myhelper: RS AGGARWAL CLASS 9 Chapter 1 Number System Exercise 1A

Exercise 1A

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Question 1:

Is zero a rational number? Justify.

Answer 1:

Yes, 0 is a rational number.

0 can be expressed in the form of the fraction $\frac{p}{q}$ $\frac{p}{q}$ , where $p = 0$ $p = 0$ and q can be any integer except 0.

Question 2:

Represent each of the following rational number line:
(i) $\frac{5}{7}$ $\frac{5}{7}$
(ii) $\frac{8}{3}$ $\frac{8}{3}$
(iii) $- \frac{23}{6}$ $- \frac{23}{6}$
(iv) 1.3
(v) – 2.4

Answer 2:

(i) $\frac{5}{7}$

(ii) $\frac{8}{3}$ $\frac{8}{3}$
$\frac{8}{3} = 2 \frac{2}{3}$

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

(iii) $- \frac{23}{6} = - 3 \frac{5}{6}$ $- \frac{23}{6} = - 3 \frac{5}{6}$

(iv) 1.3
$1.3 = \frac{13}{10} = 1 \frac{3}{10}$ $1.3 = \frac{13}{10} = 1 \frac{3}{10}$

(v) – 2.4
$- 2.4 = \frac{- 24}{10} = \frac{- 12}{5} = - 2 \frac{2}{5}$

Question 3:

Find a rational number between
(i) $\frac{3}{8} and \frac{2}{5}$ $\frac{3}{8} and \frac{2}{5}$
(ii) 1.3 and 1.4
(iii) $- 1 and \frac{1}{2}$ $- 1 and \frac{1}{2}$
(iv) $- \frac{3}{4} and - \frac{2}{5}$ $- \frac{3}{4} and - \frac{2}{5}$
(v) $\frac{1}{9} and \frac{2}{9}$ $\frac{1}{9} and \frac{2}{9}$

Answer 3:

(i) $\frac{3}{8} and \frac{2}{5}$ $\frac{3}{8} and \frac{2}{5}$
Let:
x = $\frac{3}{8}$ $\frac{3}{8}$ and y = $\frac{2}{5}$ $\frac{2}{5}$
Rational number lying between x and y:
$\frac{1}{2} (x + y) = \frac{1}{2} (\frac{3}{8} + \frac{2}{5})$ $\frac{1}{2} (x + y) = \frac{1}{2} (\frac{3}{8} + \frac{2}{5})$
= $\frac{1}{2} (\frac{15 + 16}{40}) = \frac{31}{80}$ $\frac{1}{2} (\frac{15 + 16}{40}) = \frac{31}{80}$

(ii) 1.3 and 1.4
Let:
x = 1.3 and y = 1.4
Rational number lying between x and y:
$\frac{1}{2} (x + y) = \frac{1}{2} (1.3 + 1.4)$ $\frac{1}{2} (x + y) = \frac{1}{2} (1.3 + 1.4)$
= $\frac{1}{2} (2.7) = 1.35$ $\frac{1}{2} (2.7) = 1.35$

(iii) $- 1 and \frac{1}{2}$ $- 1 and \frac{1}{2}$
Let:
x = $-$ $-$ 1 and y = $\frac{1}{2}$ $\frac{1}{2}$
Rational number lying between x and y:
$\frac{1}{2} (x + y) = \frac{1}{2} (- 1 + \frac{1}{2})$ $\frac{1}{2} (x + y) = \frac{1}{2} (- 1 + \frac{1}{2})$
= $- \frac{1}{4}$ $- \frac{1}{4}$

(iv) $- \frac{3}{4} and - \frac{2}{5}$ $- \frac{3}{4} and - \frac{2}{5}$
Let:
x = $- \frac{3}{4}$ $- \frac{3}{4}$ and y = $- \frac{2}{5}$ $- \frac{2}{5}$
Rational number lying between x and y:
$\frac{1}{2} (x + y) = \frac{1}{2} (- \frac{3}{4} - \frac{2}{5})$ $\frac{1}{2} (x + y) = \frac{1}{2} (- \frac{3}{4} - \frac{2}{5})$
= $\frac{1}{2} (\frac{- 15 - 8}{20}) = - \frac{23}{40}$ $\frac{1}{2} (\frac{- 15 - 8}{20}) = - \frac{23}{40}$

(v) $\frac{1}{9} and \frac{2}{9}$ $\frac{1}{9} and \frac{2}{9}$
A rational number lying between $\frac{1}{9} and \frac{2}{9}$ $\frac{1}{9} and \frac{2}{9}$ will be
$\frac{1}{2} (\frac{1}{9} + \frac{2}{9}) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$ $\frac{1}{2} (\frac{1}{9} + \frac{2}{9}) = \frac{1}{2} \times \frac{1}{3} = \frac{1}{6}$

Question 4:

Find three rational numbers lying between $\frac{3}{5} and \frac{7}{8}$ $\frac{3}{5} and \frac{7}{8}$ . How many rational numbers can be determined between these two numbers?

Answer 4:

$x = \frac{3}{5} and y = \frac{7}{8}$ $x = \frac{3}{5} and y = \frac{7}{8}$
n = 3
$d = \frac{(y - x)}{n + 1} = \frac{\frac{7}{8} - \frac{3}{5}}{3 + 1} = \frac{11}{40} \times \frac{1}{4} = \frac{11}{160}$ $d = \frac{(y - x)}{n + 1} = \frac{\frac{7}{8} - \frac{3}{5}}{3 + 1} = \frac{11}{40} \times \frac{1}{4} = \frac{11}{160}$
Rational numbers between $x = \frac{3}{5} and y = \frac{7}{8}$ $x = \frac{3}{5} and y = \frac{7}{8}$ will be
$(x + d), (x + 2 d), . . ., (x + n d) \Rightarrow (\frac{3}{5} + \frac{11}{160}), (\frac{3}{5} + 2 \times \frac{11}{160}), (\frac{3}{5} + 3 \times \frac{11}{160}) \Rightarrow (\frac{107}{160}), (\frac{118}{160}), (\frac{129}{160}) \Rightarrow (\frac{107}{160}), (\frac{59}{80}), (\frac{129}{160})$ $(x + d), (x + 2 d), . . ., (x + n d) \Rightarrow (\frac{3}{5} + \frac{11}{160}), (\frac{3}{5} + 2 \times \frac{11}{160}), (\frac{3}{5} + 3 \times \frac{11}{160}) \Rightarrow (\frac{107}{160}), (\frac{118}{160}), (\frac{129}{160}) \Rightarrow (\frac{107}{160}), (\frac{59}{80}), (\frac{129}{160})$
There are infinitely many rational numbers between two given rational numbers.

Question 5:

Find four rational numbers between $\frac{3}{7} and \frac{5}{7}$ $\frac{3}{7} and \frac{5}{7}$ .

Answer 5:

n = 4
n + 1 = 4 + 1 = 5
$\frac{3}{7} = \frac{3}{7} \times \frac{5}{5} = \frac{15}{35} \frac{5}{7} = \frac{5}{7} \times \frac{5}{5} = \frac{25}{35}$ $\frac{3}{7} = \frac{3}{7} \times \frac{5}{5} = \frac{15}{35} \frac{5}{7} = \frac{5}{7} \times \frac{5}{5} = \frac{25}{35}$
Thus, rational numbers between $\frac{3}{7} and \frac{5}{7}$ $\frac{3}{7} and \frac{5}{7}$ are $\frac{16}{35}, \frac{17}{35}, \frac{18}{35}, \frac{19}{35}$ $\frac{16}{35}, \frac{17}{35}, \frac{18}{35}, \frac{19}{35}$ .

Question 6:

Find six rational numbers between 2 and 3.

Answer 6:

x = 2, y = 3 and n = 6
$d = \frac{y - x}{n + 1} = \frac{3 - 2}{6 + 1} = \frac{1}{7}$ $d = \frac{y - x}{n + 1} = \frac{3 - 2}{6 + 1} = \frac{1}{7}$
Thus, the required numbers are
$(x + d), (x + 2 d), (x + 3 d), . . ., (x + n d) = (2 + \frac{1}{7}), (2 + 2 \times \frac{1}{7}), (2 + 3 \times \frac{1}{7}), (2 + 4 \times \frac{1}{7}), (2 + 5 \times \frac{1}{7}), (2 + 6 \times \frac{1}{7}) = \frac{15}{7}, \frac{16}{7}, \frac{17}{7}, \frac{18}{7}, \frac{19}{7}, \frac{20}{7}$ $(x + d), (x + 2 d), (x + 3 d), . . ., (x + n d) = (2 + \frac{1}{7}), (2 + 2 \times \frac{1}{7}), (2 + 3 \times \frac{1}{7}), (2 + 4 \times \frac{1}{7}), (2 + 5 \times \frac{1}{7}), (2 + 6 \times \frac{1}{7}) = \frac{15}{7}, \frac{16}{7}, \frac{17}{7}, \frac{18}{7}, \frac{19}{7}, \frac{20}{7}$

Question 7:

Find five rational numbers between $\frac{3}{5} and \frac{2}{3}$ $\frac{3}{5} and \frac{2}{3}$ .

Answer 7:

n = 5
n + 1 = 6
$x = \frac{3}{5}, y = \frac{2}{3}$ $x = \frac{3}{5}, y = \frac{2}{3}$
$d = \frac{y - x}{n + 1} = \frac{\frac{2}{3} - \frac{3}{5}}{6} = \frac{10 - 9}{90} = \frac{1}{90}$ $d = \frac{y - x}{n + 1} = \frac{\frac{2}{3} - \frac{3}{5}}{6} = \frac{10 - 9}{90} = \frac{1}{90}$

Thus, rational numbers between $\frac{3}{5} and \frac{2}{3}$ $\frac{3}{5} and \frac{2}{3}$ will be
$(x + d), (x + 2 d), (x + 3 d), (x + 4 d), (x + 5 d) = (\frac{3}{5} + \frac{1}{90}), (\frac{3}{5} + \frac{2}{90}), (\frac{3}{5} + \frac{3}{90}), (\frac{3}{5} + \frac{4}{90}), (\frac{3}{5} + \frac{5}{90}) = (\frac{55}{90}), (\frac{56}{90}), (\frac{57}{90}), (\frac{58}{90}), (\frac{59}{90}) = (\frac{11}{18}), (\frac{28}{45}), (\frac{19}{30}), (\frac{29}{45}), (\frac{59}{90})$ $(x + d), (x + 2 d), (x + 3 d), (x + 4 d), (x + 5 d) = (\frac{3}{5} + \frac{1}{90}), (\frac{3}{5} + \frac{2}{90}), (\frac{3}{5} + \frac{3}{90}), (\frac{3}{5} + \frac{4}{90}), (\frac{3}{5} + \frac{5}{90}) = (\frac{55}{90}), (\frac{56}{90}), (\frac{57}{90}), (\frac{58}{90}), (\frac{59}{90}) = (\frac{11}{18}), (\frac{28}{45}), (\frac{19}{30}), (\frac{29}{45}), (\frac{59}{90})$

Question 8:

Insert 16 rational numbers between 2.1 and 2.2.

Answer 8:

Let:
x = 2.1, y = 2.2 and n = 16

We know:
d = $\frac{y - x}{n + 1} = \frac{2.2 - 2.1}{16 + 1} = \frac{0.1}{17} = \frac{1}{170}$ $\frac{y - x}{n + 1} = \frac{2.2 - 2.1}{16 + 1} = \frac{0.1}{17} = \frac{1}{170}$ = 0.005 (approx.)
So, 16 rational numbers between 2.1 and 2.2 are:
(x + d), (x + 2d), ...(x + 16d)
= [2.1 + 0.005], [2.1 + 2(0.005)],...[2.1 + 16(0.005)]
= 2.105, 2.11, 2.115, 2.12, 2.125, 2.13, 2.135, 2.14, 2.145, 2.15, 2.155, 2.16, 2.165, 2.17, 2.175 and 2.18

Question 9:

State whether the following statements are true or false. Give reasons for your answer.
(i) Every natural number is a whole number.
(ii) Every whole number is a natural number.
(iii) Every integer is a whole number.
(iv) Every integer is a rational number.
(v) Every rational number is an integer.
(vi) Every rational number is a whole number.

Answer 9:

(i) Every natural number is a whole number.
True, since natural numbers are counting numbers i.e N = 1, 2,...
Whole numbers are natural numbers together with 0. i.e W = 0, 1, 2,...
So, every natural number is a whole number

(ii) Every whole number is a natural number.
False, as whole numbers contain natural numbers and 0 whereas natural numbers only contain the counting numbers except 0.

(iii) Every integer is a whole number.
False, whole numbers are natural numbers together with a zero whereas integers include negative numbers also.

(iv) Every integer is a rational number.
True, as rational numbers are of the form $\frac{p}{q} where q \neq 0$ $\frac{p}{q} where q \neq 0$ . All integers can be represented in the form $\frac{p}{q} where q \neq 0$ $\frac{p}{q} where q \neq 0$ .
(v) Every rational number is an integer.
False, as rational numbers are of the form $\frac{p}{q} where q \neq 0$ $\frac{p}{q} where q \neq 0$ . Integers are negative and positive numbers which are not in $\frac{p}{q}$ $\frac{p}{q}$ form.
For example, $\frac{1}{2}$ $\frac{1}{2}$ is a rational number but not an integer.

(vi) Every rational number is a whole number.
False, as rational numbers are of the form $\frac{p}{q} where q \neq 0$ $\frac{p}{q} where q \neq 0$ . Whole numbers are natural numbers together with a zero.
For example, $\frac{5}{7}$ $\frac{5}{7}$ is a rational number but not a whole number.

RS AGGARWAL CLASS 9 Chapter 1 Number System Exercise 1A