Request a call back

Join NOW to get access to exclusive study material for best results

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

Rational and Irrational Numbers Exercise Ex. 1(A)

Solution 1(a)

Correct option: (iii) q ≠ 0

For   a rational number, p and q should be integers and q ≠ 0.

 

Solution 1(b)

Correct option: (ii) real number

A non-terminating decimal number is a rational number and hence, a real number.

Solution 1(c)

Correct option: (ii) recurring

In 7.478478….., '478' is repeated.

Such a non-terminating decimal, in which a set of digits repeats continuously, is called recurring decimal.

Solution 1(d)

Correct option: (ii) non-terminating

  = 0.9466666666……

 

Hence, it is non-terminating.

Solution 1(e)

Correct option: (iv) None

If the denominator of a rational number can be expressed as 2m × 5n, where m and n are both whole numbers, the rational number is a terminating decimal.

Since, 85 = 5 × 17,   is not a terminating decimal.

Since, 405 = 3 × 3 × 3 × 3 × 5,   is not a terminating decimal.

Since, 524 = 2 × 2 × 131,   is not a terminating decimal.

 

Hence, none of the given numbers are terminating.

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.

Solution 3

  

Solution 4

  

Solution 5(i)

  

Solution 5(ii)

  

Solution 5(iii)

  

Solution 5(iv)

  

Solution 5(v)

  

Rational and Irrational Numbers Exercise Ex. 1(B)

Solution 1(a)

Correct option: (ii) an irrational number

Negative of an irrational number is an irrational number.

For example,   is an irrational number indicates -   is also an irrational

 

number.

Solution 1(b)

Correct option: (ii) irrational

  which is irrational.

 

Solution 1(c)

Correct option: (i)

 

 

 

In right-angled DABO,

OA2 = AB2 + OB2 = 22 + 12 = 4 + 1 = 5

OA =

 

Solution 1(d)

Correct option: (ii)

 , which is irrational.

 

Solution 1(e)

Correct option: (i)

Since

  are irrational numbers between 8 and 11.

Hence,  and   are two irrational numbers between 8 and 11. 

Solution 2

(i) open parentheses 2 plus square root of 2 close parentheses squared equals 2 squared plus 2 open parentheses 2 close parentheses open parentheses square root of 2 close parentheses plus open parentheses square root of 2 close parentheses squared

                      

 Irrational

(ii)

                      

Irrational

(iii)

Rational

(iv)

Irrational

Solution 3

 (i)

open parentheses fraction numerator 3 square root of 5 over denominator 5 end fraction close parentheses squared equals fraction numerator 3 squared open parentheses square root of 5 close parentheses squared over denominator 5 squared end fraction
space space space space space space space space space space space space space space space space space equals fraction numerator 9 cross times 5 over denominator 25 end fraction
space space space space space space space space space space space space space space space space space equals 9 over 5
space space space space space space space space space space space space space space space space space equals 1 4 over 5

(ii)

 

(iii)

open parentheses square root of 5 minus 2 close parentheses squared equals open parentheses square root of 5 close parentheses squared minus 2 open parentheses square root of 5 close parentheses open parentheses 2 close parentheses plus open parentheses 2 close parentheses squared
space space space space space space space space space space space space space space space space space equals 5 minus 4 square root of 5 plus 4
space space space space space space space space space space space space space space space equals 9 minus 4 square root of 5

(iv)

open parentheses 3 plus 2 square root of 5 close parentheses squared equals 3 squared plus 2 open parentheses 3 close parentheses open parentheses 2 square root of 5 close parentheses plus open parentheses 2 square root of 5 close parentheses squared
space space space space space space space space space space space space space space space space space space space equals 9 plus 12 square root of 5 plus 20
space space space space space space space space space space space space space space space space space space space equals 29 plus 12 square root of 5

 

 

Solution 4

(i) False

(ii) which is true

(iii) True.

(iv) False because 

2 over 7 equals 0. top enclose 285714

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.

Solution 5

  

 

(i)

(ii) 

 

(iii) 

 

(iv)  

Solution 6

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 7

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.

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.

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.

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.

x2 is a rational number. 

9 - x2 is a rational number.

 is also a rational number.

rightwards double arrow square root of 5 equals fraction numerator 9 minus x squared over denominator 4 end fractionis a rational number.

But square root of 5 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.

we arrive at a contradiction.

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

  is an irrational number.

Solution 8

are irrational numbers whose sum is irrational.

which is irrational.

Solution 9

and are two irrational numbers whose sum is rational.

Solution 10

and are two irrational numbers whose difference is irrational.

which is irrational.

Solution 11

and are irrational numbers whose difference is rational.

which is rational.

Solution 12

C o n s i d e r space t w o space i r r a t i o n a l space n u m b e r s space open parentheses 5 plus square root of 2 close parentheses space a n d space open parentheses square root of 5 minus 2 close parentheses
T h u s comma space t h e space p r o d u c t comma space open parentheses 5 plus square root of 2 close parentheses space cross times space open parentheses square root of 5 minus 2 close parentheses equals 5 square root of 5 minus 10 plus square root of 10 minus 2 square root of 2 space i s space i r r a t i o n a l.

Solution 13

Solution 14

(i)

and 45 < 48

(ii)

 

and40 < 54

(iii)

 

and 128 < 147 < 180

Solution 15

(i)

Since 162 > 96

(ii)

141 > 63

Solution 16

W e space k n o w space t h a t space 2 square root of 5 equals square root of 4 cross times 5 end root equals square root of 20 space a n d space 3 square root of 3 equals square root of 27
T h u s comma space w e space h a v e comma space square root of 20 less than square root of 21 less than square root of 22 less than square root of 23 less than square root of 24 less than square root of 25 less than square root of 26 less than square root of 27
S o space a n y space f i v e space i r r a t i o n a l space n u m b e r s space b e t w e e n space 2 square root of 5 space a n d space 3 square root of 3 space a r e :
square root of 21 comma square root of 22 comma square root of 23 comma square root of 24 space a n d space square root of 26

Solution 17

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 square root of 2.25 end root equals 1.5 space a n d space square root of 2.56 end root equals 1.6

Thus we have,

square root of 2 less than square root of 2.25 end root less than square root of 2.56 end root less than square root of 3
rightwards double arrow square root of 2 less than 1.5 less than 1.6 less than square root of 3
rightwards double arrow square root of 2 less than 15 over 10 less than 16 over 10 less than square root of 3
rightwards double arrow square root of 2 less than 3 over 2 less than 8 over 5 less than square root of 3
T h e r e f o r e space a n y space t w o space r a t i o n a l space n u m b e r s space b e t w e e n space square root of 2 space a n d space square root of 3 space a r e : space 3 over 2 space a n d space 8 over 5



Solution 18

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

square root of 3.24 end root equals 1.8 comma space square root of 3.61 end root equals 1.9 comma space square root of 4 equals 2 comma space square root of 4.41 end root equals 2.1 space a n d space square root of 4.84 end root equals 2.2

T h u s space w e space space h a v e comma
square root of 3 less than square root of 3.24 end root less than square root of 3.61 end root less than square root of 4 less than square root of 4.41 end root less than square root of 4.84 end root less than square root of 5
rightwards double arrow square root of 3 less than 1.8 less than 1.9 less than 2 less than 2.1 less than 2.2 less than square root of 5
rightwards double arrow square root of 3 less than 18 over 10 less than 19 over 10 less than 2 less than 21 over 10 less than 22 over 10 less than square root of 5
rightwards double arrow square root of 3 less than 9 over 5 less than 19 over 10 less than 2 less than 21 over 10 less than 11 over 5 less than square root of 5
T h e r e f o r e comma space a n y space t h r e e space r a t i o n a l space n u m b e r s space b e t w e e n space square root of 3 space a n d space square root of 5 space a r e :
9 over 5 comma 19 over 10 space a n d space 21 over 10

Rational and Irrational Numbers Exercise Ex. 1(C)

Solution 1(a)

Correct option: (i)

 

Rationalising,

Solution 1(b)

Correct Option:

Rationalising,

Solution 1(c)

Correct option: (iii) 3

Solution 1(d)

Correct option: (iv)

 

Solution 1(e)

Correct option: (i)

 

Solution 1(f)

Correct option: (ii)

Solution 1(g)

Correct option: (iii) 4

Rationalising,

Solution 1(h)

Correct option: (iv)

Solution 2

(i) Which is irrational

is a surd

 

(ii) Which is irrational

therefore fourth root of 27 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 square root of straight pi is irrational

 

(viii) is not a surd because square root of 3 plus square root of 2 end root is irrational.

Solution 3

(i) which is rational

lowest rationalizing factor is

 

(ii)

lowest rationalizing factor is

 

(iii) 

lowest rationalizing factor is

 

(iv)

open parentheses 7 minus square root of 7 close parentheses open parentheses 7 plus square root of 7 close parentheses equals 49 minus 7 equals 42

Therefore, lowest rationalizing factor is

 

(v) square root of 18 minus square root of 50

lowest rationalizing factor is

  (vi)

open parentheses square root of 5 minus square root of 2 close parentheses open parentheses square root of 5 plus square root of 2 close parentheses equals open parentheses square root of 5 close parentheses squared minus open parentheses square root of 2 close parentheses squared equals 3

Therefore lowest rationalizing factor is

 
 (vii)

open parentheses square root of 13 plus 3 close parentheses open parentheses square root of 13 minus 3 close parentheses equals open parentheses square root of 13 close parentheses squared minus 3 squared equals 13 minus 9 equals 4

Its lowest rationalizing factor is

 

Solution 4

(i)

(ii)

(iii)

 (iv)

 

 (v)

 

Solution 5

(i)

 (ii)

(iii)

 

Solution 6

(i)


(ii)

fraction numerator square root of 2 over denominator square root of 6 minus square root of 2 end fraction minus fraction numerator square root of 3 over denominator square root of 6 plus square root of 2 end fraction equals fraction numerator square root of 2 open parentheses square root of 6 plus square root of 2 close parentheses minus square root of 3 open parentheses square root of 6 minus square root of 2 close parentheses over denominator open parentheses square root of 6 close parentheses squared minus open parentheses square root of 2 close parentheses squared end fraction

Solution 7

(i)

 

 

(ii)

 

 

(iii) xy =

 

(iv) x2 + y2 + xy = 161 -

= 322 + 1 = 323

 

 

 

Solution 8

left parenthesis straight i right parenthesis space straight m equals fraction numerator 1 over denominator 3 minus 2 square root of 2 end fraction
space space space space equals fraction numerator 1 over denominator 3 minus 2 square root of 2 end fraction cross times fraction numerator 3 plus 2 square root of 2 over denominator 3 plus 2 square root of 2 end fraction
space space space space equals fraction numerator 3 plus 2 square root of 2 over denominator open parentheses 3 close parentheses squared minus open parentheses 2 square root of 2 close parentheses squared end fraction
space space space space equals fraction numerator 3 plus 2 square root of 2 over denominator 9 minus 8 end fraction
space space space space equals 3 plus 2 square root of 2
rightwards double arrow straight m squared equals open parentheses 3 plus 2 square root of 2 close parentheses squared
space space space space space space space space space space equals open parentheses 3 close parentheses squared plus 2 cross times 3 cross times 2 square root of 2 plus open parentheses 2 square root of 2 close parentheses squared
space space space space space space space space space space equals 9 plus 12 square root of 2 plus 8
space space space space space space space space space space equals 17 plus 12 square root of 2

 left parenthesis iii right parenthesis space space mn equals open parentheses 3 plus 2 square root of 2 close parentheses open parentheses 3 minus 2 square root of 2 close parentheses equals open parentheses 3 close parentheses squared minus open parentheses 2 square root of 2 close parentheses squared equals 9 minus 8 equals 1

Solution 9

(i)

 

(ii)

 

(iii)

 

Solution 10

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution 11

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Solution 12

straight L. straight H. straight S. equals fraction numerator 1 over denominator 3 minus 2 square root of 2 end fraction minus fraction numerator 1 over denominator 2 square root of 2 minus square root of 7 end fraction plus fraction numerator 1 over denominator square root of 7 minus square root of 6 end fraction minus fraction numerator 1 over denominator square root of 6 minus square root of 5 end fraction plus fraction numerator 1 over denominator square root of 5 minus 2 end fraction
space space space space space space space space space space space space equals fraction numerator 1 over denominator 3 minus square root of 8 end fraction minus fraction numerator 1 over denominator square root of 8 minus square root of 7 end fraction plus fraction numerator 1 over denominator square root of 7 minus square root of 6 end fraction minus fraction numerator 1 over denominator square root of 6 minus square root of 5 end fraction plus fraction numerator 1 over denominator square root of 5 minus 2 end fraction
space space space space space space space space space space space space equals fraction numerator 1 over denominator 3 minus square root of 8 end fraction cross times fraction numerator 3 plus square root of 8 over denominator 3 plus square root of 8 end fraction minus fraction numerator 1 over denominator square root of 8 minus square root of 7 end fraction cross times fraction numerator square root of 8 plus square root of 7 over denominator square root of 8 plus square root of 7 end fraction plus fraction numerator 1 over denominator square root of 7 minus square root of 6 end fraction cross times fraction numerator square root of 7 plus square root of 6 over denominator square root of 7 plus square root of 6 end fraction
space space space space space space space space space space space space space space space space space space space space space space space space space minus fraction numerator 1 over denominator square root of 6 minus square root of 5 end fraction cross times fraction numerator square root of 6 plus square root of 5 over denominator square root of 6 plus square root of 5 end fraction plus fraction numerator 1 over denominator square root of 5 minus 2 end fraction cross times fraction numerator square root of 5 plus 2 over denominator square root of 5 plus 2 end fraction
space space space space space space space space space space space space equals fraction numerator 3 plus square root of 8 over denominator open parentheses 3 close parentheses squared minus open parentheses square root of 8 close parentheses squared end fraction minus fraction numerator square root of 8 plus square root of 7 over denominator open parentheses square root of 8 close parentheses squared minus open parentheses square root of 7 close parentheses squared end fraction plus fraction numerator square root of 7 plus square root of 6 over denominator open parentheses square root of 7 close parentheses squared minus open parentheses square root of 6 close parentheses squared end fraction minus fraction numerator square root of 6 plus square root of 5 over denominator open parentheses square root of 6 close parentheses squared minus open parentheses square root of 5 close parentheses squared end fraction plus fraction numerator square root of 5 plus 2 over denominator open parentheses square root of 5 close parentheses squared minus open parentheses 2 close parentheses squared end fraction
space space space space space space space space space space space space equals fraction numerator 3 plus square root of 8 over denominator 9 minus 8 end fraction minus fraction numerator square root of 8 plus square root of 7 over denominator 8 minus 7 end fraction plus fraction numerator square root of 7 plus square root of 6 over denominator 7 minus 6 end fraction minus fraction numerator square root of 6 plus square root of 5 over denominator 6 minus 5 end fraction plus fraction numerator square root of 5 plus 2 over denominator 5 minus 4 end fraction
space space space space space space space space space space space space equals 3 plus square root of 8 minus square root of 8 minus square root of 7 plus square root of 7 plus square root of 6 minus square root of 6 minus square root of 5 plus square root of 5 plus 2
space space space space space space space space space space space space equals 3 plus 2
space space space space space space space space space space space space equals 5
space space space space space space space space space space space space equals straight R. straight H. straight S.
 

Solution 13

   

Solution 14(i)

  

Solution 14(ii)

  

Solution 14(iii)

Solution 15

Solution 16

Rational and Irrational Numbers Exercise Test Yourself

Solution 1

Solution 2

Solution 3

  

Solution 4

  

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 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 .

  

Solution 8

   

Solution 9

Solution 10

(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 = . 

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

Get Latest Study Material for Academic year 24-25 Click here
×