# SELINA Solutions for Class 9 Maths Chapter 1 - Rational and Irrational Numbers

## Chapter 1 - Rational and Irrational Numbers Exercise Ex. 1(A)

Is zero a rational number? Can it be written in the form , where p and q are integers and q≠0?

Yes, zero is a rational number.

As it can be written in the form of , where p and q are integers and q≠0?

⇒ 0 =

Are the following statements true or false? Give reasons for your answers.

i. Every whole number is a natural number.

ii. Every whole number is a rational number.

iii. Every integer is a rational number.

iv. Every rational number is a whole number.

i. False, zero is a whole number but not a natural number.

ii. True, Every whole can be written in the form of , where p and q are integers and q≠0.

iii. True, Every integer can be written in the form of , where p and q are integers and q≠0.

iv. False.

Example: is a rational number, but not a whole number.

Arrange in ascending order of their magnitudes.

Also, find the difference between the largest and smallest of these fractions. Express this difference as a decimal fraction correct to one decimal place.

Arrange in descending order of their

magnitudes.

Also, find the sum of the lowest and largest of these fractions. Express the result obtained as a decimal fraction correct to two decimal places.

## Chapter 1 - Rational and Irrational Numbers Exercise Ex. 1(B)

State, whether the following numbers are rational or not:

(i) (ii) (iii)

(iv) (v) (vi)

(i)

Irrational

(ii)

Irrational

(iii)

Rational

(iv)

Irrational

(v) Rational

(vi) Rational

Find the square of:

(i)

(ii)

(iii)

(iv)

State, in each case, whether true or false:

(i)

(ii)

(iii)

(iv) is an irrational number

(v) is a rational number.

(vi) All rational numbers are real numbers.

(vii) All real numbers are rational numbers.

(viii) Some real numbers are rational numbers.

(i) False

(ii) which is true

(iii) True.

(iv) False because

which is recurring and non-terminating and hence it is rational

(v) True because which is recurring and non-terminating

(vi) True

(vii) False

(viii) True.

Given universal set =

From the given set, find :

(i) set of rational numbers

(ii) set of irrational numbers

(iii) set of integers

(iv) set of non-negative integers

(i)

(ii)

(iii)

(iv)

Use method of contradiction to show that and are irrational numbers.

Let us suppose that and are rational numbers

and (Where a, b 7 and b, y 0 x , y)

Squaring both sides

a^{2} and x^{2} are odd as 3b^{2} and 5y^{2} are odd .

a and x are odd....(1)

Let a = 3c, x = 5z

a^{2} = 9c^{2}, x^{2} = 25z^{2}

3b^{2} = 9c^{2}, 5y^{2} = 25z^{2}(From equation )

b^{2} =3c^{2}, y^{2} = 5z^{2}

b^{2} and y^{2} are odd as 3c^{2} and 5z^{2} are odd .

b and y are odd...(2)

From equation (1) and (2) we get a, b, x, y are odd integers.

i.e., a, b, and x, y have common factors 3 and 5 this contradicts our assumption that are rational i.e, a, b and x, y do not have any common factors other than.

is not rational

and are irrational.

Prove that each of the following numbers is irrational:

Let be a rational number.

⇒ = x

Squaring on both the sides, we get

Here, x is a rational number.

⇒ x^{2 }is a rational number.

⇒ x^{2} - 5 is a rational number.

⇒ is also a rational number.

is a rational number.

But is an irrational number.

is an irrational number.

⇒ x^{2}- 5 is an irrational number.

⇒ x^{2} is an irrational number.

⇒ x is an irrational number.

But we have assume that x is a rational number.

∴ we arrive at a contradiction.

So, our assumption that is a rational number is wrong.

∴ is an irrational number.

Let be a rational number.

⇒ = x

Squaring on both the sides, we get

Here, x is a rational number.

⇒ x^{2 }is a rational number.

⇒ 11 - x^{2} is a rational number.

⇒ is also a rational number.

is a rational number.

But is an irrational number.

is an irrational number.

⇒ 11 - x^{2} is an irrational number.

⇒ x^{2} is an irrational number.

⇒ x is an irrational number.

But we have assume that x is a rational number.

∴ we arrive at a contradiction.

So, our assumption that is a rational number is wrong.

∴ is an irrational number.

Let be a rational number.

⇒ = x

Squaring on both the sides, we get

Here, x is a rational number.

⇒ x^{2 }is a rational number.

⇒ 9 - x^{2} is a rational number.

⇒ is also a rational number.

is a rational number.

But is an irrational number.

is an irrational number.

⇒ 9 - x^{2} is an irrational number.

⇒ x^{2} is an irrational number.

⇒ x is an irrational number.

But we have assume that x is a rational number.

∴ we arrive at a contradiction.

So, our assumption that is a rational number is wrong.

∴ is an irrational number.

Write a pair of irrational numbers whose sum is irrational.

are irrational numbers whose sum is irrational.

which is irrational.

Write a pair of irrational numbers whose sum is rational.

and are two irrational numbers whose sum is rational.

Write a pair of irrational numbers whose difference is irrational.

and are two irrational numbers whose difference is irrational.

which is irrational.

Write a pair of irrational numbers whose difference is rational.

and are irrational numbers whose difference is rational.

which is rational.

Write a pair of irrational numbers whose product is irrational.

Write in ascending order:

(i)

(ii)

(iii)

(i)

and 45 < 48

(ii)

and40 < 54

(iii)

and 128 < 147 < 180

Write in descending order:

(i)

(ii)

(i)

Since 162 > 96

(ii)

141 > 63

Compare.

(i) and

Make powers and same

L.C.M. of 6,4 is 12

and

(ii) and

L.C.M. of 2 and 3 is 6.

,

Insert two irrational numbers between 5 and 6.

Insert five irrational numbers between and .

Write two rational numbers between

We want rational numbers a/b and c/d such that: < a/b < c/d <

Consider any two rational numbers between 2 and 3 such that they are perfect squares.

Let us take 2.25 and 2.56 as

Thus we have,

Write three rational numbers between

Consider some rational numbers between 3 and 5 such that they are perfect squares.

Let us take, 3.24, 3.61, 4, 4.41 and 4.84 as

## Chapter 1 - Rational and Irrational Numbers Exercise Ex. 1(C)

State, with reasons, which of the following are surds and which are not:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(i) Which is irrational

is a surd

(ii) Which is irrational

is a surd

(iii)

is a surd

(iv) which is rational

is not a surd

(v)

is not a surd

(vi) = -5

is is not a surd

(vii) not a surd as is irrational

(viii) is not a surd because is irrational.

Write the lowest rationalising factor of:

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(i) which is rational

lowest rationalizing factor is

(ii)

lowest rationalizing factor is

(iii)

lowest rationalizing factor is

(iv)

Therefore, lowest rationalizing factor is

(v)

lowest rationalizing factor is

(vi)

Therefore lowest rationalizing factor is

(vii)

Its lowest rationalizing factor is

(viii)

Its lowest rationalizing factor is

(ix)

its lowest rationalizing factor is

Rationalise the denominators of :

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

Find the values of 'a' and 'b' in each of the following:

(i)

(ii)

(iii)

(iv)

(i)

(ii)

(iii)

(iv)

Simplify:

(i)

(ii)

(i)

(ii)

If x = and y = ; find:

(i) x^{2} (ii) y^{2}

(iii) xy (iv) x^{2} + y^{2} + xy.

(i)

(ii)

(iii) xy =

(iv) x^{2} + y^{2} + xy = 161 -

= 322 + 1 = 323

(i) m^{2 }

(ii) n^{2}

(iii) mn

If x = 2+ 2, find:

(i) (ii) (iii)

(i)

(ii)

(iii)

If = 1.4 and = 1.7, find the value of each of the following, correct to one decimal place:

## Chapter 1 - Rational and Irrational Numbers Exercise Ex. 1(D)

Simplify:

Simplify:

Evaluate, correct to one place of decimal, the expression, if = 2.2 and = 3.2.

If x = , find the value of:

Show that:

(i) Negative of an irrational number is irrational number.

(ii) The product of a non - zero rational number and an irrational number is a rational number.

**Proof:**

**(i)**

The given statement is TRUE.

Let us assume that negative of an irrational number is a rational number.

Let p be an irrational number,

→ - p is a rational number.

→ - (-p) = p is a rational number.

But p is an irrational number.

Therefore our assumption was wrong.

So, Negative of an irrational number is irrational number.

**(ii)**

The given statement is FALSE.

We know that 3 is a non - zero rational number and is an irrational number.

So, 3 ×= 3 is an irrational.

Therefore, the product of a non - zero rational number and an irrational number is an irrational number.

Draw a line segment of length cm.

Since

So, first we need to find and mark it on number line and then find

**Steps to draw **** on the number line are:**

- Draw a number line and mark point O.
- Mark point A on it such that OA = 1 unit.
- Draw right triangle OAB such that ∠A = 90° and AB = 1 unit.
- Join OB.
- By Pythagoras Theorem, .
- Draw a line BC perpendicular to OB such that BC = 1.
- Join OC. Thus, OBC is a right triangle. Again by Pythagoras Theorem, we have
- With centre as O and radius OC draw a circle which meets the number line at a point E.
- units. Thus, OE represents on the number line.

Draw a line segment of length cm.

Since,

We need to construct a right - angled triangle OAB, in which

∠A = 90°, OA = 2 units and AB = 2 units.

By Pythagoras theorem, we get

OB^{2} = OA^{2}
+ AB^{2}

∴ OB = units

**STEPS:**

- Draw a number line.
- Mark point A on it which is two points (units) from an initial point say O in the right/positive direction.
- Now, draw a line AB = 2 units which is perpendicular to A.
- Join OB to represent the hypotenuse of a triangle OAB, right angled at A.
- This hypotenuse of triangle OAB is showing a length of OB.
- With centre as O and radius as OB, draw an arc on the number line which cuts it at the point C.
- Here, C represents.

Show that:

Show that:

Show that x is irrational, if:

(i) x^{2} = 6 (ii) x^{2} = 0.009 (iii) x^{2} = 27

(i) x^{2} = 6

(ii) x^{2} = 0.009

(iii) x^{2} = 27

Show that x is rational, if:

(i) x^{2} = 16 (ii) x^{2} = 0.0004 (iii) x^{2} =

(i) x^{2} = 16

Taking square root on both the sides, we get

x = ± 4

As we know that 4 = and -4 =are rational numbers.

⇒ x is rational , if: x^{2} =
16 .

(ii) x^{2} = 0.0004 (iii) x^{2} =

x^{2} = 0.0004 =

Taking square root on both the sides, we get

…. is a rational number.

⇒ x is rational , if: x^{2}
=0.0004.

(iii) x^{2} =

Taking square root on both the sides, we get

…. is a rational number.

⇒ x is rational , if: x^{2} =.

Using the following figure, show that BD =.

From the image, it is cleared that

AB = x, BC = 1

⇒ AC = AB + BC = x + 1

O is a centre of the semi - circle with AC as diameter.

OA, OC and OD are the radii.

⇒ OA = OC = OD =

OB = OC - BC =

m∠OBD = 90° … given in the image

∴By Pythagoras theorem, we get

OD^{2} = OB^{2}
+ BD^{2}

⇒ BD^{2} =
OD^{2} - OB^{2}

⇒

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