Exercise 1.4 ML Aggarwal Class 9 ICSE Mathematics Chapter 1 Rational and Irrational Numbers Best Solutions

Exercise 1.1 Exercise 1.2 Exercise 1.3 Exercise 1.4 Exercise 1.5 Chapter Test-1

Exercise 1.4 ML Aggarwal Class 9 Mathematics ICSE Textbook contains a total of 9 questions. The questions are based on rational numbers and finding rational numbers between two rational numbers.

Exercise 1.4 ML Aggarwal Class 9 Mathematics Solutions

Q.1 Q.2 Q.3 Q.4 Q.5 Q.6 Q.7 Q.8 Q.9 Q.10 Q.11 Q.12

The first question of exercise 1.4 ML Aggarwal Class 9 asks us to simplify the given surd expressions.

1. Simplify the following

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

Answer

We have
$\sqrt{45} - 3 \sqrt{20} + 4 \sqrt{5}$
= $\sqrt{3 \times 3 \times 5} - 3 \sqrt{2 \times 2 \times 5} + 4 \sqrt{5}$
= $3 \sqrt{5} - 3 \times 2 \sqrt{5} + 4 \sqrt{5}$
= $\sqrt{5} (3 - 6 + 4)$
= $\sqrt{5} \times 1$
= $\sqrt{5}$ .

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

Answer

We have
$3 \sqrt{3} + 2 \sqrt{27} + \frac{7}{\sqrt{3}}$

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

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

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

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

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

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

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

(iii) $6 \sqrt{5} \times 2 \sqrt{5}$

Answer

We have
$6 \sqrt{5} \times 2 \sqrt{5}$

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

= $12 \times 5$ = $60$ .

(iv) $8 \sqrt{15} \div 2 \sqrt{3}$

Answer

We have
$8 \sqrt{15} \div 2 \sqrt{3}$

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

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

= $4 \sqrt{5}$ .

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

Answer

We have
$\frac{\sqrt{24}}{8} + \frac{\sqrt{54}}{9}$

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

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

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

= $\frac{3 \sqrt{6} + 4 \sqrt{6}}{12}$

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

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

Answer

We have
$\frac{3}{\sqrt{8}} + \frac{1}{\sqrt{2}}$

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

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

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

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

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

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

The second question of exercise 1.4 ML Aggarwal Class 9 asks us to simplify the given surd expressions.

2. Simplify the following:

(i) $(5 + \sqrt{7}) (2 + \sqrt{5})$

Answer

We have
$(5 + \sqrt{7}) (2 + \sqrt{5})$

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

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

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

(ii) $(5 + \sqrt{5}) (5 - \sqrt{5})$

Answer

We have
$(5 + \sqrt{5}) (5 - \sqrt{5})$

Applying the identity: $(a + b) (a - b) = a^{2} - b^{2}$ , we get
= $5^{2} - {(\sqrt{5})}^{2}$

= $25 - 5$ = $20$ .

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

Answer

We have
${(\sqrt{5} + \sqrt{2})}^{2}$

Applying the identity: ${(a + b)}^{2} = a^{2} + b^{2} + 2 a b$ , we get
= ${(\sqrt{5})}^{2} + {(\sqrt{2})}^{2} + 2 \times \sqrt{5} \times \sqrt{2}$

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

= $7 + 2 \sqrt{10}$ .

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

Answer

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

Applying the identity: ${(a - b)}^{2} = a^{2} + b^{2} - 2 a b$ , we get
$= {(\sqrt{3})}^{2} + {(\sqrt{7})}^{2} - 2 \times \sqrt{3} \times \sqrt{7}$

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

= $10 - 2 \sqrt{21}$ .

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

Answer

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

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

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

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

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

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

Answer

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

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

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

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

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

The third question of exercise 1.4 ML Aggarwal Class 9 asks us to find the value of the given surd expression.

3. If $\sqrt{2} = 1.414$ , then find the value of :

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

Answer

We have
$\sqrt{8} + \sqrt{50} + \sqrt{72} + \sqrt{98}$
= $\sqrt{2 \times 2 \times 2} + \sqrt{2 \times 5 \times 5} + \sqrt{2 \times 3 \times 3 \times 2 \times 2} + \sqrt{2 \times 7 \times 7}$
= $2 \sqrt{2} + 5 \sqrt{2} + 6 \sqrt{2} + 7 \sqrt{2}$
= $(2 + 5 + 6 + 7) \sqrt{2}$
= $20 \sqrt{2}$ = $20 \times 1.414$ = $28.28$ .

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

Answer

We have
$3 \sqrt{32} - 2 \sqrt{50} + 4 \sqrt{128} - 20 \sqrt{18}$
= $3 \sqrt{16 \times 2} - 2 \sqrt{25 \times 2} + 4 \sqrt{64 \times 2} - 20 \sqrt{9 \times 2}$
= $3 \sqrt{4^{2} \times 2} - 2 \sqrt{5^{2} \times 2} + 4 \sqrt{8^{2} \times 2} - 20 \sqrt{3^{2} \times 2}$
= $3 \times 4 \sqrt{2} - 2 \times 5 \sqrt{2} + 4 \times 8 \sqrt{2} - 20 \times 3 \sqrt{2}$
= $12 \sqrt{2} - 10 \sqrt{2} + 32 \sqrt{2} - 60 \sqrt{2}$
= $(12 - 10 + 32 - 60) \sqrt{2}$
= $- 26 \sqrt{2}$ = $- 26 \times 1.414$ = $- 36.764$ .

The fourth question of exercise 1.4 ML Aggarwal Class 9 asks us to find the value of given surd expression.

4. If $\sqrt{3} = 1.732$ , then find the value of:

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

Answer

We have
$\sqrt{27} + \sqrt{75} + \sqrt{108} - \sqrt{243}$

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

= $\sqrt{3^{2} \times 3} + \sqrt{5^{2} \times 3} + \sqrt{6^{2} \times 3} - \sqrt{9^{2} \times 3}$

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

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

= $5 \sqrt{3}$ = $5 \times 1.732$ = $8.66$ .

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

Answer

We have
$5 \sqrt{12} - 3 \sqrt{48} + 6 \sqrt{75} + 7 \sqrt{108}$

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

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

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

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

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

= $70 \sqrt{3}$ = $70 \times 1.732$ = $121.24$ .

The fifth question in exercise 1.4 ML Aggarwal Class 9 asks us to identify the given surds as rational or irrational.

5. State which of the following numbers are irrational:

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

Answer

The given numbers are
$\sqrt{\frac{4}{9}}, - \frac{3}{70}, \sqrt{\frac{7}{25}}, \sqrt{\frac{16}{5}}$

Let's simplify them.
$\sqrt{\frac{4}{9}} = \sqrt{\frac{2^{2}}{3^{2}}} = \frac{2}{3}$
$\frac{- 3}{70} = \frac{- 3}{70}$
$\sqrt{\frac{7}{25}} = \frac{\sqrt{7}}{\sqrt{25}} = \frac{\sqrt{7}}{5}$
$\sqrt{\frac{16}{5}} = \frac{\sqrt{16}}{\sqrt{5}} = \frac{4}{\sqrt{5}}$

Clearly, $\sqrt{\frac{7}{25}}, \sqrt{\frac{16}{5}}$ are irrational numbers. They contain prime numbers under square roots.

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

Answer

The given numbers are
$- \sqrt{\frac{2}{49}}, \frac{3}{200}, \sqrt{\frac{25}{3}}, - \sqrt{\frac{49}{16}}$ .

Let's simplify the given expressions:
$- \sqrt{\frac{2}{49}} = - \frac{\sqrt{2}}{\sqrt{49}} = - \frac{\sqrt{2}}{7}$
$\frac{3}{200} = \frac{3}{200}$
$\sqrt{\frac{25}{3}} = \frac{\sqrt{25}}{\sqrt{3}} = \frac{5}{\sqrt{3}}$
$- \sqrt{\frac{49}{16}} = - \frac{\sqrt{49}}{\sqrt{16}} = - \frac{7}{4}$

Clearly, $- \sqrt{\frac{2}{49}}, \sqrt{\frac{25}{3}}$ are irrational numbers. They contain prime numbers under square roots.

The sixth question of exercise 1.4 ML Aggarwal Class 9 asks us to identify the given surd expressions as non-terminating non-recurring decimals:

6. State which of the following numbers will change into non-terminating non-recurring decimals:

(i) $- 3 \sqrt{2}$

Answer

$- 3 \sqrt{2}$ will change into non-terminating non-recurring decimal, as product of a rational number, $- 3$ and an irrational number $\sqrt{2}$ is an irrational number and irrational number have non-terminating and non-recurring decimal expansion.

(ii) $\sqrt{\frac{256}{81}}$

Answer

$∵ \sqrt{\frac{256}{81}} = \frac{\sqrt{256}}{\sqrt{81}} = \frac{16}{9}$ , it will not change into non-terminating non-recurring decimal expansion.

(iii) $\sqrt{27 \times 16}$

Answer

$∵ \sqrt{27 \times 16}$ = $\sqrt{9 \times 3 \times 4 \times 4}$ = $\sqrt{3^{2} \times 3 \times 2^{2} \times 2^{2}}$ = $12 \sqrt{3}$ .
∴ It will change into a non-terminating non-recurring decimal expansion, as $12 \sqrt{3}$ is an irrational number.

(iv) $\sqrt{\frac{5}{36}}$

Answer

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

It will change into a non-terminating non-recurring decimal expansion, as $\frac{\sqrt{5}}{6}$ is an irrational number.

The seventh question of exercise 1.4 ML Aggarwal Class 9 asks us to state whether the given surd expressions are rational or irrational.

7. State which of the following numbers are irrational:

(i) $3 - \sqrt{\frac{7}{25}}$

Answer

We have
$3 - \sqrt{\frac{7}{25}}$ = $3 - \frac{\sqrt{7}}{\sqrt{25}}$ = $3 - \frac{\sqrt{7}}{5}$
Since $\sqrt{7}$ is an irrational number, and adding, subtracting, multiplying, or dividing by a rational number doesn’t affect it, the given number is irrational.

(ii) $- \frac{2}{3} + \sqrt[3]{2}$

Answer

Since the cube root of a prime number is also an irrational number, and adding a rational number to an irrational number preserves its irrationality, the result remains irrational. So, $- \frac{2}{3} + \sqrt[3]{2}$ is an irrational number.

(iii) $\frac{3}{\sqrt{3}}$

Answer

The given number is $\frac{3}{\sqrt{3}}$
On rationalising its denominator, we get
$\frac{3 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$ = $\frac{3 \sqrt{3}}{3}$ = $\sqrt{3}$ , which is an irrational number.

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

(iv) $- \frac{2}{7} \sqrt[3]{5}$

Answer

The given is $- \frac{2}{7} \sqrt[3]{5}$ = $- \frac{2}{7} \times \sqrt[3]{5}$ .
Since $- \frac{2}{7}$ is a rational number and $\sqrt[3]{5}$ is an irrational number, and the product of a rational number and an irrational number is also irrational, $- \frac{2}{7} \sqrt[3]{5}$ is an irrational number.

(v) $(2 - \sqrt{3}) (2 + \sqrt{3})$

Answer

We have
$(2 - \sqrt{3}) (2 + \sqrt{3})$
Applying the identity: $a^{2} - b^{2} = (a + b) (a - b)$ , we get
$2^{2} - {(\sqrt{3})}^{2}$ = $4 - 3$ = $1$ , which is a rational number.

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

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

Answer

We have
${(3 + \sqrt{5})}^{2}$
Applying the identity: $a^{2} - b^{2} = (a + b) (a - b)$ , we get
$2^{2} - {(\sqrt{3})}^{2}$ = $4 - 3$ = $1$ , which is a rational number.

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

Answer

We have
${(\frac{2}{5} \sqrt{7})}^{2}$ = $\frac{2}{3} \sqrt{7} \times \frac{2}{3} \sqrt{7}$ = ${(\frac{2}{3})}^{2} {(\sqrt{7})}^{2}$ = $\frac{4}{9} \times 7$ = $\frac{28}{9}$

Since $\frac{28}{9}$ is a rational number, ${(\frac{2}{5} \sqrt{7})}^{2}$ is also a rational number.

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

Answer

We have,
${(3 - \sqrt{6})}^{2}$ = $3^{2} + {(\sqrt{6})}^{2} - 2 \times 3 \times \sqrt{6}$ = $9 + 6 - 6 \sqrt{6}$ = $15 - 6 \sqrt{6}$ , it is an irrational number.

Therefore, ${(3 - \sqrt{6})}^{2}$ is an irrational number.

The eighth question of exercise 1.4 ML Aggarwal Class 9 asks us to prove the given numbers irrational.

8. Prove that the following numbers are irrational:

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

Answer

To prove: $\sqrt[3]{2}$ is an irrational number.

Proof: Let us assume that $\sqrt[3]{2}$ is a rational number.
$\Rightarrow \sqrt[3]{2} = \frac{p}{q}$ , where $p$ , $q$ are integers, $q \neq 0$ , $p$ and $q$ have no common factor other than 1.
$\Rightarrow {(\sqrt[3]{2})}^{3} = {(\frac{p}{q})}^{3}$ ... [Taking cube on both sides]

$\Rightarrow 2 = \frac{p^{3}}{q^{3}}$

$\Rightarrow p^{3} = 2 q^{3}$ .... (i)

We have a theorem which states that

If $a$ is any natural number and $p$ is a prime number such that $p$ divides $a^{n}$ , then $p$ divides $a$ . In this case, we choose $n = 3$ .

Using the above theorem and equation (i), we get
As $2$ divides $2 q^{3}$
$\Rightarrow$ $2$ divides $p^{3}$
$\Rightarrow 2$ divides $p$ ....... [Using above theorem]

Now, let $p = 2 k$ , where k is an integer.
Substituting this value of $p$ in (i), we get
${(2 k)}^{3} = 2 q^{3}$
$\Rightarrow 8 k^{3} = 2 q^{3}$
$\Rightarrow 4 k^{3} = q^{3}$ ......... [Dividing by 2 on both sides]

As $2$ divides $4 k^{3}$ ,
$\Rightarrow 2$ divides $q^{3}$
$\Rightarrow 2$ divides $q$ .......... [Using above theorem]

Thus, $2$ divides $p$ and $q$ both, i.e. $p$ and $q$ have a common factor other than 1. This contradicts the fact that $p$ and $q$ have no common factors other than 1.

Hence, our assumption is wrong. It follows that $\sqrt[3]{2}$ cannot be expressed as $\frac{p}{q}$ , where $p$ , $q$ are integers, $q$ ≠ $0$ , $p$ and $q$ have no common factors other than 1. Therefore, $\sqrt[3]{2}$ is an irrational number.
Hence, proved.

(ii) $\sqrt[3]{3}$

Answer

To prove: $\sqrt[3]{3}$ is an irrational number.

Proof: Let us assume that $\sqrt[3]{3}$ is a rational number.
$\Rightarrow \sqrt[3]{3} = \frac{p}{q}$ , where $p$ , $q$ are integers, $q \neq 0$ , $p$ and $q$ have no common factor other than 1.
$\Rightarrow {(\sqrt[3]{3})}^{3} = {(\frac{p}{q})}^{3}$ ... [Taking cube on both sides]

$\Rightarrow 3 = \frac{p^{3}}{q^{3}}$

$\Rightarrow p^{3} = 3 q^{3}$ .... (i)

We have a theorem which states that

If $a$ is any natural number and $p$ is a prime number such that $p$ divides $a^{n}$ , then $p$ divides $a$ . In this case, we choose $n = 3$ .

Using the above theorem and equation (i), we get
As $3$ divides $3 q^{3}$
$\Rightarrow$ $3$ divides $p^{3}$
$\Rightarrow 3$ divides $p$ ....... [Using above theorem]

Now, let $p = 3 k$ , where k is an integer.
Substituting this value of $p$ in (i), we get
${(3 k)}^{3} = 3 q^{3}$
$\Rightarrow 27 k^{3} = 3 q^{3}$
$\Rightarrow 9 k^{3} = q^{3}$ ......... [Dividing by 3 on both sides]

As $3$ divides $9 k^{3}$ ,
$\Rightarrow 3$ divides $q^{3}$
$\Rightarrow 3$ divides $q$ .......... [Using above theorem]

Thus, $3$ divides $p$ and $q$ both, i.e. $p$ and $q$ have a common factor other than 1. This contradicts the fact that $p$ and $q$ have no common factors other than 1.

Hence, our assumption is wrong. It follows that $\sqrt[3]{3}$ cannot be expressed as $\frac{p}{q}$ , where $p$ , $q$ are integers, $q$ ≠ $0$ , $p$ and $q$ have no common factors other than 1. Therefore, $\sqrt[3]{3}$ is an irrational number.
Hence, proved.

(iii) $\sqrt[4]{5}$

Answer

To prove: $\sqrt[4]{5}$ is an irrational number.

Proof: Let us assume that $\sqrt[4]{5}$ is a rational number.
$\Rightarrow \sqrt[4]{5} = \frac{p}{q}$ , where $p$ , $q$ are integers, $q \neq 0$ , $p$ and $q$ have no common factor other than 1.
$\Rightarrow {(\sqrt[4]{5})}^{4} = {(\frac{p}{q})}^{4}$ ... [Raising power 4 on both sides]

$\Rightarrow 5 = \frac{p^{4}}{q^{4}}$

$\Rightarrow p^{4} = 5 q^{4}$ .... (i)

We have a theorem which states that

If $a$ is any natural number and $p$ is a prime number such that $p$ divides $a^{n}$ , then $p$ divides $a$ . In this case, we choose $n = 4$ .

Using the above theorem and equation (i), we get
As $5$ divides $5 q^{4}$
$\Rightarrow$ $5$ divides $p^{4}$
$\Rightarrow 5$ divides $p$ ....... [Using above theorem]

Now, let $p = 5 k$ , where k is an integer.
Substituting this value of $p$ in (i), we get
${(5 k)}^{4} = 5 q^{4}$
$\Rightarrow 625 k^{4} = 5 q^{4}$
$\Rightarrow 125 k^{4} = q^{4}$ ......... [Dividing by 5 on both sides]

As $5$ divides $125 k^{4}$ ,
$\Rightarrow 5$ divides $q^{4}$
$\Rightarrow 5$ divides $q$ .......... [Using above theorem]

Thus, $5$ divides $p$ and $q$ both, i.e. $p$ and $q$ have a common factor other than 1. This contradicts the fact that $p$ and $q$ have no common factors other than 1.

Hence, our assumption is wrong. It follows that $\sqrt[4]{5}$ cannot be expressed as $\frac{p}{q}$ , where $p$ , $q$ are integers, $q$ ≠ $0$ , $p$ and $q$ have no common factors other than 1. Therefore, $\sqrt[4]{5}$ is an irrational number.
Hence, proved.

The ninth question of exercise 1.4 ML Aggarwal Class 9 asks us to find the greatest and smallest real numbers among the following real numbers:

9. Find the greatest and the smallest real numbers among the following real numbers:

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

Answer

The given numbers are
$2 \sqrt{3}, \frac{3}{\sqrt{2}}, - \sqrt{7}, \sqrt{15}$

To find the smallest and the greatest real numbers among the following real numbers, we need to write the numbers as square roots under one radical.

We can write them as follows:

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

$\frac{3}{\sqrt{2}} = \frac{\sqrt{9}}{\sqrt{2}} = \sqrt{\frac{9}{2}} = \sqrt{4 \frac{1}{2}}$

$- \sqrt{7} = - \sqrt{7}$

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

Now, on comparing, we get
$- \sqrt{7} < \sqrt{4 \frac{1}{2}} < \sqrt{12} < \sqrt{15}$

Hence, the smallest real number = $- \sqrt{7}$ and the greatest real number is $\sqrt{15}$ .

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

Answer

The given numbers are
$- 3 \sqrt{2}, \frac{9}{\sqrt{5}}, - 4, \frac{4}{3} \sqrt{5}, \frac{3}{2} \sqrt{3}$ .

To find the smallest and the greatest real numbers among the following real numbers, we need to write the numbers as square roots under one radical.

We can write them as follows:

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

$\frac{9}{\sqrt{5}}$ = $\frac{\sqrt{9 \times 9}}{\sqrt{5}}$ = $\sqrt{\frac{81}{5}}$ = $\sqrt{16 \frac{1}{5}}$

$- 4$ = $- \sqrt{4 \times 4}$ = $- \sqrt{16}$

$\frac{4}{3} \sqrt{5}$ = $\sqrt{\frac{4}{3} \times \frac{4}{3} \times 5}$ = $\sqrt{\frac{80}{9}}$ = $\sqrt{8 \frac{8}{9}}$

$\frac{3}{2} \sqrt{3}$ = $\sqrt{\frac{3}{2} \times \frac{3}{2} \times 3}$ = $\sqrt{\frac{27}{4}}$ = $\sqrt{6 \frac{3}{4}}$

Now, on comparing, we get
$- \sqrt{18} < - \sqrt{16} < \sqrt{6 \frac{3}{4}} < \sqrt{8 \frac{8}{9}} < \sqrt{16 \frac{1}{5}}$

Hence, the smallest number is $- \sqrt{18}$ = $- 3 \sqrt{2}$ and the greatest number is $\sqrt{16 \frac{1}{5}}$ = $\frac{9}{\sqrt{5}}$ .

The tenth question of exercise 1.4 ML Aggarwal Class 9 asks us to arrange the given numbers in ascending order.

10. Write the following numbers in ascending order:

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

Answer

The given number are
$3 \sqrt{2}, 2 \sqrt{3}, \sqrt{15}, 4$

To arrange the numbers in ascending order, we write the number as square roots under one radical.

We can convert the given numbers as square roots under one radical as follows:

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

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

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

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

On comparing, we get
$\sqrt{12} < \sqrt{15} < \sqrt{16} < \sqrt{18}$
$\Rightarrow 2 \sqrt{3} < \sqrt{15} < 4 < 3 \sqrt{2}$

Hence, the numbers in ascending order are
$2 \sqrt{3}, \sqrt{15}, 4, 3 \sqrt{2}$ .

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

Answer

The given number are
$3 \sqrt{2}, 2 \sqrt{8}, 4, \sqrt{50}, 4 \sqrt{3}$

To arrange the numbers in ascending order, we write the number as square roots under one radical.

We can convert the given numbers as square roots under one radical as follows:

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

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

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

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

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

On comparing, we get
$\sqrt{16} < \sqrt{18} < \sqrt{32} < \sqrt{48} < \sqrt{50}$
$\Rightarrow 4 < 3 \sqrt{2} < 2 \sqrt{8} < 4 \sqrt{3} < \sqrt{50}$

Hence, the numbers in ascending order are
$4, 3 \sqrt{2}, 2 \sqrt{8}, 4 \sqrt{3}, \sqrt{50}$ .

The eleventh question of exercise 1.4 ML Aggarwal Class 9 asks us to arrange the given numbers in descending order.

11. Write the following numbers in descending order:

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

Answer

The given number are
$\frac{9}{\sqrt{2}}, \frac{3}{2} \sqrt{5}, 4 \sqrt{3}, 3 \sqrt{\frac{6}{5}}$

To arrange the numbers in descending order, we write the number as square roots under one radical.

We can write them as follows:

$\frac{9}{\sqrt{2}} = \frac{\sqrt{9 \times 9}}{\sqrt{2}} = \frac{\sqrt{81}}{\sqrt{2}} = \sqrt{\frac{81}{2}} = \sqrt{40 \frac{1}{2}}$

$\frac{3}{2} \sqrt{5} = \sqrt{5 \times \frac{3}{2} \times \frac{3}{2}} = \sqrt{\frac{45}{4}} = \sqrt{11 \frac{1}{4}}$

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

$3 \sqrt{\frac{6}{5}} = \sqrt{\frac{6}{5} \times 3 \times 3} = \sqrt{\frac{54}{5}} = \sqrt{10 \frac{4}{5}}$

On comparing, we get,
$\sqrt{48} > \sqrt{40 \frac{1}{2}} > \sqrt{11 \frac{1}{4}} > \sqrt{10 \frac{4}{5}}$

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

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

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

Answer

The given number are
$\frac{5}{\sqrt{3}}, \frac{7}{3} \sqrt{2}, - \sqrt{3}, 3 \sqrt{5}, 2 \sqrt{7}$

To arrange the numbers in descending order, we write the number as square roots under one radical.

We can write them as follows:

$\frac{5}{\sqrt{3}} = \frac{\sqrt{5 \times 5}}{\sqrt{3}} = \frac{\sqrt{25}}{\sqrt{3}} = \sqrt{\frac{25}{3}} = \sqrt{8 \frac{1}{3}}$

$\frac{7}{3} \sqrt{2} = \sqrt{\frac{7}{3} \times \frac{7}{3} \times 2} = \sqrt{\frac{98}{9}} = \sqrt{9 \frac{7}{9}}$

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

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

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

On comparing, we get,

$\sqrt{45} > \sqrt{28} > \sqrt{9 \frac{7}{9}} > \sqrt{8 \frac{1}{3}} > - \sqrt{3}$

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

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

The twelfth question of exercise 1.4 ML Aggarwal Class 9 asks us to arrange the given numbers in ascending order.

12. Arrange the following numbers in ascending order: $\sqrt[3]{2}, \sqrt{3}, \sqrt[6]{5}$ .

Answer

The given number are
$\sqrt[3]{2}, \sqrt{3}, \sqrt[6]{5}$

⇒ $\sqrt[3]{2} = 2^{\frac{1}{3}}$ , $\sqrt{3} = 3^{\frac{1}{2}}$ , $\sqrt[6]{5} = 5^{\frac{1}{6}}$

To arrange them in ascending order, we write them as surds with same exponent. For this, we find the L.C.M. of the denominators of the exponents i.e. 3, 2, 6 which is 6.

Now, we write the numbers follows:

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

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

$\sqrt[6]{5} = 5^{\frac{1}{6}}$

As the exponents are same now, we can compare the bases to write the numbers in ascending order.

$∵ 4 < 5 < 27$
$\Rightarrow 4^{\frac{1}{6}} < 5^{\frac{1}{6}} < 27^{\frac{1}{6}}$
$\Rightarrow \sqrt[3]{2} < \sqrt[6]{5} < \sqrt{3}$

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

Exercise 1.4 ML Aggarwal Class 9 Mathematics Solutions

1. Simplify the following

(i) 45−320+45

(ii) 33+227+73

(iii) 65×25

(iv) 815÷23

(v) 248+549

(vi) 38+12

2. Simplify the following:

(i) (5+7)(2+5)

(ii) (5+5)(5−5)

(iii) (5+2)2

(iv) (3−7)2

(v) (2+3)(5+7)

(vi) (4+5)(3−7)

3. If 2=1.414, then find the value of :

(i) 8+50+72+98

(ii) 332−250+4128−2018

4. If 3=1.732, then find the value of:

(i) 27+75+108−243

(ii) 512−348+675+7108

5. State which of the following numbers are irrational:

(i) 49, −370, 725, 165

(i) −249, 3200, 253, −4916

6. State which of the following numbers will change into non-terminating non-recurring decimals:

(i) −32

(ii) 25681

(iii) 27×16

(iv) 536

7. State which of the following numbers are irrational:

(i) 3−725

(ii) −23+23

(iii) 33

(iv) −2753

(v) (2−3)(2+3)

(vi) (3+5)2

(vii) (257)2

(viii) (3−6)2

8. Prove that the following numbers are irrational:

(i) 23

(ii) 33

(iii) 54

9. Find the greatest and the smallest real numbers among the following real numbers:

(i) 23, 32, −7, 15

(ii) −32, 95, −4, 435, 323

10. Write the following numbers in ascending order:

(i) 32, 23, 15, 4

(ii) 32, 28, 4, 50,43

11. Write the following numbers in descending order:

(i) 92, 325, 43, 365

(ii) 53, 732, −3, 35, 27

12. Arrange the following numbers in ascending order: 23, 3, 56.

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

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

(iii) $6 \sqrt{5} \times 2 \sqrt{5}$

(iv) $8 \sqrt{15} \div 2 \sqrt{3}$

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

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

(i) $(5 + \sqrt{7}) (2 + \sqrt{5})$

(ii) $(5 + \sqrt{5}) (5 - \sqrt{5})$

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

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

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

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

3. If $\sqrt{2} = 1.414$ , then find the value of :

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

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

4. If $\sqrt{3} = 1.732$ , then find the value of:

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

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

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

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

(i) $- 3 \sqrt{2}$

(ii) $\sqrt{\frac{256}{81}}$

(iii) $\sqrt{27 \times 16}$

(iv) $\sqrt{\frac{5}{36}}$

(i) $3 - \sqrt{\frac{7}{25}}$

(ii) $- \frac{2}{3} + \sqrt[3]{2}$

(iii) $\frac{3}{\sqrt{3}}$

(iv) $- \frac{2}{7} \sqrt[3]{5}$

(v) $(2 - \sqrt{3}) (2 + \sqrt{3})$

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

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

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

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

(ii) $\sqrt[3]{3}$

(iii) $\sqrt[4]{5}$

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

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

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

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

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

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

12. Arrange the following numbers in ascending order: $\sqrt[3]{2}, \sqrt{3}, \sqrt[6]{5}$ .