ML Aggarwal Solution Class 9 Chapter 1 Rational and Irrational Numbers Exercise 1.4

 Exercise 1.4

Question 1

(i) $=\sqrt{45}-3 \sqrt{20}+4 \sqrt{5}$

$=\sqrt{9 \times 5}-3 \sqrt{4 \times 5}+4 \sqrt{5}$

$=\sqrt{9} \sqrt{5}-3 \sqrt{4} \sqrt{5}+4 \sqrt{5}$

$=3 \sqrt{5}-3 \times 2 \sqrt{5}+4 \sqrt{5}$

$=3 \sqrt{5}-6 \sqrt{5}+4 \sqrt{5}$

$=(3-6+4) \sqrt{5}$

$=(7-6) \sqrt{5}$

$=\sqrt{5}$


(ii) $=3 \sqrt{3}+2 \sqrt{27}+\frac{7}{\sqrt{3}}$

$=3 \sqrt{3}+2 \sqrt{9 \times 3}+\frac{7}{\sqrt{3}}$

$=3 \sqrt{3}+2 \times \sqrt{9} \sqrt{3}+\frac{7}{\sqrt{3}}$

$=3 \sqrt{3}+2 \times 3 \sqrt{3}+\frac{7}{\sqrt{3}}$

$=3 \sqrt{3}+6 \sqrt{3}+\frac{7}{\sqrt{3}}$

$=(3+6) \sqrt{3}+\frac{7}{\sqrt{3}}$

$=9 \sqrt{3}+\frac{7}{\sqrt{3}}$


Multiplying and Dividing by "√3"

$=9 \sqrt{3}+\frac{7}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$

$=9 \sqrt{3}+\frac{7 \sqrt{3}}{3}$

By cross multiplying

$=\frac{3 \times 9 \sqrt{3}+7 \sqrt{3}}{3}$

$=\frac{27 \sqrt{3}+7 \sqrt{3}}{3}$

$=\frac{(27+7) \sqrt{3}}{3}$

$=\frac{34 \sqrt{3}}{3}$


(iii) $6 \sqrt{5} \times 2 \sqrt{5}$

$6 \times 2 \times \sqrt{5} \cdot \sqrt{5}$

$=12 \times(\sqrt{5})^{2}$

$=12 \times 5$

$=60$


(iv) $8 \sqrt{15} \div 2 \sqrt{3}$

$\frac{8 \sqrt{15}}{2 \sqrt{3}}=\frac{8 \sqrt{5} \sqrt{3}}{2 \sqrt{3}}$

$=4 \sqrt{5}$


(v) $\frac{\sqrt{24}}{8}+\frac{\sqrt{54}}{9}$

Sol :

$\frac{\sqrt{6 \times 4}}{8}+\frac{\sqrt{9 \times 6}}{9}$

$=\frac{\sqrt{6} \cdot \sqrt{4}}{8}+\frac{\sqrt{9} \cdot \sqrt{6}}{9}$

$=\frac{2 \sqrt{6}}{8}+\frac{3 \sqrt{6}}{9}$

$=\frac{1 \cdot \sqrt{6}}{4}+\frac{1 \cdot \sqrt{6}}{3}$

$=\sqrt{6}\left[\frac{1}{4}+\frac{1}{3}\right]$

$=\sqrt{6}\left[\frac{3+4}{12}\right]$ [∵LCM of 4 and 3 is 12]

$=\frac{7 \sqrt{6}}{{12}}$


(vi) $\frac{3}{\sqrt{8}}+\frac{1}{\sqrt{2}}$

Sol :

$\frac{3}{\sqrt{2 \times 4}}+\frac{1}{\sqrt{2}}$

$=\frac{3}{\sqrt{2} \cdot \sqrt{4}}+\frac{1}{\sqrt{2}}$

$=\frac{3}{2 \sqrt{2}}+\frac{1}{\sqrt{2}}$

$=\frac{1}{\sqrt{2}}\left(\frac{3}{2}+1\right)$

$=\frac{1}{\sqrt{2}}\left(\frac{3+2}{2}\right)$ [∵LCM of 2 and 1 is 2]

$=\frac{1}{\sqrt{2}}\left(\frac{5}{2}\right)$

$=\frac{5}{2 \sqrt{2}}$

Multiply and Divide by '√2'

$=\frac{5}{2 \sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}}$

$=\frac{5 \sqrt{2}}{2 \cdot \sqrt{2} \times \sqrt{2}}$

$=\frac{5 \sqrt{2}}{2(\sqrt{2})^{2}}$

$=\frac{5 \sqrt{2}}{2 \cdot 2}$

$=\frac{5 \sqrt{2}}{4}$


Question 2

Sol :

(i) $(5+\sqrt{7})(2+\sqrt{5})$

Sol :

$=5 \times 2+5 \sqrt{5}+2 \sqrt{7}+\sqrt{7} \cdot \sqrt{5}$

$=10+5 \sqrt{5}+2 \sqrt{7}+\sqrt{7 \times 5}$

$=10+5 \sqrt{5}+2 \sqrt{7}+\sqrt{35}$


(ii) $(5+\sqrt{5})(5-\sqrt{5})$

$=(5)^{2}-(\sqrt{5})^{2}$

$=25-5$

$=20$


(iii) $(\sqrt{5}+\sqrt{2})^{2}$

$=(\sqrt{5})^{2}+(\sqrt{2})^{2}+2 \cdot \sqrt{5} \cdot \sqrt{2}$

$=5+2+2 \sqrt{5 \times 2}$

$=5+2+2 \sqrt{10}$

$=7+2 \sqrt{10}$


(iv) $(\sqrt{3}-\sqrt{7})^{2}$

$=(\sqrt{3})^{2}+(\sqrt{7} \cdot)^{2}-2 \cdot \sqrt{3} \cdot \sqrt{7}$

$=3+7-2 \sqrt{3 \times 7}$

$=10-2 \sqrt{21}$


(v) $(\sqrt{2}+\sqrt{3})(\sqrt{5}+\sqrt{7})$

$=\sqrt{2} \cdot \sqrt{5}+\sqrt{2} \cdot \sqrt{7}+\sqrt{3} \cdot \sqrt{5}+\sqrt{3} \cdot \sqrt{7}$

$=\sqrt{2 \times 5}+\sqrt{2 \times 7}+\sqrt{3 \times 5}+\sqrt{3 \times 7}$

$=\sqrt{10}+\sqrt{14}+\sqrt{15}+\sqrt{21}$


(vi) $(4+\sqrt{5})(\sqrt{3}-\sqrt{7})$

$=4 \sqrt{3}-4 \sqrt{7}+\sqrt{5} \cdot \sqrt{3}-\sqrt{5} \cdot \sqrt{7}$

$=4 \sqrt{3}-4 \sqrt{7}+\sqrt{5 \times 3}-\sqrt{5 \times 7}$

$=4 \sqrt{3}-4 \sqrt{7}+\sqrt{15}-\sqrt{35}$


Question 3

Sol :

(i) $\sqrt{8}+\sqrt{50}+\sqrt{72}+\sqrt{98}$

$=\sqrt{4 \times 2}+\sqrt{25 \times 2}+\sqrt{36 \times 2}+\sqrt{49 \times 2}$

$=\sqrt{4} \cdot \sqrt{2}+\sqrt{25} \cdot \sqrt{2}+\sqrt{36} \cdot \sqrt{2}+\sqrt{49} \cdot \sqrt{2}$

$=2 \sqrt{2}+5 \sqrt{2}+6 \sqrt{2}+7 \sqrt{2}$

$=(2+5+6+7) \sqrt{2}$

$=20 \times \sqrt{2}$

$=20 \times 1 \cdot 414$

=28.28


(ii) $3 \sqrt{32}-2 \sqrt{50}+4 \sqrt{128}-20 \sqrt{18}$

$=3 \sqrt{16 \times 2}-2 \sqrt{25 \times 2}+4 \sqrt{64 \times 2}-20 \sqrt{9 \times 2}$

$=3 \times 4 \sqrt{2}-2 \times 5 \sqrt{2}+4 \times 8 \sqrt{2}-20 \times 3 \sqrt{2}$

$=12 \sqrt{2}-10 \sqrt{2}+32 \sqrt{2}-60 \sqrt{2}$

$=(12-10+32-60) \sqrt{2}$

$=-26 \times \sqrt{2}$

$=-26 \times 1.414$

=-36.764


Question 4

Sol :

(i) $\sqrt{27}+\sqrt{75}+\sqrt{108}-\sqrt{243}$

$\sqrt{9 \times 3}+\sqrt{25 \times 3}+\sqrt{36 \times 3}-\sqrt{81 \times 3}$

$=3 \sqrt{3}+5 \sqrt{3}+6 \sqrt{3}-9 \sqrt{3}$

$=(3+5+6-9) \sqrt{3}$

$=(14-9) \sqrt{3}$

$=5 \times \sqrt{3}$

$=5 \times 1.732$

=8.66


(ii) $5 \sqrt{12}-3 \sqrt{48}+6 \sqrt{75}+7 \sqrt{108}$

Sol :

$=5 \sqrt{4 \times 3}-3 \sqrt{16 \times 3}+6 \sqrt{25 \times 3}+7 \sqrt{36 \times 3}$

$=5 \times 2 \sqrt{3}-3 \times 4 \sqrt{3}+6 \times 5 \sqrt{3}+7 \times 6 \sqrt{3}$

$=10 \sqrt{3}-12 \sqrt{3}+30 \sqrt{3}+42 \sqrt{3}$

$=(10-12+30+42) \sqrt{3}$

$=(82-12) \sqrt{3}$

$=70 \times \sqrt{3}$

$=70 \times 1.732$

=121.24


Question 5

(i) $\sqrt{\frac{4}{9}},-\frac{3}{70}, \sqrt{\frac{7}{25}}, \sqrt{\frac{16}{5}}$

Sol :

$\sqrt{\frac{4}{9}}=\frac{\sqrt{4}}{\sqrt{9}}$

$=\frac{2}{3}$

It is in the form of $\frac{p}{q}$and p, q are integers

Therefore $\sqrt{\frac{4}{9}}$ is a rational number

$\Rightarrow \frac{-3}{70}$ is a rational number

$\Rightarrow \sqrt{\frac{7}{25}}=\frac{\sqrt{7}}{\sqrt{25}}$

$=\frac{\sqrt{7}}{5}$

Since $\sqrt{7}$ is not an Integer

Therefore, $\sqrt{\frac{7}{25}}$ is an Irrational number

$\sqrt{\frac{16}{5}}=\frac{\sqrt{16}}{\sqrt{5}}$

$=\frac{4}{\sqrt{5}}$. Since $\sqrt{5}$ is not an Integer

Therefore, $\sqrt{\frac{16}{5}}$ is an Irrational number


(ii) $-\sqrt{\frac{2}{49}}, \frac{3}{200}, \sqrt{\frac{25}{3}},-\sqrt{\frac{49}{16}}$

Sol :

$-\sqrt{\frac{2}{49}}=-\frac{\sqrt{2}}{\sqrt{49}}$

$=-\frac{\sqrt{2}}{7}$. Since $\sqrt{2}$ is not an Integer

Therefore, $-\sqrt{\frac{2}{49}}$ is an Irrational number

$\frac{3}{200}$. It is in the form of $\frac{p}{q}$ and p, q are integers. So, it is a rational number

$\sqrt{\frac{25}{3}}=\frac{\sqrt{25}}{\sqrt{3}}$

$=\frac{5}{\sqrt{3}} \cdot$ since $\sqrt{3}$ is not an integer

Therefore, $\sqrt{\frac{25}{3}}$ is an irrational number

$-\sqrt{\frac{49}{16}}=-\frac{\sqrt{49}}{\sqrt{16}}$

$=-\frac{7}{4}$

Therefore, $-\sqrt{\frac{49}{16}}$ is a rational number


Question 6

(i) $-3 \sqrt{2}$

Sol :

Since $\sqrt{2}$ is an lrrational number

Therefore, $-3 \sqrt{2}$ will change into non-terminating , non-recurring decimal 


(ii) $\sqrt{\frac{256}{81}}$

Sol :

$\frac{\sqrt{256}}{\sqrt{81}}=\frac{16}{9}$

$=1.77777\overline{7}$

Therefore, $\sqrt{\frac{256}{81}}$ will not change into non-terminating, non - recurring decimal


(iii) $\sqrt{27 \times 16}$

Sol :

$\sqrt{27} \times \sqrt{16}=\sqrt{9 \times 3} \times 4$

$=4 \times 3 \sqrt{3}$

$=12 \sqrt{3}$

$\therefore \sqrt{3}$ is an Irrational number

Therefore, $\sqrt{27 \times 16}$ will change into non-terminating, non-recurring decimal


(iv) $\sqrt{\frac{5}{36}}$

Sol :

$\frac{\sqrt{5}}{\sqrt{36}}=\frac{\sqrt{5}}{6}$

$\because \sqrt{5}$ is an Irrational number

Therefore, $\sqrt{\frac{5}{36}}$ will change into non-terminating ,non - recurring decimal


(iv) $-\frac{2}{7} \sqrt[3]{5}$

Sol :

Since $\sqrt[3]{5}$ is an lrrational number

Therefore, $\frac{-2}{7} \sqrt[3]{5}$ is also an Irrational number

(Note: Product of rational and Irrational number is irrational)


(v) $(2-\sqrt{3})(2+\sqrt{3})$

Sol :

$=2 \times 2+2 \sqrt{3}-2 \sqrt{3}-(\sqrt{3})^{2}$

=4-3=1

Therefore $(2-\sqrt{3})(2+\sqrt{3})$ is a rational number


(vi) $(3+\sqrt{5})^{2}$

Sol :

$=(3)^{2}+(\sqrt{5})^{2}+2 \times 3 \times \sqrt{5}$

$=9+5+6 \sqrt{5}$

$=14+6 \sqrt{5}$

Since $\sqrt{5}$ is an Irrational number

$(3+\sqrt{5})^{2}$ is also an Irrational number


(vii) $\left(\frac{2}{5} \sqrt{7}\right)^{2}$

Sol :

$\left(\frac{2}{5}\right)^{2} \cdot(\sqrt{7})^{2}$

$=\frac{4}{25} \times 7$

$=\frac{28}{25}$

Therefore, $\left(\frac{2}{5} \sqrt{7}\right)^{2}$ is a rational number


(viii) $(3-\sqrt{6})^{2}$

Sol :

$(3)^{2}+(\sqrt{6})^{2}-2 \times 3 \times \sqrt{6}$

$=9+6-6 \sqrt{6}$

$=15-6 \sqrt{6}$

Since $\sqrt{6}$ is an Irrational number $(3-\sqrt{6})^{2}$ is also an irrational . number


Question 7



Question 8

(i) $\sqrt[3]{2}$

Sol :

Suppose that $\sqrt[3]{2}=\frac{p}{q}$, where p, q are integers , q>0, p and q have no common factors (except 1)

$2=\left[\frac{p}{q}\right]^{3}$

$p^{3}=2 q^{3}$...(1)

As 2 divides $2 q^{3} \Rightarrow 2$ divides $p^{3}$

2 divides p

Let p=2k, where k is an integer

Substituting this value of 'p' in (1), we get

$(2 k)^{3}=2 q^{3}$

$8k^{3}=2 q^{3}$

$4 k^{3}=q^{3}$

As 2 divides $4 k^{3} \Rightarrow 2$ divides $q^{3}$

2 divides q

Thus p and q have a common factor '2'

This contradicts that p and q have no common factor (except 1)

Therefore , $\sqrt[3]{2}$ is an irrational number


(ii) $\sqrt[3]{3}$

Sol :

Suppose that $\sqrt[3]{3}=\frac{p}{q}$, where p, q are integers, q>0 , p and q have no common factors (except 1)

$3=\left(\frac{p}{q}\right)^{3}$

$p^{3}=3 q^{3}$...(1)

As 3 divides $3 q^{3} \Rightarrow 3$ divides $p^{3}$

3 divides p

Let p=3k, Where k is an Integer

Substituting this value of  'p' in 0 , We get

$(3 k)^{3}=3 q^{3}$

$27 k^{3}=3 q^{3}$

$9 k^{3}=q^{3}$

As 3 divides $9 K^{3} \Rightarrow 3$ divides $q^{3}$

3 divide q

Thus p and q have a common factor '3' 

This contradicts that p and q have no common factor (except 1)

Therefore $\sqrt[3]{3}$ is an Irrational number


(iii) $\sqrt[4]{5}$

Sol :

Suppose that $\sqrt[4]{5}=\frac{p}{q}$, where p, q are integers , q>0, p and q have no common factors (except 1)

$5=\left(\frac{p}{q}\right)^{4}$

$p^{4}=5 q^{4}$...(1)

As 5 divides $5 q^{4} \Rightarrow 5$ divides $p^{4}$

5 divides p

Let p=5k , where k is an integer

Substituting this value of 'p' in (1) , we get

$(5k)^{4}=5 q^{4}$

$625 \mathrm{k}^{4}=5 q^{4}$

$125 k^{4}=q^{4}$

As 5 divides $125 \mathrm{~k}^{4} \Rightarrow 5$ divides $q^{4}$

5 divides q

Thus p and q have a common factor "5" 

This contradicts that p and q have no common factors (except 1)

Therefore, $\sqrt[4]{5}$ is an Irrational number.


Question 9

(i) $2 \sqrt{3}, \frac{3}{\sqrt{2}},-\sqrt{7}, \sqrt{15}$

Sol :

$\sqrt{4 \times 3}=\sqrt{12}$

$\frac{3}{\sqrt{2}}=\sqrt{\frac{9}{2}}$

$\sqrt{4 \cdot 5}$

$\therefore \sqrt{12}, \sqrt{4 \cdot 5},-\sqrt{7}, \sqrt{15}$

$-\sqrt{7}<\sqrt{4 \cdot 5}<\sqrt{12}<\sqrt{15}$

The greatest real number is $\sqrt{15}$

The smallest real number is $-\sqrt{7}$


(ii) $-3 \sqrt{2}, \frac{9}{\sqrt{5}},-4, \frac{4}{3} \sqrt{5}, \frac{3}{2} \sqrt{3}$

Sol :

$-3 \sqrt{2}=-\sqrt{9 \times 2}$

$=-\sqrt{18}$

$\frac{9}{\sqrt{5}}=\sqrt{\frac{81}{5}}$

$=\sqrt{16.2}$

$-4=-\sqrt{16}$

$\frac{4}{3} \sqrt{5}=\sqrt{\frac{16 \times 5}{9}}$

$=\sqrt{\frac{80}{9}}$

$=\sqrt{8.89}$

$\frac{3}{2} \sqrt{3}=\sqrt{\frac{9 \times 3}{4}}$

$=\sqrt{\frac{27}{4}}$

$=\sqrt{6.75}$

$\therefore-\sqrt{18}, \sqrt{16.2}, \sqrt{8.89},-\sqrt{16}, \sqrt{6.75}$

$-3 \sqrt{2}<-4<\frac{3}{2} \sqrt{3}<\frac{4}{3} \sqrt{5}<\frac{9}{\sqrt{5}}$

The greatest real number is $\frac{9}{\sqrt{5}}$

The smallest real number is $-3 \sqrt{2}$


Question 10

(i) $3 \sqrt{2}, 2 \sqrt{3}, \sqrt{15}, 4$

Sol :

White all the numbers as square roots under one radical

$3 \sqrt{2}=\sqrt{9} \times \sqrt{2}=\sqrt{18}$

$2 \sqrt{3}=\sqrt{4} \times \sqrt{3}=\sqrt{12}$

$\sqrt{15}=\sqrt{15}$

$4=\sqrt{16}$

Since 12<15<16<18

$\Rightarrow \sqrt{12}<\sqrt{15}<\sqrt{16}<\sqrt{18}$

$\Rightarrow 2 \sqrt{3}<\sqrt{15}<4<3 \sqrt{2}$

Hence, the given numbers in ascending orders are

$2 \sqrt{3}, \sqrt{15}, 4,3 \sqrt{2}$


(ii) $3 \sqrt{2}, 2 \sqrt{8}, 4, \sqrt{50}, 4 \sqrt{3}$

Sol :

Write an the numbers as square roots under one radical

$3 \sqrt{2}=\sqrt{9} \times \sqrt{2}=\sqrt{18}$

$2 \sqrt{8}=\sqrt{4} \times \sqrt{8}=\sqrt{32}$

$4=\sqrt{16}$

$\sqrt{50}=\sqrt{50}$

$4 \sqrt{3}=\sqrt{16} \times \sqrt{3}=\sqrt{48}$

Since 16<18<32<48<50

$\Rightarrow \sqrt{16}<\sqrt{18}<\sqrt{32}<\sqrt{48}<\sqrt{50}$

$\Rightarrow 4<3 \sqrt{2}<2 \sqrt{8}<4 \sqrt{3}<\sqrt{50}$


Question 11

(i) $\frac{9}{\sqrt{2}}, \frac{3}{2} \sqrt{5}, 4 \sqrt{3}, 3 \sqrt{\frac{6}{5}}$

Sol :

Write all the numbers as square roots under one radical

$\frac{9}{\sqrt{2}}=\sqrt{\frac{81}{2}}=\sqrt{40.5}$

$\frac{3}{2} \sqrt{5}=\sqrt{\frac{9}{4}} \times \sqrt{5}=\sqrt{\frac{45}{4}}=\sqrt{11 \cdot 25}$

$4 \sqrt{3}=\sqrt{16} \times \sqrt{3}=\sqrt{48}$

$3 \sqrt{\frac{6}{5}}=\sqrt{9} \times \sqrt{\frac{6}{5}}=\sqrt{\frac{54}{5}}=\sqrt{10.8}$

Since 48>40.5>11.25>10.8

$\sqrt{48}>\sqrt{40.5}>\sqrt{11.25}>\sqrt{10.8}$

$4 \sqrt{3}>\frac{9}{\sqrt{2}}>\frac{3}{2} \sqrt{5}>3 \sqrt{\frac{6}{5}}$

Hence, the given numbers in descending orders are $4 \sqrt{3}, \frac{9}{\sqrt{2}}, \frac{3}{2} \sqrt{5}, 3 \sqrt{\frac{6}{5}}$


(ii) $\frac{5}{\sqrt{3}}, \frac{7}{3} \sqrt{2},-\sqrt{3}, 3 \sqrt{5}, 2 \sqrt{7}$

Sol :

Write all the numbers as Square roots under one radical

$\frac{5}{\sqrt{3}}=\sqrt{\frac{25}{3}}=\sqrt{8.33}$

$\frac{7}{3} \sqrt{2}=\sqrt{\frac{49}{9}} \times \sqrt{2}$

$=\sqrt{\frac{98}{9}}=\sqrt{10.89}$

$-\sqrt{3}=-\sqrt{3}$

$3 \sqrt{5}=\sqrt{9} \times \sqrt{5}=\sqrt{45}$

$2 \sqrt{7}=\sqrt{4} \times \sqrt{7}=\sqrt{28}$

Since 45>28>10.89>8.33 >-3

$\sqrt{45} \times \sqrt{28}>\sqrt{10.89}>\sqrt{8.33}>-\sqrt{3}$

$3 \sqrt{5}>2 \sqrt{7}>\frac{7}{3} \sqrt{2}>\frac{5}{\sqrt{3}}>-\sqrt{3}$

Hence, the given numbers in descending orders are $3 \sqrt{5}, 2 \sqrt{7}, \frac{7}{3} \sqrt{2}, \frac{5}{\sqrt{3}},-\sqrt{3}$


Question 12

(i) $\sqrt[3]{2}, \sqrt{3}, \sqrt[6]{5}$

Sol :

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

$\sqrt[3]{2}=2^{\frac{1}{3}}=\left(2^{2}\right)^{\frac{1}{6}}=(4)^{\frac{1}{6}}$

$\sqrt{3}=3^{\frac{1}{2}}=\left(3^{3}\right)^{\frac{1}{6}}=(27)^{\frac{1}{6}}$

$\sqrt[6]{5}=5^{\frac{1}{6}}=\left(5^{1}\right)^{\frac{1}{6}}=(5)^{\frac{1}{6}}$

Since 4<5<27

$(4)^{\frac{1}{6}}<(5)^{\frac{1}{6}}<(27)^{\frac{1}{6}}$

$\sqrt[3]{2}<\sqrt[6]{5}<\sqrt{3}$

Hence, the given numbers in ascending orders are $\sqrt[3]{2}, \sqrt[6]{5}, \sqrt{3}$

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