Exercise 1.3 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 1.3 ML Aggarwal Class 9 Mathematics ICSE Textbook contains a total of 17 questions. The questions are based on real numbers, properties of real numbers, decimal expansions of real numbers.

Exercise 1.3 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 Q.13 Q.14 Q.15 Q.16 Q.17

The first question of exercise 1.3 ML Aggarwal Class 9 asks us to locate two real numbers on the number line.

1. Locate $\sqrt{10}$ and $\sqrt{17}$ on the number line.

Answer

Steps to locate $\sqrt{10}$ on the number line

Step 1: We write $10$ as the sum of squares of two natural numbers: $9 + 1$ = $3^{2} + 1^{2}$ .

Step 2: We draw a number line and take a length of the line segment OA = $3$ units.

Step 3: At A, we draw AC $⊥$ OA. From AC, we cut off AB = $1$ unit.

Step 4: We join OB. We observe that OAB is a right angled triangle at A. By Pythagoras theorem, we get
${OB}^{2} = {OA}^{2} + {AB}^{2} = 3^{2} + 1^{2} = 10$
$\Rightarrow OB = \sqrt{10}$ units.

Step 5: With O as centre and radius OB, we draw an arc of a circle to meet the number line at point P.

As OP = OB = $\sqrt{10}$ units, the point P will represent the number $\sqrt{10}$ on the number line.

Steps to locate $\sqrt{17}$ on the number line

Step 1: We write $17$ as the sum of squares of two natural numbers: $16 + 1$ = $4^{2} + 1^{2}$ .

Step 2: We draw a number line and take a length of the line segment OA = $4$ units.

Step 3: At A, we draw AC $⊥$ OA. From AC, we cut off AB = $1$ unit.

Step 4: We join OB. We observe that OAB is a right angled triangle at A. By Pythagoras theorem, we get
${OB}^{2} = {OA}^{2} + {AB}^{2} = 4^{2} + 1^{2} = 17$
$\Rightarrow OB = \sqrt{17}$ units.

Step 5: With O as centre and radius OB, we draw an arc of a circle to meet the number line at point P.

As OP = OB = $\sqrt{17}$ units, the point P will represent the number $\sqrt{17}$ on the number line.

The second question of exercise 1.3 ML Aggarwal Class 9 asks us to find the decimal expansions of the given real numbers.

2. Write the decimal expansion of each of the following numbers and say what kind of decimal expansion each has:

$(i)$ $\frac{36}{100}$

Answer

The decimal expansion of $\frac{36}{100} = 0.36$ . It is a terminating decimal expansion.

$(i i) 4 \frac{1}{8}$

Answer

The decimal expansion of $4 \frac{1}{8} = 4 + \frac{1}{8} = 4 + 0.125 = 4.125$ . It is a terminating decimal expansion.

$(i i i) \frac{2}{9}$

Answer

The decimal expansion of $\frac{2}{9} = 0.222 . . .$ = 0.2. It is a non-terminating recurring decimal expansion.

$(i v) \frac{2}{11}$

Answer

The decimal expansion of $\frac{2}{11} = 0.181818 . . .$ = 0.18. It is a non-terminating recurring decimal expansion.

$(v) \frac{3}{13}$

Answer

The decimal expansion of $\frac{3}{13} = 0.230769230769 . . .$ = 0.230769. It is a non-terminating recurring decimal expansion.

$(v i) \frac{329}{400}$

Answer

The decimal expansion of $\frac{329}{400}$ $= 0.8225$ . It is a terminating decimal expansion.

The third question of exercise 1.3 ML Aggarwal Class 9 asks us to determine if a rational number is terminating or non-terminating without actually performing the long division.

3. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

$(i)$ $\frac{13}{3125}$

Answer

First, we note that the given rational number $\frac{13}{3125}$ is in its lowest terms.

Denominator of the given rational number = $3215$ = $5 \times 643$
Clearly, the denominator is not of the form $2^{m} 5^{n}$ , where $m, n$ are are whole numbers.
∴ The given rational number i.e. $\frac{13}{3125}$ is a non-terminating repeating decimal expansion.

$(ii)$ $\frac{17}{8}$

Answer

Given rational number = $\frac{17}{8} = \frac{17}{2^{3}}$ .
We note that its denominator has prime factorisation of the form $2^{m} 5^{n}$ , where $m = 3, n = 0$ are non-negative integers.
Therefore, the given rational number $\frac{17}{8}$ has a terminating decimal expansion.

$(iii)$ $\frac{23}{75}$

Answer

Given rational number = $\frac{23}{75} = \frac{23}{3 \times 5^{2}}$ .
We note that its denominator do not have prime factorisation of the form $2^{m} 5^{n}$ .
Therefore, the given rational number $\frac{23}{75}$ has a non-terminating repeating decimal expansion.

$(iv)$ $\frac{6}{15}$

Answer

Given rational number = $\frac{6}{15} = \frac{6}{3 \times 5^{1}}$ .
We note that its denominator do not have prime factorisation of the form $2^{m} 5^{n}$ .
Therefore, the given rational number $\frac{6}{15}$ has a non-terminating repeating decimal expansion.

$(v)$ $\frac{1258}{625}$

Answer

Given rational number = $\frac{1258}{625} = \frac{1258}{5^{4}}$ .
We note that its denominator has prime factorisation of the form $2^{m} 5^{n}$ , where $m = 0, n = 4$ are non-negative integers.
Therefore, the given rational number $\frac{1258}{625}$ has a terminating decimal expansion.

$(vi)$ $\frac{77}{210}$

Answer

Given rational number = $\frac{77}{210}$ = $\frac{77}{2 \times 3 \times 5 \times 7}$
We note that its denominator do not have prime factorisation of the form $2^{m} 5^{n}$ .
Therefore, the given rational number $\frac{77}{210}$ has a non-terminating repeating decimal expansion.

The fourth question of exercise 1.3 ML Aggarwal Class 9 asks us to determine if a rational number is terminating or non-terminating without actually performing the long division.

4. Without actually performing the long division, find if $\frac{987}{10500}$ will have terminating or non-terminating repeating decimal expansion. Give reasons for your answer.

Answer

Given rational number = $\frac{987}{10500}$ = $\frac{987}{2^{2} \times 3 \times 5^{3} \times 7}$ .
We note that its denominator do not have prime factorisation of the form $2^{m} 5^{n}$ , where $m, n$ are whole numbers. It contains two extra factors 3 and 7.
Therefore, the given rational number $\frac{77}{210}$ has a non-terminating repeating decimal expansion.

The fifth question of exercise 1.3 ML Aggarwal Class 9 asks us to write the decimal expansion of the given numbers.

5. Write the decimal expansions of the following numbers which have terminating decimal expansions:

$(i)$ $\frac{17}{8}$

Answer

∵ The denominator of $\frac{17}{8}$ i.e. 8 = $2 \times 2 \times 2 \times 1 = 2^{3} \times 1 = 2^{3} \times 5^{0}$ is of the form $2^{m} 5^{n}$ , where $m, n$ are whole numbers.
∴ $\frac{17}{8}$ is a terminating decimal expansion.

Decimal expansion of $\frac{17}{8}$ can be calculated as follows:
$\frac{17}{8}$ = $\frac{17}{2^{3}}$ = $\frac{17 \times 5^{3}}{2^{3} \times 5^{3}}$ = $\frac{17 \times 125}{{(2 \times 5)}^{3}}$ = $\frac{2125}{10^{3}}$ = $\frac{2125}{1000}$ = $2.125$ .

$(ii)$ $\frac{13}{3125}$

Answer

∵ The denominator of $\frac{13}{3125}$ i.e. $3125$ = $5 \times 5 \times 5 \times 5 \times 5 = 5^{5} \times 1 = 5^{5} \times 2^{0}$ is of the form $5^{n} 2^{m}$ , where $m, n$ are whole numbers.
∴ $\frac{13}{3125}$ is a terminating decimal expansion.

Decimal expansion of $\frac{13}{3125}$ can be calculated as follows:
$\frac{13}{3125}$ = $\frac{13}{5^{5}}$ = $\frac{13 \times 2^{5}}{5^{5} \times 2^{5}}$ = $\frac{13 \times 32}{{(2 \times 5)}^{5}}$ = $\frac{416}{10^{5}}$ = $\frac{416}{100000}$ = $0.00416$ .

$(iii)$ $\frac{7}{80}$

Answer

∵ The denominator of $\frac{7}{80}$ i.e. $80$ = $2 \times 2 \times 2 \times 2 \times 5 = 2^{4} \times 5^{1}$ is of the form $2^{m} 5^{n}$ , where $m, n$ are whole numbers.
∴ $\frac{7}{80}$ is a terminating decimal expansion.

Decimal expansion of $\frac{7}{80}$ can be calculated as follows:
$\frac{7}{80}$ = $\frac{7}{2^{4} \times 5^{1}}$ = $\frac{7 \times 5^{3}}{2^{4} \times 5^{1} \times 5^{3}}$ = $\frac{7 \times 5^{3}}{2^{4} \times 5^{4}}$ = $\frac{7 \times 125}{{(2 \times 5)}^{4}}$ = $\frac{875}{10^{4}}$ = $\frac{875}{10000}$ = $0.0875$ .

$(iv)$ $\frac{6}{15}$

Answer

First, we write the rational number in its lowest terms.
$\frac{6}{15}$ = $\frac{2 \times 3}{3 \times 5}$ = $\frac{2}{5}$ .

Now,

∵ The denominator of $\frac{2}{5}$ i.e. $5 = 2^{0} \times 5^{1}$ is of the form $2^{m} 5^{n}$ , where $m, n$ are whole numbers.
∴ $\frac{2}{5}$ is a terminating decimal expansion.

The decimal expansion of $\frac{2}{5}$ can be calculated as follows:
$\frac{2}{5}$ = $\frac{2}{5^{1}}$ = $\frac{2 \times 2^{1}}{2^{1} \times 5^{1}}$ = $\frac{4}{{(2 \times 5)}^{1}}$ = $\frac{4}{10}$ = $0.4$ .

$(v)$ $\frac{2^{2} \times 7}{5^{4}}$

Answer

We can see that the denominator of $\frac{2^{2} \times 7}{5^{4}}$ i.e. $5^{4}$ = $2^{0} \times 5^{4}$ is of the form $2^{m} 5^{n}$ , where $m, n$ are non-negative integers.
Therefore, $\frac{2^{2} \times 7}{5^{4}}$ is a terminating decimal expansion.

The decimal expansion of $\frac{2^{2} \times 7}{5^{4}}$ can be calculated as follows:
$\frac{2^{2} \times 7}{5^{4}}$ = $\frac{2^{2} \times 7 \times 2^{4}}{5^{4} \times 2^{4}}$ = $\frac{4 \times 7 \times 16}{{(2 \times 5)}^{4}}$ = $\frac{448}{10^{4}}$ = $\frac{448}{10000}$ = $0.0448$ .

$(vi)$ $\frac{237}{1500}$

Answer

First, we write the rational number in its lowest terms.
$\frac{237}{1500}$ = $\frac{79 \times 3}{2 \times 2 \times 5 \times 5 \times 5 \times 3}$ = $\frac{79}{2 \times 2 \times 5 \times 5 \times 5}$ = $\frac{79}{500}$ .

Now,

∵ The denominator of $\frac{79}{500}$ i.e. $500 = 2^{2} \times 5^{3}$ is of the form $2^{m} 5^{n}$ , where $m, n$ are whole numbers.
∴ $\frac{79}{500}$ is a terminating decimal expansion.

The decimal expansion of $\frac{79}{500}$ can be calculated as follows:
$\frac{79}{500}$ = $\frac{79}{2^{2} \times 5^{3}}$ = $\frac{79 \times 2}{2^{3} \times 5^{3}}$ = $\frac{158}{{(2 \times 5)}^{3}}$ = $\frac{158}{10^{3}}$ = $\frac{158}{1000}$ = $0.158$ .

The sixth question of exercise 1.3 ML Aggarwal Class 9 asks us to find decimal expansion without actual division.

6. Write the denominator of the rational number $\frac{257}{5000}$ in the form $2^{m} \times 5^{n}$ , where $m, n$ are non-negative integers. Hence, write the decimal expansion without actual division.

Answer

The given rational number is $\frac{257}{5000}$ .
The denominator of the given rational number can be written in the form of $2^{m} 5^{n}$ as follows:

$5000 = 2 \times 2 \times 2 \times 5 \times 5 \times 5 \times 5 = 2^{3} \times 5^{4}$ , where $m = 3, n = 4$ .

The decimal expansion of the rational number $\frac{257}{5000}$ is as follows:
$\frac{257}{5000}$ = $\frac{257}{2^{3} \times 5^{4}}$ = $\frac{2 \times 257}{2 \times 2^{3} \times 5^{4}}$ = $\frac{514}{{(2 \times 5)}^{4}}$ = $\frac{514}{10^{4}}$ = $\frac{514}{10000}$ = $0.0514$ .

The seventh question of exercise 1.3 ML Aggarwal Class 9 asks us to find decimal expansion of a number.

7. Write the decimal expansion of $\frac{1}{7}$ . Hence, write the decimal expansion of $\frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}$ and $\frac{6}{7}$ .

Answer

The decimal expansion of $\frac{1}{7}$ = 0.142857 .

Now,

Decimal expansion of $\frac{2}{7}$ = $2 \times \frac{1}{7}$ = 2 $\times$ 0.142857 = 0.285714 .

Decimal expansion of $\frac{3}{7}$ = $3 \times \frac{1}{7}$ = 3 $\times$ 0.142857 = 0.428571 .

Decimal expansion of $\frac{4}{7}$ = $4 \times \frac{1}{7}$ = 4 $\times$ 0.142857 = 0.571428 .

Decimal expansion of $\frac{5}{7}$ = $5 \times \frac{1}{7}$ = 5 $\times$ 0.142857 = 0.714285 .

Decimal expansion of $\frac{6}{7}$ = $6 \times \frac{1}{7}$ = 6 $\times$ 0.142857 = 0.857142 .

The eighth question of exercise 1.3 ML Aggarwal Class 9 asks us to express the given number in the form of $\frac{p}{q}$ .

8. Express the following numbers in the form $\frac{p}{q}$ , where $p$ and $q$ are both integers and $q \neq 0$ .

(i) 0.3

Answer

Let $x$ = 0.3
$\Rightarrow x = 0.333 . . . \cdot \cdot \cdot (i)$

Concept: Number of bars in the number determine the number by which we multiply on both sides.

Multiplying both sides of (i) by 10, we get
$10 x$ = 3.333... $\cdot \cdot \cdot$ (ii)

Subtracting (i) from (ii), we get
$10 x = 3.333 . . .$
$x = 0.333 . . .$
$-$ $-$

9 x = 3

\Rightarrow x = \frac{3}{9}

= \frac{1}{3}

Hence, 0.3 = $\frac{1}{3}$ .

(ii) 5.2

Answer

Let $x$ = 5.2
$\Rightarrow x = 5.222 . . . \cdot \cdot \cdot (i)$

Concept: Number of bars in the number determine the number by which we multiply on both sides.

Multiplying both sides of (i) by 10, we get
$10 x$ = 52.222. . . $\cdot \cdot \cdot$ (ii)

Subtracting (i) from (ii), we get
$10 x = 52.222 . . .$
$x = 5.222 . . .$
$-$ $-$

9 x = 47

\Rightarrow x = \frac{47}{9}

Hence, 5.2 = $\frac{47}{9}$ .

(iii) 0.404040...

Answer

Let $x$ = 0.40
$\Rightarrow x = 0.404040 . . . \cdot \cdot \cdot (i)$

Concept: Number of bars in the number determine the number by which we multiply on both sides.

Multiplying both sides of (i) by 100, we get
$100 x$ = 40.404040. . . $\cdot \cdot \cdot$ (ii)

Subtracting (i) from (ii), we get
$100 x = 40.404040 . . .$
$x = 0.404040 . . .$
$-$ $-$

99 x = 40

\Rightarrow x = \frac{40}{99}

Hence, 0.40 = $\frac{40}{99}$ .

(iv) 0.47

Answer

Let $x$ = 0.47
$\Rightarrow x = 0.4777 . . . \cdot \cdot \cdot (i)$

Concept: Number of bars in the number determine the number by which we multiply on both sides.

Multiplying both sides of (i) by 10, we get
$10 x$ = 4.777. . . $\cdot \cdot \cdot$ (ii)

Subtracting (i) from (ii), we get
$10 x = 4.7777 . . .$
$x = 0.4777 . . .$
$-$ $-$

99 x = 4.3

\Rightarrow x = \frac{4.3}{99} = \frac{4.3 \times 10}{99 \times 10} = \frac{43}{990}

Hence, 0.47 = $\frac{43}{990}$ .

(v) 0.134

Answer

Let $x$ = 0.134
$\Rightarrow x = 0.1343434 . . . \cdot \cdot \cdot (i)$

Concept: Number of bars in the number determine the number by which we multiply on both sides.

Multiplying both sides of (i) by 100, we get
$100 x$ = 13.43434. . . $\cdot \cdot \cdot$ (ii)

Subtracting (i) from (ii), we get
$100 x = 13.4343434 . . .$
$x = 0.1343434 . . .$
$-$ $-$

99 x = 13.3

\Rightarrow x = \frac{13.3}{99} = \frac{13.3 \times 10}{99 \times 10} = \frac{133}{990}

Hence, 0.134 = $\frac{133}{990}$ .

(v) 0.001

Answer

Let $x$ = 0.001
$\Rightarrow x = 0.001001001 . . . \cdot \cdot \cdot (i)$

Concept: Number of bars in the number determine the number by which we multiply on both sides.

Multiplying both sides of (i) by 1000, we get
$1000 x$ = 1.001001001. . . $\cdot \cdot \cdot$ (ii)

Subtracting (i) from (ii), we get
$1000 x = 1.001001001 . . .$
$x = 0.001001001 . . .$
$-$ $-$

999 x = 1

\Rightarrow x = \frac{1}{999}

Hence, 0.001 = $\frac{1}{999}$ .

The ninth question of exercise 1.3 ML Aggarwal Class 9 asks us to classify the numbers as rational and irrational.

9. Classify the following numbers as rational or irrational:

(i) $\sqrt{23}$

Answer

$\sqrt{23}$ is an irrational number, as 23 is not a perfect square and cannot come out of the square root.

(ii) $\sqrt{225}$

Answer

$\sqrt{225}$ is a rational number, as 225 is a perfect square and $\sqrt{225} = \sqrt{3 \times 3 \times 5 \times 5} = 3 \times 5 = 15$ .

(iii) $0.3796$

Answer

$0.3796$ is a terminating decimal expansion of a rational number. We can write $0.3796 = \frac{3796}{10000} = \frac{1898}{5000} = \frac{949}{2500}$ . Hence, $0.3796$ is a rational number.

(iv) $7.478478 . . .$

Answer

We can see that the block of digits $478$ is repeating. So, $7.478478 . . .$ is a non-terminating repeating decimal expansion. Therefore, it is a rational number.

(v) $1.101001000100001 . . .$

Answer

$1.101001000100001 . . .$ is a non-terminating non-repeating decimal expansion, as we can see there is no block of digits repeating. Therefore, the given number is an irrational number.

(v) 347.0456

Answer

As we can see, a bar over the block of digits $456$ , the given number is a rational number .

The tenth question of exercise 1.3 ML Aggarwal Class 9 asks us to classify the numbers as rational and irrational and also find the factors of the denominator if they are rational.

10. The following real numbers have decimal expansions as given below. In each case, state whether they are rational or not. If they are rational and expressed in the form $\frac{p}{q}$ , where $p, q$ are integers and $q \neq 0$ and $p, q$ are co-prime, then what can you say about the prime factors of $q$ ?

(i) 37.09158

Answer

$37.09158$ is a rational number, as it is a terminating decimal expansion.

When $37.09158$ is expressed in the form of $\frac{p}{q}$ then prime factors of q must be either $2$ or $5$ or both, as we can see below.

$37.09158 = \frac{3709158}{100000} = \frac{1854579}{50000}$ $= \frac{1854579}{2^{4} \times 5^{5}}$ . Clearly, the denominator has $2$ and $5$ as its prime factors.

(ii) 423.04567

Answer

423.04567 is a rational number , as it is a non-terminating recurring decimal expansion.

When 423.04567 is expressed in the form of $\frac{p}{q}$ , the prime factors of q must have a prime factor other than 2 and 5, as we can see below.

423.04567 = 423 + 0.04567 = $423 + \frac{4567}{99999}$ = $\frac{42300000 + 4567}{99999}$ = $\frac{42304567}{99999}$ = $\frac{42304567}{3^{2} \times 41 \times 271}$ .
Clearly, the denominator contains prime factors other than 2 and 5.

(iii) 8.901001000100001...

Answer

8.901001000100001... is an irrational number. It can not be expressed in the form of $\frac{p}{q}$ .

(iv) 2.3476817681...

Answer

2.3476817681... = 2.347681is a rational number, as it is a non-terminating recurring decimal expansion.

When 2.347681 is expressed in the form of $\frac{p}{q}$ , the prime factors of q must include a prime number other than 2 and 5.

2.347681 = 2.34 + 0.007681 = $\frac{234}{100} + \frac{7681}{9999000}$ = $\frac{2339766000 + 7681}{9999000}$ = $\frac{2339773681}{9999000}$ = $\frac{2339773681}{2^{3} \times 3^{2} \times 5^{3} \times 11 \times 101}$ .
Clearly, the factors of 9999000 include 2, 3, 5, 11 and 101.

The eleventh question of exercise 1.3 ML Aggarwal Class 9 asks us to insert an irrational number between two rational numbers.

11. Insert an irrational number between the following:

(i) $\frac{1}{3}$ and $\frac{1}{2}$

Answer

First, we will write the numbers in decimal form.

$\frac{1}{3} = 0.333 . . .$
$\frac{1}{2} = 0.5$

To find an irrational number between $0.333 . . .$ and $0.5$ , we find a decimal number which is a non-terminating and non-recurring and lying between them.

Some of such numbers are given below:
0.333... < 0.414114111411114... < 0.5
0.333... < 0.38338333833338... < 0.5
0.333... < 0.404004000400004... < 0.5
0.333... < 0.434334333433334... < 0.5

(ii) $\frac{- 2}{5}$ and $\frac{1}{2}$

Answer

We have,
$\frac{- 2}{5} = - 0.4$
$\frac{1}{2} = 0.5$

To find an irrational number between $- 0.4$ and $0.5$ , we find a decimal number which is a non-terminating and non-recurring and lying between them.

Some of such numbers are given below:
− 0.4 < 0.151551555155551... < 0.5
− 0.4 < − 0.303003000300003... < 0.5
− 0.4 < 0.202002000200002... < 0.5
− 0.4 < 0.323223222322223... < 0.5

(iii) $0$ and $0.1$

Answer

To find an irrational number between $0$ and $0.1$ , we find a decimal number which is a non-terminating and non-recurring and lying between them.

Some of such numbers are given below:
0 < 0.07007000700007... < 0.1
0 < 0.02002000200002... < 0.1
0 < 0.03003000300003... < 0.1
0 < 0.00100110011100111100... < 0.1

The twelfth question of exercise 1.3 ML Aggarwal Class 9 asks us to insert two irrational number between two rational numbers.

12. Insert two irrational numbers between 2 and 3.

Answer

We know that
2 < 3
$\Rightarrow 2^{2} < 3^{2}$ ...[∵ Squaring on both sides]
$\Rightarrow 4 < 9$
$\Rightarrow 4 < 5 < 6 < 7 < 8 < 9$
Taking square root, we get
$\Rightarrow \sqrt{4} < \sqrt{5} < \sqrt{6} < \sqrt{7} < \sqrt{8} < \sqrt{9}$
$\Rightarrow 2 < \sqrt{5} < \sqrt{6} < \sqrt{7} < \sqrt{8} < 3$

Clearly, the numbers $\sqrt{5}$ , $\sqrt{6}$ , $\sqrt{7}$ , $\sqrt{8}$ lying between 2 and 3 are irrational numbers. Therefore, the two irrational numbers between 2 and 3 are $\sqrt{5}$ , $\sqrt{6}$ .

The thirteenth question of exercise 1.3 ML Aggarwal Class 9 asks us to write two irrational numbers between two rational numbers.

13. Write two irrational numbers between $\frac{4}{9}$ and $\frac{7}{11}$ .

Answer

First, we will write the numbers in decimal form.

$\frac{4}{9} = 0.444 . . .$

$\frac{7}{11} = 0.636363 . . .$

To find an irrational number between $0.444....$ and $0.636363...$ , we find a decimal number which is a non-terminating and non-recurring and lying between them.

Some of such numbers are given below:
0.444... < 0.505005000500005... < 0.636363...
0.444... < 0.606006000600006... < 0.636363...
0.444... < 0.515115111511115... < 0.636363...
0.444... < 0.616116111611116... < 0.636363...

The fourteenth question of exercise 1.3 ML Aggarwal Class 9 asks us to find a rational number between two irrational numbers.

14. Find a rational number between $\sqrt{2}$ and $\sqrt{3}$ .

Answer

To find a rational number between $\sqrt{2}$ and $\sqrt{3}$ , we first find a perfect square number between 2 and 3. We can find it as follows.

We know that
$\sqrt{2} < \sqrt{3}$
$\Rightarrow {(\sqrt{2})}^{2} < {(\sqrt{3})}^{2}$ ... [Squaring on both sides]
$\Rightarrow 2 < 3$

Now, we divide and multiply by a perfect square number like 100 so that we get a perfect number between 2 and 3.

$\Rightarrow \frac{2 \times 100}{100} < \frac{3 \times 100}{100}$
$\Rightarrow \frac{200}{100} < \frac{300}{100}$

$\Rightarrow \frac{200}{100} < \frac{225}{100} < \frac{256}{100} < \frac{289}{100} < \frac{300}{100}$

$\Rightarrow 2 < 2.25 < 2.56 < 2.89 < 3$
Taking square root of each number we get.
$\Rightarrow \sqrt{2} < \sqrt{2.25} < \sqrt{2.56} < \sqrt{2.89} < \sqrt{3}$

$\Rightarrow \sqrt{2} < \sqrt{{(1.5)}^{2}} < \sqrt{{(1. 6)}^{2}} < \sqrt{{(1. 7)}^{2}} < \sqrt{3}$

$\Rightarrow \sqrt{2} < 1.5 < 1.6 < 1.7 < \sqrt{3}$

Clearly, a rational number between $\sqrt{2}$ and $\sqrt{3}$ is 1.5.

The fifteenth question of exercise 1.3 ML Aggarwal Class 9 asks us to find a rational number between two irrational numbers.

15. Find two rational number between $2 \sqrt{3}$ and $\sqrt{15}$ .

Answer

The given irrational numbers are $2 \sqrt{3}$ and $\sqrt{15}$ .

We can write,
${(2 \sqrt{3})}^{2} = 12$
${(\sqrt{15})}^{2} = 15$

To find rational numbers between $2 \sqrt{3}$ and $\sqrt{15}$ , we need to perfect square numbers between 12 and 15 respectively. We can do step by step as follows:

As $2 \sqrt{3}$ < $\sqrt{15}$

$\Rightarrow {(2 \sqrt{3})}^{2} < {(\sqrt{15})}^{2}$
$\Rightarrow 12 < 15$

Multiplying and dividing by 100 in both, we get
$\Rightarrow \frac{12 \times 100}{100} < \frac{15 \times 100}{100}$
$\Rightarrow \frac{1200}{100} < \frac{1500}{100}$

Now, the perfect square numbers between 1200 and 1500 are 1225, 1296, 1369, and 1444. So, we can write,
$\Rightarrow \frac{1200}{100} < \frac{1225}{100} < \frac{1296}{100} < \frac{1369}{100} < \frac{1444}{100} < \frac{1500}{100}$
$\Rightarrow 12 < \frac{1225}{100} < \frac{1296}{100} < \frac{1369}{100} < \frac{1444}{100} < 15$
$\Rightarrow 12 < 12.25 < 12.96 < 13.69 < 14.44 < 15$

Now, taking square roots of every number, we get
$\Rightarrow \sqrt{12} < \sqrt{12.25} < \sqrt{12.96} < \sqrt{13.69} < \sqrt{14.44} < \sqrt{15}$
$\Rightarrow 2 \sqrt{3} < 3.5 < 3.6 < 3.7 < 3.8 < \sqrt{15}$

Clearly, two rational numbers between $2 \sqrt{3}$ and $\sqrt{15}$ are 3.5 and 3.6.

The sixteenth question of exercise 1.3 ML Aggarwal Class 9 asks us to find an irrational number between two rational numbers.

16. Insert an irrational number between $\sqrt{5}$ and $\sqrt{7}$ .

Answer

The given numbers are $\sqrt{5}$ and $\sqrt{7}$ .
$∵ \sqrt{5} < \sqrt{7}$
Squaring the numbers, we get
${(\sqrt{5})}^{2} < {(\sqrt{7})}^{2}$
$\Rightarrow 5 < 7$
$\Rightarrow 5 < 6 < 7$
Taking square roots of all numbers, we get,
$\Rightarrow \sqrt{5} < \sqrt{6} < \sqrt{7}$

Clearly, an irrational number between $\sqrt{5}$ and $\sqrt{7}$ is $\sqrt{6}$ .

The seventeenth question of exercise 1.3 ML Aggarwal Class 9 asks us to find two irrational number between two irrational numbers.

17. Insert two irrational number between $\sqrt{3}$ and $\sqrt{7}$ .

Answer

The given numbers are $\sqrt{3}$ and $\sqrt{7}$ .
$∵ \sqrt{3} < \sqrt{7}$
Squaring the numbers, we get
${(\sqrt{3})}^{2} < {(\sqrt{7})}^{2}$
$\Rightarrow 3 < 7$
$\Rightarrow 3 < 4 < 5 < 6 < 7$
Taking square roots of all numbers, we get,
$\Rightarrow \sqrt{3} < \sqrt{4} < \sqrt{5} < \sqrt{6} < \sqrt{7}$

Clearly, an irrational number between $\sqrt{3}$ and $\sqrt{7}$ are $\sqrt{5}$ and $\sqrt{6}$ .

Exercise 1.3 ML Aggarwal Class 9 Mathematics Solutions

1. Locate 10 and 17 on the number line.

Steps to locate 10 on the number line

Steps to locate 17 on the number line

2. Write the decimal expansion of each of the following numbers and say what kind of decimal expansion each has:

(i) 36100

(ii) 418

(iii) 29

(iv) 211

(v) 313

(vi) 329400

3. Without actually performing the long division, state whether the following rational numbers will have a terminating decimal expansion or a non-terminating repeating decimal expansion:

(i) 133125

(ii) 178

(iii) 2375

(iv) 615

(v) 1258625

(vi) 77210

4. Without actually performing the long division, find if 98710500 will have terminating or non-terminating repeating decimal expansion. Give reasons for your answer.

5. Write the decimal expansions of the following numbers which have terminating decimal expansions:

(i) 178

(ii) 133125

(iii) 780

(iv) 615

(v) 22×754

(vi) 2371500

6. Write the denominator of the rational number 2575000 in the form 2m×5n, where m, n are non-negative integers. Hence, write the decimal expansion without actual division.

7. Write the decimal expansion of 17. Hence, write the decimal expansion of 27, 37, 47, 57 and 67.

8. Express the following numbers in the form pq, where p and q are both integers and q≠0.

(i) 0.3

(ii) 5.2

(iii) 0.404040...

(iv) 0.47

(v) 0.134

(v) 0.001

9. Classify the following numbers as rational or irrational:

(i) 23

(ii) 225

(iii) 0.3796

(iv) 7.478478...

(v) 1.101001000100001...

(v) 347.0456

10. The following real numbers have decimal expansions as given below. In each case, state whether they are rational or not. If they are rational and expressed in the form pq , where p, q are integers and q≠0 and p, q are co-prime, then what can you say about the prime factors of q?

(i) 37.09158

(ii) 423.04567

(iii) 8.901001000100001...

(iv) 2.3476817681...

11. Insert an irrational number between the following:

(i) 13 and 12

(ii) −25 and 12

(iii) 0 and 0.1

12. Insert two irrational numbers between 2 and 3.

13. Write two irrational numbers between 49 and 711.

14. Find a rational number between 2 and 3.

15. Find two rational number between 23 and 15.

16. Insert an irrational number between 5 and 7.

17. Insert two irrational number between 3 and 7.

1. Locate $\sqrt{10}$ and $\sqrt{17}$ on the number line.

Steps to locate $\sqrt{10}$ on the number line

Steps to locate $\sqrt{17}$ on the number line

$(i)$ $\frac{36}{100}$

$(i i) 4 \frac{1}{8}$

$(i i i) \frac{2}{9}$

$(i v) \frac{2}{11}$

$(v) \frac{3}{13}$

$(v i) \frac{329}{400}$

$(i)$ $\frac{13}{3125}$

$(ii)$ $\frac{17}{8}$

$(iii)$ $\frac{23}{75}$

$(iv)$ $\frac{6}{15}$

$(v)$ $\frac{1258}{625}$

$(vi)$ $\frac{77}{210}$

4. Without actually performing the long division, find if $\frac{987}{10500}$ will have terminating or non-terminating repeating decimal expansion. Give reasons for your answer.

$(i)$ $\frac{17}{8}$

$(ii)$ $\frac{13}{3125}$

$(iii)$ $\frac{7}{80}$

$(iv)$ $\frac{6}{15}$

$(v)$ $\frac{2^{2} \times 7}{5^{4}}$

$(vi)$ $\frac{237}{1500}$

6. Write the denominator of the rational number $\frac{257}{5000}$ in the form $2^{m} \times 5^{n}$ , where $m, n$ are non-negative integers. Hence, write the decimal expansion without actual division.

7. Write the decimal expansion of $\frac{1}{7}$ . Hence, write the decimal expansion of $\frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}$ and $\frac{6}{7}$ .

8. Express the following numbers in the form $\frac{p}{q}$ , where $p$ and $q$ are both integers and $q \neq 0$ .

(i) $\sqrt{23}$

(ii) $\sqrt{225}$

(iii) $0.3796$

(iv) $7.478478 . . .$

(v) $1.101001000100001 . . .$

(i) $\frac{1}{3}$ and $\frac{1}{2}$

(ii) $\frac{- 2}{5}$ and $\frac{1}{2}$

(iii) $0$ and $0.1$

13. Write two irrational numbers between $\frac{4}{9}$ and $\frac{7}{11}$ .

14. Find a rational number between $\sqrt{2}$ and $\sqrt{3}$ .

15. Find two rational number between $2 \sqrt{3}$ and $\sqrt{15}$ .

16. Insert an irrational number between $\sqrt{5}$ and $\sqrt{7}$ .

17. Insert two irrational number between $\sqrt{3}$ and $\sqrt{7}$ .