Exercise 1.2 ML Aggarwal Class 9 Solutions Rational and Irrational Numbers Best Answers

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

Exercise 1.2 ML Aggarwal Class 9 Solutions of Mathematics ICSE Textbook contains solutions of all 8 questions. The questions are based on irrational numbers.

Exercise 1.2 ML Aggarwal Class 9 Solutions

Q.1 Q.2 Q.3 Q.4 Q.5 Q.6 Q.7 Q.8

Question 1 Exercise 1.2 ML Aggarwal Class 9 Solutions has been given below.

1. Prove that $\sqrt{5}$ is an irrational number.

Answer

Let $\sqrt{5}$ be a rational number, then

$\sqrt{5} = \frac{p}{q},$ where $p, q$ are integers, $q \neq 0$ and $p, q$ have no common factors other than 1.

$\Rightarrow 5 = \frac{p^{2}}{q^{2}}$ ... [∵ Squaring on both sides]

$\Rightarrow p^{2} = 5 q^{2}$ ... $(i)$

As $5 q^{2}$ is a multiple of 5,

$\Rightarrow$ 5 divides $5 q^{2}$ .

$\Rightarrow$ 5 divides $p^{2}$ ... $[∵ p^{2} = 5 q^{2}]$

According to the theorem,

If $a$ is any natural number and $p$ is a prime number such that $p$ divides $a^{2}$ , then $p$ divides $a$ .

$\Rightarrow$ 5 divides $p$ ... [∵ 5 is prime.]

Let $p = 5 m$ , where m is an integer.

Substituting this value of $p$ in $(i)$ , we get

${(5 m)}^{2} = 5 q^{2}$
$\Rightarrow 25 m^{2} = 5 q^{2}$
$\Rightarrow 5 m^{2} = q^{2}$

$\Rightarrow$ $5$ divides $5 m^{2}$ ... [∵ $5 m^{2}$ is a multiple of $5$ ]

$\Rightarrow 5$ divides $q^{2}$

$\Rightarrow 5$ divides $q$ ... [From above theorem]

Now,

Since $5$ divides $p$ and $q$ both, $p$ and $q$ have a common factor 5. This contradicts that $p$ and $q$ have no common factors other than 1.

Thus, our supposition that $\sqrt{5}$ is a rational number is incorrect.

Hence, we conclude that $\sqrt{5}$ is an irrational number.

Question 2 Exercise 1.2 ML Aggarwal Class 9 Solutions has been given below.

2. Prove that $\sqrt{7}$ is an irrational number.

Answer

Let $\sqrt{7}$ be a rational number, then

$\sqrt{7} = \frac{p}{q},$ where $p, q$ are integers, $q \neq 0$ and $p, q$ have no common factors other than 1.

$\Rightarrow 7 = \frac{p^{2}}{q^{2}}$ ... [∵ Squaring on both sides]

$\Rightarrow p^{2} = 7 q^{2}$ ... $(i)$

As $7 q^{2}$ is a multiple of 7,

$\Rightarrow$ 7 divides $7 q^{2}$ .

$\Rightarrow$ 7 divides $p^{2}$ ... $[∵ p^{2} = 7 q^{2}]$

According to the theorem,

If $a$ is any natural number and $p$ is a prime number such that $p$ divides $a^{2}$ , then $p$ divides $a$ .

$\Rightarrow$ 7 divides $p$ ... [∵ 7 is prime.]

Let $p = 7 m$ , where m is an integer.

Substituting this value of $p$ in $(i)$ , we get

${(7 m)}^{2} = 7 q^{2}$
$\Rightarrow 49 m^{2} = 7 q^{2}$
$\Rightarrow 7 m^{2} = q^{2}$

$\Rightarrow$ $7$ divides $7 m^{2}$ ... [∵ $7 m^{2}$ is a multiple of $7$ ]

$\Rightarrow 7$ divides $q^{2}$

$\Rightarrow 7$ divides $q$ ... [Using the above theorem]

Now,

Since $7$ divides $p$ and $q$ both, $p$ and $q$ have a common factor 7. This contradicts that $p$ and $q$ have no common factors other than 1.

Thus, our supposition that $\sqrt{7}$ is a rational number is incorrect.

Hence, we conclude that $\sqrt{7}$ is an irrational number.

Question 3 Exercise 1.2 ML Aggarwal Class 9 Solutions has been given below.

3. Prove that $\sqrt{6}$ is an irrational number.

Answer

Suppose that $\sqrt{6}$ is a rational number, then

$\sqrt{6} = \frac{p}{q},$ where $p, q$ are integers, $q \neq 0$ and $p, q$ have no common factors other than 1.

$\Rightarrow 6 = \frac{p^{2}}{q^{2}}$ ... [∵ Squaring on both sides]

$\Rightarrow p^{2} = 6 q^{2}$ ... $(i)$

As $6 q^{2}$ is a multiple of 2 and 3 both,

$\Rightarrow$ 2 divides $6 q^{2}$ .

$\Rightarrow$ 2 divides $p^{2}$ ... $[∵ p^{2} = 7 q^{2}]$

According to the theorem,

If $a$ is any natural number and $p$ is a prime number such that $p$ divides $a^{2}$ , then $p$ divides $a$ .

$\Rightarrow$ 2 divides $p$ ... [∵ 2 is prime.]

Let $p = 2 m$ , where m is an integer.

Substituting this value of $p$ in $(i)$ , we get

${(2 m)}^{2} = 6 q^{2}$
$\Rightarrow 4 m^{2} = 6 q^{2}$
$\Rightarrow 2 m^{2} = 3 q^{2}$

As $2$ divides $2 m^{2}$ ,

$\Rightarrow 2$ divides $3 q^{2}$ ... $[∵ 2 m^{2} = 3 q^{2}]$

$\Rightarrow 2$ divides $3$ or $2$ divides $q^{2}$

But, we know that 2 doesn't divide 3, so $2$ divides $q^{2}$ .

$\Rightarrow 2$ divides $q$ ... [From the above theorem]

∵ 2 divides both $p$ and $q$ , 2 is a common factor of $p$ and $q$ .

This contradicts the initial supposition that $p$ and $q$ have no common factors other than 1.

Hence, our supposition is wrong. Therefore, $\sqrt{6}$ is an irrational number.

Question 4 Exercise 1.2 ML Aggarwal Class 9 Solutions has been given below.

4. Prove that $\frac{1}{\sqrt{11}}$ is an irrational number.

Answer

Suppose that $\frac{1}{\sqrt{11}}$ is a rational number, then

$\frac{1}{\sqrt{11}} = \frac{p}{q}$ , where $p, q$ are integers, $q \neq 0$ and $p, q$ have no common factors other than 1.

$\Rightarrow \frac{1}{11} = \frac{p^{2}}{q^{2}}$ ... [Squaring on both sides]

$\Rightarrow q^{2} = 11 p^{2}$ ... $(i)$

As 11 divides $11 p^{2}$ ,

⇒ 11 divides $q^{2}$ ... $[∵ q^{2} = 11 p^{2}]$

According to the theorem,

If $a$ is any natural number and $p$ is a prime number such that $p$ divides $a^{2}$ , then $p$ divides $a$ .

⇒ 11 divides $q$ ... [∵ 11 is prime.]

⇒ $q$ is a multiple of 11.

Let $q = 11 m$ , where m is an integer.

Substituting this value of $q$ in $(i)$ , we get

${(11 m)}^{2} = 11 p^{2}$
$\Rightarrow 11^{2} \times m^{2} = 11 p^{2}$
$\Rightarrow 121 \times m^{2} = 11 p^{2}$
$\Rightarrow 11 m^{2} = p^{2}$ ... [Dividing by 11 on both sides]

As $11$ divides $11 m^{2}$ ,

$\Rightarrow 11$ divides $p^{2}$ ... $[∵ 11 m^{2} = p^{2}]$

$\Rightarrow 11$ divides $p$ ... [Using above theorem]

Thus, 11 divides $p$ as well as $q$ . It implies that 11 is a factor of $p$ and $q$ both.

It contradicts the fact that $p$ and $q$ have no common factors other than 1.

This contradiction arises due to the supposition that $\frac{1}{\sqrt{11}}$ is a rational number. Hence, our supposition is wrong.

$\frac{1}{\sqrt{11}}$ is not a rational number. So, we conclude that $\frac{1}{\sqrt{11}}$ is an irrational number.

Question 5 Exercise 1.2 ML Aggarwal Class 9 Solutions has been given below.

5. Prove that $\sqrt{2}$ is an irrational number. Hence, show that $3 - \sqrt{2}$ is an irrational number.

Answer

Let $\sqrt{2}$ be a rational number, then

$\sqrt{2} = \frac{p}{q},$ where $p, q$ are integers, $q \neq 0$ and $p, q$ have no common factors other than 1.

$\Rightarrow 2 = \frac{p^{2}}{q^{2}}$ ... [∵ Squaring on both sides]

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

As $2 q^{2}$ is a multiple of 2,

$\Rightarrow$ 2 divides $2 q^{2}$ .

$\Rightarrow$ 2 divides $p^{2}$ ... $[∵ p^{2} = 2 q^{2}]$

According to the theorem,

If $a$ is any natural number and $p$ is a prime number such that $p$ divides $a^{2}$ , then $p$ divides $a$ .

$\Rightarrow$ 2 divides $p$ ... [∵ 2 is prime.]

Let $p = 2 m$ , where m is an integer.

Substituting this value of $p$ in $(i)$ , we get

${(2 m)}^{2} = 2 q^{2}$
$\Rightarrow 4 m^{2} = 2 q^{2}$
$\Rightarrow 2 m^{2} = q^{2}$

$\Rightarrow$ $2$ divides $2 m^{2}$ ... [∵ $2 m^{2}$ is a multiple of $2$ ]

$\Rightarrow 2$ divides $q^{2}$

$\Rightarrow 2$ divides $q$ ... [From above theorem]

Now,

Since $2$ divides $p$ and $q$ both, $p$ and $q$ have a common factor 2. This contradicts that $p$ and $q$ have no common factors other than 1.

Thus, our supposition that $\sqrt{2}$ is a rational number is incorrect.

Hence, we conclude that $\sqrt{2}$ is an irrational number.

Now,

To Prove: $3 - \sqrt{2}$ is an irrational number.

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

$\Rightarrow (3 - \sqrt{2}) = \frac{a}{b}$ , where $a$ and $b$ are integers.

$\Rightarrow (3 - \sqrt{2}) \times b = a$

$\Rightarrow 3 b - \sqrt{2} b = a$

$\Rightarrow - \sqrt{2} b = a - 3 b$

$\Rightarrow \sqrt{2} = \frac{- (a - 3 b)}{b}$

$\Rightarrow \sqrt{2} = \frac{- a + 3 b}{b}$

$∵ \frac{- a + 3 b}{b}$ is a rational number, as $a$ and $b$ are integers.
$\Rightarrow \sqrt{2}$ is a rational number, but we have proved above that it is an irrational number.

This is a contradiction arises due to incorrect supposition that $(3 - \sqrt{2})$ is a rational number.

Hence, our supposition that $(3 - \sqrt{2})$ is a rational number is wrong. So, $(3 - \sqrt{2})$ is an irrational number.

Question 6 Exercise 1.2 ML Aggarwal Class 9 Solutions has been given below.

6. Prove that $\sqrt{3}$ is an irrational number. Hence, show that $\frac{2}{5} \sqrt{3}$ is an irrational number..

Answer

Let $\sqrt{3}$ be a rational number, then

$\sqrt{3} = \frac{p}{q},$ where $p, q$ are integers, $q \neq 0$ and $p, q$ have no common factors other than 1.

$\Rightarrow 3 = \frac{p^{2}}{q^{2}}$ ... [∵ Squaring on both sides]

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

As $3 q^{2}$ is a multiple of 3,

$\Rightarrow$ 3 divides $3 q^{2}$ .

$\Rightarrow$ 3 divides $p^{2}$ ... $[∵ p^{2} = 3 q^{2}]$

According to the theorem,

If $a$ is any natural number and $p$ is a prime number such that $p$ divides $a^{2}$ , then $p$ divides $a$ .

$\Rightarrow$ 3 divides $p$ ... [∵ 3 is prime.]

Let $p = 3 m$ , where m is an integer.

Substituting this value of $p$ in $(i)$ , we get

${(3 m)}^{2} = 3 q^{2}$
$\Rightarrow 9 m^{2} = 3 q^{2}$
$\Rightarrow 3 m^{2} = q^{2}$

$\Rightarrow$ $3$ divides $3 m^{2}$ ... [∵ $3 m^{2}$ is a multiple of $3$ ]

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

$\Rightarrow 3$ divides $q$ ... [From above theorem]

Now,

Since $3$ divides $p$ and $q$ both, $p$ and $q$ have a common factor 3. This contradicts that $p$ and $q$ have no common factors other than 1.

Thus, our supposition that $\sqrt{3}$ is a rational number is incorrect.

Hence, we conclude that $\sqrt{3}$ is an irrational number.

Now,

To Prove: $\frac{2}{5} \sqrt{3}$ is an irrational number.

Let $\frac{2}{5} \sqrt{3}$ is a rational number.

$\Rightarrow \frac{2}{5} \sqrt{3} = \frac{a}{b}$ , where $a, b$ are integers.

$\Rightarrow 2 \sqrt{3} b = 5 a$

$\Rightarrow \sqrt{3} = \frac{5 a}{2 b}$

$∵ \frac{5 a}{2 b}$ is a rational number, as $a$ and $b$ are integers.
$\Rightarrow \sqrt{3}$ is a rational number, but we have proved above that it is an irrational number.

This is a contradiction arises due to incorrect supposition that $\frac{2}{5} \sqrt{3}$ is a rational number.

Hence, our supposition that $\frac{2}{5} \sqrt{3}$ is a rational number is wrong. So, $\frac{2}{5} \sqrt{3}$ is an irrational number.

Question 7 Exercise 1.2 ML Aggarwal Class 9 Solutions has been given below.

7. Prove that $\sqrt{5}$ is an irrational number. Hence, show that $- 3 + 2 \sqrt{5}$ is an irrational number.

Answer

Let $\sqrt{5}$ be a rational number, then

$\sqrt{5} = \frac{p}{q},$ where $p, q$ are integers, $q \neq 0$ and $p, q$ have no common factors other than 1.

$\Rightarrow 5 = \frac{p^{2}}{q^{2}}$ ... [∵ Squaring on both sides]

$\Rightarrow p^{2} = 5 q^{2}$ ... $(i)$

As $5 q^{2}$ is a multiple of 5,

$\Rightarrow$ 5 divides $5 q^{2}$ .

$\Rightarrow$ 5 divides $p^{2}$ ... $[∵ p^{2} = 5 q^{2}]$

According to the theorem,

If $a$ is any natural number and $p$ is a prime number such that $p$ divides $a^{2}$ , then $p$ divides $a$ .

$\Rightarrow$ 5 divides $p$ ... [∵ 5 is prime.]

Let $p = 5 m$ , where m is an integer.

Substituting this value of $p$ in $(i)$ , we get

${(5 m)}^{2} = 5 q^{2}$
$\Rightarrow 25 m^{2} = 5 q^{2}$
$\Rightarrow 5 m^{2} = q^{2}$

$\Rightarrow$ $5$ divides $5 m^{2}$ ... [∵ $5 m^{2}$ is a multiple of $5$ ]

$\Rightarrow 5$ divides $q^{2}$

$\Rightarrow 5$ divides $q$ ... [From above theorem]

Now,

Since $5$ divides $p$ and $q$ both, $p$ and $q$ have a common factor 5. This contradicts that $p$ and $q$ have no common factors other than 1.

Thus, our supposition that $\sqrt{5}$ is a rational number is incorrect.

Hence, we conclude that $\sqrt{5}$ is an irrational number.

Now,

To Prove: $- 3 + 2 \sqrt{5}$ is an irrational number.

Let $- 3 + 2 \sqrt{5}$ be a rational number.

$\Rightarrow - 3 + 2 \sqrt{5} = \frac{a}{b}$ , where $a, b$ are integers and $b \neq 0$ .

$\Rightarrow - 3 b + 2 \sqrt{5} b = a$

$\Rightarrow 2 \sqrt{5} b = a + 3 b$

$\Rightarrow \sqrt{5} = \frac{a + 3 b}{2 b}$

$∵ \frac{a + 3 b}{2 b}$ is a rational number, as $a$ and $b$ are integers.
$\Rightarrow \sqrt{5}$ is a rational number, but we have proved above that it is an irrational number.

This is a contradiction arises due to incorrect supposition that $- 3 + 2 \sqrt{5}$ is a rational number.

Hence, our supposition that $- 3 + 2 \sqrt{5}$ is a rational number is wrong. So, $- 3 + 2 \sqrt{5}$ is an irrational number.

Question 8 Exercise 1.2 ML Aggarwal Class 9 Solutions has been given below.

8. Prove that the following numbers are irrational:

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

Answer

Let us assume that $5 + \sqrt{2}$ be a rational number, say r.

Then, $5 + \sqrt{2} = r$

$\Rightarrow \sqrt{2} = r - 5$

As $r$ is rational, $r - 5$ is also rational.

$\Rightarrow \sqrt{2}$ is rational. .... $[∵ \sqrt{2} = r - 5]$

But this contradicts the fact that $\sqrt{2}$ is irrational.

Hence, our assumption is wrong. Therefore, $5 + \sqrt{2}$ is an irrational number.

(ii) $3 - 5 \sqrt{3}$

Answer

Let us assume that $3 - 5 \sqrt{3}$ be a rational number, say $r$ .

Then, $3 - 5 \sqrt{3} = r$

$\Rightarrow - 5 \sqrt{3} = r - 3$

$\Rightarrow \sqrt{3} = \frac{r - 3}{- 5}$

$∵ r$ is rational, $\Rightarrow r - 3$ is rational.

$\Rightarrow \frac{r - 3}{- 5}$ is rational.

$\Rightarrow \sqrt{3}$ is rational. .... $[∵ \sqrt{3} = \frac{r - 3}{- 5}]$

But this contradicts the fact that $\sqrt{3}$ is irrational.

Hence, our assumption is wrong. Therefore, $3 - 5 \sqrt{3}$ is an irrational number.

(iii) $2 \sqrt{3} - 7$

Answer

Let us assume that $2 \sqrt{3} - 7$ be a rational number, say $r$ .

Then, $2 \sqrt{3} - 7 = r$

$\Rightarrow 2 \sqrt{3} = r + 7$

$\Rightarrow \sqrt{3} = \frac{r + 7}{2}$

$∵ r$ is rational, $\Rightarrow r + 7$ is rational.

$\Rightarrow \frac{r + 7}{2}$ is rational.

$\Rightarrow \sqrt{3}$ is rational. .... $[∵ \sqrt{3} = \frac{r + 7}{2}]$

But this contradicts the fact that $\sqrt{3}$ is irrational.

Hence, our assumption is wrong. Therefore, $2 \sqrt{3} - 7$ is an irrational number.

(iv) $\sqrt{2} + \sqrt{5}$

Answer

Let us assume that $\sqrt{2} + \sqrt{5}$ be a rational number, say $r$ . (Here, $r$ ≠ 0)

Then, $\sqrt{2} + \sqrt{5} = r$

$\Rightarrow \sqrt{5} = r - \sqrt{2}$

$\Rightarrow {(\sqrt{5})}^{2} = {(r - \sqrt{2})}^{2}$ ... [Squaring on both sides]

$\Rightarrow 5 = r^{2} + 2 - 2 \sqrt{2} r$

$\Rightarrow 2 \sqrt{2} r = r^{2} + 2 - 5$

$\Rightarrow 2 \sqrt{2} r = r^{2} - 3$

$\Rightarrow \sqrt{2} = \frac{r^{2} - 3}{2 r}$

As $r$ is rational and $r \neq 0$

$\Rightarrow \frac{r^{2} - 3}{2 r}$ is rational.

$\Rightarrow \sqrt{2}$ is rational.

But, this contradicts that $\sqrt{2}$ is irrational.

Hence, our assumption is wrong. Therefore, $\sqrt{2} + \sqrt{5}$ is an irrational number.

Exercise 1.2 ML Aggarwal Class 9 Solutions

1. Prove that $\sqrt{5}$ is an irrational number.

2. Prove that $\sqrt{7}$ is an irrational number.

3. Prove that $\sqrt{6}$ is an irrational number.

4. Prove that $\frac{1}{\sqrt{11}}$ is an irrational number.

5. Prove that $\sqrt{2}$ is an irrational number. Hence, show that $3 - \sqrt{2}$ is an irrational number.

6. Prove that $\sqrt{3}$ is an irrational number. Hence, show that $\frac{2}{5} \sqrt{3}$ is an irrational number..

7. Prove that $\sqrt{5}$ is an irrational number. Hence, show that $- 3 + 2 \sqrt{5}$ is an irrational number.

8. Prove that the following numbers are irrational: