SELINA Solution Class 9 Chapter 1 Rational and Irrational Numbers Exercise 1B

Question 1.1

State, whether the following numbers is rational or not : ( 2 + √2 )²

Sol :

( 2 + √2 )²  = 2² + 2 ( 2 ) ( √2 ) + ( √2 )²
                   = 4 + 4√2 + 2 
                   = 6 + 4√2 
Irrational number.

Question 1.2

State, whether the following numbers is rational or not : ( 3 - √3 )²

Sol :

( 3 - √3 )²  = 3² - 2 ( 3 ) ( √3 ) + ( √3 )²
                   = 9 - 6√3 + 3 
                   = 12 - 6√3  = 6 ( 2 - √3 )
Irrational number.

Question 1.3

State, whether the following numbers is rational or not : ( 5 + √5 )( 5 - √5 )

Sol :

( 5 + √5 )( 5 - √5 ) = ( 5 )² - ( √5 )²
= 25 - 5 = 20 
Rational Number

Question 1.4

State, whether the following numbers is rational or not : ( √3 - √2 )²

Sol:

( √3 - √2 )²  = ( √3 )² - 2 ( √3 )( √2 ) + ( √2 )²
= 3 - 2√6 + 2 
= 5 - 2√6 
Irrational Number

Question 1.5

State, whether the following numbers is rational or not : 
${\displaystyle {\left(\frac{3}{2\sqrt{2}}\right)}^2}$

Sol:

${\displaystyle {\left(\frac{3}{2\sqrt{2}}\right)}^2=\frac{{\left(3\right)}^2}{{\left(2\sqrt{2}\right)}^2}=\frac{9}{4\times 2}=\frac{9}{8}}$

Rational number

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Question 1.6

State, whether the following number is rational or not :
${\displaystyle {\left(\frac{\surd 7}{6\sqrt{2}}\right)}^2}$

Sol:

${\displaystyle {\left(\frac{\surd 7}{6\sqrt{2}}\right)}^2}$

= ${\displaystyle \frac{{(\surd 7)}^2}{{(6^2\times \surd 2)}^2} ...[\because {\left(\frac{a}{b}\right)}^2=\frac{a^2}{b^2}]}$

= ${\displaystyle \frac{7}{36\times 2}}$

= ${\displaystyle \frac{7}{72}}$

${\displaystyle \frac{7}{72}}$ is rational Number

∴ ${\displaystyle {\left(\frac{\surd 7}{6\sqrt{2}}\right)}^2}$ is a rational Number.

Question 2.1

Find the square of : ${\displaystyle \frac{3\sqrt{5}}{5}}$

Sol:

${\displaystyle {\left(\frac{3\sqrt{5}}{5}\right)}^2=\frac{3^2{(\surd 5)}^2}{5^{20}}=\frac{9\times 5}{25}=\frac{9}{5}=1\frac{4}{5}}$

Question 2.2

Find the square of : √3 + √2

Sol:

( √3 + √2 )² = ( √3 )² + 2( √3 )( √2 ) + ( √2 )²
= 3 + 2√6 + 2
= 5 + 2√6

Question 2.3

Find the square of : √5 - 2

Sol:

( √5 - 2 )² = ( √5 )² - 2( √5 )( 2 ) + ( 2 )²
= 5 - 4√5 + 4
= 9 - 4√5

Question 2.4

Find the square of : 3 + 2√5

Sol:

( 3 + 2√5 )² = 3² + 2( 3 )( 2√5) + ( 2√5 )²
= 9 + 12√5 + 20
= 29 + 12√5

Question 3.1

State, in each case, whether true or false : 
√2 + √3 = √5

Sol:

False

Question 3.2

State, in each case, whether true or false : 
2√4 + 2 = 6

Sol:

2√4 + 2 = 2 x 2 + 2 = 4 + 2 = 6 which is True.

Question 3.3

State, in each case, whether true or false : 
3√7 - 2√7 = √7 

Sol:

3√7 - 2√7 = √7 - True.

Question 3.4

State, in each case, whether true or false : 
${\displaystyle \frac{2}{7}}$ ia an irrational number.

Sol:

False Because ${\displaystyle \frac{2}{7}=0.\overline{285714}}$ which is recurring and non-terminating and hence it is rational.

Question 3.5

State, in each case, whether true or false :
${\displaystyle \frac{5}{11}}$ is a rational number.

Sol:

True, because ${\displaystyle \frac{5}{11}=0.\overline{45}}$ which is recurring and non-terminating.

Question 3.6

State, in each case, whether true or false : 
All rational numbers are real numbers.

Sol: True

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Question 4.2

Given universal set =
${\displaystyle \{-6,-5\frac{3}{4},-\sqrt{4},-\frac{3}{5},-\frac{3}{8},0,\frac{4}{5},1,1\frac{2}{3},\sqrt{8},3.01,\pi ,8.47\}}$
From the given set, find : set of irrational numbers

Sol:

Given universal set =
${\displaystyle \{-6,-5\frac{3}{4},-\sqrt{4},-\frac{3}{5},-\frac{3}{8},0,\frac{4}{5},1,1\frac{2}{3},\sqrt{8},3.01,\pi ,8.47\}}$

We need to find the set of irrational numbers.
Irrational numbers are numbers which are not rational.
From the above subpart, the set of rational
Numbers is Q,
and Q = ${\displaystyle \{-6,-5\frac{3}{4},-\frac{3}{5},-\frac{3}{8},0,\frac{4}{5},1,1\frac{2}{3},3.01,8.47\}}$

Set of irrational numbers is the set of complement of the rational numbers over real numbers.
Here the set of irrational numbers is U - Q = { √8 , π }

Question 4.3

Given universal set =
${\displaystyle \{-6,-5\frac{3}{4},-\sqrt{4},-\frac{3}{5},-\frac{3}{8},0,\frac{4}{5},1,1\frac{2}{3},\sqrt{8},3.01,\pi ,8.47\}}$
From the given set, find : set of integers

Sol: Given Universal set is (Image to be added)

We need to find the set of integers.
Set of integers consists of zero, the natural numbers and their additive inverses.
The set of integers is Z.
Z = ${\displaystyle \{......-3,-2,-1,0,1,2,3,.......\}}$
Here the set of integers is U ∩ Z = ${\displaystyle \{-6,\surd 4,0,1\}}$

Question 4.4

Given universal set =
${\displaystyle \{-6,-5\frac{3}{4},-\sqrt{4},-\frac{3}{5},-\frac{3}{8},0,\frac{4}{5},1,1\frac{2}{3},\sqrt{8},3.01,\pi ,8.47\}}$
From the given set, find : set of non-negative integers

Given universal set =
${\displaystyle \{-6,-5\frac{3}{4},-\sqrt{4},-\frac{3}{5},-\frac{3}{8},0,\frac{4}{5},1,1\frac{2}{3},\sqrt{8},3.01,\pi ,8.47\}}$
From the given set, find : set of non-negative integers

Sol:

Given universal set =
${\displaystyle \{-6,-5\frac{3}{4},-\sqrt{4},-\frac{3}{5},-\frac{3}{8},0,\frac{4}{5},1,1\frac{2}{3},\sqrt{8},3.01,\pi ,8.47\}}$

We need to find the set of non-negative integers.
Set of non-negative integers consists of zero and the natural numbers.
The set of non-negative integers is Z⁺ and Z⁺ = { 0, 1, 2, 3,...... }
Here the set of integers is U ∩ Z⁺ = {0, 1}

Question 5

Use method of contradiction to show that √3 and √5 are irrational numbers.

Sol:

Let us suppose that √3 and √5 are rational numbers.

∴ √3 = ${\displaystyle \frac{a}{b}}$ and √5 = ${\displaystyle \frac{x}{y}}$ (Where a, b ∈ 7 and b, y  ≠ 0 x , y)

Squaring on both sides

3 = ${\displaystyle a^{/}{b}^2}$,             5 = ${\displaystyle \frac{x^2}{y^2}}$

${\displaystyle 3b^2=a^2}$,            ${\displaystyle 5y^2=x^2}$

⇒ a² and x² are odd as 3b² and 5y² are odd .

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

Let a = 3c, x = 5z

a² = 9c², x² = 25z²

3b² = 9c², 5y² = 25z²       (From equation )

⇒ b² =3c², y² = 5z²

⇒ b² and y² are odd as 3c² and 5z² 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 ${\displaystyle \frac{a}{b}}$ and ${\displaystyle \frac{x}{y}}$ are rational i.e, a, b and x, y do not have any common factors other than.

⇒ ${\displaystyle \frac{a}{b}}$ and ${\displaystyle \frac{x}{y}}$ is not rational.

⇒  √3 and √5 and are irrational.

Question 6.1

Prove that the following number is irrational: √3 + √2

Sol:

√3 + √2
Let √3 + √2 be a rational number.
⇒  √3 + √2 = x
Squaring on both the sides, we get
( √3 + √2 )² = x²
⇒ 3 + 2 + 2 x √3 x √2 = x^2
⇒ x² - 5 = 2√6
⇒ √6 = ${\displaystyle \frac{x^2-5}{2}}$
Here, x is a rational number.
⇒ x² is a rational number. 

⇒ x² - 5 is a rational number.

⇒  ${\displaystyle \frac{x^2-5}{2}}$  is also a rational number.

But √6 is an irrational number.

⇒  ${\displaystyle \frac{x^2-5}{2}}$  is also a irrational number.

⇒ x² - 5 is an irrational number.

⇒ x² - 5 is an irrational number.

⇒ x² is an irrational number. 

But we have assume that x is a rational numebr. We arrive at a contradiction.

So, our assumption that √3 + √2 is a rational number is wrong.

∴ √3 + √2 is an irrational number.

Question 6.2

Prove that the following number is irrational:  3 - √2

Sol:

3 - √2
Let  3 - √2 be a rational number.
⇒  3 - √2 = x
Squaring on both the sides, we get
(  3 - √2 )² = x²
⇒ 9 + 2 - 2 x  3 x √2 = x^2
⇒  11 - x² = 6√2
⇒ √2 = ${\displaystyle \frac{11-x^2}{6}}$
Here, x is a rational number.
⇒ x² is a rational number. 

⇒ 11 - x² is a rational number.

⇒  ${\displaystyle \frac{11-x^2}{6}}$ is also a rational number.

⇒ ${\displaystyle \sqrt{2}=\frac{11-x^2}{6}}$  is a rational number.

But √2 is an irrational number.

⇒  ${\displaystyle \frac{11-x^2}{6}=\sqrt{2}}$  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  3 - √2 is a rational number is wrong.
∴ 3 - √2 is an irrational number.

Question 6.3

Prove that the following number is irrational: √5 - 2

Sol:

√5 - 2
Let √5 - 2 be a rational number.
⇒ √5 - 2 = x
Squaring on both the sides, we get
${\displaystyle {(\surd 5-2)}^2=x^2}$

⇒ 5 + 4 - 2 x 2 x √5 = x²

⇒ 9 - x² = 4√5

⇒ √5 = ${\displaystyle \frac{9-x^2}{4}}$

Here, x is rational number

⇒ x² is a rational number. 

⇒ 9 - x² is a rational number.

⇒ ${\displaystyle \frac{9-x^2}{4}}$ is also a rational number.

⇒ √2 = ${\displaystyle \frac{9-x^2}{4}}$ is a rational number

But √2 is an irrational number.

⇒ √5 = ${\displaystyle \frac{9-x^2}{4}}$ 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 √5 - 2 is a rational number is wrong.
∴ √5 - 2 is an irrational number.

Question 7

Write a pair of irrational numbers whose sum is irrational.

Sol:

√3 + 5 and √5 - 3 are irrational numbers whose sum is irrational.
( √3 + 5  ) + ( √5 - 3 ) = √3 + √5 + 5 - 3 = √3 + √5 + 2 which is irrational.

Question 8

Write a pair of irrational numbers whose sum is rational.

Sol:

√3 + 5 and  4 - √3 are two irrational numbers whose sum is rational.
( √3 + 5  ) + ( 4 - √3 ) = √3 + 5+ 4 - √3 = 9

Question 9

Write a pair of irrational numbers whose difference is irrational.

Sol:

√3 + 2 and √2 - 3 are two irrational numbers whose difference is irrational.
( √3 + 2  ) - ( √2 - 3 ) = √3 - √2 + 2 + 3 = √3 - √2 + 5 which is irrational.

Question 10

Write a pair of irrational numbers whose difference is rational.

Sol:

√5 - 3 and √5 + 3 are irrational numbers whose difference is rational.
( √5 - 3 ) - ( √5 + 3  ) = √5 - 3 - √5 - 3 = -6 which is rational.

Question 11

Write a pair of irrational numbers whose product is irrational.

Sol:

Consider two irrational numbers ( 5 + √2 ) and ( √5 - 2 )
Thus, the product, ( 5 + √2 ) x ( √5 - 2 ) = 5√5 - 10 + √10 - 2√2 is irrational. 

Question 12

Write a pair of irrational numbers whose product is rational. 

Sol:

Consider √2 as an irrational number.
√2×√2= √4= 2 which is a rational number.

Question 13.1

Write in ascending order: 3√5 and 4√3

Sol:

3√5 = ${\displaystyle \sqrt{3^2\times 5}=\sqrt{45},4\sqrt{3}=\sqrt{4^2\times 3}=\sqrt{48}}$ 
and 45 < 48 
∴ ${\displaystyle \sqrt{45}<\sqrt{48}\Rightarrow 3\sqrt{5}<4\sqrt{3}}$

Question 13.2

Write in ascending order :  ${\displaystyle 2\sqrt[3]{5} \text{and}3\sqrt[3]{2}}$

Sol:

${\displaystyle 2\sqrt[3]{5}=\sqrt[3]{2^3\times 5}=\sqrt[3]{40}}$

${\displaystyle 3\sqrt[3]{2}=\sqrt[3]{3^3\times 2}=\sqrt[3]{54}}$

and 40 < 54 ⇒ ${\displaystyle \sqrt[3]{40}<\sqrt[3]{54}}$

⇒ ${\displaystyle 2\sqrt[3]{5} <3\sqrt[3]{2}}$

Question 13.3

Write in ascending order :  6√5, 7√3 and 8√2

Sol:

${\displaystyle 6\sqrt{5}=\sqrt{6^2\times 5}=\sqrt{180}}$

${\displaystyle 7\sqrt{3}=\sqrt{7^2\times 3}=\sqrt{147}}$

${\displaystyle 8\sqrt{2}=\sqrt{8^2\times 2}=\sqrt{128}}$

and 128 < 147 < 180

∴ ${\displaystyle \sqrt{128}<\sqrt{147}<\sqrt{180}}$

⇒ ${\displaystyle 8\sqrt{2}<7\sqrt{3}<6\sqrt{5}}$

Question 14.1

Write in descending order: ${\displaystyle 2\sqrt[3]{6}\text{and}3\sqrt[4]{2}}$

Sol:

${\displaystyle 2\sqrt[4]{6}=\sqrt[4]{2^4\times 6}=\sqrt[4]{96}}$

${\displaystyle 3\sqrt[4]{2}=\sqrt[4]{3^4}=\sqrt[4]{162}}$

Since 162 > 96
⇒ ${\displaystyle \sqrt[4]{162}>\sqrt[4]{96}}$

⇒ ${\displaystyle 3\sqrt[4]{2}>2\sqrt[4]{6}}$

Question 14.2

Write in descending order: 7√3 and 3√7

Sol:

7√3 = ${\displaystyle \sqrt{7^2\times 3}=\sqrt{147}}$

3√7 = ${\displaystyle \sqrt{3^2\times 7}=\sqrt{63}}$

147 > 63 
⇒ ${\displaystyle \sqrt{147}>\sqrt{63}}$

⇒ 7√3 > 3√7

Question 15.1

Compare : ${\displaystyle \sqrt[6]{15}\text{and}\sqrt[4]{12}}$

Sol:

${\displaystyle \sqrt[6]{15}={\left(15\right)}^{\frac{1}{6}}\text{and}\sqrt[4]{12}={\left(12\right)}^{\frac{1}{4}}}$

Make powers ${\displaystyle \frac{1}{6}\text{and}\frac{1}{4}}$ same
L.C.M. of 6,4 is 12

${\displaystyle \frac{1}{6}\times \frac{2}{2}=\frac{2}{12} \text{and}\frac{1}{4}\times \frac{3}{3}=\frac{3}{12}}$
⇒ ${\displaystyle \sqrt[6]{15}={15}^{\frac{1}{6}}={15}^{\frac{2}{12}}={\left({15}^2\right)}^{\frac{1}{12}}={225}^{\frac{1}{12}}}$

and ${\displaystyle \sqrt[4]{12}={12}^{\frac{1}{4}}={12}^{\frac{3}{12}}={\left({12}^3\right)}^{\frac{1}{12}}={\left(1728\right)}^{\frac{1}{12}}}$

⇒ 1272 > 225
⇒ ${\displaystyle {\left(1728\right)}^{\frac{1}{12}}>{225}^{\frac{1}{12}}}$ 

⇒ ${\displaystyle \sqrt[4]{12}>\sqrt[6]{15}}$

Question 15.2

Compare : ${\displaystyle \sqrt{24}\text{and}\sqrt[3]{35}}$

Sol:

${\displaystyle \sqrt{24}={\left(24\right)}^{\frac{1}{2}}\text{and}\sqrt[3]{35}={35}^{\frac{1}{3}}}$
L.C.M. of 2 and 3 is 6.

${\displaystyle \frac{1}{2}\times \frac{3}{3}=\frac{3}{6},\frac{1}{3}\times \frac{2}{2}=\frac{2}{6}}$

⇒ ${\displaystyle {24}^{\frac{1}{2}}={24}^{\frac{3}{6}}={\left({24}^3\right)}^{\frac{1}{6}}={\left(13824\right)}^{\frac{1}{6}}}$

⇒ ${\displaystyle {35}^{\frac{1}{3}}={35}^{\frac{2}{6}}={\left({35}^2\right)}^{\frac{1}{6}}={\left(1225\right)}^{\frac{1}{6}}}$

⇒ 13824 > 1225

⇒ ${\displaystyle {13824}^{\frac{1}{6}}>\sqrt[3]{35}}$ 
⇒ ${\displaystyle \sqrt{24}>\sqrt[3]{35}}$

Question 16

Insert two irrational numbers between 5 and 6.

Sol:

We know that 5 = √25 and 6 = √36.

Thus consider the numbers.
√25 < √26 < √27 < √28 < √29 < √30 < √31 < √32 < √33 < √34 < √35 < √36.

Therefore, any two irrational numbers between 5 and 6 is √27 and √28. 

Question 17

Insert five irrational numbers between 2√5 and 3√3.

Sol:

We know that 2√5 = ${\displaystyle \sqrt{4\times 5}}$ = √20 and 3√3 = √20

Thus, We have, √20 < √21 < √22 < √23 < √24 < √25 < √26 < √27.
So any five irrational numbers between 2√5 and 3√3 are :
√21, √22, √23, √24, and √26.

Question 18

Write two rational numbers between √2 and √3.

Sol:

We want rational numbers a/b and c/d such that : √2 < ${\displaystyle \frac{a}{b}<\frac{c}{d}}$ < √3
Consider any two rational numbers between 2 and 3 such that they are perfect squares.
Let us take 2.25 and 2.56 as √2.25 = 1.5 and √2.56 = 1.6

Thus we have,
√2 < √2.25 < √2.56 < √3
⇒ √2 < 1.5 < 1.6 < √3

⇒ √2 < ${\displaystyle \frac{15}{10}<\frac{16}{10}}$ < √3

⇒ √2 < ${\displaystyle \frac{3}{2}<\frac{8}{5}}$ < √3
Therefore any two rational numbers between √2 and √3 are : ${\displaystyle \frac{3}{2}\text{and}\frac{8}{5}}$

Question 19

Write three rational numbers between √3 and √5.

Sol:

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 √3.24 = 1.8, √3.61 = 1.9, √4 = 2, √4.41 = 2.1 and √4.84 = 2.2

Thus we have,
√3 < √3.24 < √3.61 < √4 < √4.41 < √4.84 < √5
⇒ √3 < 1.8 < 1.9 < 2 < 2.1 < 2.2 < √5

⇒ √3 < ${\displaystyle \frac{18}{10}<\frac{19}{10}<2<\frac{21}{10}<\frac{22}{10}}$ < √5

⇒ √3 < ${\displaystyle \frac{9}{5}<\frac{19}{10}<2<\frac{21}{10}<\frac{11}{5}}$ < √5

Therefore, any three rational numbers between √3 and √5 are :
${\displaystyle \frac{9}{5},\frac{19}{10}\text{and}\frac{21}{10}}$

Question 20.1

Simplify : ${\displaystyle \sqrt[5]{16}\times \sqrt[5]{2}}$

Sol:

${\displaystyle \sqrt[5]{16}\times \sqrt[5]{2}}$

= ${\displaystyle {16}^{\frac{1}{5}}\times 2^{\frac{1}{5}}}$

= ${\displaystyle 2^{4\times \frac{1}{5}}\times 2^{\frac{1}{5}}}$

= ${\displaystyle 2^{\frac{4}{5}}\times 2^{\frac{1}{5}}}$

= ${\displaystyle 2^{\frac{5}{5}}}$

= ${\displaystyle 2^1}$

= 2

Question 20.2

Simplify : ${\displaystyle \frac{\sqrt[4]{243}}{\sqrt[4]{3}}}$

Sol:

${\displaystyle \frac{\sqrt[4]{243}}{\sqrt[4]{3}}}$

= ${\displaystyle \frac{\sqrt[4]{3^5}}{\sqrt[4]{3}}}$

= ${\displaystyle \frac{3^{5\times \frac{1}{4}}}{3^{\frac{1}{4}}}}$

= ${\displaystyle \frac{3^{\frac{5}{4}}}{3^{\frac{1}{4}}}}$

= ${\displaystyle 3^{\frac{5}{4}-\frac{1}{4}}}$

= ${\displaystyle 3^{\frac{4}{4}}}$
= ${\displaystyle 3^1}$
= 3

Question 20.3

Simplify : ( 3 + √2 )( 4 + √7 )

Sol:

( 3 + √2 )( 4 + √7 )
= 3 x 4 + 3 x √7 + 4 x √2 + √2  x √7
= 12 + 3√7 + 4√2 + √14

Question 20.4

Simplify : (√3 - √2 )²

Sol:

(√3 - √2 )²

= ( √3 )² + ( √2 )² - 2 x √3 x √2

= 3 + 2 - 2√6

= 5 - 2√6