB as radius draw an arc meeting OX at P. Then, OP = OB =

Now draw BC ⊥ OB and set off BC = 1 unit Join OC. Then,

With O as centre and OC as radius, draw an arc, meeting OX at Q. Then, OQ = OC =

Thus, Q represents on the real line. Now, draw CD ⊥ OC as set off CD = 1 units Join OD. Then,

With O as centre and OD as radius, draw an arc, meeting OX at R. Then OR = OD =

Thus, R represents on the real line.

## Question 5:

(i)

Since 4 is a rational number and is an irrational number.

So, is irrational because sum of a rational number and irrational number is always an irrational number.

(ii)

Since – 3 is a rational number and is irrational.

So, is irrational because sum of a rational number and irrational number is always an irrational number.

(iii)

Since 5 is a rational number and is an irrational number.

So, is irrational because product of a rational number and an irrational number is always irrational.

(iv)

Since -3 is a rational number and is an irrational number.

So, is irrational because product of a rational number and an irrational number is always irrational.

(v)

is irrational because it is the product of a rational number and the irrational number .

(vi)

is an irrational number because it is the product of rational number and irrational number .

## Question 6:

- True
- False
- True
- False
- True
- False
- False
- True
- True

# Exercise 1D

## Question 1:

(i)

We have:

(ii)

We have:

## Question 2:

**Question 3:**

(i) by

(ii) by

(iii) by

## Question 4:

**Question 5:**

Draw a line segment AB = 3.2 units and extend it to C such that BC = 1 units. Find the midpoint O of AC.

With O as centre and OA as radius, draw a semicircle. Now, draw BD AC, intersecting the semicircle at D. Then, BD = units.

With B as centre and BD as radius, draw an arc meeting AC produced at E. Then, BE = BD = units.

## Question 6:

Draw a line segment AB = 7.28 units and extend it to C such that BC = 1 unit. Find the midpoint O of AC.

With O as centre and OA as radius, draw a semicircle.

Now, draw BD AC, intersecting the semicircle at D. Then, BD = units.

With D as centre and BD as radius, draw an arc, meeting AC produced at E. Then, BE = BD = units.

## Question 7:

Closure Property: The sum of two real numbers is always a real number. Associative Law: (a + b) + c = a + (b + c) for al real numbers a, b, c.

Commutative Law: a + b = b + a, for all real numbers a and b.

Existence of identity: 0 is a real number such that 0 + a = a + 0, for every real number a. Existence of inverse of addition: For each real number a, there exists a real number (-a) such that

a + (-a) = (-a) + a= 0

a and (-a) are called the additive inverse of each other. Existence of inverse of multiplication:

For each non zero real number a, there exists a real number such that

a and are called the multiplicative inverse of each other.

## Question 1:

**Exercise 1E**

On multiplying the numerator and denominator of the given number by , we get

## Question 2:

On multiplying the numerator and denominator of the given number by , we get

## Question 3:

**Question 4:**

**Question 5:**

**Question 6:**

**Question 7:**

**Question 8:**

**Question 9:**

**Question 10:**

**Question 11:**

**Question 12:**

**Question 13:**

**Question 14:**

**Question 15:**

**Question 16:**

**Question 17:**

**Question 18:**

**Exercise 1F**

**Question 1:**

(i)

(ii)

(iii)

## Question 2:

(i)

(ii)

(iii)

## Question 3:

(i)

(ii)

(iii)

## Question 4:

(i)

(ii)

(iii)

## Question 5:

(i)

(ii)

(iii)

## Question 6:

(i)

(ii)

(iii)

## Question 7:

(i)

(ii)

(iii)