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

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

### Solution 1

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 =

### Solution 2

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.

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

### Solution 5

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.

### Solution 12

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 =

mOBD = 90° … given in the image

By Pythagoras theorem, we get

OD2 = OB2 + BD2

⇒ BD2 = OD2 - OB2

### Solution 6

Since

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

Steps to draw  on the number line are:

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

### Solution 7

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

OB2 = OA2 + AB2

OB =  units

STEPS:

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

(i) x2 = 6

(ii) x2 = 0.009

(iii) x2 = 27

### Solution 11

(i) x2 = 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: x2 = 16 .

(ii) x2 = 0.0004 (iii) x2 =

x2 = 0.0004 =

Taking square root on both the sides, we get

…. is a rational number.

x is rational , if: x2 =0.0004.

(iii) x2 =

Taking square root on both the sides, we get

…. is a rational number.

x is rational , if: x2 =.

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

### Solution 5

1. Take OA = 1 unit and draw OAB = 90o such that AB = 1 unit. Hence by Pythagoras, OB =

2. Now draw OBC = 90o such that BC = 1 unit. Hence by Pythagoras, OC =

3. With point O as center and OC as radius, draw an arc which meets the number line at point P, so OP =

4. Now with P as center and OP as radius, draw an arc which meets the number line at point Q, so OQ = 2

1. Take OA = 1 unit and draw OAB = 90o such that AB = 1 unit. Hence by Pythagoras, OB =

2. Now with point O as center and OB as radius, draw an arc which meets the number line at point P, so OP =

3. Now with P as center and OP as radius, draw an arc which meets the number line at point Q, so OQ = 2

1. Draw the figure as shown below

2. Clearly, O'P = O'O + OP =

1. Draw the figure as shown below

2. Clearly, O'P' = O'O - OP' =

1. Take OA = 2 units and draw OAB = 90o such that AB = 1 unit. Hence by Pythagoras OB =

2. With point O as center and OB as radius, draw an arc which meets the number line at point P, so OP =

3. Now with P as center and OP as radius, draw an arc which meets the number line at point Q, so OQ = 2

### Solution 1

(i)

Irrational

(ii)

Irrational

(iii)

Rational

(iv)

Irrational

(v) Rational

(vi) Rational

(i)

(ii)

(iii)

(iv)

### Solution 3

(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.

(i)

(ii)

(iii)

(iv)

### Solution 6

Let be a rational number.

= x

Squaring on both the sides, we get

Here, x is a rational number.

x2 is a rational number.

x2 - 5 is a rational number.

is also a rational number.

is a rational number.

But  is an irrational number.

is an irrational number.

x2- 5 is an irrational number.

x2 is an irrational number.

x is an irrational number.

But we have assume that x is a rational number.

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.

x2 is a rational number.

11 - x2 is a rational number.

is also a rational number.

is a rational number.

But  is an irrational number.

is an irrational number.

11 - x2 is an irrational number.

x2 is an irrational number.

x is an irrational number.

But we have assume that x is a rational number.

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.

x2 is a rational number.

9 - x2 is a rational number.

is also a rational number.

is a rational number.

But  is an irrational number.

is an irrational number.

9 - x2 is an irrational number.

x2 is an irrational number.

x is an irrational number.

But we have assume that x is a rational number.

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

is an irrational number.

### Solution 7

are irrational numbers whose sum is irrational.

which is irrational.

### Solution 8

and are two irrational numbers whose sum is rational.

### Solution 9

and are two irrational numbers whose difference is irrational.

which is irrational.

### Solution 10

and are irrational numbers whose difference is rational.

which is rational.

### Solution 13

(i)

and 45 < 48

(ii)

and40 < 54

(iii)

and 128 < 147 < 180

(i)

Since 162 > 96

(ii)

141 > 63

### Solution 15

(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.

,

### Solution 18

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,

### Solution 19

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

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

### Solution 1

(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.

### Solution 2

(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

(i)

(ii)

(iii)

(iv)

(v)

(vi)

(vii)

(viii)

(ix)

(i)

(ii)

(iii)

(iv)

(i)

(ii)

### Solution 6

(i)

(ii)

(iii) xy =

(iv) x2 + y2 + xy = 161 -

= 322 + 1 = 323

(i)

(ii)

(iii)