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}\mathrm{and}\frac{2}{5}$$\frac{3}{8} \mathrm{and} \frac{2}{5}$
(ii) 1.3 and 1.4
(iii) $-1\mathrm{and}\frac{1}{2}$$-1 \mathrm{and} \frac{1}{2}$
(iv) $-\frac{3}{4}\mathrm{and}-\frac{2}{5}$$-\frac{3}{4} \mathrm{and}-\frac{2}{5}$
(v) $\frac{1}{9}\mathrm{and}\frac{2}{9}$$\frac{1}{9} \mathrm{and} \frac{2}{9}$

Answer 3:

(i) $\frac{3}{8}\mathrm{and}\frac{2}{5}$$\frac{3}{8} \mathrm{and} \frac{2}{5}$
Let:
x = $\frac{3}{8}$$\frac{3}{8}$ and y = $\frac{2}{5}$$\frac{2}{5}$
Rational number lying between x and y:
$\frac{1}{2}\left(x+y\right)=\frac{1}{2}\left(\frac{3}{8}+\frac{2}{5}\right)$$\frac{1}{2}\left(x + y\right) = \frac{1}{2}\left(\frac{3}{8} + \frac{2}{5}\right)$
= $\frac{1}{2}\left(\frac{15+16}{40}\right)=\frac{31}{80}$$\frac{1}{2}\left(\frac{15+16}{40}\right) = \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}\left(x+y\right)=\frac{1}{2}\left(1.3+1.4\right)$$\frac{1}{2}\left(x + y\right) = \frac{1}{2}\left(1.3+1.4\right)$
= $\frac{1}{2}\left(2.7\right)=1.35$$\frac{1}{2}\left(2.7\right)= 1.35$

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

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

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

Question 4:

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

Answer 4:

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

Find four rational numbers between $\frac{3}{7}\mathrm{and}\frac{5}{7}$$\frac{3}{7} \mathrm{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}$$$\begin{gathered} \frac{3}{7}=\frac{3}{7}\times \frac{5}{5}=\frac{15}{35} \\ \frac{5}{7}=\frac{5}{7}\times \frac{5}{5}=\frac{25}{35} \end{gathered}$$
Thus, rational numbers between $\frac{3}{7}\mathrm{and}\frac{5}{7}$$\frac{3}{7} \mathrm{and} \frac{5}{7}$ are $\frac{16}{35},\frac{17}{35},\frac{18}{35},\frac{19}{35}$$\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}$$d=\frac{y-x}{n+1}=\frac{3-2}{6+1}=\frac{1}{7}$
Thus, the required numbers are 
$\left(x+d\right),\left(x+2d\right),\left(x+3d\right),...,\left(x+nd\right)=\left(2+\frac{1}{7}\right),\left(2+2\times \frac{1}{7}\right),\left(2+3\times \frac{1}{7}\right),\left(2+4\times \frac{1}{7}\right),\left(2+5\times \frac{1}{7}\right),\left(2+6\times \frac{1}{7}\right)=\frac{15}{7},\frac{16}{7},\frac{17}{7},\frac{18}{7},\frac{19}{7},\frac{20}{7}$$$\begin{gathered} \left(x+d\right),\left(x+2d\right),\left(x+3d\right),...,\left(x+nd\right) \\ =\left(2+\frac{1}{7}\right),\left(2+2\times \frac{1}{7}\right),\left(2+3\times \frac{1}{7}\right),\left(2+4\times \frac{1}{7}\right),\left(2+5\times \frac{1}{7}\right),\left(2+6\times \frac{1}{7}\right) \\ =\frac{15}{7},\frac{16}{7},\frac{17}{7},\frac{18}{7},\frac{19}{7},\frac{20}{7} \end{gathered}$$
 

Question 7:

Find five rational numbers between $\frac{3}{5}\mathrm{and}\frac{2}{3}$$\frac{3}{5} \mathrm{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{{\displaystyle \frac{2}{3}}-{\displaystyle \frac{3}{5}}}{6}=\frac{10-9}{90}=\frac{1}{90}$

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

 

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}\mathrm{where}q\ne 0$$\frac{p}{q} \mathrm{where} q\ne 0$. All integers can be represented in the form $\frac{p}{q}\mathrm{where}q\ne 0$$\frac{p}{q} \mathrm{where} q\ne 0$.
(v) Every rational number is an integer.
False, as rational numbers are of the form $\frac{p}{q}\mathrm{where}q\ne 0$$\frac{p}{q} \mathrm{where} q\ne 0$. Integers are negative and positive numbers which are not in $\frac{p}{q}$$\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}\mathrm{where}q\ne 0$$\frac{p}{q} \mathrm{where} q\ne 0$. Whole numbers are natural numbers together with a zero.
For example, $\frac{5}{7}$$\frac{5}{7}$ is a rational number but not a whole number.