====== Exercise 2.1 (Solutions) ======

====Question 1====
Identify which of the following are rational and irrational numbers:

(i) $\sqrt{3}$	(ii) $\frac{1}{6}$	(iii) $\pi$	(iv) $\frac{15}{2}$	(v) $7.25$	(vi)$\sqrt{29}$ 

**Solution**\\

  * Rational: $\frac{1}{6}$, $\frac{15}{2}$, $7.25$
  * Irrational: $\sqrt{3}$, $\pi$, $\sqrt{29}$

====Question 2====
Convert the following fraction into decimal fraction.

(i) $\frac{17}{25}$		(ii) $\frac{19}{4}$		(iii)$\frac{57}{8}$	
	
(iv) $\frac{205}{18}$		(v) $\frac{5}{8}$			(vi) $\frac{25}{38}$

**Soluation**\\

  * (i) o.68
  * (ii) 4.75
  * (iii) 7.125
  * (iv) 11.3889
  * (v) 0.625
  * (vi) .65789

====Question 3====
Which of the statements are true and which of the false?
 
  * (i) $\frac{2}{3}$ is an irrational number.
  * (ii) $\pi$ is an irrational number.
  * (iii) $\frac{1}{9}$ is a terminating fraction.
  * (iv) $\frac{3}{4}$ is a terminating fraction.
  * (v) $\frac{4}{5}$ is a recurring fraction.

**Soluaton**\\

  * (i) $\frac{2}{3}$ is an irrational number.	**False**
  * (ii) $\pi$ is an irrational number.		**True**
  * (iii) $\frac{1}{9}$ is a terminating fraction.	**False**
  * (iv) $\frac{3}{4}$ is a terminating fraction.	**True**
  * (v) $\frac{4}{5}$ is a recurring fraction.	**False**

====Question 4====
Represent the following numbers on the number line.

(i) $\frac{2}{3}$		(ii) $-\frac{4}{5}$		(iii) $\frac{3}{4}$

(iv) $-2\frac{5}{8}$		(v) $2\frac{3}{4}$		(vi) $\sqrt{5}$

**Soluation**\\

 (i)  
  - To represent the rational number $\frac{2}{3}$ , divide unit length between 0 and 1 into 3 equal parts.
  - (ii) Take 2 parts on right of 0
  - (iii) Point  M represents $\frac{2}{3}$ on the number line in the following figure.










(ii)

  - To represent the rational number $-\frac{4}{5}$ , divide unit length between 0 and -1 into 5 equal parts.
  - Take 4 parts on right of 0
  - Point  M represents $-\frac{4}{5}$ on the number line in the following figure.












(iii)
  - Rational number $\frac{3}{4}$ lies between 1 and 2 on the number line.
  - The distance between 1 and 2 is divided into 4 equal parts , from L we take 3 parts.
  - Point M represent $\frac{3}{4}$ on the number line.














(iv)
  - Rational number $-2\frac{5}{8}$ lies between 2 and -3 on the number line.
  - The distance between 2 and -3 is divided into 8 equal parts , from 2 we take 5 parts.











(v)
  - Rational number $2\frac{3}{4}$ lies between 2 and 3 on the number line.
  - The distance between 2 and 3 is divided into 4 equal parts , from 2 we take 3 parts.

 














(vi)

  - (i)	Construct a $\Delta OAB$ which $m \overline{OA} = 2$ unit and $\perp BA = 1$
  - (ii) Take O as centre and draw an arc of radius $\overline{OB}$. It cuts the number line at M. $m \overline{OM} = \sqrt{5}$

$$(m OB)^2 = (m OA)^2 + (m AB)^2 \quad	(Pythagorean\,\, Theorem)$$
$$(m \overline{OB})^2 = 4 + 1 = 5 $$
therefore
$$m \overline{OB} = \sqrt{5}$$
thus 
$$m \overline{OB} = m \overline{OM} = \sqrt{5}$$












====Question 5====
Give a rational number between $\frac{3}{4}$ and $\frac{5}{9}$.

**Soluation**\\

The mean of the numbers is between given numbers.
Therefore $\frac{3/4+5/9}{2}= \frac{47}{72}$ is a number between required numbers.

====Question 6====
Express the  following recurring decimals as the rational number $\frac{p}{q}$, where p , q are integers and $q \neq 0$.

(i) $0.\overline{5}$		(ii)	$0.\overline{13}$		(iii)	$0.\overline{67}$

**Soluation**\\

(i)
Let		
$x = 0.\overline{5}$.
That is
$$x = 0.5555....\qquad (1)$$

Only one digit 5 is being repeated, multiply 10 on both sides

$$10x = (0.5555.....) \times 10$$
$$10x = 5.5555.....	\qquad	(2)$$
Subtract (1) from (2), we get

$$9x = 5$$
$$\therefore \,\,  x = \frac{5}{9}$$
$$0.\overline{5} = \frac{5}{9}$$

(ii) Let $x = 0.\overline{13}$. That is
$$x = 0.13131313....	\qquad	(1)$$
Only one digit 13 is being repeated, multiply 100 on both sides
$$100x = (0.13131313.....) \times 100$$
$$100x = 13.13131313.....	\qquad	(2)$$
Subtract (1) from (2)
$$99x = 13$$
$$\therefore \,\,  x = \frac{13}{99}$$
$$0.\overline{13} = \frac{13}{99}$$

(iii) Let $x = 0.\overline{67}$. That is
$$x = 0.67676767....	\qquad	(1)$$
Only one digit 67 is being repeated, multiply 100 on both sides
$$100x = (0.67676767.....) \times 100$$
$$100x = 67.67676767.....	\qquad	(2)$$
Subtract (1) from (2)
$$99x = 67$$
$$\therefore\,\,   x = \frac{67}{99}$$
$$0.\overline{67} = \frac{67}{99}$$