Rational & Irrational Numbers

In rational numbers, if the denominator of the fraction can be expressed in the form of 2^m. × 5^n., then it is a terminating decimal.

(i) So, 25 can be expressed as 2^0. × 5^2., which is in the form of 2^m. × 5^n..

Hence, $\frac{9}{25}$ is a terminating decimal number.

(ii) So, 12 can be expressed as 3 × 2^2. × 5^0., which is not in the form of 2^m. × 5^n..

Hence, $\frac{7}{12}$ is not a terminating decimal number.

(iii) So, 16 can be expressed as 2^4. × 5^0., which is in the form of 2^m. × 5^n..

Hence, $\frac{13}{16}$ is a terminating decimal number.

(iv) So, 128 can be expressed as 2^7. × 5^0., which is in the form of 2^m. × 5^n..

Hence, $\frac{25}{128}$ is a terminating decimal number.

(v) So, 50 can be expressed as 2^1. × 5^2., which is in the form of 2^m. × 5^n..

Hence, $\frac{9}{50}$ is a terminating decimal number.

(vi) So, 125 can be expressed as 2^0. × 5^3., which is in the form of 2^m. × 5^n..

Hence, $\frac{121}{125}$ is a terminating decimal number.

(vii) So, 55 can be expressed as 11 × 2^0. × 5^1., which is not in the form of 2^m. × 5^n..

Hence, $\frac{19}{55}$ is not a terminating decimal number.

(viii) So, 78 can be expressed as 39 × 2^1. × 5^0., which is not in the form of 2^m. × 5^n..

Hence, it is not a terminating decimal number.

(ix) So, 80 can be expressed as 2^4. × 5^1., which is in the form of 2^m. × 5^n..

Hence, $\frac{23}{80}$ is a terminating decimal number.

(x) So, 30 can be expressed as 3 × 2^1. × 5^1., which is not in the form of 2^m. × 5^n..

Hence, $\frac{19}{30}$ is not a terminating decimal number.

(i) Given,

$\Rightarrow 0.65 = \frac{65}{100}$

The H.CF. of 65 and 100 is 5, dividing the numerator and denominator by 5,

$$\begin{aligned}\Rightarrow \frac{65 ÷ 5}{100 ÷ 5} \\[1em] \Rightarrow \frac{13}{20}.\end{aligned}$$

Hence, 0.65 = $\frac{13}{20}$.

(ii) Given,

$\Rightarrow 1.08 = \frac{108}{100}$

The H.CF. of 108 and 100 is 4, dividing the numerator and denominator by 4,

$$\begin{aligned}\Rightarrow \frac{108 ÷ 4}{100 ÷ 4} \\[1em] \Rightarrow \frac{27}{25}\end{aligned}$$

Hence, 1.08 = $\frac{27}{25}$.

(iii) Given,

$\Rightarrow 0.075 = \frac{75}{1000}$

The H.CF. of 75 and 1000 is 25, dividing the numerator and denominator by 25,

$$\begin{aligned}\Rightarrow \frac{75 ÷ 25}{1000 ÷ 25} \\[1em] \Rightarrow \frac{3}{40}\end{aligned}$$

Hence, 0.075 = $\frac{3}{40}$.

(iv) Given,

$\Rightarrow 2.016 = \frac{2016}{1000}$

The H.CF. of 2016 and 1000 is 8, dividing the numerator and denominator by 8,

$$\begin{aligned}\Rightarrow \frac{2016 ÷ 8}{1000 ÷ 8} \\[1em] \Rightarrow \frac{252}{125}\end{aligned}$$

Hence, 2.016 = $\frac{252}{125}$.

(v) Given,

$\Rightarrow 1.732 = \frac{1732}{1000}$

The H.CF. of 1732 and 1000 is 4, dividing the numerator and denominator by 4

$$\begin{aligned}\Rightarrow \frac{1732 ÷ 4}{1000 ÷ 4} \\[1em] \Rightarrow \frac{433}{250}\end{aligned}$$

Hence, 1.732 = $\frac{433}{250}$.

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

On dividing,

$\frac{1}{8} = 0.125$

Hence, $\frac{1}{8} = 0.125$.

(ii) $\frac{3}{32}$

On dividing,

$\frac{3}{32} = 0.09375$

Hence, $\frac{3}{32} = 0.09375$.

(iii) $\frac{44}{9}$

On dividing,

$\frac{44}{9} = 4.\overline{8}$

Hence, $\frac{44}{9} = 4.\overline{8}$.

(iv) $\frac{11}{24}$

On dividing,

$\frac{11}{24} = 0.458\overline{3}$

Hence, $\frac{11}{24} = 0.458\overline{3}$.

(v) $\frac{12}{13}$

On dividing,

$\frac{12}{13} = 0.\overline{923076}$

Hence, $\frac{12}{13} = 0.\overline{923076}$.

(vi) $\frac{27}{44}$

On dividing,

$\frac{27}{44} = 0.61\overline{36}$

Hence, $\frac{27}{44} = 0.61\overline{36}$.

(vii) $2\frac{5}{12}$

On dividing,

$2\frac{5}{12} = \frac{29}{12} = 2.41\overline{6}$

Hence, $\frac{29}{12} = 2.41\overline{6}$.

(viii) $1\frac{31}{55}$

On dividing,

$1\frac{31}{55} = \frac{86}{55} = 1.5\overline{63}$

Hence, $1\frac{31}{55} = 1.5\overline{63}.$.

On dividing,

$\frac{15}{56} = 0.26785$

Hence, $\frac{15}{56} = 0.2679$.

On dividing,

$\frac{13}{34} = 0.3823$

Hence, $\frac{13}{34} = 0.382$.

(i) On dividing,

$\frac{11}{9} = 1.222......$

Hence, proved that $\frac{11}{9} = 1.\overline{2}$.

(ii) On dividing,

$\frac{43}{11} = 3.9090...$

Hence, proved that $\frac{43}{11} = 3.\overline{90}$.

(iii) On dividing,

$\frac{107}{45} = 2.3777...$

Hence, proved that $\frac{107}{45} = 2.3\overline{7}$.

(iv) On dividing,

$\frac{21}{55} = 0.38181...$

Hence, proved that $\frac{21}{55} = 0.3\overline{81}$.

(i) $0.\overline{5}$

Let x = $0.\overline{5}$.

⇒ x = 0.555..     ...........(1)

⇒ 10x = 5.555..     ..........(2)

On subtracting (1) from (2), we get :

⇒ 10x - x = 5.555... - 0.555...

⇒ 9x = 5

⇒ x = $\frac{5}{9}$

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

(ii) $0.\overline{43}$

Let x = $0.\overline{43}$.

⇒ x = 0.434343..     .........(1)

⇒ 100x = 43.4343..     ........(2)

On subtracting (1) from (2), we get :

⇒ 100x - x = 43.4343.... - 0.4343......

⇒ 99x = 43

⇒ x = $\frac{43}{99}$

Hence, $0.\overline{43} = \frac{43}{99}$.

(iii) $0.\overline{158}$

Let x = $0.\overline{158}$

⇒ x = 0.158158..     ......(1)

⇒ 1000x = 158.158158..     ......(2)

On subtracting (1) from (2), we get :

⇒ 1000x - x = 158.158158.... - 0.158158.....

⇒ 999x = 158

⇒ x = $\frac{158}{999}$

Hence, $0.\overline{158} = \frac{158}{999}$.

(iv) $1.\overline{3}$

Let x = $1.\overline{3}$

⇒ x = 1.3333..     ......(1)

⇒ 10x = 13.333..     ......(2)

On subtracting (1) from (2), we get :

⇒ 10x - x = 13.333.. - 1.333..

⇒ 9x = 12

⇒ x = $\frac{12}{9}$

⇒ x = $\frac{4}{3}$

Hence, $1.\overline{3} = \frac{4}{3}$.

(v) $4.\overline{17}$

Let x = $4.\overline{17}$

⇒ x = 4.1717..     ........(1)

⇒ 100x = 417.1717..     .......(2)

On subtracting (1) from (2), we get :

⇒ 100x - x = 417.1717..... - 4.1717.....

⇒ 99x = 413

⇒ x = $\frac{413}{99}$

Hence, $4.\overline{17} = \frac{413}{99}$.

(vi) $0.\overline{12}$

Let x = $0.\overline{12}$

⇒ x = 0.1222..     ...........(1)

⇒ 100x = 12.222...(ii)

On subtracting (i) from (ii), we get

⇒ 99x = 12

⇒ x = $\frac{12}{99}$

⇒ x = $\frac{12÷3}{99÷3}$

⇒ x = $\frac{4}{33}$

Hence, 0.12... = $\frac{4}{33}$.

(vii) $0.1\overline{36}$

Let x = $0.1\overline{36}$

⇒ x = 0.13636..     ...........(1)

⇒ 10x = 1.3636..     ...........(2)

⇒ 1000x = 136.3636..     ...........(3)

On subtracting (2) from (3), we get

⇒ 1000x - 10x = 136.3636.. - 1.3636..

⇒ 990x = 135

⇒ x = $\frac{135}{990}$

⇒ x = $\frac{135 ÷ 45}{990 ÷ 45}$

⇒ x = $\frac{3}{22}$

Hence, $0.1\overline{36}\frac{3}{22}$.

(viii) $1.5\overline{7}$

Let x = $1.5\overline{7}$.

⇒ x = 1.5777..     ...........(1)

⇒ 10x = 15.777..     ...........(2)

⇒ 100x = 157.7777..     ...........(3)

On subtracting (i) from (iii), we get

⇒ 100x - 10x = 157.7777.. - 15.777......

⇒ 90x = 142

⇒ x = $\frac{142}{90}$

⇒ x = $\frac{142 ÷ 2}{90 ÷ 2}$

⇒ x = $\frac{71}{45}$

Hence, $1.5\overline{7} = \frac{71}{45}$.

The additive inverse of a number is the number which, when added to the original number, results in zero.

(i) Let x be the additive inverse of 5, then :

⇒ 5 + x = 0

⇒ x = -5.

Hence, additive inverse of 5 = -5.

(ii) Let x be the additive inverse of -7, then :

⇒ -7 + x = 0

⇒ x = 7.

Hence, additive inverse of -7 = 7.

(iii) Let x be the additive inverse of $\frac{5}{9}$, then :

⇒ $\frac{5}{9}$ + x = 0

⇒ x = $-\frac{5}{9}$.

Hence,additive inverse of $\frac{5}{9} = -\frac{5}{9}$.

(iv) Let x be the additive inverse of $-\frac{3}{17}$, then :

⇒ $-\frac{3}{17}$ + x = 0

⇒ x = $\frac{3}{17}$.

Hence,additive inverse of $-\frac{3}{17} = \frac{3}{17}$.

(v) Let x be the additive inverse of 0, then :

⇒ 0 + x = 0

⇒ x = 0.

Hence, additive inverse of 0 is 0.

(vi) Let x be the additive inverse of $11\frac{5}{17}$, then :

⇒ $11\frac{5}{17}$ + x = 0

⇒ $\frac{192}{17}$ + x = 0

⇒ x = $-\frac{192}{17}$.

Hence, additive inverse of $11\frac{5}{17} = -\frac{192}{17}$.

(vii) Let x be the additive inverse of $-5\frac{3}{8}$, then :

⇒ $-5\frac{3}{8}$ + x = 0

⇒ $-\frac{43}{8}$ + x = 0

⇒ x = $\frac{43}{8}$.

Hence, additive inverse of $-5\frac{3}{8} = \frac{43}{8}$.

(viii) Let x be the additive inverse of -37, then :

⇒ -37 + x = 0

⇒ x = 37.

Hence, additive inverse of -37 = 37.

(ix) Let x be the additive inverse of 1, then :

⇒ 1 + x = 0

⇒ x = -1.

Hence, additive inverse of 1 = -1.

The multiplicative inverse of a number is defined as a number that when multiplied by the original number gives the product as 1.

(i) Let the multiplicative inverse of 9, be x.

⇒ 9 × x = 1

⇒ x = $\frac{1}{9}$.

Hence, multiplicative inverse of 9 = $\frac{1}{9}$.

(ii) Let the multiplicative inverse of -1, be x.

⇒ -1 × x = 1

⇒ x = $-\frac{1}{1}$ = -1.

Hence, multiplicative inverse of -1 = -1.

(iii) Let the multiplicative inverse of $\frac{11}{16}$, be x.

⇒ $\frac{11}{16}$ × x = 1

⇒ x = $\frac{16}{11}$.

Hence, multiplicative inverse of $\frac{11}{16} = \frac{16}{11}$.

(iv) Let the multiplicative inverse of $5\frac{1}{4}$, be x.

⇒ $5\frac{1}{4}$ × x = 1

⇒ $\frac{21}{4}$ × x = 1

⇒ x = $\frac{4}{21}$.

Hence, multiplicative inverse of $5\frac{1}{4} = \frac{4}{21}$.

(v) Let the multiplicative inverse of $-\frac{2}{3}$, be x.

⇒ $-\frac{2}{3}$ × x = 1

⇒ x = $-\frac{3}{2}$.

Hence, multiplicative inverse of $-\frac{2}{3} = -\frac{3}{2}$.

(vi) Let the multiplicative inverse of $17\frac{3}{20}$, be x.

⇒ $17\frac{3}{20}$ × x = 1

⇒ $\frac{343}{20}$ × x = 1

⇒ x = $\frac{20}{343}$.

Hence, multiplicative inverse of $17\frac{3}{20} = \frac{20}{343}$.

(vii) Let the multiplicative inverse of $–18\frac{1}{2}$, be x.

⇒ $–18\frac{1}{2}$ × x = 1

⇒ $-\frac{37}{2}$ × x = 1

⇒ x = $–\frac{2}{37}$.

Hence, multiplicative inverse of $-18\frac{1}{2} = -\frac{2}{37}$.

(viii) Let the multiplicative inverse of –5, be x.

⇒ –5 × x = 1

⇒ x = $-\frac{1}{5}$.

Hence, multiplicative inverse of $-5 = -\frac{1}{5}$.

(ix) Let the multiplicative inverse of $\frac{-20}{41}$, be x.

⇒ $\frac{-20}{41}$ × x = 1

⇒ x = $-\frac{41}{20}$.

Hence, multiplicative inverse of $-\frac{20}{41} = -\frac{41}{20}$.

(i) On dividing,

$\frac{3}{7}$ = 0.428

(ii) On dividing,

$\frac{16}{5}$ = 3.2

(iii) On dividing,

$-\frac{4}{9}$ = -0.4444..

(iv) On dividing,

$-\frac{18}{11}$ = -1.6363..

(v) On dividing,

$-3\frac{1}{6} = -\frac{19}{6}$ = -3.166..

Let x be a rational number between $\frac{3}{5}$ and $\frac{7}{9}$.

$$\begin{aligned}\Rightarrow x = \frac{1}{2}\Big(\frac{3}{5} + \frac{7}{9}\Big) \\[1em] \Rightarrow x = \frac{1}{2}\Big(\frac{27 + 35}{45} \Big) \\[1em] \Rightarrow x = \frac{1}{2}\Big(\frac{62}{45} \Big) \\[1em] \Rightarrow x = \frac{31}{45} \\[1em]\end{aligned}$$

Hence, a rational number between $\frac{3}{5} \text{ and } \frac{7}{9} \text{ is } \frac{31}{45}$.

(i) Let the first rational number between 2 and 3 be x.

$$\begin{aligned}\Rightarrow x = \frac{1}{2}\Big(2 + 3\Big) \\[1em] \Rightarrow x = \frac{1}{2} \times 5 \\[1em] \Rightarrow x = \frac{5}{2} \\[1em]\end{aligned}$$

Let the second rational number be y.

$$\begin{aligned}\Rightarrow y = \frac{1}{2} \Big(\frac{5}{2} + 3\Big) \\[1em] \Rightarrow y = \frac{1}{2} \Big(\frac{5+6}{2}\Big) \\[1em] \Rightarrow y = \frac{1}{2} \Big(\frac{11}{2}\Big) \\[1em] \Rightarrow y = \frac{11}{4} \\[1em]\end{aligned}$$

Hence, two rational numbers between 2 and 3 are $\frac{5}{2} \text{ and } \frac{11}{4}$.

(ii) Let the first rational number between $\frac{1}{3}$ and $\frac{2}{5}$ be x.

$$\begin{aligned}\Rightarrow x = \frac{1}{2} \Big(\frac{1}{3} + \frac{2}{5}\Big) \\[1em] \Rightarrow x = \frac{1}{2} \Big(\frac{5 + 6}{15}\Big) \\[1em] \Rightarrow x = \frac{1}{2} \Big(\frac{11}{15} \Big) \\[1em] \Rightarrow x = \frac{11}{30}\end{aligned}$$

Let the second rational number be y.

$$\begin{aligned}\Rightarrow y = \frac{1}{2} \Big(\frac{11}{30} + \frac{2}{5}\Big) \\[1em] \Rightarrow y = \frac{1}{2} \Big(\frac{11 + 12}{30}\Big) \\[1em] \Rightarrow y = \frac{1}{2} \Big(\frac{23}{30}\Big) \\[1em] \Rightarrow y = \frac{23}{60} \\[1em]\end{aligned}$$

Hence, two rational numbers between $\frac{1}{3}$ and $\frac{2}{5}$ are $\frac{11}{30} \text{ and } \frac{23}{60}$ .

(iii) Let the first rational number between $\frac{3}{4}$ and $1\frac{1}{5}$ be x.

$$\begin{aligned}\Rightarrow x = \frac{1}{2}\Big(\frac{3}{4} + 1\frac{1}{5}\Big) \\[1em] \Rightarrow x = \frac{1}{2}\Big(\frac{3}{4} + \frac{6}{5}\Big) \\[1em] \Rightarrow x = \frac{1}{2}\Big(\frac{15 + 24}{20}\Big) \\[1em] \Rightarrow x = \frac{1}{2}\Big(\frac{39}{20} \Big) \\[1em] \Rightarrow x = \frac{39}{40}\end{aligned}$$

Let the second rational number be y.

$$\begin{aligned}\Rightarrow y = \frac{1}{2}\Big(\frac{39}{40} + \frac{6}{5}\Big) \\[1em] \Rightarrow y = \frac{1}{2}\Big(\frac{39 + 48}{40}\Big) \\[1em] \Rightarrow y = \frac{1}{2}\Big(\frac{87}{40}\Big) \\[1em] \Rightarrow y = \frac{87}{80} \\[1em]\end{aligned}$$

Hence, two rational numbers between $\frac{3}{4}$ and $1\frac{1}{5}$ are $\frac{39}{40} \text{ and } \frac{87}{80}$ .

(iv) Let the first rational number between -2 and 1 be x.

$$\begin{aligned}\Rightarrow x = \frac{1}{2}(-2 + 1) \\[1em] \Rightarrow x = \frac{1}{2} \times -1 \\[1em] \Rightarrow x = -\frac{1}{2} \\[1em]\end{aligned}$$

Let the second rational number be y.

$$\begin{aligned}\Rightarrow y = \frac{1}{2}\Big[-\frac{1}{2} + (-2)\Big] \\[1em] \Rightarrow y = \frac{1}{2}\Big(\frac{-1-4}{2}\Big) \\[1em] \Rightarrow y = \frac{1}{2} \times -\frac{5}{2} \\[1em] \Rightarrow y = -\frac{5}{4}.\end{aligned}$$

Hence, two rational numbers between -2 and 1 are $-\frac{1}{2} \text{ and } -\frac{5}{4}$.

(i) Let the first rational number between 4 and 5 be x.

$$\begin{aligned}\Rightarrow x = \frac{1}{2}(4 + 5) \\[1em] \Rightarrow x = \frac{1}{2} \times 9 \\[1em] \Rightarrow x = \frac{9}{2}.\end{aligned}$$

Let the second rational number be y.

$$\begin{aligned}\Rightarrow y = \frac{1}{2}\Big(\frac{9}{2} + 5\Big) \\[1em] \Rightarrow y = \frac{1}{2}\Big(\frac{9 + 10}{2}\Big) \\[1em] \Rightarrow y = \frac{1}{2}\Big(\frac{19}{2}\Big) \\[1em] \Rightarrow y = \frac{19}{4}\end{aligned}$$

Let the third rational number be z.

$$\begin{aligned}\Rightarrow z = \frac{1}{2}\Big(\frac{9}{2} + 4\Big) \\[1em] \Rightarrow z = \frac{1}{2}\Big(\frac{9 + 8}{2}\Big) \\[1em] \Rightarrow z = \frac{1}{2}\Big(\frac{17}{2}\Big) \\[1em] \Rightarrow z = \frac{17}{4}\end{aligned}$$

Hence, three rational numbers between 4 and 5 are $\frac{17}{4}, \frac{9}{2} \text{ and }\frac{19}{4}$.

(ii) Let the first rational number between $\frac{1}{2}$ and $\frac{3}{5}$ be x.

$$\begin{aligned}\Rightarrow x = \frac{1}{2}\Big(\frac{1}{2} + \frac{3}{5}\Big)\\[1em] \Rightarrow x = \frac{1}{2}\Big(\frac{5 + 6}{10}\Big)\\[1em] \Rightarrow x = \frac{1}{2} \times \frac{11}{10} \\[1em] \Rightarrow x = \frac{11}{20}.\end{aligned}$$

Let the second rational number be y.

$$\begin{aligned}\Rightarrow y = \frac{1}{2}\Big(\frac{11}{20} + \frac{3}{5}\Big) \\[1em] \Rightarrow y = \frac{1}{2}\Big(\frac{11 + 12}{20}\Big) \\[1em] \Rightarrow y = \frac{1}{2} \Big(\frac{23}{20}\Big) \\[1em] \Rightarrow y = \frac{23}{40}.\end{aligned}$$

Let the third rational number be z.

$$\begin{aligned}\Rightarrow z = \frac{1}{2}\Big(\frac{11}{20} + \frac{1}{2}\Big) \\[1em] \Rightarrow z = \frac{1}{2}\Big(\frac{11 + 10}{20}\Big) \\[1em] \Rightarrow z = \frac{1}{2} \times \frac{21}{20} \\[1em] \Rightarrow z = \frac{21}{40} \\[1em]\end{aligned}$$

Hence, three rational numbers between $\frac{1}{2}$ and $\frac{3}{5}$ are $\frac{21}{40}, \frac{11}{20} \text{ and } \frac{23}{40}$.

(iii) Let the first rational number between -1 and 1 be x.

$$\begin{aligned}\Rightarrow x = \frac{1}{2}(-1 + 1) \\[1em] \Rightarrow x = \frac{1}{2} \times 0 \\[1em] \Rightarrow x = 0.\end{aligned}$$

Let the second rational number be y.

$$\begin{aligned}\Rightarrow y = \frac{1}{2}(0 + 1) \\[1em] \Rightarrow y = \frac{1}{2} \times 1 \\[1em] \Rightarrow y = \frac{1}{2}.\end{aligned}$$

Let the third rational number be z.

$$\begin{aligned}\Rightarrow z = \frac{1}{2}[0 + (-1)] \\[1em] \Rightarrow z = \frac{1}{2} \times -1 \\[1em] \Rightarrow z = -\frac{1}{2}\end{aligned}$$

Hence, three rational numbers between -1 and 1 are $-\frac{1}{2}, 0 \text{ and } \frac{1}{2}$.

(iv) Let the first rational number between $2\frac{1}{3}$ and $3\frac{2}{3}$ be x.

$$\begin{aligned}\Rightarrow x = \frac{1}{2}\Big(2\frac{1}{3} + 3\frac{2}{3}\Big) \\[1em] \Rightarrow x = \frac{1}{2}\Big(\frac{7}{3} + \frac{11}{3}\Big) \\[1em] \Rightarrow x = \frac{1}{2} \times \frac{18}{3} \\[1em] \Rightarrow x = \frac{1}{2} \times 6 \\[1em] \Rightarrow x = 3\end{aligned}$$

Let the second rational number be y.

$$\begin{aligned}\Rightarrow y = \frac{1}{2}\Big(3 + \frac{11}{3}\Big) \\[1em] \Rightarrow y = \frac{1}{2}\Big(\frac{9 + 11}{3}\Big) \\[1em] \Rightarrow y = \frac{1}{2} \times \frac{20}{3} \\[1em] \Rightarrow y = \frac{10}{3}.\end{aligned}$$

Let the third rational number be z.

$$\begin{aligned}\Rightarrow z = \frac{1}{2}\Big(3 + \frac{7}{3}\Big) \\[1em] \Rightarrow z = \frac{1}{2}\Big(\frac{9 + 7}{3}\Big) \\[1em] \Rightarrow z = \frac{1}{2} \times \frac{16}{3} \\[1em] \Rightarrow z = \frac{8}{3}.\end{aligned}$$

Hence, three rational numbers between $\frac{7}{3}$ and $\frac{11}{3}$ are $\frac{8}{3}, 3 \text{ and } \frac{10}{3}$.

(v) Let the first rational number between $-\frac{1}{2}$ and $\frac{1}{3}$ be x.

$$\begin{aligned}\Rightarrow x = \frac{1}{2}\Big(-\frac{1}{2} + \frac{1}{3}\Big) \\[1em] \Rightarrow x = \frac{1}{2}\Big(\frac{-3 + 2}{6}\Big) \\[1em] \Rightarrow x = \frac{1}{2}\Big(\frac{-1}{6} \Big) \\[1em] \Rightarrow x = \frac{-1}{12}\end{aligned}$$

Let the second rational number be y.

$$\begin{aligned}\Rightarrow y = \frac{1}{2}\Big(\frac{-1}{12} + \frac{-1}{2}\Big) \\[1em] \Rightarrow y = \frac{1}{2}\Big(\frac{-1 - 6}{12}\Big) \\[1em] \Rightarrow y = \frac{1}{2}\Big(\frac{-7}{12}\Big) \\[1em] \Rightarrow y = -\frac{7}{24} \\[1em]\end{aligned}$$

Let the third rational number be z.

$$\begin{aligned}\Rightarrow z = \frac{1}{2}\Big(\frac{-1}{12} + \frac{1}{3}\Big) \\[1em] \Rightarrow z = \frac{1}{2}\Big(\frac{-1 + 4}{12}\Big) \\[1em] \Rightarrow z = \frac{1}{2}\Big(\frac{3}{12}\Big) \\[1em] \Rightarrow z = \frac{1}{8} \\[1em]\end{aligned}$$

Hence, three rational numbers between $\frac{-1}{2}$ and $\frac{1}{3}$ are $\frac{-7}{24}, \frac{-1}{12} \text{ and } \frac{1}{8}$.

(vi) Let the first rational number between $-\frac{1}{3}$ and $\frac{1}{4}$ be x.

$$\begin{aligned}\Rightarrow x = \frac{1}{2}\Big(-\frac{1}{3} + \frac{1}{4}\Big) \\[1em] \Rightarrow x = \frac{1}{2}\Big(\frac{-4 + 3}{12}\Big) \\[1em] \Rightarrow x = \frac{1}{2}\Big(\frac{-1}{12} \Big) \\[1em] \Rightarrow x = -\frac{1}{24}.\end{aligned}$$

Let the second rational number be y.

$$\begin{aligned}\Rightarrow y = \frac{1}{2} \Big(\frac{-1}{24} + \frac{-1}{3}\Big) \\[1em] \Rightarrow y = \frac{1}{2}\Big(\frac{-1 - 8}{24}\Big) \\[1em] \Rightarrow y = \frac{1}{2} \times \frac{-9}{24} \\[1em] \Rightarrow y = -\frac{9}{48} = -\frac{3}{16}.\end{aligned}$$

Let the third rational number be z.

$$\begin{aligned}\Rightarrow z = \frac{1}{2}\Big(\frac{-1}{24} + \frac{1}{4}\Big) \\[1em] \Rightarrow z = \frac{1}{2}\Big(\frac{-1+6}{24} \Big) \\[1em] \Rightarrow z = \frac{1}{2} \times \frac{5}{24} \\[1em] \Rightarrow z = \frac{5}{48}\end{aligned}$$

Hence, three rational numbers between $-\frac{1}{3}$ and $\frac{1}{4}$ are $-\frac{3}{16}, -\frac{1}{24} \text{ and } \frac{5}{48}$.

Let a = 4, b = 4.5 and n = 4

Difference between consecutive rational numbers =

$\frac{b - a}{n + 1} = \frac{4.5 - 4}{4 + 1} = \frac{0.5}{5} = 0.1$

Rational numbers between 4 and 4.5 are :

⇒ a + d, a + 2d, a + 3d, a + 4d

⇒ 4 + 0.1, 4 + 0.2, 4 + 0.3, 4 + 0.4

⇒ 4.1, 4.2, 4.3, 4.4

Hence, four rational numbers between 4 and 4.5 are 4.1, 4.2, 4.3 and 4.4.

Let a = 3, b = 4 and n = 9

Difference between consecutive rational numbers =

$\frac{b - a}{n + 1} = \frac{4 - 3}{9 + 1} = \frac{1}{10} = 0.1$

Rational numbers between 3 and 4 are :

⇒ a + d, a + 2d, a + 3d, a + 4d, a + 5d, a + 6d

⇒ 3 + 0.1. 3 + 0.2, 3 + 0.3, 3 + 0.4, 3 + 0.5, 3 + 0.6

⇒ 3.1, 3.2, 3.3, 3.4, 3.5, 3.6

Hence, six rational numbers between 3 and 4 are 3.1, 3.2, 3.3, 3.4, 3.5, 3.6.

(i) 5 can be expressed in the form $\frac{p}{q}$, where p and q are integers and q ≠ 0.

Hence, 5 is a rational number.

(ii) $\frac{9}{14}$ can be expressed in the form $\frac{p}{q}$, where p and q are integers and q ≠ 0.

Hence, $\frac{9}{14}$ is a rational number.

(iii) $\sqrt{3}$ is square root of non-perfect square i.e. 3.

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

(iv) π is a non-terminating and non-repeating decimal.

Hence, π is a irrational number.

(v) 3.1416 is a terminating decimal, so it can be expressed in the form of $\frac{p}{q}$, where p and q are integers and q ≠ 0.

Hence, 3.1416 is a rational number.

(vi) $\sqrt{4}$ is square root of perfect square i.e. 4.

$\sqrt{4} = 2 = \frac{2}{1}$.

Hence, $\sqrt{4}$ is a rational number.

(vii) $-\sqrt{5}$ is square root of non-perfect square.

Hence, $-\sqrt{5}$ is an irrational number.

(viii) Given,

$\sqrt[3]{8} = 2 = \frac{2}{1}$.

∴ $\sqrt[3]{8}$ can be expressed in the form of $\frac{p}{q}$, where p and q are integers and q ≠ 0.

Hence, $\sqrt[3]{8}$ is a rational number.

(ix) $\sqrt[3]{3}$ is cube root of non-perfect cube.

Hence, $\sqrt[3]{3}$ is an irrational number.

(x) $2\sqrt{6}$

Here, $\sqrt{6}$ is square root of a non-perfect square i.e. 6, thus it is an irrational number.

The product of a non-zero rational number and an irrational number is always an irrational number.

Hence, $2\sqrt{6}$ is an irrational number.

(xi) $0.\overline{36}$ is a repeating decimal.

Thus, $0.\overline{36}$ can be expressed as a fraction with an integer numerator and a non-zero integer denominator.

Hence, $0.\overline{36}$ is a rational number.

(xii) 0.2022022220... is a non-terminating and non-repeating decimal.

Hence, 0.2022022220... is an irrational number.

(xiii) $\frac{2}{\sqrt{3}}$.

2 is rational number and $\sqrt{3}$ is an irrational number.

Since, on dividing a rational number by irrational number the solution is always an irrational number.

Hence, $\frac{2}{\sqrt{3}}$ is an irrational number.

(xiv) $\frac{22}{7}$ can be expressed in the form of $\frac{p}{q}$, where p and q are integers and q ≠ 0.

Hence, $\frac{22}{7}$ is a rational number.

(i) -8 can be expressed in the form of $\frac{p}{q}$, where p and q are integers and q ≠ 0.

Hence, -8 is a rational number.

(ii) $\sqrt{25} = 5 = \frac{5}{1}$

Thus, $\sqrt{25}$ can be expressed in the form of $\frac{p}{q}$, where p and q are integers and q ≠ 0.

Hence, $\sqrt{25}$ is a rational number.

(iii) $\frac{-3}{5}$ can be expressed in the form of $\frac{p}{q}$, where p and q are integers and q ≠ 0.

Hence, $\frac{-3}{5}$ is a rational number.

(iv) $\sqrt{8}$ is square root of non-perfect square i.e. 8.

Hence, $\sqrt{8}$ is an irrational number.

(v) 0 can be expressed in the form of $\frac{p}{q}$, where p and q are integers and q ≠ 0.

Hence, 0 is a rational number.

(vi) π is a non-terminating and non-repeating decimal.

Hence, π is an irrational number.

(vii) $\sqrt[3]{5}$ is cube root of non-perfect cube i.e. 5.

Hence, $\sqrt[3]{5}$ is an irrational number.

(viii) $2.\overline{4}$ is a repeating decimal.

Hence, $2.\overline{4}$ is a rational number.

(ix) $-\sqrt{3}$ is square root of non-perfect square i.e. 3.

Hence, $-\sqrt{3}$ is an irrational number.

(i) $\sqrt{3}$ = 1.732..

(ii) $\sqrt{5}$ = 2.236..

(iii) $\sqrt{6}$ = 2.449..

(iv) $\sqrt{10}$ = 3.162..

(i) Solving,

$\Rightarrow (2\sqrt{3})^2$

$\Rightarrow 2\sqrt{3} \times 2\sqrt{3}$

⇒ 4 × 3

⇒ 12.

Hence, $(2\sqrt{3})^2$ = 12.

(ii) Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{3}{2}\sqrt{2}\Big)^2 \\[1em] \Rightarrow \frac{3}{2}\sqrt{2} \times \frac{3}{2}\sqrt{2} \\[1em] \Rightarrow \frac{9}{4} \times 2 \\[1em] \Rightarrow \frac{9}{2}.\end{aligned}$$

Hence, $\Big(\frac{3}{2}\sqrt{2}\Big)^2 = \frac{9}{2}$.

(iii) Solving,

$$\begin{aligned}\Rightarrow (5 + \sqrt{3})^2 \\[1em] \Rightarrow (5)^2 + (\sqrt{3})^2 + 2 \times 5 \times \sqrt{3} \\[1em] \Rightarrow 25 + 3 + 10\sqrt{3} \\[1em] \Rightarrow 28 + 10\sqrt{3}\end{aligned}$$

Hence, $(5 + \sqrt{3})^2 = 28 + 10\sqrt{3}$.

(iv) Solving,

$$\begin{aligned}\Rightarrow (\sqrt{6} - 3)^2 \\[1em] \Rightarrow (\sqrt{6})^2 + (3)^2 - 2 \times 3 \times \sqrt{6} \\[1em] \Rightarrow 6 + 9 - 6\sqrt{6} \\[1em] \Rightarrow 15 - 6\sqrt{6}\end{aligned}$$

Hence, $(\sqrt{6} - 3)^2 = 15 - 6\sqrt{6}$.

(v) Solving,

$$\begin{aligned}\Rightarrow (3 + 2\sqrt{5})^2 \\[1em] \Rightarrow (3)^2 + (2\sqrt{5})^2 + 2 \times 3 \times 2\sqrt{5} \\[1em] \Rightarrow 9 + 4 \times 5 + 12\sqrt{5} \\[1em] \Rightarrow 9 + 20 + 12\sqrt{5} \\[1em] \Rightarrow 29 + 12\sqrt{5}\end{aligned}$$

Hence, $(3 + 2\sqrt{5})^2 = 29 + 12\sqrt{5}$.

(vi) Solving,

$$\begin{aligned}\Rightarrow (\sqrt{5} + \sqrt{6})^2 \\[1em] \Rightarrow (\sqrt{5})^2 + (\sqrt{6})^2 + 2 \times \sqrt{5} \times \sqrt{6} \\[1em] \Rightarrow 5 + 6 + 2\sqrt{30} \\[1em] \Rightarrow 11 + 2\sqrt{30}\end{aligned}$$

Hence, $(\sqrt{5} + \sqrt{6})^2 = 11 + 2\sqrt{30}$.

(vii) Solving,

$$\begin{aligned}\Rightarrow \Big(\frac{3}{2\sqrt{2}}\Big)^2 \\[1em] \Rightarrow \frac{3}{2\sqrt{2}} \times \frac{3}{2\sqrt{2}} \\[1em] \Rightarrow \frac{9}{4 \times 2} \\[1em] \Rightarrow \frac{9}{8}\end{aligned}$$

Hence, $\Big(\frac{3}{2\sqrt{2}}\Big)^2 = \frac{9}{8}$.

(viii) Solving,

$$\begin{aligned}\Rightarrow (5 - 6\sqrt{3})^2 \\[1em] \Rightarrow (5)^2 + (6\sqrt{3})^2 - 2 \times 5 \times 6\sqrt{3} \\[1em] \Rightarrow 25 + 108 - 60\sqrt{3} \\[1em] \Rightarrow 133 - 60\sqrt{3}\end{aligned}$$

Hence, $(5 - 6\sqrt{3})^2 = 133 + 60\sqrt{3}$.

(i) Given,

$(3 + \sqrt{5})$

3 is a rational number as it can be expressed in the form of $\frac{p}{q}$, where p and q are integers and q ≠ 0.

$\sqrt{5}$ is an irrational number as it is a square root of a non-perfect square i.e. 5.

The sum of a rational number and an irrational number is always irrational.

Hence, $(3 + \sqrt{5})$ is a irrational number.

(ii) Given,

$(-1 + \sqrt{3})$

-1 is a rational number as it can be expressed in the form of $\frac{p}{q}$, where p and q are integers and q ≠ 0.

$\sqrt{3}$ is an irrational number as it is a square root of a non-perfect square i.e. 3.

The sum of a rational number and an irrational number is always irrational.

Hence, $(-1 + \sqrt{3})$ is a irrational number.

(iii) Given,

$5\sqrt{6}$

5 is a rational number as it can be expressed in the form of $\frac{p}{q}$, where p and q are integers and q ≠ 0.

$\sqrt{6}$ is an irrational number as it is a square root of a non-perfect square i.e. 6.

The product of a rational number and an irrational number is always irrational.

Hence, $5\sqrt{6}$ is an irrational number.

(iv) Given,

$\sqrt{7}$, is an irrational number as it is a square root of a non-perfect square i.e. 7.

∴ $-\sqrt{7}$ is an irrational number.

Hence, $-\sqrt{7}$ is an irrational number.

(v) Given,

$\frac{\sqrt{6}}{4}$

4 is a rational number as it can be expressed in the form of $\frac{p}{q}$, where p and q are integers and q ≠ 0.

$\sqrt{6}$ is an irrational number as it is a square root of a non-perfect square i.e. 6.

The division of a rational number and an irrational number is always irrational.

Hence, $\frac{\sqrt{6}}{4}$ is an irrational number.

(vi) Given,

$\frac{3}{\sqrt{2}}$

3 is a rational number as it can be expressed in the form of $\frac{p}{q}$, where p and q are integers and q ≠ 0.

$\sqrt{2}$ is an irrational number as it is a square root of a non-perfect square i.e. 2.

The division of a rational number and an irrational number is always irrational.

Hence, $\frac{3}{\sqrt{2}}$ is an irrational number.

(vii) Given,

$$\begin{aligned}\Rightarrow (3 + \sqrt{3}) (3 - \sqrt{3}) \\[1em] \Rightarrow (3)^2- (\sqrt{3})^2 \\[1em] \Rightarrow 9 - 3 \\[1em] \Rightarrow 3\end{aligned}$$

3 is a rational number as it can be expressed in the form of $\frac{p}{q}$, where p and q are integers and q ≠ 0.

Hence, $(3 + \sqrt{3})(3 - \sqrt{3})$ is a rational number.

(i) Given,

$$\begin{aligned}\Rightarrow \Big(2 + \sqrt{5}\Big)^2 \\[1em] \Rightarrow (2)^2 + (\sqrt{5})^2 + 2 \times 2 \times \sqrt{5} \\[1em] \Rightarrow 4 + 5 + 4\sqrt{5} \\[1em] \Rightarrow 9 + 4\sqrt{5}.\end{aligned}$$

9 is a rational number as it can be expressed in the form of $\frac{p}{q}$, where p and q are integers and q ≠ 0.

4 is a rational number as it can be expressed in the form of $\frac{p}{q}$, where p and q are integers and q ≠ 0.

$\sqrt{5}$ is an irrational number as it is a square root of a non-perfect square i.e. 5.

The product of a rational number and an irrational number is always irrational. i.e. $4\sqrt{5}$

The sum of a rational number and an irrational number is always irrational. i.e. $9 + 4\sqrt{5}$

Hence, $\Big(2 + \sqrt{5}\Big)^2$ is an irrational number.

(ii) Given,

$$\begin{aligned}\Rightarrow \Big(3 - \sqrt{3}\Big)^2 \\[1em] \Rightarrow (3)^2 + (\sqrt{3})^2 - 2 \times 3 \times \sqrt{3} \\[1em] \Rightarrow 9 + 3 - 6 \sqrt{3} \\[1em] \Rightarrow 12 - 6\sqrt{3} \\[1em]\end{aligned}$$

12 is a rational number as it can be expressed in the form of $\frac{p}{q}$, where p and q are integers and q ≠ 0.

6 is a rational number as it can be expressed in the form of $\frac{p}{q}$, where p and q are integers and q ≠ 0.

$\sqrt{3}$ is an irrational number as it is a square root of a non-perfect square i.e. 3.

The product of a rational number and an irrational number is always irrational. i.e. $6\sqrt{3}$

The difference between a rational number and an irrational number is always irrational. i.e. $12 - 6\sqrt{3}$.

Hence, $\Big(3 - \sqrt{3}\Big)^2$ is an irrational number.

(iii) Given,

$$\begin{aligned}\Rightarrow \Big(\sqrt{5} + \sqrt{3}\Big)^2 \\[1em] \Rightarrow (\sqrt{5})^2 + (\sqrt{3})^2 + 2 \times \sqrt{5} \times \sqrt{3} \\[1em] \Rightarrow 5 + 3 + 2\sqrt{15} \\[1em] \Rightarrow 8 + 2\sqrt{15} \\[1em]\end{aligned}$$

8 is a rational number as it can be expressed in the form of $\frac{p}{q}$, where p and q are integers and q ≠ 0.

2 is a rational number as it can be expressed in the form of $\frac{p}{q}$, where p and q are integers and q ≠ 0.

$\sqrt{15}$ is an irrational number as it is a square root of a non-perfect square i.e. 15.

The product of a rational number and an irrational number is always irrational. i.e. $2\sqrt{15}$

The sum of a rational number and an irrational number is always irrational. i.e. $8 + 2\sqrt{15}$

Hence, $\Big(\sqrt{5} + \sqrt{3}\Big)^2$ is an irrational number.

(iv) Given,

$\frac{6}{\sqrt{3}}$

6 is a rational number as it can be expressed in the form of $\frac{p}{q}$, where p and q are integers and q ≠ 0.

$\sqrt{3}$ is an irrational number as it is a square root of a non-perfect square i.e. 3.

The division of a rational number and an irrational number is always irrational. i.e. $\frac{6}{\sqrt{3}}$

Hence, $\frac{6}{\sqrt{3}}$ is an irrational number.

Let $\sqrt{5}$ be rational.

Thus, $\sqrt{5}$ can be expressed in the form of $\frac{p}{q}$.

$$\begin{aligned}\Rightarrow \sqrt{5} = \frac{p}{q} \\[1em] \Rightarrow \sqrt{5}q = p \\[1em] \text{Squaring both sides, we get : } \\[1em] \Rightarrow (\sqrt{5}q)^2 = p^2 \\[1em] \Rightarrow 5q^2 = p^2 \text{ .......(1)}\end{aligned}$$

As 5 divides 5q², so 5 divides p² but 5 is prime,

Thus, 5 divides p.

Let p = 5m for some positive integer m.

Then, p = 5m

Substituting this value of p in (1), we get :

$$\begin{aligned}\Rightarrow 5q^2 = (5m)^2 \\[1em] \Rightarrow 5q^2 = 25m^2 \\[1em] \Rightarrow q^2 = 5m^2\end{aligned}$$

As 5 divides 5m², so 5 divides q² but 5 is prime.

Thus, 5 divides q.

This shows that 5 is a common factor of p and q. This contradicts the hypothesis that p and q have no common factor, other than 1.

$\therefore \sqrt{5}$ is not a rational number.

Hence, proved that $\sqrt{5}$ is a irrational number.

We know that,

Square root of non-perfect squares are irrational numbers.

Since, 2, 3, 5 and 6 are non-perfect squares.

∴ $\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{6}$ are irrational numbers.

Hence, $\sqrt{2}, \sqrt{3}, \sqrt{5}, \sqrt{6}$ are 4 distinct irrational numbers.

Let us assume $\sqrt{3} + \sqrt{7}$ is a rational number.

Let, $\sqrt{3} + \sqrt{7} = x$

Squaring both sides, we get :

$$\begin{aligned}\Rightarrow (\sqrt{3} + \sqrt{7})^2 = x^2 \\[1em] \Rightarrow (\sqrt{3})^2 + (\sqrt{7})^2 + 2 \times \sqrt{3} \times \sqrt{7} = x^2 \\[1em] \Rightarrow 3 + 7 + 2\sqrt{21} = x^2 \\[1em] \Rightarrow 10 + 2\sqrt{21} = x^2 \\[1em] \Rightarrow \sqrt{21} = \frac{x^2 -10}{2}.\end{aligned}$$

Here, x is rational,

∴ x² is rational .........(1)

⇒ x² - 10 is rational (As difference between two rational numbers is always rational)

⇒ $\frac{x^2 - 10}{2}$ is rational (Dividing two rational numbers results in a rational number)

But, $\sqrt{21}$ is irrational.

$\therefore \frac{x^2 - 10}{2}$ is irrational.

Thus, x² - 10 is irrational and so x² is irrational ........(2)

(1) and (2) do not match with each other.

∴ We arrive at a contradiction.

So, our assumption that $\sqrt{3} + \sqrt{7}$ is a rational number is wrong.

∴ $\sqrt{3} + \sqrt{7}$ is irrational.

Hence, proved that $\sqrt{3} + \sqrt{7}$ is an irrational number.

Let us assume $\sqrt{2} + \sqrt{3}$ is a rational number.

Let, $(\sqrt{2} + \sqrt{3}) = x$

Squaring on both sides, we get :

$$\begin{aligned}\Rightarrow (\sqrt{2}+\sqrt{3})^2 = x^2 \\[1em] \Rightarrow (\sqrt{2})^2 + (\sqrt{3})^2 + 2 \times \sqrt{2} \times \sqrt{3} = x^2 \\[1em] \Rightarrow 2 + 3 + 2\sqrt{6} = x^2 \\[1em] \Rightarrow 5 + 2\sqrt{6} = x^2 \\[1em] \Rightarrow 2\sqrt{6} = x^2 - 5 \\[1em] \Rightarrow \sqrt{6} = \frac{x^2 - 5}{2}.\end{aligned}$$

Here, x is rational,

∴ x² is rational .........(1)

⇒ x² - 5 is rational (Difference between two rational numbers is always rational)

So, $\frac{x^2 - 5}{2}$ is rational (Dividing two rational numbers results in a rational number)

But $\sqrt{6}$ is irrational,

$\therefore \frac{x^2 - 5}{2}$ is irrational

Thus, x² - 5 is irrational and so x² is irrational ........(2)

(1) and (2) do not match with each other.

∴ We arrive at a contradiction.

So, our assumption that $\sqrt{2} + \sqrt{3}$ is a rational number is wrong.

∴ $\sqrt{2} + \sqrt{3}$ is irrational.

Hence, proved that $\sqrt{2} + \sqrt{3}$ is an irrational number.

We want two irrational numbers between $\sqrt{14} \text{ and } \sqrt{19}$.

Consider any two numbers between 14 and 19 such that they are not perfect squares.

Let us take 15 and 17 as they are not perfect squares.

We know that square root of a non-perfect square is an irrational number.

∴ $\sqrt{15} \text{ and } \sqrt{17}$ are irrational numbers.

Thus, we have :

$\Rightarrow \sqrt{14} \lt \sqrt{15} \lt \sqrt{17} \lt \sqrt{19}$

Hence, two irrational numbers between

$\sqrt{14} \text{ and } \sqrt{19} \text{ are } \sqrt{15} \text{ and } \sqrt{17}$.

We want three irrational numbers between $\sqrt{2} \text{ and } \sqrt{7}$.

Consider any three numbers between 2 and 7 such that they are not perfect squares.

Let us take 3, 5 and 6 as they are not perfect squares.

We know that square root of a non-perfect square is an irrational number.

∴ $\sqrt{3}, \sqrt{5} \text{ and } \sqrt{6}$ are irrational numbers.

Thus, we have :

$\Rightarrow \sqrt{2} \lt \sqrt{3} \lt \sqrt{5} \lt \sqrt{6} \lt\sqrt{7}$

Hence, three irrational numbers between

$\sqrt{2} \text{ and } \sqrt{7} \text{ are } \sqrt{3}, \sqrt{5} \text{ and } \sqrt{6}$.

(i) Adding two rational numbers will always result in rational number.

Hence, above statement is true.

(ii) Adding two irrational numbers can be rational as well as irrational.

Hence, above statement is false.

(iii) Multiplying two rational numbers will always result in rational number.

Hence, above statement is true.

(iv) Multiplying two irrational numbers can be rational as well as irrational.

Hence, above statement is false.

(v) Adding a rational and an irrational will always result irrational number.

Hence, above statement is true.

(iv) Multiplying a rational and an irrational can be rational as well as irrational.

Hence, above statement is false.

The numbers of the form $\frac{p}{q}$, where p and q are integers and q is not equal to zero, are called rational numbers.

Examples:

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

(ii) $-\frac{5}{8}$

(iii) 7

(iv) 0

(v) -9

(vi) $\frac{11}{2}$

(vii) -1.25

(viii) 3.75

(ix) 10000

(x) $\frac{-16}{7}$

A number which when expressed in decimal form is expressible as a non-terminating and non-repeating decimal, is called an irrational number.

Examples:

1. π

2. $\sqrt{2}$

3. $\sqrt{3}$

4. $\sqrt{5}$

5. $\sqrt{6}$

6. $\sqrt{7}$

7. $\sqrt{8}$

8. $\sqrt{10}$

9. $\sqrt{11}$

10. $\sqrt{12}$

(i) Rationalizing the denominator,

$$\begin{aligned}\Rightarrow \frac{2}{\sqrt{6}} \times \frac{ \sqrt{6}}{\sqrt{6}} \\[1em] \Rightarrow \frac{2(\sqrt{6})}{(\sqrt{6})^2} \\[1em] \Rightarrow \frac{2\sqrt{6}}{6} \\[1em] \Rightarrow \frac{\sqrt{6}}{3}\end{aligned}$$

Hence, on rationalizing $\frac{2}{\sqrt{6}} = \frac{\sqrt{6}}{3}$.

Rationalizing the denominator, $$\begin{aligned}\Rightarrow \frac{\sqrt{2}}{2\sqrt{3}} \times \frac{2\sqrt{3}}{2\sqrt{3}} \\[1em] \Rightarrow \frac{2(\sqrt{6})}{(2\sqrt{3})^2} \\[1em] \Rightarrow \frac{2\sqrt{6}}{(4 \times 3)} \\[1em] \Rightarrow \frac{2\sqrt{6}}{12} \\[1em] \Rightarrow \frac{\sqrt{6}}{6}.\end{aligned}$$

Hence, on rationalizing = $\frac{\sqrt{2}}{2\sqrt{3}} = \frac{\sqrt{6}}{6}$.

Rationalizing the denominator,

$$\begin{aligned}\Rightarrow \frac{1}{(3 + \sqrt{5})} \times \frac{(3 - \sqrt{5})} {(3 - \sqrt{5})} \\[1em] \Rightarrow \frac{(3 - \sqrt{5})} {(3)^2 - (\sqrt{5})^2} \\[1em] \Rightarrow \frac{(3 - \sqrt{5})} {9 - 5} \\[1em] \Rightarrow \frac{(3 - \sqrt{5})} {4} \\[1em]\end{aligned}$$

Hence, on rationalizing $\frac{1}{(3 + \sqrt{5})} = \frac{(3 - \sqrt{5})} {4}$.

Rationalizing on denominator,

$$\begin{aligned}\Rightarrow \frac{1}{(\sqrt{3} - 1)} \times \frac{(\sqrt{3} + 1)}{(\sqrt{3} + 1)} \\[1em] \Rightarrow \frac{(\sqrt{3} + 1)}{(\sqrt{3})^2 - (1)^2} \\[1em] \Rightarrow \frac{(\sqrt{3} + 1)}{3 - 1} \\[1em] \Rightarrow \frac{(\sqrt{3} + 1)}{2} \\[1em]\end{aligned}$$

Hence, on rationalizing $\frac{1}{(\sqrt{3} - 1)} = \frac{(\sqrt{3} + 1)}{2}$.

Rationalizing the denominator,

$$\begin{aligned}\Rightarrow \frac{1}{(4 + 2\sqrt{3})} \times \frac{(4 - 2\sqrt{3})} {(4 - 2\sqrt{3})} \\[1em] \Rightarrow \frac{(4 - 2\sqrt{3})} {(4)^2 - (2\sqrt{3})^2} \\[1em] \Rightarrow \frac{(4 - 2\sqrt{3})} {16 - 12} \\[1em] \Rightarrow \frac{2(2 - \sqrt{3})} {4} \\[1em] \Rightarrow \frac{(2 - \sqrt{3})} {2} \\[1em]\end{aligned}$$

Hence, on rationalizing $\frac{1}{(4 + 2\sqrt{3})} = \frac{(2 - \sqrt{3})} {2}$.

Rationalizing the denominator,

$$\begin{aligned}\Rightarrow \frac{1}{(\sqrt{6} - \sqrt{3})} \times \frac{(\sqrt{6} + \sqrt{3})}{(\sqrt{6} + \sqrt{3})} \\[1em] \Rightarrow \frac{(\sqrt{6} + \sqrt{3})}{(\sqrt{6})^2 - (\sqrt{3})^2} \\[1em] \Rightarrow \frac{(\sqrt{6} + \sqrt{3})}{6 - 3} \\[1em] \Rightarrow \frac{(\sqrt{6} + \sqrt{3})}{3} \\[1em]\end{aligned}$$

Hence, on rationalizing $\frac{1}{(\sqrt{6} - \sqrt{3})} = \frac{(\sqrt{6} + \sqrt{3})}{3}$.

Rationalizing the denominator,

$$\begin{aligned}\Rightarrow \frac{\sqrt{3} - 1}{\sqrt{3} + 1} \times \frac{\sqrt{3} - 1}{\sqrt{3} - 1} \\[1em] \Rightarrow \frac{(\sqrt{3} - 1)^2}{(\sqrt{3})^2 - (1)^2} \\[1em] \Rightarrow \frac{(\sqrt{3})^2 + (1)^2 - 2 \times \sqrt{3} \times 1 }{3 - 1} \\[1em] \Rightarrow \frac{3 + 1 - 2\sqrt{3}}{2} \\[1em] \Rightarrow \frac{(4 - 2\sqrt{3})}{2} \\[1em] \Rightarrow \frac{2(2 - \sqrt{3})}{2} \\[1em] \Rightarrow (2 - \sqrt{3})\end{aligned}$$

Hence, on rationalizing = $\frac{\sqrt{3} - 1}{\sqrt{3} + 1} = (2 - \sqrt{3})$.

Rationalizing the denominator,

$$\begin{aligned}\Rightarrow \frac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}} \times \frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} \\[1em] \Rightarrow \frac{(3 - 2\sqrt{2})^2} {(3)^2 - ( 2\sqrt{2})^2} \\[1em] \Rightarrow \frac{(3)^2 + ( 2\sqrt{2})^2 -2 \times 3 \times 2\sqrt{2} } {9-8} \\[1em] \Rightarrow \frac{(9 + 8 - 12\sqrt{2})} {1} \\[1em] \Rightarrow (17 - 12\sqrt{2}) \\[1em]\end{aligned}$$

Hence, on rationalizing $\frac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}} = (17 - 12\sqrt{2})$.

Rationalizing the denominator,

$$\begin{aligned}\Rightarrow \frac{1}{(2\sqrt{5} - \sqrt{3})} \times \frac{(2\sqrt{5} + \sqrt{3})} {(2\sqrt{5} + \sqrt{3})} \\[1em] \Rightarrow \frac{(2\sqrt{5} + \sqrt{3})} {(2\sqrt{5})^2 - (\sqrt{3})^2} \\[1em] \Rightarrow \frac{(2\sqrt{5} + \sqrt{3})} {20 - 3} \\[1em] \Rightarrow \frac{(2\sqrt{5} + \sqrt{3})} {17} \\[1em]\end{aligned}$$

Hence, on rationalizing $\frac{1}{(2\sqrt{5} - \sqrt{3})} = \frac{(2\sqrt{5} + \sqrt{3})}{17}$.

Rationalizing the denominator,

$$\begin{aligned}\Rightarrow \frac{1}{(1 + \sqrt{5} + \sqrt{3})} \times \frac{(1 + \sqrt{5} - \sqrt{3})} {(1 + \sqrt{5} - \sqrt{3})} \\[1em] \Rightarrow \frac{(1 + \sqrt{5} - \sqrt{3})}{(1 + \sqrt{5} + \sqrt{3}) \times (1 + \sqrt{5} - \sqrt{3})} \\[1em] \Rightarrow \frac{(1 + \sqrt{5} - \sqrt{3})}{(1 + \sqrt{5})^2 - (\sqrt{3})^2} \\[1em] \Rightarrow \frac{(1 + \sqrt{5} - \sqrt{3})}{1 + 5 + 2\sqrt{5} - 3} \\[1em] \Rightarrow \frac{(1 + \sqrt{5} - \sqrt{3})}{(3 + 2\sqrt{5})}\end{aligned}$$

Rationalizing the denominator again,

$$\begin{aligned}\Rightarrow \frac{(1 + \sqrt{5} - \sqrt{3})}{(3 + 2\sqrt{5})} \times \frac {(3 - 2\sqrt{5})}{(3 - 2\sqrt{5})} \\[1em] \Rightarrow \frac{3 - 2\sqrt{5} + 3\sqrt{5} - 10 - 3\sqrt{3} + 2\sqrt{15}}{(3)^2 - (2\sqrt{5})^2} \\[1em] \Rightarrow \frac{-7 + \sqrt{5} - 3\sqrt{3} + 2\sqrt{15}}{9 - 20} \\[1em] \Rightarrow \frac{-(7 - \sqrt{5} + 3\sqrt{3} - 2\sqrt{15})}{-11} \\[1em] \Rightarrow \frac{(7 - \sqrt{5} + 3\sqrt{3} - 2\sqrt{15})}{11}\end{aligned}$$

Hence, on rationalizing $\frac{1}{(1 + \sqrt{5} + \sqrt{3})} = \frac{(7 - \sqrt{5} + 3\sqrt{3} - 2\sqrt{15})} {11}$.

Rationalizing the denomaintor,

$$\begin{aligned}\Rightarrow \frac{\sqrt{2}}{(\sqrt{2} + \sqrt{3} - \sqrt{5})} \times \frac{(\sqrt{2} + \sqrt{3} + \sqrt{5})}{(\sqrt{2} + \sqrt{3} + \sqrt{5})} \\[1em] \Rightarrow \frac{ \sqrt{2} \times (\sqrt{2}+ \sqrt{3} + \sqrt{5})} {(\sqrt{2}+ \sqrt{3} - \sqrt{5}) \times (\sqrt{2} + \sqrt{3} + \sqrt{5})} \\[1em] \Rightarrow \frac{(2 + \sqrt{6} + \sqrt{10})} {(2 + \sqrt{6} + \sqrt{10} + \sqrt{6} + 3 + \sqrt{15} - \sqrt{10} - \sqrt{15} - 5)} \\[1em] \Rightarrow \frac{(2 + \sqrt{6} + \sqrt{10})} {(2\sqrt{6})}\end{aligned}$$

Rationalizing again,

$$\begin{aligned}\Rightarrow \frac{(2 + \sqrt{6} + \sqrt{10})} {(2\sqrt{6})} \times \frac{\sqrt{6}}{\sqrt{6}} \\[1em] \Rightarrow \frac{(2 + \sqrt{6} + \sqrt{10}) \times \sqrt{6}} {(2\sqrt{6})\times \sqrt{6}} \\[1em] \Rightarrow \frac{(2\sqrt{6} + 6 + \sqrt{60})} {12} \\[1em] \Rightarrow \frac{(2\sqrt{6} + 6 + \sqrt{15 \times 4})}{12} \\[1em] \Rightarrow \frac{2\sqrt{6} + 6 + 2\sqrt{15}}{12} \\[1em] \Rightarrow \frac{2(\sqrt{6} + 3 + \sqrt{15})}{12} \\[1em] \Rightarrow \frac{(\sqrt{6} + 3 + \sqrt{15})}{6}.\end{aligned}$$

Hence, on rationalizing $\frac{(\sqrt{6} + 3 + \sqrt{15})}{6} = \frac{(\sqrt{6} + 3 + \sqrt{15})}{6}$.

Given,

Equation : $\frac{\sqrt{3} + 1}{\sqrt{3} - 1} = a + b\sqrt{3}$

Rationalizing the denominator of L.H.S. of the above equation :

$$\begin{aligned}\Rightarrow \frac{\sqrt{3} + 1}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} \\[1em] \Rightarrow \frac{(\sqrt{3} + 1)^2}{(\sqrt{3})^2 - 1^2} \\[1em] \Rightarrow \frac{(\sqrt{3})^2 + 1^2 + 2 \times \sqrt{3} \times 1}{3 - 1} \\[1em] \Rightarrow \frac{3 + 1 + 2\sqrt{3}}{2} \\[1em] \Rightarrow \frac{4 + 2\sqrt{3}}{2} \\[1em] \Rightarrow \frac{2(2 + \sqrt{3})}{2} \\[1em] \Rightarrow 2 + \sqrt{3}.\end{aligned}$$

Comparing $2 + \sqrt{3} \text{ with } a + b\sqrt{3}$, we get :

a = 2 and b = 1.

Hence, a = 2 and b = 1.

Given,

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

Rationalizing L.H.S. of the above equation :

$$\begin{aligned}\Rightarrow \frac{3 + \sqrt{2}}{3 - \sqrt{2}} \times \frac{3 + \sqrt{2}}{3 + \sqrt{2}} \\[1em] \Rightarrow \frac{(3 + \sqrt{2})^2}{(3)^2 - (\sqrt{2})^2} \\[1em] \Rightarrow \frac{(3)^2 + (\sqrt{2})^2 + 2 \times 3 \times \sqrt{2}}{9 - 2} \\[1em] \Rightarrow \frac{9 + 2 + 6\sqrt{2}}{7} \\[1em] \Rightarrow \frac{11 + 6\sqrt{2}}{7} \\[1em] \Rightarrow \frac{11}{7} + \frac{6\sqrt{2}}{7}\end{aligned}$$

Comparing $\frac{11}{7} + \frac{6}{7}\sqrt{2} \text{ with } a + b\sqrt{2}$, we get :

$a = \frac{11}{7} \text{ and } b = \frac{6}{7}.$

Hence, $a = \frac{11}{7} \text{ and } b = \frac{6}{7}$.

Given,

Equation : $\frac{5 - \sqrt{6}}{5 + \sqrt{6}} = a - b\sqrt{6}$

Rationalizing the denomiantor of L.H.S. of the above equation :

$$\begin{aligned}\Rightarrow \frac{5 - \sqrt{6}}{5 + \sqrt{6}} \times \frac{5 - \sqrt{6}}{5 - \sqrt{6}} \\[1em] \Rightarrow \frac{(5 - \sqrt{6})^2}{(5)^2 - (\sqrt{6})^2} \\[1em] \Rightarrow \frac{(5)^2 + (\sqrt{6})^2 - 2 \times 5 \times \sqrt{6}}{25 - 6} \\[1em] \Rightarrow \frac{25 + 6 - 10\sqrt{6}}{19} \\[1em] \Rightarrow \frac{31 - 10\sqrt{6}}{19} \\[1em] \Rightarrow \frac{31}{19} - \frac{10\sqrt{6}}{19}\end{aligned}$$

Comparing $\frac{31}{19} - \frac{10}{19}\sqrt{6} \text{ with } a - b\sqrt{6}$, we get :

$a = \frac{31}{19} \text{ and } b = \frac{10}{19}.$

Hence, $a = \frac{31}{19} \text{ and } b = \frac{10}{19}$.

Given,

Equation : $\frac{5 + 2\sqrt{3}}{7 + 4\sqrt{3}} = a - b\sqrt{3}$

Rationalizing the denominator of L.H.S. of the above equation :

$$\begin{aligned}\Rightarrow \frac{5 + 2\sqrt{3}}{7 + 4\sqrt{3}} \times \frac{7 - 4\sqrt{3}}{7 - 4\sqrt{3}} \\[1em] \Rightarrow \frac{(5 + 2\sqrt{3}) \times (7 - 4\sqrt{3})}{(7)^2 - (4\sqrt{3})^2} \\[1em] \Rightarrow \frac{35 - 20\sqrt{3} + 14\sqrt{3} - 8 \times 3}{49 - 48} \\[1em] \Rightarrow \frac{35 - 6\sqrt{3} - 24}{1} \\[1em] \Rightarrow 11 - 6\sqrt{3} \\[1em]\end{aligned}$$

Comparing, $11 - 6\sqrt{3} \text{ with } a - b\sqrt{3}$, we get :

a = 11 and b = 6.

Hence, a = 11 and b = 6.

Given,

Equation : $\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}} + \frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}$

Simplifying the above equation :

$$\begin{aligned}\Rightarrow \frac{(\sqrt{5} + \sqrt{3})^2 + (\sqrt{5} - \sqrt{3})^2}{(\sqrt{5} - \sqrt{3})(\sqrt{5} + \sqrt{3})} \\[1em] \Rightarrow \frac{(\sqrt{5})^2 + (\sqrt{3})^2 + 2 \times \sqrt{5} \times \sqrt{3} + (\sqrt{5})^2 + (\sqrt{3})^2 - 2 \times \sqrt{5} \times \sqrt{3}}{(\sqrt{5})^2 - (\sqrt{3})^2} \\[1em] \Rightarrow \frac{5 + 3 + 2\sqrt{15} + 5 + 3 - 2\sqrt{15}}{5 - 3} \\[1em] \Rightarrow \frac{16}{2} \\[1em] \Rightarrow 8\end{aligned}$$

Hence, $\frac{\sqrt{5} + \sqrt{3}}{\sqrt{5} - \sqrt{3}}+\frac{\sqrt{5} - \sqrt{3}}{\sqrt{5} + \sqrt{3}}$ = 8.

Given,

Equation : $\frac{7 + 3\sqrt{5}}{3 + \sqrt{5}} - \frac{7 - 3\sqrt{5}}{3 - \sqrt{5}}$

Simplifying the above equation :

$$\begin{aligned}\Rightarrow \frac{(7 + 3\sqrt{5}) \times (3 - \sqrt{5}) - (7 - 3\sqrt{5}) \times (3 + \sqrt{5})}{(3 + \sqrt{5}) \times (3 - \sqrt{5})} \\[1em] \Rightarrow \frac{(21 - 7\sqrt{5} + 9\sqrt{5} - 15)-(21 + 7\sqrt{5} - 9\sqrt{5} - 15)}{3^2 - (\sqrt{5})^2} \\[1em] \Rightarrow \frac{21 - 7\sqrt{5} + 9\sqrt{5} - 15 - 21 - 7\sqrt{5} + 9\sqrt{5} + 15}{9 - 5} \\[1em] \Rightarrow \frac{4\sqrt{5}}{4} \\[1em] \Rightarrow \sqrt{5}\end{aligned}$$

Hence, $\frac{7 + 3\sqrt{5}}{3 + \sqrt{5}}- \frac{7 - 3\sqrt{5}}{3 - \sqrt{5}} = \sqrt{5}$.

Given,

Equation : $\frac{1}{(3 - \sqrt{8})} + \frac{1}{(\sqrt{7} - \sqrt{6})}+ \frac{1}{(\sqrt{5} - 2)} - \frac{1}{(\sqrt{8} - \sqrt{7})} - \frac{1}{(\sqrt{6} - \sqrt{5})}$

Simplifying L.H.S. of the above equation :

$$\begin{aligned}\Rightarrow \frac{1}{(3 - \sqrt{8})} \times \frac{(3 + \sqrt{8})}{(3 + \sqrt{8})} + \frac{1}{(\sqrt{7} - \sqrt{6})} \times \frac{(\sqrt{7} + \sqrt{6})}{(\sqrt{7} + \sqrt{6})}+ \frac{1}{(\sqrt{5} - 2)} \times \frac{(\sqrt{5} + 2)}{(\sqrt{5} + 2)} - \frac{1}{(\sqrt{8} - \sqrt{7})} \times \frac{(\sqrt{8} + \sqrt{7})}{(\sqrt{8} + \sqrt{7})} - \frac{1}{(\sqrt{6} - \sqrt{5})} \times \frac{(\sqrt{6} + \sqrt{5})}{(\sqrt{6} + \sqrt{5})} \\[1em] \Rightarrow \frac{3 + \sqrt{8}}{3^2 - (\sqrt{8})^2} + \frac{\sqrt{7} + \sqrt{6}}{(\sqrt{7})^2 - (\sqrt{6})^2} + \frac{\sqrt{5} + 2}{(\sqrt{5})^2 - (2)^2} - \frac{\sqrt{8} + \sqrt{7}}{(\sqrt{8})^2 - (\sqrt{7})^2} - \frac{\sqrt{6} + \sqrt{5}}{(\sqrt{6})^2 - (\sqrt{5})^2} \\[1em] \Rightarrow \frac{3 + \sqrt{8}}{9 - 8} + \frac{\sqrt{7} + \sqrt{6}}{7 - 6} + \frac{\sqrt{5} + 2}{5 - 4} - \frac{\sqrt{8} + \sqrt{7}}{8 - 7} - \frac{\sqrt{6} + \sqrt{5}}{6 - 5} \\[1em] \Rightarrow 3 + \sqrt{8} + \sqrt{7} + \sqrt{6} + \sqrt{5} + 2 - (\sqrt{8} + \sqrt{7}) - (\sqrt{6} + \sqrt{5}) \\[1em] \Rightarrow 3 + 2 + \sqrt{8} - \sqrt{8} + \sqrt{7} - \sqrt{7} + \sqrt{6} - \sqrt{6} + \sqrt{5} - \sqrt{5} \\[1em] \Rightarrow 5\end{aligned}$$

Hence, proved that

$\frac{1}{(3 - \sqrt{8})} + \frac{1}{(\sqrt{7} - \sqrt{6})}+ \frac{1}{(\sqrt{5} - 2)} - \frac{1}{(\sqrt{8} - \sqrt{7})} - \frac{1}{(\sqrt{6} - \sqrt{5})} = 5$.

Given,

x = $(3 + \sqrt{8})$

$\therefore \frac{1}{x} = \frac{1}{(3 + \sqrt{8})}$

Rationalizing,

$$\begin{aligned}\Rightarrow \frac{1}{x} = \frac{1}{(3 + \sqrt{8})} \times \frac{(3 - \sqrt{8})}{(3 - \sqrt{8})} \\[1em] = \frac{3 - \sqrt{8}}{3^2 - (\sqrt{8})^2} \\[1em] = \frac{3 - \sqrt{8}}{9 - 8} \\[1em] = \frac{3 - \sqrt{8}}{1} \\[1em] = 3 - \sqrt{8}.\end{aligned}$$

By formula,

$x^2 + \frac{1}{x^2} = \Big(x + \frac{1}{x}\Big)^2 - 2$

Substituting values we get :

$$\begin{aligned}\Rightarrow x^2 + \frac{1}{x^2} = (3 + \sqrt{8} + 3 - \sqrt{8})^2 - 2 \\[1em] = 6^2 - 2 \\[1em] = 36 - 2 \\[1em] = 34.\end{aligned}$$

Hence, $x^2 + \frac{1}{x^2}$ = 34.

Given,

x = $(4 - \sqrt{15})$

$\therefore \frac{1}{x} = \frac{1}{(4 - \sqrt{15})}$

Rationalizing,

$$\begin{aligned}\Rightarrow \frac{1}{x} = \frac{1}{(4 - \sqrt{15})} \times \frac{(4 + \sqrt{15})}{(4 + \sqrt{15})} \\[1em] = \frac{4 + \sqrt{15}}{4^2 - (\sqrt{15})^2} \\[1em] = \frac{4 + \sqrt{15}}{16 - 15} \\[1em] = \frac{4 + \sqrt{15}}{1} \\[1em] = 4 + \sqrt{15}.\end{aligned}$$

By formula,

$x^2 + \frac{1}{x^2} = \Big(x + \frac{1}{x}\Big)^2 - 2$

Substituting values we get :

$$\begin{aligned}\Rightarrow x^2 + \frac{1}{x^2} = (4 - \sqrt{15} + 4 + \sqrt{15})^2 - 2 \\[1em] = 8^2 - 2 \\[1em] = 64 - 2 \\[1em] = 62.\end{aligned}$$

Hence, $x^2 + \frac{1}{x^2}$ = 62.

3.4 is a terminating decimal, thus it is a rational number.

Hence, Option 3 is the correct option.

$\sqrt{11}$ is square root of non-perfect square i.e 11, thus it is an irrational number.

Hence, Option 3 is the correct option.

71 has only two factors i.e is 1 and 71 itself, thus it is a prime number.

Hence, Option 3 is the correct option.

Let x = $8.\overline{32}$

⇒ x = 8.32323232..     ..........(1)

Multiplying both sides by 100 (since there are two recurring digits)

⇒ 100x = 832.3232..     ..........(2)

Subtracting equation (1) from (2), we get :

⇒ 100x - x = 832.3232.. - 8.3232..

⇒ 99x = 824

⇒ x = $\frac{824}{99}$

Hence, Option 2 is the correct option.

$\frac{103}{330} = 0.3\overline{12}$

$\frac{52}{165} = 0.3\overline{15}$

$\frac{103}{111} = 0.\overline{927}$

$\frac{104}{333} = 0.\overline{312}$

Since, $0.3\overline{12}$ is nearest to 0.3

Hence, Option 1 is the correct option.

A rational number will have a terminating decimal expansion if and only if the prime factorization of its denominator contains only powers of 2 and/or 5.

The prime factor of 16 is only 2.

∴ $\frac{3}{16}$ will have a terminating decimal.

Hence, Option 3 is the correct option.

A rational number will have a recurring decimal expansion if and only if the prime factorization of its denominator contains any prime factor other than 2 or 5.

The prime factors of 51 are 17 and 3.

Hence, Option 2 is the correct option.

Let the number to be subtarcted be x.

$$\begin{aligned}\Rightarrow \sqrt{72} - x = \sqrt{32} \\[1em] \Rightarrow x = \sqrt{72} - \sqrt{32} \\[1em] \Rightarrow x = \sqrt{36 \times 2} - \sqrt{16 \times 2} \\[1em] \Rightarrow x = 6\sqrt{2} - 4\sqrt{2} \\[1em] \Rightarrow x = 2\sqrt{2}.\end{aligned}$$

Hence, Option 4 is the correct option.

Given,

⇒ x = 3 + $2\sqrt{2}$

$\Rightarrow \frac{1}{x} = \frac{1}{(3 + 2\sqrt{2})}$

Rationalizing,

$$\begin{aligned}\Rightarrow \frac{1}{(3 + 2\sqrt{2})} \times \frac{(3 - 2\sqrt{2})}{(3 - 2\sqrt{2})} \\[1em] \Rightarrow \frac{3 - 2\sqrt{2}}{3^2 - (2\sqrt{2})^2} \\[1em] \Rightarrow \frac{3 - 2\sqrt{2}}{9 - 8} \\[1em] \Rightarrow 3 - 2\sqrt{2}\end{aligned}$$

Adding x and $\frac{1}{x}$ we get

$$\begin{aligned}\Rightarrow x + \frac{1}{x} = 3 + 2\sqrt{2} + 3 - 2\sqrt{2} \\[1em] \Rightarrow 6\end{aligned}$$

Hence, Option 3 is the correct option.

Given,

⇒ x = 2 - $\sqrt{2}$

$\Rightarrow \frac{1}{x} = \frac{1}{(2 - \sqrt{2})}$

Rationalizing the denominator, we get :

$$\begin{aligned}\Rightarrow \frac{1}{(2 - \sqrt{2})} \times \frac{(2 + \sqrt{2})}{(2 + \sqrt{2})} \\[1em] \Rightarrow \frac{2 + \sqrt{2}}{2^2 - (\sqrt{2})^2} \\[1em] \Rightarrow \frac{2 + \sqrt{2}}{4 - 2} \\[1em] \Rightarrow \frac{2 + \sqrt{2}}{2}\end{aligned}$$

Substituting values in $x - \frac{1}{x}$, we get :

$$\begin{aligned}\Rightarrow 2 - \sqrt{2} - \frac{2 + \sqrt{2}}{2} \\[1em] \Rightarrow \frac{2(2 - \sqrt{2})-(2 + \sqrt{2})}{2} \\[1em] \Rightarrow \frac{4 - 2\sqrt{2} - 2 - \sqrt{2}}{2} \\[1em] \Rightarrow \frac{2 - 3\sqrt{2}}{2} \\[1em]\end{aligned}$$

$x - \frac{1}{x} = \frac{2 - 3\sqrt{2}}{2}$

Hence, Option 3 is the correct option.

Given,

x = $(5 + 2\sqrt{6})$

$\therefore \frac{1}{x} = \frac{1}{(5 + 2\sqrt{6})}$

Rationalizing,

$$\begin{aligned}\Rightarrow \frac{1}{x} = \frac{1}{5 + 2\sqrt{6}} \times \frac{(5 - 2\sqrt{6})}{(5 - 2\sqrt{6})} \\[1em] = \frac{5 - 2\sqrt{6}}{5^2 - (2\sqrt{6})^2} \\[1em] = \frac{5 - 2\sqrt{6}}{25 - 24} \\[1em] = \frac{5 - 2\sqrt{6}}{1} \\[1em] = 5 - 2\sqrt{6}.\end{aligned}$$

By formula,

$x^2 + \frac{1}{x^2} = \Big(x + \frac{1}{x}\Big)^2 - 2$

Substituting values we get :

$$\begin{aligned}\Rightarrow x^2 + \frac{1}{x^2} = (5 + 2\sqrt{6} + 5 - 2\sqrt{6})^2 - 2 \\[1em] = 10^2 - 2 \\[1em] = 100 - 2 \\[1em] = 98.\end{aligned}$$

Hence, Option 1 is the correct option.

Let the first rational number between $-\frac{3}{7} \text{ and } -\frac{1}{7}$ be x.

$$\begin{aligned}\Rightarrow x = \frac{1}{2}\Big[-\frac{3}{7} + \Big(-\frac{1}{7}\Big)\Big] \\[1em] \Rightarrow x = \frac{1}{2}\Big(-\frac{4}{7}\Big)\\[1em] \Rightarrow x = -\frac{4}{14}\end{aligned}$$

Let the second rational number be y.

$$\begin{aligned}\Rightarrow y = \frac{1}{2}\Big[-\frac{4}{14} + \Big(-\frac{1}{7}\Big)\Big] \\[1em] \Rightarrow y = \frac{1}{2}\Big(\frac{-4 - 2}{14}\Big) \\[1em] \Rightarrow y = \frac{1}{2}\Big(-\frac{6}{14}\Big)\\[1em] \Rightarrow y = -\frac{6}{28} = -\frac{3}{14}\end{aligned}$$

Hence, Option 4 is the correct option.

On solving,

⇒ $\sqrt{3}$ = 1.732

⇒ $\sqrt[3]{6}$ = 1.817

⇒ $\sqrt[4]{7}$ = 1.626

∴ $\sqrt[4]{7} \lt \sqrt{3} \lt \sqrt[3]{6}$

Hence, Option 2 is the correct option.

Given,

$$\begin{aligned}\Rightarrow \sqrt[3]{432} \\[1em] \Rightarrow \sqrt[3]{216 \times 2} \\[1em] \Rightarrow \sqrt[3]{216} \times \sqrt[3]{2} \\[1em] \Rightarrow 6\sqrt[3]{2}\end{aligned}$$

Hence, Option 2 is the correct option.

Given,

$$\begin{aligned}\Rightarrow 5\sqrt[3]{2} \\[1em] \Rightarrow \sqrt[3]{5^3 \times 2} \\[1em] \Rightarrow \sqrt[3]{125 \times 2} \\[1em] \Rightarrow \sqrt[3]{250}\end{aligned}$$

Hence, Option 3 is the correct option.

1. Real numbers include all rational and irrational numbers.

Rational numbers are a subset of real numbers.

Hence, Option (d) is the correct option.

2. Real numbers include all rational and irrational numbers.

Hence, Option (d) is the correct option.

3. The sum of two irrational numbers can be either rational or irrational.

Hence, Option (c) is the correct option.

4. If the rational number is zero, the product is rational.

If the rational number is non-zero, the product is irrational.

Hence, Option (c) is the correct option.

5. The number of irrational number is infinite as the number of real numbers is also infinite.

Hence, Option (b) is the correct option.

1. On dividing,

$\frac{3}{11} = 0.\overline{27}$

Hence, Option (d) is the correct option.

2. $0.\overline{52}$ = 0.525252...

Hence, Option (b) is the correct option.

3. $\frac{3}{11}$ = 0.272727.... = $0.\overline{27}$

Hence, Option (c) is the correct option.

4. Given,

Let x = 0.525252..     .........(1)

Multiply both side by 100(as two digits are recurring)

⇒ 100x = 52.5252..     .........(2)

Subtracting eqn (1) from eqn (2), we get :

⇒ 100x - x = 52.5252.. - 0.525252

⇒ 99x = 52

⇒ x = $\frac{52}{99}$.

Hence, Option (a) is the correct option.

5. From part 4,

$0.\overline{52} = \frac{52}{99}$

Adding $\frac{52}{99}$ and $\frac{3}{11}$

$$\begin{aligned}\Rightarrow \frac{52}{99} + \frac{3}{11} \\[1em] \Rightarrow \frac{52 + 27}{99} \\[1em] \Rightarrow \frac{79}{99}.\end{aligned}$$

Hence, Option (a) is the correct option.

Given,

$\frac{1}{\sqrt{5} - 2}$

Rationalizing the denominator of $\frac{1}{\sqrt{5} - 2}$,

$$\begin{aligned}\Rightarrow \frac{1}{\sqrt{5} - 2} \times \frac{\sqrt{5} + 2}{\sqrt{5} + 2} \\[1em] \Rightarrow \frac{\sqrt{5} + 2}{(\sqrt{5})^2 - (2)^2}\\[1em] \Rightarrow \frac{\sqrt{5} + 2}{5 - 4} \\[1em] \Rightarrow \frac{\sqrt{5} + 2}{1} \\[1em] \Rightarrow 2 + \sqrt{5}\end{aligned}$$

∴ Assertion (A) is true.

We know that,

If the product of two irrational numbers is rational, then each one is called the rationalizing factor of the other.

∴ Reason (R) is true.

Hence, Option 3 is the correct option.

$\sqrt[3]{2}, \sqrt[3]{3}, \sqrt[3]{4}, \sqrt[3]{5}, \sqrt[3]{6}, \sqrt[3]{7}$ are irrational, because these are cube roots of not perfect cubes.

∴ Assertion (A) is true.

The cube roots of all perfect cubes are rational. Thus, we cannot say that the cube roots of all natural numbers is irrational.

∴ Reason (R) is false.

Hence, Option 1 is the correct option.

Between any two real numbers like 0 and 0.1, there are infinitely many rational numbers. So the sum is infinite.

Hence, Option 2 is the correct option.

Given,

$$\begin{aligned}\Rightarrow q = \frac{1}{p} \\[1em] \Rightarrow \frac{1}{q} = p \\[1em] \Rightarrow s = \frac{1}{r} \\[1em] \Rightarrow \frac{1}{s} = r.\end{aligned}$$

Substituting value from above in given equation,

$$\begin{aligned}\Rightarrow \Big([p + p] ÷ [r + r]\Big)÷ \Big([s + s] ÷ [q + q]\Big) \\[1em] \Rightarrow \Big(\frac{2p}{2r}\Big) ÷ \Big(\frac{2s}{2q}\Big) \\[1em] \Rightarrow \frac{p}{r} ÷ \frac{s}{q} \\[1em] \Rightarrow \Big(\frac{p}{r}\Big) \times \Big(\frac{q}{s}\Big) \\[1em] \Rightarrow \frac{p}{r} \times q \times \frac{1}{s} \\[1em] \Rightarrow \frac{p}{r} \times \frac{1}{p} \times r \\[1em] \Rightarrow 1.\end{aligned}$$

Hence, Option 1 is the correct option.

Convert repeating decimals to fraction,

Let, x = $0.\overline{7}$

⇒ x = 0.7777....     .......(1)

⇒ 10x = 7.777...     .......(2)

Subtracting equation (1) from (2), we get :

⇒ 10x - x = 7.777..... - 0.777.....

⇒ 9x = 7

⇒ x = $\frac{7}{9}$

Let, y = $0.4\overline{7}$

⇒ 10y = 4.7777....     .......(3)

⇒ 100y = 47.777...     .......(4)

Subtracting equation (3) from (4), we get :

⇒ 100y - 10y = 47.777..... - 4.777.....

⇒ 90y = 43

⇒ y = $\frac{43}{90}$

$$\begin{aligned}\Rightarrow 0.6 + 0.\overline{7} + 0.4\overline{7} = 0.6 + \frac{7}{9} + \frac{43}{90} \\[1em] = \frac{6}{10} + \frac{7}{9} + \frac{43}{90} \\[1em] = \frac{54 + 70 + 43}{90} \\[1em] = \frac{167}{90}.\end{aligned}$$

Hence, Option 3 is the correct option.

We know that,

$\Rightarrow \sqrt[n]{x} = x^\frac{1}{n}$

Solving,

$$\begin{aligned}\Rightarrow \sqrt[4]{\sqrt[3]{3^2}} \\[1em] \Rightarrow \sqrt[4]{3^\frac{2}{3}} \\[1em] \Rightarrow 3^\frac{2}{3 \times 4} \\[1em] \Rightarrow 3^\frac{2}{12} \\[1em] = {3^\frac{1}{6}}\end{aligned}$$

Hence, Option 4 is the correct option.

Let x = $1.\overline{9}$

⇒ x = 1.999...     .......(1)

⇒ 10x = 19.999...     .......(2)

Subtracting equation (1) from (2), we get :

⇒ 10x - x = 19.999..... - 1.999....

⇒ 9x = 18

⇒ x = $\frac{18}{9}$ = 2.

⇒ $1.\overline{9} - 1.9$

⇒ 2 - 1.9

⇒ 0.1

Hence, Option 4 is the correct option.

A non-terminating and non-repeating decimal is the definition of an irrational number.

Since 2 is not a perfect 9th power of any rational number, its 9th root will be an irrational number.

Hence, Option 2 is the correct option.

We get the maximum possible value of $\frac{c - b}{a}$, if c is the largest and a, b are the smallest values.

$$\begin{aligned}\Rightarrow \frac{c - b}{a} \\[1em] \Rightarrow \frac{25 - 3}{2} \\[1em] \Rightarrow \frac{22}{2} \\[1em] \Rightarrow 11.\end{aligned}$$

Hence, the maximum possible value of $\frac{c - b}{a}$ = 11.

As, the complete line is divided in into 16 equal parts. Thus, A = 1, B = 2, C = 3.

Rational number less than 1 and greater than 0, will be the interval in which reciprocal of any rational number is greater than the number itself.