Rational and Irrational Numbers

The L.C.M of 9 and 8 is 72.

$$\begin{aligned}\frac{2}{9} = \frac{2 \times 8}{9 \times 8} = \frac{16}{72} \\[0.5em] \frac{3}{8} = \frac{3 \times 9}{8 \times 9} = \frac{27}{72} \\[0.5em] \text{Since } 16 \lt 27, \frac{2}{9} \lt \frac{3}{8}\end{aligned}$$

A rational number between $\frac{2}{9}$ and $\frac{3}{8}$

$$\begin{aligned}= \frac{\frac{2}{9} + \frac{3}{8}}{2} \\[0.5em] = \frac{\frac{16 + 27}{72}}{2} \\[0.5em] = \bold{\frac{43}{144}} \\[0.5em]\end{aligned}$$

Numbers in descending order are:

$\bold{\frac{3}{8}}, \bold{\frac{43}{144}}, \bold{\frac{2}{9}}$

The L.C.M of 3 and 4 is 12.

$$\begin{aligned}\frac{1}{3} = \frac{1 \times 4}{3 \times 4} = \frac{4}{12} \\[0.5em] \frac{1}{4} = \frac{1 \times 3}{4 \times 3} = \frac{3}{12} \\[0.5em] \text{Since } 3 \lt 4, \frac{1}{4} \lt \frac{1}{3}\end{aligned}$$

A rational number between $\frac{1}{4}$ and $\frac{1}{3}$

$$\begin{aligned}= \frac{\frac{1}{4} + \frac{1}{3}}{2} \\[0.5em] = \frac{\frac{4 + 3}{12}}{2} \\[0.5em] = \bold{\frac{7}{24}} \\[0.5em]\end{aligned}$$

A rational number between $\frac{1}{4}$ and $\frac{7}{24}$

$$\begin{aligned}= \frac{\frac{1}{4} + \frac{7}{24}}{2} \\[0.5em] = \frac{\frac{6 + 7}{24}}{2} \\[0.5em] = \bold{\frac{13}{48}} \\[0.5em]\end{aligned}$$

Numbers in ascending order are:

$\bold{\frac{1}{4}}, \bold{\frac{13}{48}}, \bold{\frac{7}{24}}, \bold{\frac{1}{3}}$

The L.C.M of 2 and 3 is 6.

$$\begin{aligned}-\frac{1}{3} = -\frac{1 \times 2}{3 \times 2} = -\frac{2}{6} \\[0.5em] -\frac{1}{2} = -\frac{1 \times 3}{2 \times 3} = -\frac{3}{6} \\[0.5em] \text{Since } -3 \lt -2, -\frac{1}{2} \lt -\frac{1}{3}\end{aligned}$$

A rational number between -$\frac{1}{2}$ and -$\frac{1}{3}$

$$\begin{aligned}= \frac{-\frac{1}{2} + \Big(-\frac{1}{3}\Big)}{2} \\[0.5em] = \frac{\frac{-3 + (-2)}{6}}{2} \\[0.5em] = \frac{\frac{-3 -2}{6}}{2} \\[0.5em] = \bold{-\frac{5}{12}} \\[0.5em]\end{aligned}$$

A rational number between -$\frac{1}{2}$ and -$\frac{5}{12}$

$$\begin{aligned}= \frac{-\frac{1}{2} + \Big(-\frac{5}{12}\Big)}{2} \\[0.5em] = \frac{\frac{-6 + (-5)}{12}}{2} \\[0.5em] = \frac{\frac{-6 -5}{12}}{2} \\[0.5em] = \bold{-\frac{11}{24}} \\[0.5em]\end{aligned}$$

Numbers in ascending order are:

$\bold{-\frac{1}{2}}, \bold{-\frac{11}{24}}, \bold{-\frac{5}{12}}, \bold{-\frac{1}{3}}$

The L.C.M of 3 and 5 is 15.

$$\begin{aligned}\frac{1}{3} = \frac{1 \times 5}{3 \times 5} = \frac{5}{15} \\[0.5em] \frac{4}{5} = \frac{4 \times 3}{5 \times 3} = \frac{12}{15} \\[0.5em] \text{Since } 5 \lt 12, \frac{1}{3} \lt \frac{4}{5}\end{aligned}$$

A rational number between $\frac{1}{3}$ and $\frac{4}{5}$

$$\begin{aligned}= \frac{\frac{1}{3} + \frac{4}{5}}{2} \\[0.5em] = \frac{\frac{5 + 12}{15}}{2} \\[0.5em] = \bold{\frac{17}{30}} \\[0.5em]\end{aligned}$$

A rational number between $\frac{1}{3}$ and $\frac{17}{30}$

$$\begin{aligned}= \frac{\frac{1}{3} + \frac{17}{30}}{2} \\[0.5em] = \frac{\frac{10 + 17}{30}}{2} \\[0.5em] = \bold{\frac{27}{60}} \\[0.5em]\end{aligned}$$

A rational number between $\frac{17}{30}$ and $\frac{4}{5}$

$$\begin{aligned}= \frac{\frac{17}{30} + \frac{4}{5}}{2} \\[0.5em] = \frac{\frac{17 + 24}{30}}{2} \\[0.5em] = \bold{\frac{41}{60}} \\[0.5em]\end{aligned}$$

Numbers in descending order are:

$\bold{\frac{4}{5}}, \bold{\frac{41}{60}}, \bold{\frac{17}{30}}, \bold{\frac{27}{60}}, \bold{\frac{1}{3}}$

(i) We want three rational numbers between 3 and 3.5.

We know that,

Terminating decimals are rational numbers.

Let us take 3.1, 3.2, 3.3

Thus, we have :

⇒ 3 < 3.1 < 3.2 < 3.3 <3.5

Hence, three rational numbers between 3 and 3.5 are 3.1, 3.2 and 3.3.

(ii) Converting fraction in decimal form,

$\frac{1}{4}$ = 0.25

$\frac{2}{5}$ = 0.40

We know that,

Terminating decimals are rational numbers.

Let us take 0.28, 0.30, 0.32, 0.34

Thus, we have :

⇒ 0.25 < 0.28 < 0.30 < 0.32 < 0.34 < 0.40

Hence, four rational numbers between $\frac{1}{4} \text{ and } \frac{2}{5}$ = 0.28, 0.30, 0.32 and 0.34

(iii) We want five rational numbers between $1\frac{1}{2}$ and $1\frac{3}{4}$.

Numbers : $1\frac{1}{2} = \frac{3}{2} = 1.50$ and $1\frac{3}{4} = \frac{7}{4} = 1.75$

We know that,

Terminating decimals are rational numbers.

Let us take 1.61, 1.62, 1.63, 1.64, 1.65

Thus, we have :

⇒ 1.50 < 1.61 < 1.62 < 1.63 <1.64 < 1.65 <1.75

Hence, five rational numbers between $1\frac{1}{2} \text{ and } 1\frac{3}{4}$ = 1.61, 1.62, 1.63, 1.64 and 1.65.

The numbers 3 and 4 can be written as $\frac{3}{1}$ and $\frac{4}{1}$.

Since we want to find six rational numbers between the given numbers, multiplying the numerator and denominator of the above numbers by 6 + 1 i.e. by 7, we get $\frac{21}{7}$ and $\frac{28}{7}$, which are equivalent to the given numbers.

$$\begin{aligned}\text{As } 21 \lt 22 \lt 23 \lt 24 \lt 25 \lt 26 \lt 27 \lt 28, \\[0.7em] \frac{21}{7} \lt \frac{22}{7} \lt \frac{23}{7} \lt \frac{24}{7} \lt \frac{25}{7} \lt \frac{26}{7} \lt \frac{27}{7} \lt \frac{28}{7}\end{aligned}$$

Therefore, six rational numbers between 3 and 4 are:

$\bold{\frac{22}{7}}, \bold{\frac{23}{7}}, \bold{\frac{24}{7}}, \bold{\frac{25}{7}}, \bold{\frac{26}{7}}, \bold{\frac{27}{7}}$

Since we want to find five rational numbers between the given numbers, multiplying the numerator and denominator of the above numbers by 5 + 1 i.e. by 6, we get $\frac{18}{30}$ and $\frac{24}{30}$, which are equivalent to the given numbers.

$$\begin{aligned}\text{As } 18 \lt 19 \lt 20 \lt 21 \lt 22 \lt 23 \lt 24, \\[0.7em] \frac{18}{30} \lt \frac{19}{30} \lt \frac{20}{30} \lt \frac{21}{30} \lt \frac{22}{30} \lt \frac{23}{30} \lt \frac{24}{30} \\[0.7em] \Rightarrow \frac{3}{5} \lt \frac{19}{30} \lt \frac{2}{3} \lt \frac{7}{10} \lt \frac{11}{15} \lt \frac{23}{30} \lt \frac{4}{5}\end{aligned}$$

Therefore, six rational numbers between $\frac{3}{5}$ and $\frac{4}{5}$ are:

$\bold{\frac{19}{30}}, \bold{\frac{2}{3}}, \bold{\frac{7}{10}}, \bold{\frac{11}{15}}, \bold{\frac{23}{30}}$

Writing the given numbers with same denominator 35 (L.C.M. of 5 and 7), we get:

$$\begin{aligned}-\frac{2}{5} = -\frac{14}{35} \\[0.5em] \frac{1}{7} = \frac{5}{35}\end{aligned}$$

$$\begin{aligned}\text{As } -14 \lt -13 \lt -12 \lt -11 \lt -10 \lt -9 \lt -8 \lt -7 \lt 0 \lt 1 \lt 2 \lt 5, \\[0.7em] -\frac{14}{35} \lt -\frac{13}{35} \lt -\frac{12}{35} \lt -\frac{11}{35} \lt -\frac{10}{35} \lt -\frac{9}{35} \lt -\frac{8}{35} \lt -\frac{7}{35} \lt 0 \lt \frac{1}{35} \lt \frac{2}{35} \lt \frac{5}{35} \\[0.7em] \Rightarrow -\frac{2}{5} \lt -\frac{13}{35} \lt -\frac{12}{35} \lt -\frac{11}{35} \lt -\frac{2}{7} \lt -\frac{9}{35} \lt -\frac{8}{35} \lt -\frac{1}{5} \lt 0 \lt \frac{1}{35} \lt \frac{2}{35} \lt \frac{5}{35}\end{aligned}$$

Therefore, ten rational numbers between $-\frac{2}{5}$ and $\frac{1}{7}$ are:

$$\begin{aligned}\bold{-\frac{13}{35}}, \bold{-\frac{12}{35}}, \bold{-\frac{11}{35}}, \bold{-\frac{2}{7}}, \bold{-\frac{9}{35}}, \\[0.5em] \bold{-\frac{8}{35}}, \bold{-\frac{1}{5}}, 0, \bold{\frac{1}{35}}, \bold{\frac{2}{35}}\end{aligned}$$

Writing the given numbers with same denominator 6 (L.C.M. of 2 and 3), we get:

$$\begin{aligned}\frac{1}{2} = \frac{3}{6} \\[0.5em] \frac{2}{3} = \frac{4}{6}\end{aligned}$$

Since we want to find six rational numbers between the given numbers, multiplying the numerator and denominator of the above numbers by 6 + 1 i.e. by 7, we get $\frac{21}{42}$ and $\frac{28}{42}$, which are equivalent to the given numbers.

$$\begin{aligned}\text{As } 21 \lt 22 \lt 23 \lt 24 \lt 25 \lt 26 \lt 27 \lt 28, \\[0.7em] \frac{21}{42} \lt \frac{22}{42} \lt \frac{23}{42} \lt \frac{24}{42} \lt \frac{25}{42} \lt \frac{26}{42} \lt \frac{27}{42} \lt \frac{28}{42} \\[0.7em] \Rightarrow \frac{1}{2} \lt \frac{11}{21} \lt \frac{23}{42} \lt \frac{4}{7} \lt \frac{25}{42} \lt \frac{13}{21} \lt \frac{9}{14} \lt \frac{2}{3}\end{aligned}$$

Therefore, six rational numbers between $\frac{1}{2}$ and $\frac{2}{3}$ are:

$\bold{\frac{11}{21}}, \bold{\frac{23}{42}}, \bold{\frac{4}{7}}, \bold{\frac{25}{42}}, \bold{\frac{13}{21}}, \bold{\frac{9}{14}}$

Let $\sqrt{5}$ be a rational number, then

$\sqrt{5} = \frac{p}{q},$

where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)

$$\begin{aligned}\Rightarrow 5 = \frac{p^2}{q^2} \\[0.5em] \Rightarrow p^2 = 5q^2 \qquad \text{....(i)}\end{aligned}$$

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

$\Rightarrow 5 \text{ divides } p \qquad \text{(Theorem 1)}$

Let p = 5m, where m is an integer.

Substituting this value of p in (i), we get

$$\begin{aligned}(5m)^2 = 5q^2 \\[0.5em] \Rightarrow 25m^2 = 5q^2 \\[0.5em] \Rightarrow 5m^2 = q^2 \\[0.5em]\end{aligned}$$

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

$\Rightarrow 5 \text{ divides } q \qquad \text{(Theorem 1)}$

Thus, p and q have a common factor 5. This contradicts that p and q have no common factors (except 1).

Hence, $\sqrt{5}$ is not a rational number. So, we conclude that $\sqrt{5}$ is an irrational number.

Let $\sqrt{7}$ be a rational number, then

$\sqrt{7} = \frac{p}{q},$

where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)

$$\begin{aligned}\Rightarrow 7 = \frac{p^2}{q^2} \\[0.5em] \Rightarrow p^2 = 7q^2 \qquad \text{....(i)}\end{aligned}$$

As 7 divides 7q², so 7 divides p² but 7 is prime

$\Rightarrow 7 \text{ divides } p \qquad \text{(Theorem 1)}$

Let p = 7m, where m is an integer.

Substituting this value of p in (i), we get

$$\begin{aligned}(7m)^2 = 7q^2 \\[0.5em] \Rightarrow 49m^2 = 7q^2 \\[0.5em] \Rightarrow 7m^2 = q^2 \\[0.5em]\end{aligned}$$

As 7 divides 7m², so 7 divides q² but 7 is prime

$\Rightarrow 7 \text{ divides } q \qquad \text{(Theorem 1)}$

Thus, p and q have a common factor 7. This contradicts that p and q have no common factors (except 1).

Hence, $\sqrt{7}$ is not a rational number. So, we conclude that $\sqrt{7}$ is an irrational number.

Suppose that $\sqrt{6}$ is a rational number, then

$\sqrt{6} = \frac{p}{q},$

where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)

$$\begin{aligned}\Rightarrow 6 = \frac{p^2}{q^2} \\[0.5em] \Rightarrow p^2 = 6q^2 \qquad \text{....(i)}\end{aligned}$$

As 2 divides 6q², so 2 divides p² but 2 is prime

$\Rightarrow 2 \text{ divides } p \qquad \text{(Theorem 1)}$

Let p = 2k, where k is some integer.

Substituting this value of p in (i), we get

$$\begin{aligned}(2k)^2 = 6q^2 \\[0.5em] \Rightarrow 4k^2 = 6q^2 \\[0.5em] \Rightarrow 2k^2 = 3q^2 \\[0.5em]\end{aligned}$$

As 2 divides 2k², so 2 divides 3q²

$\Rightarrow$ 2 divides 3 or 2 divides q²

But 2 does not divide 3, therefore, 2 divides q²

$\Rightarrow$ 2 divides q      (Theorem 1)

Thus, p and q have a common factor 2. This contradicts that p and q have no common factors (except 1).

Hence, our supposition is wrong. Therefore, $\sqrt{6}$ is not a rational number. So, we conclude that $\sqrt{6}$ is an irrational number.

Let $\frac{1}{\sqrt{11}}$ be a rational number, then

$\frac{1}{\sqrt{11}} = \frac{p}{q},$

where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)

$$\begin{aligned}\Rightarrow \frac{1}{11} = \frac{p^2}{q^2} \\[0.5em] \Rightarrow q^2 = 11p^2 \qquad \text{....(i)}\end{aligned}$$

As 11 divides 11p², so 11 divides q² but 11 is prime

$\Rightarrow 11 \text{ divides } q \qquad \text{(Theorem 1)}$

Let q = 11m, where m is an integer.

Substituting this value of q in (i), we get

$$\begin{aligned}(11m)^2 = 11p^2 \\[0.5em] \Rightarrow 121m^2 = 11p^2 \\[0.5em] \Rightarrow 11m^2 = p^2 \\[0.5em]\end{aligned}$$

As 11 divides 11m², so 11 divides p² but 11 is prime

$\Rightarrow 11 \text{ divides } p \qquad \text{(Theorem 1)}$

Thus, p and q have a common factor 11. This contradicts that p and q have no common factors (except 1).

Hence, $\frac{1}{\sqrt{11}}$ is not a rational number. So, we conclude that $\frac{1}{\sqrt{11}}$ is an irrational number.

Let $\sqrt{2}$ be a rational number, then

$\sqrt{2} = \frac{p}{q},$

where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)

$$\begin{aligned}\Rightarrow 2 = \frac{p^2}{q^2} \\[0.5em] \Rightarrow p^2 = 2q^2 \qquad \text{....(i)}\end{aligned}$$

As 2 divides 2q², so 2 divides p² but 2 is prime

$\Rightarrow 2 \text{ divides } p \qquad \text{(Theorem 1)}$

Let p = 2m, where m is an integer.

Substituting this value of p in (i), we get

$$\begin{aligned}(2m)^2 = 2q^2 \\[0.5em] \Rightarrow 4m^2 = 2q^2 \\[0.5em] \Rightarrow 2m^2 = q^2 \\[0.5em]\end{aligned}$$

As 2 divides 2m², so 2 divides q² but 2 is prime

$\Rightarrow 2 \text{ divides } q \qquad \text{(Theorem 1)}$

Thus, p and q have a common factor 2. This contradicts that p and q have no common factors (except 1).

Hence, $\sqrt{2}$ is not a rational number. So, we conclude that $\sqrt{2}$ is an irrational number.

Suppose that $3 - \sqrt{2}$ is a rational number, say r.

Then, $3 - \sqrt{2}$ = r (note that r ≠ 0)

$$\begin{aligned}\Rightarrow - \sqrt{2} = r - 3 \\[0.5em] \Rightarrow \sqrt{2} = 3 - r \\[0.5em]\end{aligned}$$

As r is rational and r ≠ 0, so 3 - r is rational

$\Rightarrow \sqrt{2}$ is rational

But this contradicts that $\sqrt{2}$ is irrational. Hence, our supposition is wrong.

∴ $3 - \sqrt{2}$ is an irrational number.

Let $\sqrt{3}$ be a rational number, then

$\sqrt{3} = \frac{p}{q},$

where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)

$$\begin{aligned}\Rightarrow 3 = \frac{p^2}{q^2} \\[0.5em] \Rightarrow p^2 = 3q^2 \qquad \text{....(i)}\end{aligned}$$

As 3 divides 3q², so 3 divides p² but 3 is prime

$\Rightarrow 3 \text{ divides } p \qquad \text{(Theorem 1)}$

Let p = 3m, where m is an integer.

Substituting this value of p in (i), we get

$$\begin{aligned}(3m)^2 = 3q^2 \\[0.5em] \Rightarrow 9m^2 = 3q^2 \\[0.5em] \Rightarrow 3m^2 = q^2 \\[0.5em]\end{aligned}$$

As 3 divides 3m², so 3 divides q² but 3 is prime

$\Rightarrow 3 \text{ divides } q \qquad \text{(Theorem 1)}$

Thus, p and q have a common factor 3. This contradicts that p and q have no common factors (except 1).

Hence, $\sqrt{3}$ is not a rational number. So, we conclude that $\sqrt{3}$ is an irrational number.

Suppose that $\frac{2}{5}\sqrt{3}$ is a rational number, say r.

Then, $\frac{2}{5}\sqrt{3}$ = r (note that r ≠ 0)

$$\begin{aligned}\Rightarrow \sqrt{3} = \frac{5}{2}r \\[0.5em]\end{aligned}$$

As r is rational and r ≠ 0, so $\frac{5}{2}r$ is rational
[∵ System of rational numbers is closed under all four fundamental arithmetic operations (except division by zero)]

$\Rightarrow \sqrt{3}$ is rational

But this contradicts that $\sqrt{3}$ is irrational. Hence, our supposition is wrong.

∴ $\frac{2}{5}\sqrt{3}$ is an irrational number.

Let $\sqrt{5}$ be a rational number, then

$\sqrt{5} = \frac{p}{q},$

where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)

$$\begin{aligned}\Rightarrow 5 = \frac{p^2}{q^2} \\[0.5em] \Rightarrow p^2 = 5q^2\end{aligned}$$

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

$\Rightarrow 5 \text{ divides } p \qquad \text{(Theorem 1)}$

Let p = 5m, where m is an integer.

Substituting this value of p in (i), we get

$$\begin{aligned}(5m)^2 = 5q^2 \\[0.5em] \Rightarrow 25m^2 = 5q^2 \\[0.5em] \Rightarrow 5m^2 = q^2 \\[0.5em]\end{aligned}$$

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

$\Rightarrow 5 \text{ divides } q \qquad \text{(Theorem 1)}$

Thus, p and q have a common factor 5. This contradicts that p and q have no common factors (except 1).

Hence, $\sqrt{5}$ is not a rational number. So, we conclude that $\sqrt{5}$ is an irrational number.

Suppose that $-3 + 2\sqrt{5}$ is a rational number, say r.

Then, $-3 + 2\sqrt{5}$ = r (note that r ≠ 0)

$$\begin{aligned}\Rightarrow 2\sqrt{5} = r + 3 \\[0.5em] \Rightarrow \sqrt{5} = \frac{r + 3}{2} \\[0.5em]\end{aligned}$$

As r is rational and r ≠ 0, so $\frac{r + 3}{2}$ is rational

$\Rightarrow \sqrt{5}$ is rational

But this contradicts that $\sqrt{5}$ is irrational. Hence, our supposition is wrong.

∴ $-3 + 2\sqrt{5}$ is an irrational number.

$\text{(i) } 5 + \sqrt{2}$

Let us assume that $5 + \sqrt{2}$ is a rational number, say r.

Then,

$$\begin{aligned}5 + \sqrt{2} = r \\[0.5em] \Rightarrow \sqrt{2} = r - 5\end{aligned}$$

As r is rational, r - 5 is rational

$\Rightarrow \sqrt{2}$ is rational

But this contradicts the fact that $\sqrt{2}$ is irrational.

Hence, our assumption is wrong.

∴ $5 + \sqrt{2}$ is an irrational number.

$\text{(ii) } 3 - 5\sqrt{3}$

Let us assume that $3 - 5\sqrt{3}$ is a rational number, say r.

Then,

$$\begin{aligned}3 - 5\sqrt{3} = r \\[0.5em] \Rightarrow 5\sqrt{3} = 3 - r \\[0.5em] \Rightarrow \sqrt{3} = \frac{3 - r}{5} \\[0.5em]\end{aligned}$$

As r is rational, 3 - r is rational

$\Rightarrow \frac{3 - r}{5}$ is rational

$\Rightarrow \sqrt{3}$ is rational

But this contradicts the fact that $\sqrt{3}$ is irrational.

Hence, our assumption is wrong.

∴ $3 - 5\sqrt{3}$ is an irrational number.

$\text{(iii) } 2\sqrt{3} - 7$

Let us assume that $2\sqrt{3} - 7$ is a rational number, say r.

Then,

$$\begin{aligned}2\sqrt{3} - 7 = r \\[0.5em] \Rightarrow 2\sqrt{3} = r + 7 \\[0.5em] \Rightarrow \sqrt{3} = \frac{r + 7}{2} \\[0.5em]\end{aligned}$$

As r is rational, r + 7 is rational

$\Rightarrow \frac{r + 7}{2}$ is rational

$\Rightarrow \sqrt{3}$ is rational

But this contradicts the fact that $\sqrt{3}$ is irrational.

Hence, our assumption is wrong.

∴ $2\sqrt{3} - 7$ is an irrational number.

$\text{(iv) } \sqrt{2} + \sqrt{5}$

Let us assume that $\sqrt{2} + \sqrt{5}$ is a rational number, say r.

Then,

$$\begin{aligned}\sqrt{2} + \sqrt{5} = r \\[0.5em] \Rightarrow \sqrt{5} = r - \sqrt{2} \\[0.5em] \Rightarrow (\sqrt{5})^2 = (r - \sqrt{2})^2 \\[0.5em] \Rightarrow 5 = r^2 + 2 - 2\sqrt{2}r \\[0.5em] \Rightarrow 2\sqrt{2}r = r^2 + 2 - 5 \\[0.5em] \Rightarrow 2\sqrt{2}r = r^2 - 3 \\[0.5em] \Rightarrow \sqrt{2} = \frac{r^2 - 3}{2r} \\[0.5em]\end{aligned}$$

As r is rational,

$\Rightarrow \frac{r^2 - 3}{2r}$ is rational

$\Rightarrow \sqrt{2}$ is rational

But this contradicts the fact that $\sqrt{2}$ is irrational.

Hence, our assumption is wrong.

∴ $\sqrt{2} + \sqrt{5}$ is an irrational number.

Locating $\sqrt{10}$:

Representing 10 as the sum of squares of two natural numbers:

10 = 9 + 1 = 3² + 1²

Let l be the number line. If point O represents number 0 and point A represents number 3, then draw a line segment OA = 3 units.

At A, draw AC ⟂ OA. From AC, cut off AB = 1 unit.

We observe that OAB is a right angled triangle at A. By Pythagoras theorem, we get:

$$\begin{aligned}OB^2 = OA^2 + AB^2 \\[0.5em] \Rightarrow OB^2 = 3^2 + 1^2 \\[0.5em] \Rightarrow OB^2 = 9 + 1 \\[0.5em] \Rightarrow OB^2 = 10 \\[0.5em] \Rightarrow OB = \sqrt{10} \text{ units} \\[0.5em]\end{aligned}$$

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

As OP = OB = $\sqrt{10}$ units, the point P will represent the number $\sqrt{10}$ on the number line as shown in the figure below:

Locating $\sqrt{17}$:

Representing 17 as the sum of squares of two natural numbers:

17 = 16 + 1 = 4² + 1²

Let l be the number line. If point O represents number 0 and point A represents number 4, then draw a line segment OA = 4 units.

At A, draw AC ⟂ OA. From AC, cut off AB = 1 unit.

We observe that OAB is a right angled triangle at A. By Pythagoras theorem, we get:

$$\begin{aligned}OB^2 = OA^2 + AB^2 \\[0.5em] \Rightarrow OB^2 = 4^2 + 1^2 \\[0.5em] \Rightarrow OB^2 = 16 + 1 \\[0.5em] \Rightarrow OB^2 = 17 \\[0.5em] \Rightarrow OB = \sqrt{17} \text{units} \\[0.5em]\end{aligned}$$

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

As OP = OB = $\sqrt{17}$ units, the point P will represent the number $\sqrt{17}$ on the number line as shown in the figure below:

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

$\therefore \frac{36}{100} = 0.36$

Remainder becomes zero.

Decimal expansion of $\bold{\frac{36}{100}}$ is terminating.

$\text{(ii) } 4\frac{1}{8}$

$\therefore 4\frac{1}{8} = 4.125$

Remainder becomes zero.

Decimal expansion of $\bold{4\frac{1}{8}}$ is terminating.

$\text{(iii) } \frac{2}{9}$

$\therefore \frac{2}{9} = 0.2222..... = 0.\overline{2}$

Remainder is repeating.

Decimal expansion of $\bold{\frac{2}{9}}$ is non-terminating repeating.

$\text{(iv) } \frac{2}{11}$

$\therefore \frac{2}{11} = 0.1818..... = 0.\overline{18}$

Remainder is repeating.

Decimal expansion of $\bold{\frac{2}{11}}$ is non-terminating repeating.

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

$\therefore \frac{3}{13} = 0.2307692307..... = 0.\overline{230769}$

Remainder is repeating.

Decimal expansion of $\bold{\frac{3}{13}}$ is non-terminating repeating.

$\text{(vi) } \frac{329}{400}$

$\therefore \frac{329}{400} = 0.8225$

Remainder becomes zero.

Decimal expansion of $\bold{\frac{329}{400}}$ is terminating.

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

The given number $\frac{13}{3125}$ is in its lowest form.

Prime factorization of denominator 3125:

$$\begin{array}{l|l} 5 & 3125 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

3125 = 5 x 5 x 5 x 5 x 5 x 1
= 5⁵ x 1
= 1 x 5⁵
= 2⁰ x 5⁵    [∵ 2⁰ = 1]

Denominator is of the form 2^m x 5ⁿ, where m, n are non-negative integers.

∴ The given number $\frac{13}{3125}$ has a terminating decimal expansion.

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

The given number $\frac{17}{8}$ is in its lowest form.

Prime factorization of denominator 8:

$$\begin{array}{l|l} 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$$

8 = 2 x 2 x 2 x 1
= 2³ x 1
= 2³ x 5⁰    [∵ 5⁰ = 1]

Denominator is of the form 2^m x 5ⁿ, where m, n are non-negative integers.

∴ The given number $\frac{17}{8}$ has a terminating decimal expansion.

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

The given number $\frac{23}{75}$ is in its lowest form.

Prime factorization of denominator 75:

$$\begin{array}{l|l} 3 & 75 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

75 = 3 x 5 x 5 x 1
= 3 x 5² x 1
= 3 x 5² x 2⁰    [∵ 2⁰ = 1]

Denominator has a prime factor 3 other than 2 or 5.

∴ The given number $\frac{23}{75}$ has a non-terminating repeating decimal expansion.

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

Both numerator and denominator contain common factor 3. Reducing the number to its lowest form:

$$\begin{aligned}\frac{6}{15} = \frac{\cancel{3} \times 2}{\cancel{3} \times 5} \\[0.5em] = \frac{2}{5}\end{aligned}$$

The Denominator 5 = 2⁰ x 5¹

Denominator is of the form 2^m x 5ⁿ, where m, n are non-negative integers.

∴ The given number $\frac{6}{15}$ has a terminating decimal expansion.

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

The given number $\frac{1258}{625}$ is in its lowest form.

Prime factorization of denominator 625:

$$\begin{array}{l|l} 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

625 = 5 x 5 x 5 x 5 x 1
= 5⁴ x 1
= 1 x 5⁴
= 2⁰ x 5⁴    [∵ 2⁰ = 1]

Denominator is of the form 2^m x 5ⁿ, where m, n are non-negative integers.

∴ The given number $\frac{1258}{625}$ has a terminating decimal expansion.

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

Both numerator and denominator contain common factor 7. Reducing the number to its lowest form:

$$\begin{aligned}\frac{77}{210} = \frac{\cancel{7} \times 11}{\cancel{7} \times 30} \\[0.5em] = \frac{11}{30}\end{aligned}$$

Prime factorization of denominator 30:

$$\begin{array}{l|l} 2 & 30 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

30 = 2 x 3 x 5

Denominator has a prime factor 3 other than 2 or 5.

∴ The given number $\frac{77}{210}$ has a non-terminating repeating decimal expansion.

GCD of numerator and denominator is 21. Reducing the number to its lowest form:

$$\begin{aligned}\frac{987}{10500} = \frac{\cancel{21} \times 47}{\cancel{21} \times 500} \\[0.5em] = \frac{47}{500}\end{aligned}$$

Prime factorization of denominator 500:

$$\begin{array}{l|l} 2 & 500 \\ \hline 2 & 250 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

500 = 2 x 2 x 5 x 5 x 5 = 2² x 5³

Denominator is of the form 2^m x 5ⁿ, where m, n are non-negative integers.

∴ The given number $\frac{987}{10500}$ has a terminating decimal expansion.

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

The given number $\frac{17}{8}$ is in its lowest form.

Prime factorization of denominator 8:

$$\begin{array}{l|l} 2 & 8 \\ \hline 2 & 4 \\ \hline 2 & 2 \\ \hline & 1 \end{array}$$

8 = 2 x 2 x 2 x 1
= 2³ x 5⁰    [∵ 5⁰ = 1]

$$\begin{aligned}\frac{17}{8} = \frac{17}{2^3} \\[0.5em] = \frac{17 \times 5^3}{2^3 \times 5^3} \\[0.5em] = \frac{17 \times 125}{(2 \times 5)^3} \\[0.5em] = \frac{2125}{10^3} \\[0.5em] = 2.125 \\[0.5em] \bold{\therefore \frac{17}{8} = 2.125}\end{aligned}$$

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

The given number $\frac{13}{3125}$ is in its lowest form.

Prime factorization of denominator 3125:

$$\begin{array}{l|l} 5 & 3125 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

3125 = 5 x 5 x 5 x 5 x 5 x 1
= 5⁵ x 1
= 1 x 5⁵
= 2⁰ x 5⁵    [∵ 2⁰ = 1]

$$\begin{aligned}\frac{13}{3125} = \frac{13}{5^5} \\[0.5em] = \frac{13 \times 2^5}{2^5 \times 5^5} \\[0.5em] = \frac{13 \times 32}{(2 \times 5)^5} \\[0.5em] = \frac{416}{10^5} \\[0.5em] = 0.00416 \\[0.5em] \bold{\therefore \frac{13}{3125} = 0.00416}\end{aligned}$$

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

The given number $\frac{7}{80}$ is in its lowest form.

Prime factorization of denominator 80:

$$\begin{array}{l|l} 2 & 80 \\ \hline 2 & 40 \\ \hline 2 & 20 \\ \hline 2 & 10 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

80 = 2 x 2 x 2 x 2 x 5
= 2⁴ x 5
= 2⁴ x 5¹

$$\begin{aligned}\frac{7}{80} = \frac{7}{2^4 \times 5^1} \\[0.5em] = \frac{7 \times 5^3}{2^4 \times 5^1 \times 5^3} \\[0.5em] = \frac{7 \times 125}{2^4 \times 5^4} \\[0.5em] = \frac{7 \times 125}{(2 \times 5)^4} \\[0.5em] = \frac{875}{10^4} \\[0.5em] = 0.0875 \\[0.5em] \bold{\therefore \frac{7}{80} = 0.0875}\end{aligned}$$

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

GCD of numerator and denominator is 3. Reducing the number to its lowest form:

$$\begin{aligned}\frac{6}{15} = \frac{\cancel{3} \times 2}{\cancel{3} \times 5} \\[0.5em] = \frac{2}{5}\end{aligned}$$

$$\begin{aligned}\frac{6}{15} = \frac{2}{5} \\[0.5em] = \frac{2 \times 2}{2 \times 5} \\[0.5em] = \frac{4}{10} \\[0.5em] = 0.4 \\[0.5em] \bold{\therefore \frac{6}{15} = 0.4}\end{aligned}$$

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

$$\begin{aligned}\frac{2^2 \times 7}{5^4} = \frac{2^2 \times 7 \times 2^4}{5^4 \times 2^4} \\[0.5em] = \frac{4 \times 7 \times 16}{(2 \times 5)^4} \\[0.5em] = \frac{448}{10^4} \\[0.5em] = 0.0448 \\[0.5em] \bold{\therefore \frac{2^2 \times 7}{5^4} = 0.0448}\end{aligned}$$

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

GCD of numerator and denominator is 3. Reducing the number to its lowest form:

$$\begin{aligned}\frac{237}{1500} = \frac{\cancel{3} \times 79}{\cancel{3} \times 500} \\[0.5em] = \frac{79}{500}\end{aligned}$$

$$\begin{aligned}\frac{237}{1500} = \frac{79}{500} \\[0.5em] = \frac{79 \times 2}{500 \times 2} \\[0.5em] = \frac{158}{10^3} \\[0.5em] = 0.158 \\[0.5em] \bold{\therefore \frac{237}{1500} = 0.158}\end{aligned}$$

The given number $\frac{257}{5000}$ is in its lowest form.

Prime factorization of denominator 5000:

$$\begin{array}{l|l} 2 & 5000 \\ \hline 2 & 2500 \\ \hline 2 & 1250 \\ \hline 5 & 625 \\ \hline 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

5000 = 2 x 2 x 2 x 5 x 5 x 5 x 5
= 2³ x 5⁴

Hence, denominator of rational number $\frac{257}{5000}$ in the form 2^m × 5ⁿ is 2³ x 5⁴ where m = 3 and n = 4.

$$\begin{aligned}\frac{257}{5000} = \frac{257}{2^3 \times 5^4} \\[0.5em] = \frac{257 \times 2}{2^3 \times 5^4 \times 2} \\[0.5em] = \frac{514}{2^4 \times 5^4} \\[0.5em] = \frac{514}{(2 \times 5)^4} \\[0.5em] = \frac{514}{10^4} \\[0.5em] = 0.0514 \\[0.5em] \bold{\therefore \frac{257}{5000} = 0.0514}\end{aligned}$$

The fraction : $\frac{1}{7}$

Decimal expansion of $\frac{1}{7}$ = $\bold{0.\overline{142857}}$

Since, it is recurring

$\frac{2}{7}=2\times\frac{1}{7}=2\times0.\overline{142857}=\bold{0.\overline{285714}}$

$\frac{3}{7}=3\times\frac{1}{7}=3\times0.\overline{142857}=\bold{0.\overline{428571}}$

$\frac{4}{7}=4\times\frac{1}{7}=4\times0.\overline{142857}=\bold{0.\overline{571428}}$

$\frac{5}{7}=5\times\frac{1}{7}=5\times0.\overline{142857}=\bold{0.\overline{714285}}$

$\frac{6}{7}=6\times\frac{1}{7}=6\times0.\overline{142857}=\bold{0.\overline{857142}}$

(i) Let x = $0.\overline{3}$ = 0.333333... $\qquad \text{....(i)}$

As there is one repeating digit after decimal point,

So multiplying both sides of (i) by 10

we get,

10x = 3.3333...$\qquad \text{....(ii)}$

Subtracting (i) from (ii), we get

9x = 3

x = $\frac{3}{9}$ = $\bold{\frac{1}{3}}$,

which is in the form of $\frac{p}{q}$, q ≠ 0.

(ii) Let x = $5.\overline{2}$ = 5.2222... $\qquad \text{....(i)}$

As there is one repeating digit after decimal point,

So multiplying both sides of (i) by 10

we get,

10x = 52.2222...$\qquad \text{....(ii)}$

Subtracting (i) from (ii), we get

9x = 47

x = $\bold{\frac{47}{9}}$,

Which is in the form of $\frac{p}{q}$, q ≠ 0.

(iii) Let x = $0.\overline{40}$ = 0.4040... $\qquad \text{....(i)}$

As there are two repeating digit after decimal point,

So multiplying both sides of (i) by 100

we get,

100x = 40.4040...$\qquad \text{....(ii)}$

Subtracting (i) from (ii), we get

99x = 40

x = $\bold{\frac{40}{99}}$,

Which is in the form of $\frac{p}{q}$, q ≠ 0

(iv) Let x = $0.4\overline{7}$ = 0.477777... $\qquad \text{....(i)}$

As there is one repeating digit after decimal point ,

So multiplying both sides of (i) by 10

we get,

10x=4.7777...$\qquad \text{....(ii)}$

Multiply by 100 on both sides

100x=47.77777.....$\qquad \text{....(iii)}$

subtracting (ii) from (iii), we get

100x-10x=47.7777.. -4.777...

90x= 43

x = $\bold{\frac{43}{90}}$

Which is in the form of $\frac{p}{q}$, q ≠ 0

(v) Let x = $0.1\overline{34}$ = 0.13434 ... $\qquad \text{....(i)}$

So multiplying both sides of (i) by 10

we get,

10x=1.343434...$\qquad \text{....(ii)}$

Again multiply by 100 on both sides ,

1000x =134.3434.....$\qquad \text{....(iii)}$

Subtracting (ii) from (iii), we get

1000x - 10x = 134.3434... - 1.3434...

990x = 133

x = $\bold{\frac{133}{990}}$

which is in the form of $\frac{p}{q}$, q ≠ 0

(vi) Let x = $0.\overline{001}$ = 0.001001001... $\qquad \text{....(i)}$

So multiplying both sides of (i) by 1000,

we get,

1000x = 1.001001...$\qquad \text{....(ii)}$

Subtracting (i) from (ii), we get

1000x - x = 1.001001... - 0.001001...

999x = 1

x = $\bold{\frac{1}{999}}$

which is in the form of $\frac{p}{q}$, q ≠ 0.

Rational numbers are in the form c, q ≠ 0, and p and q are integers.

(i) $\bold{\sqrt{23}}$ is an irrational number , as it is not a perfect square so it cannot be written in the form $\frac{p}{q}$ , q ≠ 0.

(ii) $\sqrt{225}$ = $\sqrt{15×15}$ = 15 = $\frac{15}{1}$,

As it can be written in the form $\frac{p}{q}$ , q ≠ 0.

∴ $\bold{\sqrt{225}}$ is a rational number .

(iii) 0.3796 = $\frac{3796}{1000}$

Since, the decimal expansion is terminating decimal.

∴ 0.3796 is a rational number .

(iv) 7.478478

Let x = 7.478478 $\text{....(i)}$

Since there is three repeating digit after decimal point,

Multiplying both sides by 1000, we get

1000x = 7478.478478... $\text{....(ii)}$

Subtracting (i) from (ii) we get,

999x = 7471

x = $\bold{\frac{7471}{999}}$

∴ It is non terminating , repeating rational number .

(v) 1.101001000100001...

Since number of 0's are increasing between two consecutive terms as we move further , So it is non terminating, non repeating decimal.

∴ 1.101001000100001... is an irrational number.

(vi) $345.0\overline{456}$

$345.0\overline{456}$ = 345.0456456...

Let x = 345.0456456...

Multiply both sides by 10, we get

10x = 3450.456456.. $\text{....(i)}$

Since, after decimal there are three repeating digit:

Multiply both sides by 1000, we get

10000x = 3450456.456456... $\text{....(ii)}$

Subtracting (i) from (ii) ,

9990x = 3447006

x = $\bold{\frac{3447006}{9990}}$

since, it is non-terminating , repeating decimal.

∴ $\bold{345.0\overline{456}}$ is a Rational number .

(i) 37.07158

This can be written as 37.09158 = $\frac{3709158}{100000}$

Since , it is terminating decimal

It is Rational number and the prime factors of its denominator q will be 2 or 5 or both .

(ii) $423.\overline{04567}$

since it has non-terminating recurring decimal,

$423.\overline{04567}$ = 423.0456704567...

It is a rational number which is non-terminating and repeating. Its denominator q will have prime factors other than 2 or 5.

(iii) 8.9010010001...

Since, it is non-terminating non-repeating decimal number

∴ It is not a Rational number.

(iv) 2.3476817681... = $2.34\overline{7681}$

Since, it is a non-terminating repeating decimal number,

∴ It is a Rational number and its denominator q will have prime factors other than 2 or 5.

(i) One irrational number between $\frac{1}{3}$ and $\frac{1}{2}$

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

$\frac{1}{2}$ = 0.5

So there are infinite irrational number between $\frac{1}{3}$ and $\frac{1}{2}$

One irrational number among them can be $\bold{0.4040040004...}$

(ii) One irrational number between $-\frac{2}{5}$ and $\frac{1}{2}$

$-\frac{2}{5}$ = -0.4

$\frac{1}{2}$ = 0.5

So, there are infinite irrational numbers between $-\frac{2}{5}$ and $\frac{1}{2}$

One irrational number among them can be $\bold{0.2020020002...}$

(iii) One irrational number among 0 and 0.1 , can be $\bold{0.050050005...}$

Consider, the squares $(2)^2$ = 4 and $(3)^2$ = 9

Two irrational numbers can be te squares root of any natural number between 4 and 9.

As, As, $4 \lt 5 \lt 6\lt 9$ , it follows that

$\sqrt{4} \lt \sqrt{5} \lt \sqrt{6} \lt \sqrt{9}$

therefore , $\sqrt{5}$ and $\sqrt{6}$ lie between 2 and 3

$$\begin{aligned}2 \lt \sqrt{5} \lt \sqrt{6} \lt \\ 3\end{aligned}$$

Hence, two irrational number between 2 and 3 or $\sqrt{4}$ and $\sqrt{9}$ are $\sqrt{5}$ , $\sqrt{6}$ .

$\frac{4}{9}$ is expressed as 0.4444...

$\frac{7}{11}$ is expressed as 0.636363..

So, two irrational number between $\frac{4}{9}$ and $\frac{7}{11}$ are 0.5050050005... and 0.6060060006...

Consider the squares of $\sqrt{2}$ and $\sqrt{3}$

${(\sqrt2)^2}$ = 2 and ${(\sqrt3)^2}$ = 3

Take any rational number between 2 and 3 which is a perfect squares of a rational number,

One such number is 2.25 and

2.25 = $(1.5)^2$

$\sqrt{2.25}$ = 1.5

As, $2 \lt 2.25 \lt 3$ , it follows that

$\sqrt{2} \lt \sqrt{2.25} \lt \sqrt{3}$

$\sqrt{2} \lt 1.5 \lt \sqrt{3}$

Hence , one rational number between $\sqrt{2}$ and $\sqrt{3}$ is 1.5 .

$2\sqrt{3}$ = $\sqrt{4 × 3}$ = $\sqrt{12}$

So, we need to find two irrational number between $\sqrt{12}$ and $\sqrt{15}$

$$\begin{aligned}\text{Since}, 12 \lt 12.25 \lt 12.96 \lt 15 \\[0.5em] \Rightarrow \sqrt{12} \lt \sqrt{12.25} \lt \sqrt{12.96} \lt \sqrt{15}\end{aligned}$$

Hence, two rational number between $\sqrt{12}$ and $\sqrt{15}$ are $\bold{\sqrt{12.25}}$ and $\bold{\sqrt{12.96}}$ .

consider the squares of $\sqrt{5}$ and $\sqrt{7}$

${(\sqrt5)^2}$ = 5 and ${(\sqrt7)^2}$ = 7

As, $5 \lt 6\lt 7$ , it follows that

$\sqrt{5} \lt \sqrt{6} \lt \sqrt{7}$ therefore , $\sqrt{6}$ lie between $\sqrt{5}$ and $\sqrt{7}$

Hence, irrational number between $\sqrt{5}$ and $\sqrt{7}$ is $\bold{\sqrt{6}}$ .

consider the squares of $\sqrt{3}$ and $\sqrt{7}$

${(\sqrt3)^2}$ = 3 and ${(\sqrt7)^2}$ = 7

As, $3 \lt 5 \lt 6\lt 7$ , it follows that

$\sqrt{3} \lt \sqrt{5} \lt \sqrt{6} \lt \sqrt{7}$ therefore , $\sqrt{5}$ and $\sqrt{6}$ lie between $\sqrt{3}$ and $\sqrt{7}$

Hence, two irrational number between $\sqrt{3}$ and $\sqrt{7}$ is $\bold{\sqrt{5}}$ and $\bold{\sqrt{6}}$ .

$$\begin{aligned}\text{(i) } \sqrt{45} - 3\sqrt{20} + 4\sqrt{5} \\[1.5em] =\sqrt{3 × 3 × 5} - 3\sqrt{5 × 4} + 4\sqrt{5} \\[1.5em] = 3\sqrt{5} - 3 × 2\sqrt{5} + 4\sqrt5 \\[1.5em] = 3\sqrt{5} - 6\sqrt{5} + 4\sqrt5 \\[1.5em] = \sqrt{5}(3 - 6 + 4) \\[1.5em] = \bold{\sqrt{5}}\end{aligned}$$

$$\begin{aligned}\text{(ii) } 3\sqrt{3} + 2\sqrt{27} + \frac{7}{\sqrt{3}} \\[1.5em] = 3\sqrt{3} + 2\sqrt{3 × 3 × 3} + \frac{7}{\sqrt{3}} \\[1.5em] = 3\sqrt{3} + 2 × 3\sqrt{3} + \frac{7}{\sqrt{3}} \\[1.5em] = 3\sqrt{3} + 6\sqrt{3} + \frac{7}{\sqrt{3}} × \frac{\sqrt3}{\sqrt3} \\[1.5em] = \sqrt{3} × (3 + 6 + \frac{7}{3}) = \bold{\frac{34}{3}{\sqrt{3}}} \\[1.5em]\end{aligned}$$

$$\begin{aligned}\text{(iii) } 6\sqrt{5} × 2\sqrt{5} \\[1.5em] = 12 × (\sqrt{5} × \sqrt{5}) \\[1.5em] = 12 × (\sqrt{5})^2 \\[1.5em] = 12 × 5 \\[1.5em] = \bold{60} \\[1.5em]\end{aligned}$$

$$\begin{aligned}\text{(iv) } 8\sqrt{15} ÷ 2\sqrt{3} \\[1.5em] = \frac{8\sqrt{15}}{2\sqrt{3}} \\[1.5em] = \frac{8\sqrt{3×5}}{2\sqrt{3}} \\[1.5em] = \frac{8\sqrt{3}\sqrt{5}}{2\sqrt{3}} \\[1.5em] = \frac{8\sqrt{5}}{2} \\[1.5em] = \bold{4\sqrt{5}} \\[1.5em]\end{aligned}$$

$$\begin{aligned}\text{(v) } \frac{\sqrt{24}}{8} + \frac{\sqrt{54}}{9} \\[1.5em] = \frac{\sqrt{2 × 2 × 6}}{8} + \frac{\sqrt{3 × 3 × 6}}{9} \\[1.5em] = \frac{2\sqrt{6}}{8} + \frac{3\sqrt6}{9}\\[1.5em] = \frac{\sqrt{6}}{4} + \frac{\sqrt6}{3} \\[1.5em] = \sqrt{6} × {(\frac{1}{4} + \frac{1}{3})} \\[1.5em] = \sqrt{6} × (\frac{3 + 4}{12}) \\[1.5em] = \bold{\frac{7}{12}{\sqrt{6}}}\end{aligned}$$

$$\begin{aligned}\text{(vi) } \frac{3}{\sqrt{8}} + \frac{1}{\sqrt{2}} \\[1.5em] = \frac{3}{\sqrt{2 × 2 × 2}} + \frac{1}{\sqrt{2}} \\[1.5em] = \frac{3}{2\sqrt{2}} + \frac{1}{\sqrt2} \\[1.5em] = \frac{1}{\sqrt2} × (\frac{3}{2} + 1) \\[1.5em] = \frac{1}{\sqrt2} × (\frac{3 + 2}{2}) \\[1.5em] = \frac{1}{\sqrt2} × \frac{5}{2} \\[1.5em] = \frac{1}{\sqrt2} × \frac{\sqrt{2}}{\sqrt{2}} × \frac{5}{2} \\[1.5em] = \bold{\frac{5\sqrt{2}}{4}} \\[1.5em]\end{aligned}$$

$$\begin{aligned}\text{(i) } (5 + \sqrt{7})(2 + \sqrt{5}) \\[1.5em] = 10 + 5\sqrt{5} + 2\sqrt{7} + \sqrt{7}\sqrt{5} \\[1.5em] = \bold{10 + 5\sqrt{5} + 2\sqrt{7} + \sqrt{35}} \\[1.5em]\end{aligned}$$

$$\begin{aligned}\text{(ii) } (5 + \sqrt{5})(5 - \sqrt{5}) \\[1.5em] \text{Using identity} : (a + b)(a - b) = a^2 - b^2 \\[1.5em] = 5^2 - (\sqrt{5})^2 \\[1.5em] = \bold{25 - 5 = 20} \\[1.5em]\end{aligned}$$

$$\begin{aligned}\text{(iii) } (\sqrt{5} + \sqrt{2})^2 \\[1.5em] \text{Using identity} : (a + b)^2 = a^2 + 2ab + b^2 \\[1.5em] = (\sqrt{5} + \sqrt{2})^2 = {(\sqrt{5})}^2 + 2 × \sqrt{5} ×\sqrt{2} + {(\sqrt{2})}^2 \\[1.5em] = 5 + 2\sqrt{10} + 2 \\[1.5em] = \bold{7 + 2\sqrt{10}} \\[1.5em]\end{aligned}$$

$$\begin{aligned}\text{(iv) } (\sqrt{3} - \sqrt{7})^2 \\[1.5em] \text{Using identity} : (a - b)^2 = a^2 - 2ab + b^2 \\[1.5em] = (\sqrt{3} - \sqrt{7})^2 = {(\sqrt{3})}^2 - 2 × \sqrt{3} ×\sqrt{7} + {(\sqrt{7})}^2 \\[1.5em] = 3 - 2\sqrt{21} + 7 \\[1.5em] = \bold{10 - 2\sqrt{21}} \\[1.5em]\end{aligned}$$

$$\begin{aligned}\text{(v) } (\sqrt{2} + \sqrt{3})(\sqrt{5} + \sqrt{7}) \\[1.5em] = \sqrt{2} × \sqrt{5} + \sqrt{2} × \sqrt{7} + \sqrt{3} × \sqrt{5} + \sqrt{3} × \sqrt{7} \\[1.5em] = \bold{\sqrt{10} + \sqrt{14} + \sqrt{15} + \sqrt{21}} \\[1.5em]\end{aligned}$$

$$\begin{aligned}\text{(vi) } (4 + \sqrt{5})(\sqrt{3} - \sqrt{7}) \\[1.5em] = 4\sqrt{3} - 4\sqrt{7} + \sqrt{5} × \sqrt{3} - \sqrt{5} × \sqrt{7} \\[1.5em] = \bold{4\sqrt{3} - 4\sqrt{7} + \sqrt{15} - \sqrt{35}} \\[1.5em]\end{aligned}$$

$$\begin{aligned}\text{(i) } \sqrt{8} + \sqrt{50} + \sqrt{72} + \sqrt{98} \\[1.5em] = \sqrt{2 × 2 × 2} + \sqrt{5 × 5 × 2} + \sqrt{6 × 6 × 2} + \sqrt{2 × 7 × 7} \\[1.5em] = 2\sqrt{2} + 5\sqrt{2} + 6\sqrt{2} + 7\sqrt{2} \\[1.5em] = (2 + 5 + 6 + 7) × \sqrt{2} \\[1.5em] = (20) × \sqrt{2} \\[1.5em] = \bold{20 × 1.414 = 28.28 } \\[1.5em]\end{aligned}$$

$$\begin{aligned}\text{(ii) } 3\sqrt{32} - 2\sqrt{50} + 4\sqrt{128} - 20\sqrt{18} \\[1.5em] = 3\sqrt{2 × 4 × 4 } - 2\sqrt{5 × 5 × 2} + 4\sqrt{8 × 8 × 2} - 20\sqrt{2 × 3 × 3} \\[1.5em] = 12\sqrt{2} - 10\sqrt{2} + 32\sqrt{2} - 60\sqrt{2} \\[1.5em] = (12 - 10 + 32 - 60) × \sqrt{2} \\[1.5em] = (44 - 70) × \sqrt{2} \\[1.5em] = (-26) × \sqrt{2} \\[1.5em] = \bold{-26 × 1.414 = -36.764 } \\[1.5em]\end{aligned}$$

$$\begin{aligned}\text{(i) } \sqrt{27} + \sqrt{75} + \sqrt{108} - \sqrt{243} \\[1.5em] = \sqrt{3 × 3 × 3 } + \sqrt{3 × 5 × 5} + \sqrt{2 × 2 × 3 × 3 ×3 } - \sqrt{3 × 3 × 3 × 3 × 3} \\[1.5em] = 3\sqrt{3} + 5\sqrt{3} + 6\sqrt{3} - 9\sqrt{3} \\[1.5em] = (3 + 5 + 6 - 9) × \sqrt{3} \\[1.5em] = (14 - 9) × \sqrt{3} \\[1.5em] = \bold{5 × 1.732 = 8.660} \\[1.5em]\end{aligned}$$

$$\begin{aligned}\text{(ii) } 5\sqrt{12} - 3\sqrt{48} + 6\sqrt{75} + 7\sqrt{108} \\[1.5em] = 5\sqrt{2 × 2 × 3 } - 3\sqrt{2 × 2 × 2 × 2 × 3} + 6\sqrt{5 × 5 × 3 } + 7\sqrt{2 × 2 × 3 × 3 × 3} \\[1.5em] = 5 × 2\sqrt{3} - 4 × 3\sqrt{3} + 6 × 5\sqrt{3} + 7 × 6\sqrt{3} \\[1.5em] = 10\sqrt{3} - 12\sqrt{3} + 30\sqrt{3} + 42\sqrt{3} \\[1.5em] = (10 - 12 + 30 +42) × \sqrt{3} \\[1.5em] = (82 - 12) × \sqrt{3} \\[1.5em] = \bold{70 × 1.732 = 121.24} \\[1.5em]\end{aligned}$$

$$\begin{aligned}\text{(i) } \sqrt{\frac{4}{9}} = \frac{2}{3} \\[1.5em] -\frac{3}{70} = -\frac{3}{70} \\[1.5em] \sqrt{\frac{7}{25}} = \frac{\sqrt{7}}{5} \\[1.5em] \sqrt{\frac{16}{5}} = \frac{4}{\sqrt{5}} \\[1.5em]\end{aligned}$$

∴ $\bold{\sqrt{\frac{7}{25}}}$ and $\bold{\sqrt{\frac{16}{5}}}$ are irrational numbers as they cannot be written in the form $\frac{{p}}{q}$ where p and q are integers.

$\bold{\sqrt{\frac{4}{9}}}$ and $\bold{- \frac{3}{70}}$ are rational numbers and they can be written in the form $\frac{{p}}{q}$ where p and q are integers .

$$\begin{aligned}\text{(ii) } - \sqrt{\frac{2}{49}} = - \frac{\sqrt{2}}{7} \\[1.5em] \frac{3}{200} = \frac{3}{200} \\[1.5em] \sqrt{\frac{25}{3}} = \frac{5}{\sqrt{3}} \\[1.5em] -\sqrt{\frac{49}{16}} = -\frac{7}{{4}} \\[1.5em]\end{aligned}$$

∴ $\bold{- \sqrt{\frac{2}{49}}}$ and $\bold{\sqrt{\frac{25}{3}}}$ are irrational numbers as they cannot be written in the form $\frac{{p}}{q}$ where p and q are integers.

$-\bold{\sqrt{\frac{49}{16}}}$ and $\bold{\frac{3}{200}}$ are rational numbers and they can be written in the form $\frac{{p}}{q}$ where p and q are integers.

$\text{(i) } -3\sqrt{2}$

it is an irrational number.

We know that $\sqrt2$ is a non-terminating, non-recurring decimal.
So, $\bold{-3\sqrt{2}}$ is also non-terminating, non-recurring decimal .

$\text{(ii) } \sqrt{\frac{256}{81}}$

Here, $\sqrt{\frac{256}{81}} = \frac{16}{9}$ it is a rational number.

$\text{(iii) } \sqrt{(27 × 16)} = \sqrt{27} × \sqrt{16} = 3\sqrt{3} × 4 = 12\sqrt{3}$

It is an irrational number.

We know that $\sqrt{3}$ is a non-terminating, non-recurring decimal.

So , $12\sqrt{3}$ is non-terminating, non-recurring decimal.

Hence, $\bold{\sqrt{27 × 16}}$ is also non-terminating, non-recurring decimal .

$\text{(iv) }\sqrt{\frac{5}{36}}$

Here, $\sqrt{\frac{5}{36}}$ = $\frac{\sqrt{5}}{6}$ , It is an irrational number.

As, $\frac{\sqrt{5}}{6}$ is non-terminating , non-recurring decimal So , $\sqrt{\frac{5}{36}}$ is also non-terminating, non-recurring decimal .

$\text{(i) } 3 - \sqrt{\frac{7}{25}} = 3 - \frac{\sqrt7}{(\sqrt{5 × 5})} = 3 - \frac{\sqrt7}{5}$

As , $3 - \frac{\sqrt{7}}{5}$ is an irrational number,

∴ $\bold{3 - \sqrt{\frac{7}{25}}}$ is also an irrational number .

$\text{(ii) } -\frac{2}{3} + \sqrt[3]{2}$

Here, 2 is not perfect cube

∴ $-\frac{2}{3} + \sqrt[3]{2}$ is an irrational number .

$\text{(iii) } \frac{3}{\sqrt{3}} = \frac{3}{\sqrt{3}} × \frac{\sqrt{3}}{\sqrt{3}} = \frac{3\sqrt{3}}{3} = \sqrt{3}$

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

∴ $\frac{3}{\sqrt{3}}$ is an irrational number .

$\text{(iv) } -\frac{2}{7}\sqrt[3]{5}$

Here, 5 is not perfect cube

∴ $-\frac{2}{7}\sqrt[3]{5}$ is an irrational number .

$\text{(v) } (2-\sqrt{3})(2+\sqrt{3})$

Using identity : $(a + b)(a - b) = a^2 - b^2$

$(2-\sqrt{3})(2+\sqrt{3}) = 2^2 - (\sqrt3)^2 = 4 - 3 = 1$

Hence , $(2-\sqrt{3})(2+\sqrt{3})$ is a rational number .

$\text{(vi) }(3 + \sqrt{5})^2$

Using identity : (a + b)² = a² + 2ab + b²

$$\begin{aligned}(3 + \sqrt{5})^2 = 3^2 + 2 × 3 × \sqrt{5} + (\sqrt{5})^2 \\[1.5em] = 9 + 6\sqrt{5} + 5 \\[1.5em] = 9 + 5 + 6\sqrt{5} \\[1.5em] = 14 + 6\sqrt{5}\end{aligned}$$

As, 14 + $6\sqrt{5}$ is an irrational number

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

$\text{(vii) } \Big(\frac{2}{5}\sqrt{7}\Big)^2$

$$\begin{aligned}\Big(\frac{2}{5}\sqrt{7}\Big)^2 = \frac{2}{5}\sqrt{7} × \frac{2}{5}\sqrt{7} \\[1.5em] = \frac{4}{25} × (\sqrt7)^2 \\[1.5em] = \frac{4}{25} × 7 \\[1.5em] = \frac{28}{25}\end{aligned}$$

As, $\frac{28}{25}$ is a rational number,

∴ $(\frac{2}{5}\sqrt{7})^2$ is a rational number.

$\text{(viii) } (3 - \sqrt{6})^2$

Using identity : (a + b)² = a² + 2ab + b²

$$\begin{aligned}(3 - \sqrt{6})^2 = 3^2 - 2 × 3 × \sqrt{6} + (\sqrt{6})^2 \\[1.5em] = 9 - 6\sqrt{6} + 6 \\[1.5em] = 9 + 6 - 6\sqrt{6} \\[1.5em] = 15 - 6\sqrt{6}\end{aligned}$$

As, 15 - $6\sqrt{6}$ is an irrational number.

∴ $(3 - \sqrt{6})^2$ is an irrational number .

(i) Suppose that $\sqrt[3]{2}$ = $\frac{p}{q}$, where p, q are integers , q ≠ 0 , p and q have no common factors (except 1)

$$\begin{aligned}\Rightarrow 2 = \Big(\frac{p}{q}\Big)^3 \\[1.5em] \Rightarrow p^3 = 2q^3 \qquad \text{....(i)}\end{aligned}$$

As 2 divides 2q³ $\Rightarrow$ 2 divides p³
$\Rightarrow$ 2 divides p    (using generalisation of theorem 1)

Let p = 2k , where k is an integer.

Substituting this value of p in (i), we get

$\phantom{\Rightarrow}$(2k)³ = 2q³
$\Rightarrow$ 8k³ = 2q³
$\Rightarrow$ 4k³ = q³

As 2 divides 4k³ $\Rightarrow$ 2 divides q³

$\Rightarrow$ 2 divides q    (using generalisation of theorem 1)

Thus, p and q have a common factor 2. This contradicts that p and q have no common factors (except 1).

Hence, our supposition is wrong. It follows that $\sqrt[3]{2}$ cannot be expressed as $\frac{p}{q}$, where p, q are integers, q > 0, p and q have no common factors (except 1).

∴ $\bold{\sqrt[3]{2}}$ is an irrational number.

(ii) Suppose that $\sqrt[3]{3}$ = $\frac{p}{q}$, where p, q are integers, q ≠ 0, p and q have no common factors (except 1)

$$\begin{aligned}\Rightarrow 3 = \Big(\frac{p}{q}\Big)^3 \\[1.5em] \Rightarrow p^3 = 3q^3 \qquad \text{....(i)}\end{aligned}$$

As 3 divides 3q³ $\Rightarrow$ 3 divides p³
$\Rightarrow$ 3 divides p    (using generalisation of theorem 1)

Let p = 3k, where k is an integer.

Substituting this value of p in (i), we get

$\phantom{\Rightarrow}$(3k)³ = 3q³
$\Rightarrow$ 27k³ = 3q³
$\Rightarrow$ 9k³ = q³

As 3 divides 9k³ $\Rightarrow$ 3 divides q³
$\Rightarrow$ 3 divides q    (using generalisation of theorem 1)

Thus, p and q have a common factor 3. This contradicts that p and q have no common factors (except 1).

Hence, our supposition is wrong . It follows that $\sqrt[3]{3}$ cannot be expressed as $\frac{p}{q}$, where p, q are integers, q > 0, p and q have no common factors (except 1).

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

(iii) Suppose that $\sqrt[4]{5}$ = $\frac{p}{q}$, where p, q are integers, q ≠ 0, p and q have no common factors (except 1)

$$\begin{aligned}\Rightarrow 5 = \Big(\frac{p}{q}\Big)^4 \\[1.5em] \Rightarrow p^4 = 5q^4 \qquad \text{....(i)}\end{aligned}$$

As 5 divides 5q⁴ $\Rightarrow$ 5 divides p⁴
$\Rightarrow$ 5 divides p    (using generalisation of theorem 1)

Let p= 5k, where k is an integer.

Substituting this value of p in (i), we get

$\phantom{\Rightarrow}$(5k)⁴ = 5q⁴
$\Rightarrow$ 625k⁴ = 5q⁴
$\Rightarrow$ 125k⁴ = q⁴

As 5 divides 125k⁴ $\Rightarrow$ 5 divides q⁴
$\Rightarrow$ 5 divides q    (using generalisation of theorem 1)

Thus, p and q have a common factor 5. This contradicts that p and q have no common factors (except 1).

Hence, our supposition is wrong . It follows that $\sqrt[4]{5}$ cannot be expressed as $\frac{p}{q}$, where p, q are integers, q > 0, p and q have no common factors (except 1).

Therefore, $\bold{\sqrt[4]{5}}$ is an irrational number .

(i) We will write all the numbers as square roots under one radical.

$$\begin{aligned}2\sqrt{3}=\sqrt{4 × 3} = \sqrt{12} \\[1.5em] \frac{3}{\sqrt{2}} = \sqrt{\frac{9}{2}} = \sqrt{4.5} \\[1.5em] -\sqrt{7} \\[1.5em] \sqrt{15} \\[1.5em]\end{aligned}$$

∴ The greatest real number is $\bold{\sqrt{15}}$ and smallest real number is $\bold{-\sqrt{7}}$ .

(ii) We will write all the numbers as square roots under one radical.

$$\begin{aligned}-3\sqrt{2}=-\sqrt{(9×2)} = -\sqrt{18} \\[1.5em] \frac{9}{\sqrt{5}} = \sqrt{\frac{81}{5}} = \sqrt{16.2} \\[1.5em] -4 = -\sqrt{16} \\[1.5em] {\frac{4}{3}}\sqrt{5} = \frac{\sqrt{16}×\sqrt{5}}{\sqrt{9}} = \sqrt{\frac{80}{9}} = \sqrt{8.88} \\[1.5em] {\frac{3}{2}}\sqrt{3} = \frac{\sqrt{9} × \sqrt{3}}{\sqrt{4}} = \sqrt{\frac{27}{4}} = \sqrt{6.75} \\[1.5em]\end{aligned}$$

Here, $\sqrt{16.2}$ is the greatest and $-\sqrt{18}$ is the smallest.

∴ The greatest real number is $\bold{\frac{9}{\sqrt{5}}}$ and smallest real number is $\bold{-3\sqrt{2}}.$

(i) Write all the numbers as square root under one radical :

$$\begin{aligned}3\sqrt{2} = \sqrt{9} × \sqrt{2} = \sqrt{9 × 2} = \sqrt{18} \\[1.5em] 2\sqrt{3} = \sqrt{4} × \sqrt{3} = \sqrt{4 × 3} = \sqrt{12} \\[1.5em] \sqrt{15} = \sqrt{15} \\[1.5em] 4 = \sqrt{16} \\[1.5em] \text{Since} , 12 \lt 15 \lt 16 \lt 18 \\[1.5em] \Rightarrow \sqrt{12} \lt \sqrt{15} \lt \sqrt{16} \lt \sqrt{18} \\[1.5em] \Rightarrow 2\sqrt3 \lt \sqrt{15} \lt 4 \lt 3\sqrt{2}\end{aligned}$$

Hence, the given numbers in ascending order are, $2\bold{\sqrt{3}} , \bold{\sqrt{15}} , \bold{4} , \bold{3\sqrt{2}}$.

(ii) Write all the numbers as square root under one radical :

$$\begin{aligned}3\sqrt{2} = \sqrt{9} × \sqrt{2} = \sqrt{9 × 2} = \sqrt{18} \\[1.5em] 2\sqrt{8} = \sqrt{4} × \sqrt{8} = \sqrt{4 × 8} = \sqrt{32} \\[1.5em] 4 = \sqrt{16} \\[1.5em] \sqrt{50} = \sqrt{50} \\[1.5em] 4\sqrt{3} = \sqrt{16} × \sqrt{3} = \sqrt{16 × 3} = \sqrt{48} \\[1.5em] \text{Since} , 16 \lt 18 \lt 32 \lt 48 \lt 50 \\[1.5em] \sqrt{16} \lt \sqrt{18} \lt \sqrt{32} \lt \sqrt{48} \lt \sqrt{50} \\[1.5em] \Rightarrow 4 \lt 3\sqrt2 \lt 2\sqrt{8} \lt 4\sqrt{3} \lt \sqrt{50}\end{aligned}$$

Hence , $3\bold{\sqrt{2}} , 2\bold{\sqrt{8}} , 4\bold{\sqrt{3}} , 4\bold{\sqrt{50}}$ are in ascending order .

(i) Write all the numbers as square root under one radical :

$$\begin{aligned}\frac{9}{\sqrt{2}} = \frac{\sqrt{81}}{\sqrt{2}} = \sqrt{\frac{81}{2}} = \sqrt{40.5} \\[1.5em] {\frac{3}{2}}{\sqrt{5}} = \sqrt{\frac{9}{4}}×{\sqrt{5}} = \sqrt{\frac{9 × 5}{4}} = \sqrt{\frac{45}{4}} = \sqrt{11.25} \\[1.5em] 4\sqrt{3} = \sqrt{16} × \sqrt{3} = \sqrt{16 × 3} = \sqrt{48} \\[1.5em] 3\sqrt{\frac{6}{5}} = \sqrt{9} × \sqrt{\frac{6}{5}} = \sqrt{\frac{9 × 6}{5}} = \sqrt{\frac{54}{5}} = \sqrt{10.8} \\[1.5em] \text{Since} , 48 \gt 40.5 \gt 11.25 \gt 10.8 \\[1.5em] \Rightarrow \sqrt{48} \gt \sqrt{40.5} \gt \sqrt{11.25} \gt \sqrt{10.8} \\[1.5em] \Rightarrow 4\sqrt3 \gt \frac{9}{\sqrt{2}} \gt {\frac{3}{2}}{\sqrt{5}} \gt 3\sqrt{\frac{6}{5}}\end{aligned}$$

Hence, the given numbers in descending order are $\bold{4\sqrt{3}} , \bold{\frac{9}{\sqrt{2}}} , \bold{{\frac{3}{2}}{\sqrt{5}}} , \bold{3\sqrt{\frac{6}{5}}}$.

(ii) Write all the numbers as square root under one radical :

$$\begin{aligned}\frac{5}{\sqrt{3}} = \frac{\sqrt{25}}{\sqrt{3}} = \sqrt{\frac{25}{3}} = \sqrt{8.33} \\[1.5em] {\frac{7}{3}}{\sqrt{2}} = \sqrt{\frac{49}{9}} ×{\sqrt{2}} = \sqrt{\frac{49 × 2}{9}} = \sqrt{\frac{98}{9}} = \sqrt{10.88} \\[1.5em] -\sqrt{3} = -\sqrt{3} \\[1.5em] 3\sqrt{5} = \sqrt{9 × 5} = \sqrt{45} \\[1.5em] 2\sqrt{7} = \sqrt{4 × 7} = \sqrt{28} \\[1.5em] \text{Since} , 45 \gt 28 \gt 10.88 \gt 8.33 \gt -3 \\[1.5em] \sqrt{45} \gt \sqrt{28} \gt \sqrt{10.88} \gt \sqrt{8.33} \gt -\sqrt{3} \\[1.5em] \Rightarrow 3\sqrt{5} \gt 2\sqrt{7} \gt {\frac{7}{3}}{\sqrt{2}} \gt \frac{5}{\sqrt{3}} \gt -\sqrt{3}\end{aligned}$$

Hence, $\bold{3\sqrt{5}}, \bold{2\sqrt{7}}, \bold{{{\frac{7}{3}}{\sqrt{2}}}}, \bold{\frac{5}{\sqrt{3}}}, -\bold{\sqrt{3}}$ are in descending order.

L.C.M of 3, 2, 6 is 6 :

$$\begin{aligned}\sqrt[3]{2} = 2^\frac{1}{3} = (2^2)^\frac{1}{6} = (4)^\frac{1}{6} \\[1.5em] \sqrt{3} = 3^\frac{1}{2} = (3^3)^\frac{1}{6} = (27)^\frac{1}{6} \\[1.5em] \sqrt[6]{5} = (5)^\frac{1}{6} \\[1.5em] \text{As, } 4 \lt 5 \lt 27 \\[1.5em] \Rightarrow (4)^\frac{1}{6} \lt (5)^\frac{1}{6} \lt (27)^\frac{1}{6} \\[1.5em] \Rightarrow \sqrt[3]{2} \lt \sqrt[6]{5} \lt \sqrt3 \\[1.5em]\end{aligned}$$

Hence , the given number in ascending order are $\sqrt[3]{2} , \sqrt[6]{5} , \sqrt{3}$ .

$\text{(i)}$

$\frac{3}{4\sqrt{5}}$

Let us rationalise the denominator,

Then,

$$\begin{aligned}\frac{3}{4\sqrt{5}} = \frac{3×\sqrt{5}}{4\sqrt{5} × \sqrt{5}} \\[1.5em] \Rightarrow\frac{3\sqrt{5}}{4 × 5} \\[1.5em] \bold{\Rightarrow\frac{3\sqrt{5}}{20}}\end{aligned}$$

$\text{(ii)}$

$\frac{5\sqrt{7}}{\sqrt{3}}$

Let us rationalise the denominator,

Then,

$$\begin{aligned}\frac{5\sqrt{7}}{\sqrt{3}} = \frac{5\sqrt{7}×\sqrt{3}}{\sqrt{3} × \sqrt{3}} \\[1.5em] \bold{\Rightarrow\frac{5\sqrt{21}}{3}}\end{aligned}$$

$\text{(iii)}$

$\frac{3}{4 - \sqrt{7}}$

Let us rationalise the denominator,

Then,

$$\begin{aligned}\frac{3}{4 - \sqrt{7}} = \frac{3}{4 - \sqrt{7}} × \frac{4 + \sqrt{7}}{4 + \sqrt{7}} \\[1.5em] \Rightarrow\frac{3(4 + \sqrt{7})}{(4)^2 - (\sqrt{7})^2} \\[1.5em] \Rightarrow\frac{3(4 + \sqrt{7})}{(16) - (7)} \\[1.5em] \Rightarrow\frac{3(4 + \sqrt{7})}{9} \\[1.5em] \bold{\Rightarrow\frac{(4 + \sqrt{7})}{3}} \\[1.5em]\end{aligned}$$

$\text{(iv)}$

$\frac{17}{3\sqrt{2}+1}$

Let us rationalise the denominator,

Then,

$$\begin{aligned}\frac{17}{3\sqrt{2} + 1} = \frac{17}{3\sqrt{2} + 1}×\frac{3\sqrt{2} - 1}{3\sqrt{2} - 1} \\[1.5em] \Rightarrow\frac{17(3\sqrt{2} - 1)}{(3\sqrt{2})^2 - 1} \\[1.5em] \Rightarrow\frac{17(3\sqrt{2} - 1)}{17} \\[1.5em] \bold{\Rightarrow(3\sqrt{2} - 1)} \\[1.5em]\end{aligned}$$

$\text{(v)}$

$\frac{16}{\sqrt{41}-5}$

Let us rationalise the denominator,

Then,

$$\begin{aligned}\frac{16}{\sqrt{41} - 5} = \frac{16}{\sqrt{41} - 5}×\frac{\sqrt{41} + 5}{\sqrt{41} + 5} \\[1.5em] \Rightarrow\frac{16({\sqrt{41} + 5})}{(\sqrt{41})^2 - 5^2} \\[1.5em] \Rightarrow\frac{16({\sqrt{41} + 5})}{41 - 25} \\[1.5em] {\Rightarrow\frac{16({\sqrt{41} + 5})}{16}} \\[1.5em] \bold{\sqrt{41} + 5}\end{aligned}$$

$\text{(vi)}$

$\frac{1}{\sqrt{7} - \sqrt{6}}$

Let us rationalise the denominator,

Then,

$$\begin{aligned}\frac{1}{\sqrt{7} - \sqrt{6}} = \frac{1}{\sqrt{7} - \sqrt{6}} × \frac{\sqrt{7} + \sqrt{6}}{\sqrt{7} + \sqrt{6}} \\[1.5em] \Rightarrow\frac{({\sqrt{7} + \sqrt{6}})}{(\sqrt{7})^2 - (\sqrt{6})^2} \\[1.5em] \Rightarrow\frac{\sqrt{7} + \sqrt{6}}{7 - 6 } \\[1.5em] \bold{{\Rightarrow\sqrt{7} + \sqrt{6}}} \\[1.5em]\end{aligned}$$

$\text{(vii)}$

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

Let us rationalise the denominator,

Then,

$$\begin{aligned}\frac{1}{\sqrt{5} + \sqrt{2}} = \frac{1}{\sqrt{5} +\sqrt{2}} × \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} - \sqrt{2}} \\[1.5em] \Rightarrow\frac{{\sqrt{5} - \sqrt{2}}}{(\sqrt{5})^2 - (\sqrt{2})^2} \\[1.5em] \bold{\Rightarrow\frac{{\sqrt{5} - \sqrt{2}}}{3}} \\[1.5em]\end{aligned}$$

$\text{(viii)}$

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

Let us rationalise the denominator,

Then,

$$\begin{aligned}\frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} = \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} - \sqrt{3}} × \frac{\sqrt{2} + \sqrt{3}}{\sqrt{2} + \sqrt{3}} \\[1.5em] = \frac{{(\sqrt{2} + \sqrt{3})^2}}{(\sqrt{2})^2 - (\sqrt{3})^2} \\[1.5em] = \frac{{(\sqrt{2})^2 + (\sqrt{3})^2} + 2 × \sqrt{2} × \sqrt{3}}{(\sqrt{2})^2 - (\sqrt{3})^2} \\[1.5em] = \frac{2 + 3 + 2\sqrt{2}\sqrt{3}}{2 - 3} \\[1.5em] = \bold{-(5 + 2\sqrt{6})} \\[1.5em]\end{aligned}$$

$\text(i)$

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

Let us rationalise the denominator,

Then,

$$\begin{aligned}\frac{7 + 3\sqrt{5}}{7 - 3\sqrt{5}} = \frac{7 + 3\sqrt{5}}{7 - 3\sqrt{5}} × \frac{7 + 3\sqrt{5}}{7 + 3\sqrt{5}} \\[1.5em] \Rightarrow\frac{(7 + 3\sqrt{5})^2}{7^2 - (3\sqrt{5})^2} \\[1.5em] \Rightarrow\frac{7^2 + (3\sqrt{5})^2 + 2 × 7 × 3\sqrt{5}}{49 - 45} \\[1.5em] \Rightarrow\frac{49 + 45 + 42\sqrt{5}}{49 - 45} \\[1.5em] \Rightarrow\frac{94 + 42\sqrt{5}}{49 - 45} \\[1.5em] \Rightarrow\frac{2 × (47 + 21\sqrt{5})}{4} \\[1.5em] \bold{\Rightarrow\frac{(47 + 21\sqrt{5})}{2}} \\[1.5em]\end{aligned}$$

$\text(ii)$

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

$$\begin{aligned}\frac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}} = \frac{3 - 2\sqrt{2}}{3 + 2\sqrt{2}} × \frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} \\[1.5em] \Rightarrow\frac{(3 - 2\sqrt{2})^2}{3^2 - (2\sqrt{2})^2} \\[1.5em] \Rightarrow\frac{3^2 + (2\sqrt{2})^2 - 2 × 3 × 2\sqrt{2}}{9-8} \\[1.5em] \Rightarrow\frac{9 + 8 - 12\sqrt{2}}{1} \\[1.5em] \Rightarrow\frac{(17 - 12\sqrt{2})}{1} \\[1.5em] \bold{\Rightarrow{17 - 12\sqrt{2}}} \\[1.5em]\end{aligned}$$

$\text(iii)$

$\frac{5 - 3\sqrt{14}}{7 + 2\sqrt{14}}$

$$\begin{aligned}\frac{5 - 3\sqrt{14}}{7 + 2\sqrt{14}} = \frac{5 - 3\sqrt{14}}{7 + 2\sqrt{14}} × \frac{7 - 2\sqrt{14}}{7 - 2\sqrt{14}} \\[1.5em] \Rightarrow\frac{(5 - 3\sqrt{14})(7 - 2\sqrt{14})}{7^2 - (2\sqrt{14})^2} \\[1.5em] \Rightarrow\frac{35 - 10\sqrt{14} - 21\sqrt{14} + (6 × 14)}{49 - 56} \\[1.5em] \Rightarrow\frac{119 - 31\sqrt{14}}{-7} \\[1.5em] \bold{\Rightarrow\frac{-119+31\sqrt{14}}{7}} \\[1.5em]\end{aligned}$$

$\frac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} - \frac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} - \frac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}}$ $\qquad \text{....(i)}$

Simplifying each term individually,

$\frac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}}$

Let us rationalize its denominator,

$$\begin{aligned}\frac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} = \frac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} × \frac{\sqrt{10} - \sqrt{3}}{\sqrt{10} - \sqrt{3}} \\[1.5em] \Rightarrow\frac{(7\sqrt{3})(\sqrt{10} - \sqrt{3}) }{(\sqrt{10})^2 - (\sqrt{3})^2} \\[1.5em]\Rightarrow\frac{7\sqrt{3} × \sqrt{10} - 7\sqrt{3} × \sqrt{3} }{(\sqrt{10})^2 - (\sqrt{3})^2} \\[1.5em] \Rightarrow\frac{7\sqrt{3 × 10} - 7\sqrt{3 × 3} }{(\sqrt{10})^2 - (\sqrt{3})^2} \\[1.5em] \Rightarrow\frac{7\sqrt{30}-(7×3)}{10-3} \\[1.5em] \Rightarrow\Big(\frac{7\sqrt{30}-(7×3)}{7}\Big) \\[1.5em] \Rightarrow 7 ×\Big(\frac{\sqrt{30} - 3}{7}\Big) \\[1.5em] \bold{\Rightarrow{\sqrt{30}-3}} \qquad \text{....(ii)} \\[1.5em]\end{aligned}$$

$\frac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}}$

Let us rationalize its denominator,

$$\begin{aligned}\frac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} = \frac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} × \frac{\sqrt6 - \sqrt{5}}{\sqrt{6} - \sqrt{5}} \\[1.5em] \Rightarrow\frac{(2\sqrt{5})(\sqrt{6} - \sqrt{5}) }{(\sqrt{6})^2 - (\sqrt{5})^2} \\[1.5em] \Rightarrow\frac{2\sqrt{5} × \sqrt{6} - 2\sqrt{5} ×\sqrt{5} }{(\sqrt{6})^2 - (\sqrt{5})^2} \\[1.5em] \Rightarrow\frac{2\sqrt{5 × 6} - 2\sqrt{5 × 5} }{6 - 5} \\[1.5em] \Rightarrow\frac{2\sqrt{30} - 10}{1} \\[1.5em] \bold{\Rightarrow{2\sqrt{30} - 10}} \qquad \text{....(iii)} \\[1.5em]\end{aligned}$$

$\frac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}}$

Let us rationalize its denominator,

$$\begin{aligned}\frac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}} = \frac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}} × \frac{\sqrt{15} - 3\sqrt{2}}{\sqrt{15} - 3\sqrt{2}} \\[1.5em] \Rightarrow\frac{(3\sqrt{2})(\sqrt{15} - 3\sqrt{2}) }{(\sqrt{15})^2 - (3\sqrt{2})^2} \\[1.5em] \Rightarrow\frac{3\sqrt{2} × \sqrt{15} - 3\sqrt{2} × 3\sqrt{2} }{(\sqrt{15})^2 - (3\sqrt{2})^2} \\[1.5em] \Rightarrow\frac{3\sqrt{30} - 18}{15 - 18} \\[1.5em] \Rightarrow\frac{3\sqrt{30} - 18}{-3} \\[1.5em] \Rightarrow\Big[\frac{-3(-\sqrt{30} + 6)}{-3}\Big] \\[1.5em] \Rightarrow-\Big(\frac{-\sqrt{30} + 6}{1}\Big) \\[1.5em] \bold{\Rightarrow{6 - \sqrt{30}}} \qquad \text{....(iv)} \\[1.5em]\end{aligned}$$

Using (ii) , (iii) , (iv) in equation (i):

$$\begin{aligned}\frac{7\sqrt{3}}{\sqrt{10} + \sqrt{3}} - \frac{2\sqrt{5}}{\sqrt{6} + \sqrt{5}} - \frac{3\sqrt{2}}{\sqrt{15} + 3\sqrt{2}} =(\sqrt{30} - 3)-(2\sqrt{30} - 10) - (6 -\sqrt{30}) \\[1.5em]\end{aligned}$$ $$\begin{aligned}\Rightarrow \sqrt{30} - 3 - 2\sqrt{30} + 10 - 6 + \sqrt{30} \\[1.5em]\end{aligned}$$ $$\begin{aligned}\Rightarrow 2\sqrt{30} - 2\sqrt{30} -3 + 10 - 6 \\[1.5em]\end{aligned}$$ $\bold{\Rightarrow 1}$

$\frac{1}{\sqrt{4} + \sqrt{5}} + \frac{1}{\sqrt{5} + \sqrt{6}} + \frac{1}{\sqrt{6} + \sqrt{7}} + \frac{1}{\sqrt{7} + \sqrt{8}} +\frac{1}{\sqrt{8} + \sqrt{9}} \qquad \text{....(i)}$

Simplifying each term individually,

$\frac{1}{\sqrt{4} + \sqrt{5}}$

Let us rationalise its denominator,

Then,

$$\begin{aligned}\frac{1}{\sqrt{4} + \sqrt{5}} = \frac{1}{\sqrt{4} + \sqrt{5}} × \frac{\sqrt{4} - \sqrt{5}}{\sqrt{4} - \sqrt{5}} \\[1.5em] \Rightarrow\frac{{\sqrt{4} - \sqrt{5}}}{(\sqrt{4})^2 - (\sqrt{5})^2} \\[1.5em] \Rightarrow\frac{{\sqrt{4} - \sqrt{5}}}{4 - 5} \\[1.5em] \Rightarrow{-(\sqrt{4} - \sqrt{5})} \\[1.5em] \Rightarrow{(\sqrt{5} - \sqrt{4})} \qquad \text{....(ii)} \\[1.5em]\end{aligned}$$

$\frac{1}{\sqrt{5} + \sqrt{6}}$

Let us rationalise its denominator,

Then,

$$\begin{aligned}\frac{1}{\sqrt{5} + \sqrt{6}} = \frac{1}{\sqrt{5}+\sqrt{6}}×\frac{\sqrt{5} - \sqrt{6}}{\sqrt{5} - \sqrt{6}} \\[1.5em] \Rightarrow\frac{{\sqrt{5} - \sqrt{6}}}{(\sqrt{5})^2 - (\sqrt{6})^2} \\[1.5em] \Rightarrow\frac{{\sqrt{5} - \sqrt{6}}}{5 - 6} \\[1.5em] \Rightarrow{-(\sqrt{5} - \sqrt{6})} \\[1.5em] \Rightarrow{(\sqrt{6} - \sqrt{5})} \qquad \text{....(iii)} \\[1.5em]\end{aligned}$$

$\frac{1}{\sqrt{6} + \sqrt{7}}$

Let us rationalise its denominator,

Then,

$$\begin{aligned}\frac{1}{\sqrt{6} + \sqrt{7}} = \frac{1}{\sqrt{6}+ \sqrt{7}} × \frac{\sqrt{6} - \sqrt{7}}{\sqrt{6} - \sqrt{7}} \\[1.5em] \Rightarrow\frac{{\sqrt{6} - \sqrt{7}}}{(\sqrt{6})^2 - (\sqrt{7})^2} \\[1.5em] \Rightarrow\frac{{\sqrt{6} - \sqrt{7}}}{6-7} \\[1.5em] \Rightarrow{-(\sqrt{6} - \sqrt{7})} \\[1.5em] \Rightarrow{(\sqrt{7} - \sqrt{6})} \qquad \text{....(iv)} \\[1.5em]\end{aligned}$$

$\frac{1}{\sqrt7+\sqrt8}$

Let us rationalise its denominator,

Then,

$$\begin{aligned}\frac{1}{\sqrt{7} + \sqrt{8}} = \frac{1}{\sqrt{7}+ \sqrt{8}} × \frac{\sqrt{7} - \sqrt{8}}{\sqrt{7} - \sqrt{8}} \\[1.5em] \Rightarrow\frac{{\sqrt{7} - \sqrt{8}}}{(\sqrt{7})^2 - (\sqrt{8})^2} \\[1.5em] \Rightarrow\frac{{\sqrt{7} - \sqrt{8}}}{7-8} \\[1.5em] \Rightarrow{-(\sqrt{7} - \sqrt{8})} \\[1.5em] \Rightarrow{(\sqrt{8} - \sqrt{7})} \qquad \text{....(v)} \\[1.5em]\end{aligned}$$

$\frac{1}{\sqrt{4} + \sqrt{5}}$

Let us rationalise its denominator,

Then,

$$\begin{aligned}\frac{1}{\sqrt{8} + \sqrt{9}} = \frac{1}{\sqrt{8}+ \sqrt{9}} × \frac{\sqrt{8} - \sqrt{9}}{\sqrt{8} - \sqrt{9}} \\[1.5em] \Rightarrow\frac{{\sqrt {8} - \sqrt{9}}}{(\sqrt{8})^2 - (\sqrt{9})^2} \\[1.5em] \Rightarrow\frac{{\sqrt{8} - \sqrt{9}}}{8 - 9} \\[1.5em] \Rightarrow{-(\sqrt{8} - \sqrt{9})} \\[1.5em] \Rightarrow{(\sqrt{9}-\sqrt{8})} \qquad \text{....(vi)} \\[1.5em]\end{aligned}$$

Using (ii) , (iii) , (iv) , (v) , (vi) in equation (i):

$$\begin{aligned}\frac{1}{\sqrt4+\sqrt5} + \frac{1}{\sqrt5+\sqrt6} + \frac{1}{\sqrt6+\sqrt7} + \frac{1}{\sqrt7+\sqrt8} +\frac{1}{\sqrt8+\sqrt9} \\[1.5em] = \sqrt{5} - \sqrt{4} + \sqrt{6} - \sqrt{5} + \sqrt{7} - \sqrt{6} + \sqrt{8} - \sqrt{7} + \sqrt{9} - \sqrt{8} \\[1.5em] = \sqrt{9} - \sqrt{4} \\[1.5em] = \bold{3 - 2 = 1 } \\[1.5em]\end{aligned}$$

(i) Since, it is given that

$\frac{3 - \sqrt{5}}{3 + 2\sqrt{5}}$ = $-\frac{19}{11} + a\sqrt{5}$

On solving,

$$\begin{aligned}\frac{(3 - \sqrt{5})}{(3 + 2\sqrt{5})} × \frac{(3-2\sqrt{5})}{(3-2\sqrt{5})} \\[1.5em] \Rightarrow \frac{9 - 6\sqrt{5} - 3\sqrt{5} + 10 }{3^2 - (2\sqrt{5})^2} \\[1.5em] \Rightarrow \frac{19 - 9{\sqrt{5}}}{9 - 20} \\[1.5em] \Rightarrow \frac{-9{\sqrt{5}} + 19}{-11} \\[1.5em] \Rightarrow -\frac{19}{11} + \frac{9\sqrt{5}}{11} \\[1.5em] \therefore -\frac{19}{11} + \frac{9\sqrt5}{11} = -\frac{19}{11} + a\sqrt{5}\end{aligned}$$

Hence, a = $\frac{9}{11}$.

(ii) Since, it is given that

$\frac{\sqrt{2} + \sqrt{3}}{3\sqrt{2} - 2\sqrt{3}} = a - b\sqrt6$

On solving,

$$\begin{aligned}\frac{\sqrt{2} + \sqrt3}{3\sqrt{2} - 2\sqrt{3}} × \frac{(3\sqrt{2} + 2\sqrt{3})}{(3\sqrt{2} + 2\sqrt{3})} \\[1.5em] \Rightarrow \frac{6 + 2\sqrt{6} + 3\sqrt6 + 6 }{(3\sqrt{2})^2 -(2\sqrt{3})^2} \\[1.5em] \Rightarrow \frac{12 + 5{\sqrt6}}{18-12} \\[1.5em] \Rightarrow \frac{12 + 5{\sqrt{6}}}{6} \\[1.5em] \Rightarrow \frac{12}{6} + \frac{5\sqrt{6}}{6} \\[1.5em] \Rightarrow 2 + \frac{5\sqrt{6}}{6} \\[1.5em] \therefore 2 - \frac{(-5\sqrt{6})}{6} = a - b\sqrt{6}\end{aligned}$$

Hence, a = 2 and b = $\frac{-5}{6}$.

(iii) Since, it is given that

$\frac{(7 + \sqrt5)}{(7 - \sqrt{5})} - \frac{(7 - \sqrt5)}{(7 + \sqrt{5})} = a+\frac{7}{11}b\sqrt{5}$

On solving,

$$\begin{aligned}\frac{(7 + \sqrt{5})(7 + \sqrt{5}) - (7-\sqrt{5})(7 - \sqrt{5})}{(7 - \sqrt{5})(7 + \sqrt{5})} \\[1.5em] \frac{(7 + \sqrt{5})^2 - (7 - \sqrt{5})^2}{(7 - \sqrt{5})(7+\sqrt{5})} \\[1.5em] \Rightarrow \frac{(7^2 + 2 × 7 × \sqrt5 + (\sqrt{5})^2)-(7^2 - 2 × 7 × \sqrt5 + (\sqrt{5})^2)}{7^2-(\sqrt5)^2} \\[1.5em] \Rightarrow \frac{(49 + 2 × 7\sqrt{5} + 5)-(49 - 2 × 7\sqrt{5}+ 5)}{7^2 - (\sqrt{5})^2} \\[1.5em] \Rightarrow \frac{49+14\sqrt{5} + 5 - 49 + 14\sqrt{5} - 5}{7^2 - (\sqrt5)^2} \\[1.5em] \Rightarrow \frac{28{\sqrt{5}}}{49 - 5} \\[1.5em] \Rightarrow \frac{28{\sqrt{5}}}{44} \\[1.5em] \Rightarrow \frac{7{\sqrt{5}}}{11} \\[1.5em] \therefore a+\frac{7}{11}b\sqrt{5} = \frac{7{\sqrt5}}{11} \\[1.5em] \Rightarrow a + \frac{7}{11}b\sqrt{5} = 0 + \frac{7}{11} × 1 × \sqrt{5}\end{aligned}$$

Hence, value of a = 0 and b = 1.

(iv) Since, it is given that

$\frac{2\sqrt{2} - \sqrt{3}}{2\sqrt{2} + 2\sqrt{3}} = a + b\sqrt{24}$

Rationalizing, $\frac{2\sqrt{2} - \sqrt{3}}{2\sqrt{2} + 2\sqrt{3}}$, we get :

$$\begin{aligned}\Rightarrow \frac{2\sqrt{2} - \sqrt{3}}{2\sqrt{2} + 2\sqrt{3}} \times \frac{2\sqrt{2} - 2\sqrt{3}}{2\sqrt{2} - 2\sqrt{3}} \\[1em] \Rightarrow \frac{(2\sqrt{2} - \sqrt{3})(2\sqrt{2} - 2\sqrt{3})}{(2\sqrt{2})^2 - (2\sqrt{3})^2} \\[1em] \Rightarrow \frac{8 - 4\sqrt{6} - 2\sqrt{6} + 6}{8 - 12} \\[1em] \Rightarrow \frac{14 - 6\sqrt{6}}{-4} \\[1em] \Rightarrow -\frac{14}{4} + \frac{6\sqrt{6}}{4} \\[1em] \Rightarrow -\frac{7}{2} + \frac{3}{2}\sqrt{6} \\[1em] \Rightarrow -\frac{7}{2} + \frac{3\sqrt{6}}{2} \times \frac{2}{2} \\[1em] \Rightarrow -\frac{7}{2} + \frac{3\sqrt{6 \times 2^2}}{4} \\[1em] \Rightarrow -\frac{7}{2} + \frac{3\sqrt{24}}{4}.\end{aligned}$$

Comparing $-\frac{7}{2} + \frac{3\sqrt{24}}{4}$ with a + $b\sqrt{24}$, we get :

a = $-\frac{7}{2}, b = \frac{3}{4}$.

Hence, a = $-\frac{7}{2}, b = \frac{3}{4}$.

Since, it is given that

$\frac{(7 + 3\sqrt{5})}{3 + \sqrt{5}}$ - $\frac{(7 - 3\sqrt{5})} {(3 - \sqrt{5})}$ = p + q$\sqrt{5}$

On solving,

$$\begin{aligned}\frac{(7 + 3\sqrt{5})(3 - \sqrt{5}) - (7 - 3\sqrt{5})(3 + \sqrt{5})}{(3 + \sqrt{5})(3 - \sqrt{5})} \\[1.5em] \Rightarrow \frac{(21 - 7\sqrt{5} + 9\sqrt{5} - 15) - (21 + 7\sqrt{5} - 9\sqrt{5} - 15)}{3^2 - {(\sqrt{5})}^2} \\[1.5em] \Rightarrow \frac{(21 - 7\sqrt{5} + 9\sqrt{5} - 15) - (21 + 7\sqrt{5} - 9\sqrt{5} - 15)}{9-5} \\[1.5em] \Rightarrow \frac{21 - 7\sqrt{5} + 9\sqrt{5} - 15 - 21 - 7\sqrt{5} + 9\sqrt{5} + 15}{9-5} \\[1.5em] \Rightarrow \frac{18{\sqrt{5}} - 14{\sqrt{5}}}{4} \\[1.5em] \Rightarrow \frac{4 × {\sqrt{5}}}{4} = \sqrt{5} \\[1.5em] \therefore\sqrt{5} = p + q{\sqrt{5}} \\[1.5em] \Rightarrow 0 + 1 × \sqrt{5} = p + q{\sqrt{5}}\end{aligned}$$

Hence, value of p = 0 and q = 1.

(i) Rationalise the denominator ,

$$\begin{aligned}\frac{\sqrt{2}}{2 + \sqrt{2}} = \frac{\sqrt{2}}{2+ \sqrt{2}} × \frac{2 - \sqrt{2}}{2 - \sqrt{2}} \\[1.5em] \Rightarrow \sqrt{2} × \frac{(2 - \sqrt{2})}{2^2 - (\sqrt{2})^2} \\[1.5em] \Rightarrow \frac{2{\sqrt{2}} - 2}{4 - 2} \\[1.5em] \Rightarrow (\sqrt{2} - 1) \\[1.5em]\end{aligned}$$

Since, $\sqrt{2}$ = 1.414

$$\begin{aligned}\Rightarrow {1.414 - 1} \\[1.5em] \Rightarrow\bold{ 0.414} \\[1.5em]\end{aligned}$$

$(\text{ii})$ Rationalise the denominator ,

$$\begin{aligned}\frac{1}{\sqrt{3} + \sqrt{2}} = \frac{1}{\sqrt{3} + \sqrt{2}} × \frac{{\sqrt{3} - \sqrt{2}}}{{\sqrt{3} - \sqrt{2}}} \\[1.5em] \Rightarrow \frac{(\sqrt{3} - \sqrt{2})}{(\sqrt{3})^2 -(\sqrt{2})^2} \\[1.5em] \Rightarrow \frac{\sqrt{3} - \sqrt{2}}{3-2} \\[1.5em] \Rightarrow (\sqrt{3} - \sqrt{2}) \\[1.5em]\end{aligned}$$

Since, $\sqrt{3}$ = 1.732

$$\begin{aligned}\Rightarrow {1.732 - 1.414} \\[1.5em] \Rightarrow\bold{0.318} \\[1.5em]\end{aligned}$$

Given,

$$\begin{aligned}a = 2 + \sqrt{3} \\[1.5em] \therefore \frac{1}{a} = \frac{1}{2 + \sqrt{3}} ×\frac{2 - \sqrt{3}}{2 - \sqrt{3}} \\[1.5em] \Rightarrow \frac{2 - \sqrt{3}}{2^2 - (\sqrt{3})^2} \\[1.5em] \Rightarrow \frac{2 - \sqrt{3}}{4 - 3} \\[1.5em] \Rightarrow (2 - \sqrt{3}) \\[1.5em] \therefore a - \frac{1}{a} = 2 + \sqrt{3} - 2 + \sqrt{3} \\[1.5em] \Rightarrow\bold{ a - \frac{1}{a} = 2\sqrt{3}} \\[1.5em]\end{aligned}$$

Given,

$$\begin{aligned}x = 1 - \sqrt{2} \\[1.5em] \therefore \frac{1}{x} = \frac{1}{1 - \sqrt{2}} × \frac{1 + \sqrt{2}}{1 + \sqrt{2}} \\[1.5em] \Rightarrow \frac{1 + \sqrt{2}}{1 - (\sqrt{2})^2} \\[1.5em] \Rightarrow \frac{1 + \sqrt{2}}{1 - 2} \\[1.5em] \Rightarrow - (1 +\sqrt{2}) \\[1.5em] \therefore x - \frac{1}{x} = 1 - \sqrt{2} + 1 + \sqrt{2} \\[1.5em] \Rightarrow x - \frac{1}{x} = 2 \\[1.5em] \therefore \bold{{\Big(x-\frac{1}{x}\Big)}^4 = 2 ^4 = 16}\end{aligned}$$

Given x = $5 - 2\sqrt{6}$

$$\begin{aligned}\therefore \frac{1}{x} = \frac{1}{5 - 2\sqrt{6}} = \frac{1}{5 - 2\sqrt{6}} × \frac{5 + 2\sqrt{6}}{5 + 2\sqrt{6}} \\[1.5em] \Rightarrow \frac{5 + 2\sqrt{6}}{(5)^2 - (2\sqrt{6})^2} \\[1.5em] \Rightarrow \frac{5 + 2\sqrt{6}}{25 - 24} = \frac{5 + 2\sqrt{6}}{1} \\[1.5em] \Rightarrow\frac{1}{x} = 5+2\sqrt6 \\[1.5em] \therefore (x + \frac{1}{x}) = (5 - 2\sqrt6) + (5 + 2\sqrt6) = 10 \qquad \text{....(i)}\end{aligned}$$

We know that ${\Big(x + \frac{1}{x}\Big)}^2 = x^2 + \frac{1}{x^2} + 2$

$$\begin{aligned}\Rightarrow x^2 + \frac{1}{x^2} = {\Big(x + \frac{1}{x}\Big)}^2 -2 \\[1.5em] \Rightarrow x^2 + \frac{1}{x^2} = 10^2 - 2 ..... \text{using(i)} \\[1.5em] \Rightarrow x^2 + \frac{1}{x^2} = 100 - 2 \\[1.5em] \bold{x^2 + \frac{1}{x^2} = 98}\end{aligned}$$

(i)

$$\begin{aligned}p+q = \frac{2-\sqrt{5}}{2+\sqrt{5}} +\frac{2+\sqrt{5}}{2-\sqrt{5}} \\[1.5em] \Rightarrow\frac{(2-\sqrt{5})^2 + (2+\sqrt{5})^2}{(2-\sqrt{5})(2+\sqrt{5})} \\[1.5em] \Rightarrow\frac{4 + 5 - 4\sqrt{5} + 4 + 5 + 4\sqrt{5}}{(2)^2 -(\sqrt{5})^2} \\[1.5em] \bold{p+q = -18} \qquad \text{....(i)} \\[1.5em]\end{aligned}$$

$(\text{ii})$

$$\begin{aligned}p-q = \frac{2-\sqrt{5}}{2+\sqrt{5}} -\frac{2+\sqrt{5}}{2-\sqrt{5}} \\[1.5em] \Rightarrow\frac{(2-\sqrt{5})^2 - (2+\sqrt{5})^2}{(2-\sqrt{5})(2+\sqrt{5})} \\[1.5em] \Rightarrow\frac{4 + 5 - 4\sqrt{5} - 4 - 5 - 4\sqrt{5}}{(2)^2 - (\sqrt{5})^2} \\[1.5em] \Rightarrow\bold {p - q = 8\sqrt{5}} \qquad \text{....(ii)}\end{aligned}$$

$(\text{iii})$

$$\begin{aligned}(p+q)^2 = (p)^2 + (q)^2 +2pq \\[1.5em] \Rightarrow(p)^2+(q)^2 =(p+q)^2 -2pq \qquad \text{....(iii)} \\[1.5em] \Rightarrow pq = \frac{2-\sqrt{5}}{2+\sqrt{5}} ×\frac{2+\sqrt{5}} {2-\sqrt{5}} = 1 \qquad \text{....(iv)} \\[1.5em]\end{aligned}$$

substituting value of (i) and (iv) in (iii) :

$\bold{p^2+q^2 = (-18)^2 - 2 = 324-2 = 322}$

$(\text{iv})$

$(p)^2 - (q)^2 = (p+q)(p-q) \qquad \text{....(v)}$

Using (i) and (ii) in (v) we get,

$\bold{p^2 - q^2 = -18 × 8\sqrt{5} = -144\sqrt{5}}$

The difference of a rational number and an irrational number is an irrational number.

For example, 2 is rational number, $\sqrt{3}$ is irrational number and 2 - $\sqrt{3}$ is an irrational number.

∴ Option 4, is the correct option.

Every rational number is a real number.

∴ Option 3, is the correct option.

Between two rational numbers there are infinitely many rational numbers.

∴ Option 3, is the correct option.

Decimal representation of a rational number is terminating or non-terminating repeating but not non-terminating non-repeating.

∴ Option 4, is the correct option.

The product of any two irrational numbers is sometimes rational, sometimes irrational

For example, $2\sqrt{3}$ and $3\sqrt{3}$ are two irrational number .

Their product $2\sqrt{3}$ × $3\sqrt{3}$ = 6 × $(\sqrt{3})^2$ = 6 × 3 = 18 which is rational number.

Again, let $2\sqrt{2}$ and $3\sqrt{3}$ be two irrational numbers.

Their product $2\sqrt{2}$ × $3\sqrt{3}$ = 6 × $\sqrt{2}$ × $\sqrt{3}$ = $6\sqrt{6}$ which is an irrational number.

∴ Option 4, is the correct option.

The division of two irrational numbers is either a rational number or an irrational number.

For example, let $2\sqrt{3}$ and $3\sqrt{3}$ be two irrational numbers.

$\frac{2\sqrt{3}}{3\sqrt{3}}$ = $\frac{2}{3}$ is rational number.

Let $2\sqrt{3}$ and $3\sqrt{5}$ be another two irrational numbers.

$\frac{2\sqrt{3}}{3\sqrt{5}}$ is an irrational number.

∴ Option 3, is the correct option.

$\sqrt{\frac{4}{9}}$ = $\frac{\sqrt{4}}{\sqrt{9}}$ = $\frac{(\sqrt{2})^2}{(\sqrt{3})^2}$ = $\frac{2}{3}$ is a rational number .

$\frac{\sqrt{12}}{\sqrt{3}}$ = $\frac{\sqrt{2 × 2 × 3}}{\sqrt{3}}$ = $\frac{2\sqrt{3}}{\sqrt{3}}$ = 2 is a rational number.

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

$\sqrt{81}$ = $\sqrt{9 × 9}$ = 9 is a rational number.

∴ Option 3, is the correct option.

$\frac{3}{7}$ = 0.428571429.. non-terminating decimal representation.

$\frac{3}{5}$ = 0.6 is terminating decimal representation.

$\frac{1}{3}$ = 0.33333... non-terminating decimal representation.

$\frac{3}{11}$ = 0.272727273.. non-terminating decimal representation.

∴ Option 2, is the correct option.

0.14 is terminating decimal number hence, rational number.

$0.14\overline{16}$ = 0.14161616... is non-terminating repeating decimal number hence, rational number.

$0.\overline{1416}$ = 0.14161416... is non-terminating repeating decimal number hence, rational number.

0.4014001400014... is non-terminating, non-repeating decimal number hence, it is an irrational number.

∴ Option 4, is the correct option.

$\frac{11}{80}$ = 0.1375 has terminating decimal expansion.

$\frac{17}{160}$ = 0.10625 has terminating decimal expansion.

$\frac{63}{240}$ = 0.2625 has terminating decimal expansion.

$\frac{93}{420}$ = 0.221428571..has non-terminating repeating decimal expansion.

∴ Option 4, is the correct option.

Consider the squares of $\sqrt{2}$ and $\sqrt{3}$

${(\sqrt2)^2}$ = 2 and ${(\sqrt3)^2}$ = 3

Take any rational number between 2 and 3 which is a perfect square of a rational number,

One such number is 2.25 and

2.25 = $(1.5)^2$

$\sqrt{2.25}$ = 1.5

As, $2 \lt 2.25 \lt 3$, it follows that

$\sqrt{2} \lt \sqrt{2.25} \lt \sqrt{3}$

$\sqrt{2} \lt 1.5 \lt \sqrt{3}$

Hence, one rational number between $\sqrt{2}$ and $\sqrt{3}$ is 1.5 .

∴ Option 3, is the correct option.

The decimal expansion of 2 - $\sqrt{3}$ is non-terminating, non-repeating as $\sqrt{3}$ is an irrational number, 2 - $\sqrt{3}$ is also an irrational number and decimal expansion of an irrational number is non-terminating, non-repeating.

∴ Option 3, is the correct option.

$\frac{33}{2^2 × 5}$ = $\frac{33}{4 × 5}$ = $\frac{33}{20}$ = 1.65

Hence, decimal expansion of the rational number $\frac{33}{2^2 × 5}$ will terminate after two decimal place.

∴ Option 2, is the correct option.

$\sqrt{10} × \sqrt{15}$ = $\sqrt{2 × 5} × \sqrt{3 × 5}$ = $\sqrt{2 × 5 × 3 × 5}$ = $5\sqrt{6}$

∴ Option 2, is the correct option.

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

∴ Option 3, is the correct option.

$\sqrt{8} + \sqrt{18}$ = $\sqrt{2 × 2 × 2} + \sqrt{2 × 3 × 3}$ = $2\sqrt{2} + 3\sqrt{2}$ = $(2 + 3)\sqrt{2}$ = $5\sqrt{2}$

∴ Option 3, is the correct option.

$(2 - \sqrt{3})^2$ = $(2)^2 + (\sqrt{3})^2 - 2 × 2 × \sqrt{3}$ = 4 + 3 - $4\sqrt{3}$ = 7 - $4\sqrt{3}$

Since, 7 - $4\sqrt{3}$ is an irrational number therefore, $(2 - \sqrt{3})^2$ is also an irrational number.

∴ Option 4, is the correct option.

x is a positive rational number and x is not a perfect square.

Then, $\sqrt{x}$ is an irrational number,

Therefore , $-5\sqrt{x}$ is also an irrational number.

∴ Option 4, is the correct option.

$$\begin{aligned}(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y}) = \sqrt{x} × \sqrt{x} - \sqrt{x} × \sqrt{y} - \sqrt{x} × \sqrt{y} - \sqrt{y} × \sqrt{y} \\[1.5em] \Rightarrow (\sqrt{x})^2 - \sqrt{xy} + \sqrt{xy} -(\sqrt{y})^2 = x - y \\[1.5em]\end{aligned}$$

Since, x, y are both positive rational numbers, so the difference of two positive rational numbers is also a rational number .

Therefore, $x - y$ is also a rational number. Hence, $(\sqrt{x} + \sqrt{y})(\sqrt{x} - \sqrt{y})$ is a rational number.

∴ Option 1, is the correct option.

$\frac{7}{3\sqrt3 - 2\sqrt{2}}$

Let us rationalise the denominator,

Then,

$$\begin{aligned}\frac{7}{3\sqrt3 - 2\sqrt{2}} = \frac{7}{3\sqrt{3} - 2\sqrt{2}} × \frac{3\sqrt{3} + 2\sqrt{2}}{3\sqrt{3} + 2\sqrt{2}} \\[1.5em] \Rightarrow \frac{7({3\sqrt{3} + 2\sqrt{2}})}{(3\sqrt{3})^2 - (2\sqrt{2})^2} \\[1.5em] \Rightarrow \frac{7 × 3\sqrt{3} + 14 × \sqrt{2}}{27 - 8} \\[1.5em] \Rightarrow \frac{21\sqrt{3}+ 14\sqrt{2}}{19} \\[1.5em]\end{aligned}$$

∴ Option 2, is the correct option.

Given,

${\frac{1}{\sqrt{7} - 2}}$

Let us rationalise the denominator,

Then,

$$\begin{aligned}{\frac{1}{\sqrt{7} - 2}} = {\frac{1}{\sqrt{7} - 2}} × \frac{\sqrt{7} + 2}{\sqrt{7} + 2} \\[1.5em] \Rightarrow \frac{\sqrt{7} + 2}{(\sqrt{7})^2 - {2}^2} \\[1.5em] \Rightarrow \frac{\sqrt{7} + 2}{7 - 4} \\[1.5em] \Rightarrow \frac{\sqrt{7} + 2}{3} \\[1.5em]\end{aligned}$$

∴ Option 1, is the correct option.

Let x = $0.\overline{25}$

x = 0.252525.....      .........(1)

Multiplying both side by 100, we get

100x = 25.252525...     .........(2)

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

⇒ 100x - x = 25.252525..... - 0.252525......

⇒ 99x = 25

⇒ x = $\frac{25}{99}$.

Hence, option 4 is the correct option.

Let x = $1.99\overline{9}$

x = 1.999999.....      .........(1)

Multiplying both side with 10, we get

10x = 19.99999...     .........(2)

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

⇒ 10x - x = 19.99999..... - 1.99999......

⇒ 9x = 18

⇒ x = $\frac{18}{9}$ = 2.

Hence, option 3 is the correct option.

According to statement 1 :

2^m x 3ⁿ = (2 + 3)^{m + n}, where m, n are positive integers.

This statement does not reflect any general law of exponents.

∴ Statement 1 is false.

According to statement 2 :

If a is a rational number, and m, n are integers, then a^m.aⁿ = a^{m + n}

This is one of the fundamental laws of exponents. It holds for any rational base, and all integer exponents.

∴ Statement 2 is true.

Hence, option 4 is the correct option.

Any number that can be expressed in the form $\frac{p}{q}$ is a rational number.

The conditions are that p and q must be integers, and q must not be equal to zero.

∴ Reason (R) is false.

In the number $-\frac{2}{7}$, p = -2(which is an integer) and q = 7 (which is a non-zero integer).

Therefore, $-\frac{2}{7}$ fits the definition of a rational number.

∴ Assertion (A) is true.

Hence, option 1 is the correct option.

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

This is a fundamental property of irrational and rational numbers.

∴ Reason (R) is true.

π is an irrational number, -10 is a rational number.

So, their sum -10 + π, will be an irrational number.

∴ Assertion (A) is true.

∴ Both Assertion (A) and Reason (R) are true, and Reason (R) is the correct reason for Assertion (A).

Hence, option 3 is the correct option.

Let x = $0.\overline{36}$

x = 0.363636.....      .........(1)

Multiplying both side by 100, we get

100x = 36.363636...     .........(2)

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

⇒ 100x - x = 36.363636..... - 0.363636......

⇒ 99x = 36

⇒ x = $\frac{36}{99} = \frac{4}{11}$.

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

∴ Assertion (A) is false.

By definition,

Any real number that can be expressed in the form of $\frac{p}{q}$ where p, q are integers, q ≠ 0 and p, q have no common factor except 1 is a rational number.

∴ Reason (R) is true.

∴ Assertion (A) is false, Reason (R) is true.

Hence, option 2 is the correct option.

A surd is a number that cannot be expressed as a simple fraction and is the root of a rational number (like $\sqrt{2}, \sqrt[3]{5}$). These roots are always irrational.

∴ Assertion (A) is true.

While all surds are irrational, not all irrational numbers are surds.

For example, pi (π) is irrational number but are not roots of rational numbers and therefore are not surds.

∴ Reason (R) is false.

∴ Assertion (A) is true, Reason (R) is false.

Hence, option 1 is the correct option.

Multiplying 2 + $\sqrt{3}$ and 2 - $\sqrt{3}$

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

Since, 1 is a rational number.

Thus, we can say that the rationalising factor of 2 + $\sqrt{3}$ is 2 - $\sqrt{3}$.

∴ Assertion (A) is true.

A surd is an irrational root of a rational number.

Thus, 2 + $\sqrt{3}$ and 2 - $\sqrt{3}$ are irrational numbers containing surds $(\sqrt{3})$, but not surd itself.

∴ Reason (R) is false.

Hence, option 1 is the correct option.

$\text{(i) } \frac{13}{45}$

The given number $\frac{13}{45}$ is in its lowest form.

Prime factorization of denominator 45:

$$\begin{array}{l|l} 3 & 45 \\ \hline 3 & 15 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

45 = 3 x 3 x 5 x 1
= 3² x 5 x 1

Denominator is not of the form 2^m x 5ⁿ, where m, n are non-negative integers.

∴ The given number $\frac{13}{45}$ is recurring decimal.

$\text{(ii) } -\frac{5}{56}$

The given number $-\frac{5}{56}$ is in its lowest form.

Prime factorization of denominator 56:

$$\begin{array}{l|l} 2 & 56 \\ \hline 2 & 28 \\ \hline 2 & 14 \\ \hline 7 & 7 \\ \hline & 1 \end{array}$$

56 = 2 x 2 x 2 x 7 x 1
= 2³ x 7 x 1

Denominator is not of the form 2^m x 5ⁿ, where m, n are non-negative integers.

∴ The given number $-\frac{5}{56}$ is recurring decimal.

$\text{(iii) } \frac{7}{125}$

The given number $\frac{7}{125}$ is in its lowest form.

Prime factorization of denominator 125:

$$\begin{array}{l|l} 5 & 125 \\ \hline 5 & 25 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

125= 5 x 5 x 5 x 1
= 5³ x 1
= 5³ x 2⁰

Denominator is of the form 2^m x 5ⁿ, where m, n are non-negative integers.

$$\begin{aligned}\frac{7}{125} = \frac{7}{2^0 × 5^3} \\[1.5em] = \frac{7 × 2^3}{2^3 × 5^3} \\[1.5em] = \frac{56}{(2 × 5)^3} \\[1.5em] = \frac{56}{(10)^3} \\[1.5em] = \frac{56}{1000} \\[1.5em] = 0.056\end{aligned}$$

∴ The given number $\frac{7}{125}$ is a terminating decimal and its decimal expansion is 0.056.

$\text{(iv) } \frac{-23}{80}$

The given number $-\frac{23}{80}$ is in its lowest form.

Prime factorization of denominator 80:

$$\begin{array}{l|l} 2 & 80 \\ \hline 2 & 40 \\ \hline 2 & 20 \\ \hline 2 & 10 \\ \hline 5 & 5 \\ \hline & 1 \end{array}$$

80 = 2 x 2 x 2 x 2 x 5 x 1
= 2⁴ x 5¹

Denominator is of the form 2^m x 5ⁿ, where m, n are non-negative integers.

$$\begin{aligned}-\frac{23}{80} = -\frac{23}{2^4 × 5^1} \\[1.5em] = -\frac{23 × 5^3} {2^4 × 5^4} \\[1.5em] = -\frac{23 × 125}{(2 × 5)^4} \\[1.5em] = -\frac{2875}{(10)^4} \\[1.5em] = -\frac{2875}{10000} = -0.2875\end{aligned}$$

∴ The given number $-\frac{23}{80}$ is a terminating decimal and its decimal expansion is -0.2875.

$\text{(v) } -\frac{15}{66}$

The given number $-\frac{15}{66}$ is in its lowest form.

Prime factorization of denominator 66:

$$\begin{array}{l|l} 2 & 66 \\ \hline 3 & 33 \\ \hline 11 & 11 \\ \hline & 1 \end{array}$$

66 = 2 x 3 x 11 x 1
= 2 x 3 x 11

Denominator is not of the form 2^m x 5ⁿ, where m, n are non-negative integers.

∴ The given number $-\frac{15}{66}$ is recurring decimal.

(i) Let x = $1.3\overline{45}$ = 1.3454545 ... $\qquad \text{....(i)}$

So multiplying both sides of (i) by 10

we get,

10x = 13.4545...$\qquad \text{....(ii)}$

Again multiply by 100 on both sides ,

1000x =1345.4545.....$\qquad \text{....(iii)}$

Subtracting (ii) from (iii), we get

1000x - 10x = 1345.4545... - 13.4545...

990x = 1332

x = $\frac{1332}{990}$ = $\bold{\frac{74}{55}}$

which is in the form of $\frac{p}{q}$, q ≠ 0

(ii) Let x = $2.\overline{357}$ = 2.357357... $\qquad \text{....(i)}$

So multiplying both sides of (i) by 1000,

we get,

1000x = 2357.357357...$\qquad \text{....(ii)}$

Subtracting (i) from (ii), we get

1000x - x = 2357.357357... - 2.357357...

999x = 2355

x = $\bold{\frac{2355}{999}}$

which is in the form of $\frac{p}{q}$, q ≠ 0.

The L.C.M. of 9 and 13 is 117.

$$\begin{aligned}\frac{5}{9} = \frac{5 \times 13}{9 \times 13} = \frac{65}{117} \\[1.5em] \frac{7}{13} = \frac{7 \times 9}{13 \times 9} = \frac{63}{117} \\[1.5em] \text{Since } 9 \lt 13, \frac{1}{13} \lt \frac{1}{9}\end{aligned}$$

A rational number between $\frac{5}{9}$ and $\frac{7}{13}$

$$\begin{aligned}= \frac{\frac{5}{9} + \frac{7}{13}}{2} \\[1.5em] = \frac{\frac{65 + 63}{117}}{2} \\[1.5em] = \frac{128}{117 × 2} \\[1.5em] = \bold{\frac{64}{117}} \\[1.5em]\end{aligned}$$

$\therefore\frac{7}{13} \lt \frac{64}{117} \lt \frac{5}{9}$

Hence, numbers in ascending order are:

$\bold{\frac{7}{13}}, \bold{\frac{64}{117}}, \bold{\frac{5}{9}}$

The L.C.M of 5 and 6 is 30.

$$\begin{aligned}\frac{4}{5} = \frac{4 \times 6}{5 \times 6} = \frac{24}{30} \\[1.5em] \frac{5}{6} = \frac{5 \times 5}{6 \times 5} = \frac{25}{30} \\[1.5em] \text{Since } 24 \lt 25, \\[1.5em] \therefore \frac{4}{5} \lt \frac{5}{6}\end{aligned}$$

To find four rational number between $\frac{4}{5}$ and $\frac{5}{6}$ multiply the numerator and denominator of $\frac{24}{30}$ and $\frac{25}{30}$ by 4 + 1 i.e. by 5 we get, $\frac{120}{150}$ and $\frac{125}{150}$

$$\begin{aligned}\text{Since}, 120 \lt 121 \lt 122 \lt 123 \lt 124 \lt 125 \\[1.5em] \Rightarrow\frac{120}{150} \lt \frac{121}{150} \lt \frac{122}{150} \lt \frac{123}{150} \lt \frac{124}{150} \lt \frac{125}{150} \\[1.5em] \Rightarrow\frac{4}{5} \lt \frac{121}{150} \lt \frac{61}{75} \lt \frac{123}{150} \lt \frac{62}{75} \lt \frac{5}{6} \\[1.5em]\end{aligned}$$

Four rational number between $\frac{4}{5}$ and $\frac{5}{6}$ are:

$\bold{\frac{121}{150}}, \bold{\frac{61}{75}}, \bold{\frac{123}{150}}, \bold{\frac{62}{75}}$

Let us consider, x as an irrational number.

Reciprocal of x is $\frac{1}{x}$.

Let us consider $\frac{1}{x}$ to be a non-zero rational number.

Then, $x × \frac{1}{x}$ will also be an irrational number as product of non-zero rational number and an irrational number is also an irrational number.

But $x × \frac{1}{x}$ = 1 is a rational number.

Hence, our supposition is wrong. So, $\frac{1}{x}$ is an irrational number.

(i) $\sqrt{8}$ can be written as $2\sqrt{2}$, now we are going to show that $\sqrt{2}$ is an irrational number.

Let $\sqrt{2}$ be a rational number, then

$\sqrt{2} = \frac{p}{q},$

where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)

$$\begin{aligned}\Rightarrow 2 = \frac{p^2}{q^2} \\[1.5em] \Rightarrow p^2 = 2q^2 \qquad \text{....(i)}\end{aligned}$$

As 2 divides 2q², so 2 divides p² but 2 is prime

$\Rightarrow 2 \text{ divides } p \qquad \text{(Theorem 1)}$

Let p = 2m, where m is an integer.

Substituting this value of p in (i), we get

$$\begin{aligned}(2m)^2 = 2q^2 \\[1.5em] \Rightarrow 4m^2 = 2q^2 \\[1.5em] \Rightarrow 2m^2 = q^2 \\[1.5em]\end{aligned}$$

As 2 divides 2m², so 2 divides q² but 2 is prime

$\Rightarrow 2 \text{ divides } q \qquad \text{(Theorem 1)}$

Thus, p and q have a common factor 2. This contradicts that p and q have no common factors (except 1).

Hence, $\sqrt{2}$ is not a rational number. So, we conclude that $\sqrt{2}$ is an irrational number.

Since, product of non-zero rational number and an irrational number is an irrational number.

And $\sqrt{2}$ is an irrational number this implies that $2\sqrt{2}$ = $\bold{\sqrt{8}}$ is an irrational number.

(ii) Suppose that $\sqrt{14}$ is a rational number, then

$\sqrt{14} = \frac{p}{q},$

where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)

$$\begin{aligned}\Rightarrow 14 = \frac{p^2}{q^2} \\[1.5em] \Rightarrow p^2 = 14q^2 \qquad \text{....(i)}\end{aligned}$$

As 2 divides 14q², so 2 divides p² but 2 is prime

$\Rightarrow 2 \text{ divides } p \qquad \text{(Theorem 1)}$

Let p = 2k, where k is some integer.

Substituting this value of p in (i), we get

$$\begin{aligned}(2k)^2 = 14q^2 \\[1.5em] \Rightarrow 4k^2 = 14q^2 \\[1.5em] \Rightarrow 2k^2 = 7q^2 \\[1.5em]\end{aligned}$$

As 2 divides 2k², so 2 divides 7q²

$\Rightarrow$ 2 divides 7 or 2 divides q²

But 2 does not divide 7, therefore, 2 divides q²

$\Rightarrow$ 2 divides q      (Theorem 1)

Thus, p and q have a common factor 2. This contradicts that p and q have no common factors (except 1).

Hence, our supposition is wrong. Therefore, $\sqrt{14}$ is not a rational number. So, we conclude that $\bold{\sqrt{14}}$ is an irrational number.

(iii) Suppose that $\sqrt[3]{2}$ = $\frac{p}{q}$, where p, q are integers , q ≠ 0 , p and q have no common factors (except 1)

$$\begin{aligned}\Rightarrow 2 = \Big(\frac{p}{q}\Big)^3 \\[1.5em] \Rightarrow p^3 = 2q^3 \qquad \text{....(i)}\end{aligned}$$

As 2 divides 2q³ $\Rightarrow$ 2 divides p³

$\Rightarrow$ 2 divides p    (using generalisation of theorem 1)

Let p = 2k , where k is an integer.

Substituting this value of p in (i), we get

$\phantom{\Rightarrow}$(2k)³ = 2q³
$\Rightarrow$ 8k³ = 2q³
$\Rightarrow$ 4k³ = q³

As 2 divides 4k³ $\Rightarrow$ 2 divides q³

$\Rightarrow$ 2 divides q    (using generalisation of theorem 1)

Thus, p and q have a common factor 2. This contradicts that p and q have no common factors (except 1).

Hence, our supposition is wrong. It follows that $\sqrt[3]{2}$ cannot be expressed as $\frac{p}{q}$, where p, q are integers, q > 0, p and q have no common factors (except 1).

∴ $\bold{\sqrt[3]{2}}$ is an irrational number.

Let $\sqrt{3}$ be a rational number, then

$\sqrt{3} = \frac{p}{q},$

where p, q are integers, q ≠ 0 and p, q have no common factors (except 1)

$$\begin{aligned}\Rightarrow 3 = \frac{p^2}{q^2} \\[1.5em] \Rightarrow p^2 = 3q^2 \qquad \text{....(i)}\end{aligned}$$

As 3 divides 3q², so 3 divides p² but 3 is prime

$\Rightarrow 3 \text{ divides } p \qquad \text{(Theorem 1)}$

Let p = 3m, where m is an integer.

Substituting this value of p in (i), we get

$$\begin{aligned}(3m)^2 = 3q^2 \\[1.5em] \Rightarrow 9m^2 = 3q^2 \\[1.5em] \Rightarrow 3m^2 = q^2 \\[1.5em]\end{aligned}$$

As 3 divides 3m², so 3 divides q² but 3 is prime

$\Rightarrow 3 \text{ divides } q \qquad \text{(Theorem 1)}$

Thus, p and q have a common factor 3. This contradicts that p and q have no common factors (except 1).

Hence, $\sqrt{3}$ is not a rational number. So, we conclude that $\sqrt{3}$ is an irrational number.

Suppose that $5 - \sqrt{3}$ is a rational number, say r.

Then, $5 - \sqrt{3}$ = r (note that r ≠ 0)

$$\begin{aligned}\Rightarrow - \sqrt{3} = r - 5 \\[1.5em] \Rightarrow \sqrt{3} = 5 - r \\[1.5em]\end{aligned}$$

As r is rational and r ≠ 0, so 5 - r is rational

$\Rightarrow \sqrt{3}$ is rational

But this contradicts that $\sqrt{3}$ is irrational. Hence, our supposition is wrong.

∴ $\bold{5 - \sqrt{3}}$ is an irrational number.

$\text{(i) } 3 + \sqrt{5}$

Let us assume that $3 + \sqrt{5}$ is a rational number, say r.

Then,

$$\begin{aligned}3 + \sqrt{5} = r \\[1.5em] \Rightarrow \sqrt{5} = r - 3\end{aligned}$$

As r is rational, r - 3 is rational

$\Rightarrow \sqrt{5}$ is rational

But this contradicts the fact that $\sqrt{5}$ is irrational.

Hence, our assumption is wrong.

∴ $\bold{3 + \sqrt{5}}$ is an irrational number.

$\text{(ii) } 15 - 2\sqrt{7}$

Let us assume that $15 - 2\sqrt{7}$ is a rational number, say r.

Then,

$$\begin{aligned}15 - 2\sqrt{7} = r \\[1.5em] \Rightarrow 2\sqrt{7} = 15 - r \\[1.5em] \Rightarrow \sqrt{7} = \frac{15 - r}{2} \\[1.5em]\end{aligned}$$

As r is rational, 15 - r is rational

$\Rightarrow \frac{15 - r}{2}$ is rational

$\Rightarrow \sqrt{7}$ is rational

But this contradicts the fact that $\sqrt{7}$ is irrational.

Hence, our assumption is wrong.

∴ $\bold{15 - 2\sqrt{7}}$ is an irrational number.

$\text{(iii) }\frac{1}{3 - \sqrt5}$

Let us rationalise the denominator

$$\begin{aligned}\frac{1}{3 - \sqrt5} = \frac{1}{3 - \sqrt5} × \frac{3 + \sqrt5}{3 + \sqrt5} \\[1.5em] = \frac{3 + \sqrt5}{(3)^2 - (\sqrt5)^2} \\[1.5em] = \frac{3 + \sqrt5}{9 - 5} \\[1.5em] = \frac{3 + \sqrt5}{4} \\[1.5em]\end{aligned}$$

Let us assume that $\frac{3 + \sqrt5}{4}$ is a rational number, say r.

Then,

$$\begin{aligned}\frac{3 + \sqrt5}{4} = r \\[1.5em] \Rightarrow 3 +\sqrt{5} = 4r \\[1.5em] \Rightarrow \sqrt{5} = 4r - 3 \\[1.5em]\end{aligned}$$

As r is rational, 4r is rational

$\Rightarrow$ 4r - 3 is also rational

$\Rightarrow \sqrt{5}$ is rational

But this contradicts the fact that $\sqrt{5}$ is irrational.

Hence, our assumption is wrong.

$\Rightarrow \frac{3 + \sqrt5}{4}$ is an irrational number.

∴ $\bold{\frac{1}{3 - \sqrt5}}$ is an irrational number.

$\text{(i)}$

$\frac{10}{2\sqrt{2} + \sqrt{3}}$

Let us rationalise the denominator,

Then,

$$\begin{aligned}\frac{10}{2\sqrt{2} + \sqrt{3}} = \frac{10}{2\sqrt{2} + \sqrt{3}} × \frac{2\sqrt{2} - \sqrt{3}}{2\sqrt{2} - \sqrt{3}} \\[1.5em] = \frac{10({2\sqrt{2} - \sqrt{3}})}{(2\sqrt{2})^2 - (\sqrt{3})^2} \\[1.5em] = \frac{10 × 2\sqrt{2} - 10 × \sqrt{3}}{(2\sqrt{2})^2 - (\sqrt{3})^2} \\[1.5em] = \frac{10(2\sqrt{2} - \sqrt{3})}{8 - 3} \\[1.5em] = \frac{10(2\sqrt{2} - \sqrt{3})}{5} \\[1.5em] \bold{= 2(2\sqrt{2} - \sqrt{3}) } \\[1.5em]\end{aligned}$$

$\text{(ii)}$ Since, it is given that

$\frac{7\sqrt{3} - 5\sqrt{2}}{\sqrt{48} + \sqrt{18}}$

Let us rationalise the denominator,

$$\begin{aligned}\frac{7\sqrt{3} - 5\sqrt{2}}{\sqrt{48} + \sqrt{18}} × \frac{\sqrt{48} - \sqrt{18}}{\sqrt{48} - \sqrt{18}} \\[1.5em] = \frac{7\sqrt{3} × \sqrt{48} - 7\sqrt{3} × \sqrt{18} - 5\sqrt{2} × \sqrt{48} + 5\sqrt{2} × \sqrt{18} }{(\sqrt{48})^2 -(\sqrt{18})^2} \\[1.5em] = \frac{7\sqrt{144} - 7\sqrt{54} - 5\sqrt{96} +5\sqrt{36}}{48 - 18} \\[1.5em] = \frac{7 × 12 - 7\sqrt{2 × 3 × 3 × 3 } - 5\sqrt{2 × 2 × 2 × 2 × 2 × 3} +5\sqrt{2 × 2 × 3 × 3}}{30} \\[1.5em] = \frac{84 - 21\sqrt{6} - 20\sqrt{6} + 30}{30} \\[1.5em] = \frac{84 - 21\sqrt{6} - 20\sqrt{6} + 30}{30} \\[1.5em] = \frac{114 - 41{\sqrt6}}{30} \\[1.5em] = \frac{114}{30} - \frac{41\sqrt{6}}{30} \\[1.5em] = \bold{\frac{57}{15} - \frac{41\sqrt{6}}{30}} \\[1.5em]\end{aligned}$$

$\text{(iii)}$

$\frac{1}{\sqrt{3} - \sqrt{2} + 1}$

Let us rationalise the denominator,

Then,

$$\begin{aligned}\frac{1}{\sqrt{3} - \sqrt{2} + 1} = \frac{1}{\sqrt{3} - (\sqrt{2} - 1)} × \frac{\sqrt{3} + (\sqrt{2} - 1)}{\sqrt{3} + (\sqrt{2} - 1)} \\[1.5em] =\frac{{\sqrt{3} + \sqrt{2}} - 1}{{(\sqrt{3})^2 }- (\sqrt{2} - 1)^2} \\[1.5em] =\frac{\sqrt{3} + \sqrt{2} - 1}{(\sqrt{3})^2 - ( (\sqrt{2})^2 - 2 × \sqrt{2} + 1)} \\[1.5em] =\frac{\sqrt{3} + \sqrt{2} - 1}{3 - (2 - 2\sqrt{2} + 1)} \\[1.5em] =\frac{\sqrt{3} + \sqrt{2} - 1}{3 - 2 + 2\sqrt{2} -1} \\[1.5em] =\frac{\sqrt{3} + \sqrt{2} - 1}{2\sqrt{2}} \\[1.5em] =\frac{\sqrt{3} + \sqrt{2} - 1}{2\sqrt{2}} × \frac{\sqrt{2}}{\sqrt{2}} \\[1.5em] =\frac{\sqrt{2}(\sqrt{3} + \sqrt{2} - 1)}{2\sqrt{2} × \sqrt{2}} \\[1.5em] =\frac{\sqrt{2} × \sqrt{3} + \sqrt{2} × \sqrt{2} - \sqrt{2}}{2\sqrt{2} × \sqrt{2}} \\[1.5em] = \frac{\sqrt{6} + 2 -\sqrt{2}}{4} \\[1.5em] \bold{{=} \frac{2 + \sqrt{6} -\sqrt{2}}{4}} \\[1.5em]\end{aligned}$$

Since, it is given that

$\frac{2\sqrt{3} - \sqrt{5}}{4\sqrt{3} - 3\sqrt{5}} = p - \sqrt{15} q$

On solving,

$$\begin{aligned}\frac{2\sqrt{3} - \sqrt{5}}{4\sqrt{3} - 3\sqrt{5}} × \frac{4\sqrt{3} + 3\sqrt{5}}{4\sqrt{3} + 3\sqrt{5}} \\[1.5em] = \frac{2\sqrt{3} × 4\sqrt{3}+ 2\sqrt{3} × 3\sqrt{5} - \sqrt{5} × 4\sqrt{3} - \sqrt{5} × 3\sqrt{5} }{(4\sqrt{3})^2 -(3\sqrt{5})^2} \\[1.5em] = \frac{24 + 6\sqrt{15} - 4\sqrt{15} - 15 }{(4\sqrt{3})^2 -(3\sqrt{5})^2} \\[1.5em] = \frac{24 - 15 + 6\sqrt{15} - 4\sqrt{15} }{48 - 45} \\[1.5em] = \frac{9 + 2{\sqrt{15}}}{3} \\[1.5em] = \frac{9}{3} + \frac{2\sqrt{15}}{3} \\[1.5em] = 3 + \frac{2}{3}{\sqrt{15}} \\[1.5em] \therefore 3 - \Big(-\frac{2}{3}\Big){\sqrt{15}} = p - q\sqrt{15}\end{aligned}$$

Hence, p = 3 and q = $-\frac{2}{3}$.

Given,

$$\begin{aligned}x =\frac{1}{3 + 2\sqrt{2}} \qquad \text{....(i)} \\[1.5em]\end{aligned}$$

Let us rationalise the denominator,

$$\begin{aligned}x = \frac{1}{3 + 2\sqrt{2}} ×\frac{3 - 2\sqrt{2}}{3 - 2\sqrt{2}} \\[1.5em] = \frac{3 - 2\sqrt{2}}{3^2 - (2\sqrt{2})^2} \\[1.5em] = \frac{3 - 2\sqrt{2}}{9 - 8} \\[1.5em] = (3 - 2\sqrt{2}) \\[1.5em] \therefore x = (3 - 2\sqrt{2})\end{aligned}$$

From (i) we get,

$$\begin{aligned}\frac{1}{x} = 3 + 2\sqrt{2} \\[1.5em] \therefore x - \frac{1}{x} = (3 - 2\sqrt{2}) - (3 + 2\sqrt{2}) = 3 - 2\sqrt{2} - 3 - 2\sqrt{2}\\[1.5em] \Rightarrow\bold{ x - \frac{1}{x} = -4\sqrt{2}} \\[1.5em]\end{aligned}$$

(i) Given x = $\frac{7 + 3\sqrt{5}}{7- 3\sqrt{5}}$

Rationalising the denominator,

$$\begin{aligned}\frac{7 + 3\sqrt{5}}{7- 3\sqrt{5}} = \frac{7 + 3\sqrt{5}}{7- 3\sqrt{5}} × \frac{7 + 3\sqrt{5}}{7 + 3\sqrt{5}} \\[1.5em] = \frac{({7 + 3\sqrt{5}})^2}{(7)^2 - (3\sqrt{5})^2} \\[1.5em] = \frac{7^2 + 2 × 7 × 3\sqrt{5} + (3\sqrt{5})^2}{49 - 45} = \frac{49 + 42\sqrt{5} + 45}{4} = \frac{94 + 42\sqrt{5}}{4} \\[1.5em] = \frac{47 + 21\sqrt{5}}{2} \\[1.5em] \therefore x = \frac{47 + 21\sqrt{5}}{2} \\[1.5em] \Rightarrow \frac{1}{x} = \frac{2}{47 + 21\sqrt{5}} \\[1.5em]\end{aligned}$$

Rationalising denominator of $\frac{1}{x}$,

$$\begin{aligned}\frac{1}{x} = \frac{2}{47 + 21\sqrt{5}} × \frac{47 - 21\sqrt{5}}{47 - 21\sqrt{5}} \\[1.5em] = \frac{{2}(47 - 21\sqrt{5})} {(47)^2 - (21\sqrt{5})^2} \\[1.5em] = \frac{{2}(47 - 21\sqrt{5})} {2209 - 2205} \\[1.5em] = \frac{{2}(47 - 21\sqrt{5})} {4} \\[1.5em] = \frac{(47 - 21\sqrt{5})} {2} \\[1.5em]\end{aligned}$$

Now,

$$\begin{aligned}{\Big(x + \frac{1}{x}\Big)} = \frac{47 + 21\sqrt{5}}{2} + \frac{(47 - 21\sqrt{5})}{2} \\[1.5em] = \frac{47 + 21\sqrt{5} + 47 -21\sqrt{5}}{2} \\[1.5em] = \frac{94}{2} = 47 \\[1.5em] \therefore {\Big(x + \frac{1}{x}\Big)} = 47 \qquad \text{....(i)}\end{aligned}$$

We know that ${\Big(x + \frac{1}{x}\Big)}^2 = x^2 + \frac{1}{x^2} + 2$

$$\begin{aligned}\Rightarrow x^2 + \frac{1}{x^2} = {\Big(x + \frac{1}{x}\Big)}^2 -2 \\[1.5em] \Rightarrow x^2 + \frac{1}{x^2} = (47)^2 - 2 \qquad ..... \text{using(i)} \\[1.5em] \Rightarrow x^2 + \frac{1}{x^2} = 2209 - 2 = 2207 \\[1.5em] \bold{x^2 + \frac{1}{x^2} = 2207} \\[1.5em]\end{aligned}$$

$$\begin{aligned}\text{(ii) } x = \frac{\sqrt{5} - \sqrt{2} }{\sqrt{5} + \sqrt{2}} \text{ and y} = \frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}} \\[1.5em] \therefore x+y = \frac{\sqrt{5} - \sqrt{2}}{\sqrt{5} + \sqrt{2}} +\frac{\sqrt{5} + \sqrt{2}}{\sqrt{5} - \sqrt{2}} \\[1.5em] \Rightarrow\frac{(\sqrt{5} - \sqrt{2})^2 + (\sqrt{5} + \sqrt{2})^2}{(\sqrt{5} + \sqrt{2})(\sqrt{5} - \sqrt{2})} \\[1.5em] \Rightarrow\frac{(\sqrt{5})^2 - 2 × \sqrt{2} × \sqrt{5} + (\sqrt{2})^2+ (\sqrt{5})^2 + 2 × \sqrt{2} × \sqrt{5} + (\sqrt{2})^2 }{(\sqrt{5})^2 - (\sqrt{2})^2 } \\[1.5em] \Rightarrow\frac{5 - 2\sqrt{10} + 2 + 5 + 2\sqrt{10} + 2}{5 - 2} = \frac{14}{3} \\[1.5em] \Rightarrow x + y = \frac{14}{3} \qquad \text{....(i)} \\[1.5em] \text{Also } xy = \frac{\sqrt{5} - \sqrt{2} }{\sqrt{5} + \sqrt{2}} ×\frac{\sqrt{5} + \sqrt{2} }{\sqrt{5} - \sqrt{2}} = 1 \qquad \text{....(ii)} \\[1.5em]\end{aligned}$$

We need to find the value of ${x^2 + xy + y^2}$ $$\begin{aligned}{x^2 + xy + y^2} = x^2 + y^2 + 2xy - xy \\[1.5em] \Rightarrow {x^2 + xy + y^2} = (x+y)^2 - xy \qquad \text{....(iii)} \\[1.5em]\end{aligned}$$

Substituting the values from (i) and (ii) in (iii),

$$\begin{aligned}{x^2 + xy + y^2} = \Big(\frac{14}{3}\Big)^2 - 1 = \frac{196}{9} - 1 = \frac{196 - 9}{9} = \frac{187}{9} \\[1.5em] \therefore \bold{x^2 + xy + y^2} = \bold{\frac{187}{9}} \\[1.5em]\end{aligned}$$

$$\begin{aligned}\text{(iii) } x = \frac{\sqrt{3} - \sqrt{2} }{\sqrt{3} + \sqrt{2}} \text{ and y} = \frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \\[1.5em] \therefore x+y = \frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}} +\frac{\sqrt{3} + \sqrt{2}}{\sqrt{3} - \sqrt{2}} \\[1.5em] = \frac{(\sqrt{3} - \sqrt{2})^2 + (\sqrt{3} + \sqrt{2})^2}{(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2})} \\[1.5em] = \frac{(\sqrt{3})^2 - 2 × \sqrt{2} × \sqrt{3} + (\sqrt{2})^2+ (\sqrt{3})^2 + 2 × \sqrt{2} × \sqrt{3} + (\sqrt{2})^2 }{(\sqrt{3})^2 - (\sqrt{2})^2 } \\[1.5em] = \frac{3 - 2\sqrt{6} + 2 + 3 + 2\sqrt{6} + 2}{3 - 2} = \frac{10}{1} = 10 \\[1.5em] \therefore x + y = 10 \qquad \text{....(i)} \\[1.5em] \text{Also } xy = \frac{\sqrt{3} - \sqrt{2} }{\sqrt{3} + \sqrt{2}} ×\frac{\sqrt{3} + \sqrt{2} }{\sqrt{3} - \sqrt{2}} = 1 \qquad \text{....(ii)} \\[1.5em]\end{aligned}$$

We need to find the value of ${x^3 + y^3}$ $$\begin{aligned}{x^3 + y^3} = (x+y)^3 - 3xy(x + y) \qquad \text{....(iii)} \\[1.5em]\end{aligned}$$

Substituting the values from (i) and (ii) in (iii),

$$\begin{aligned}{x^3+ y^3} = (10)^3 - 3 × 1 × 10 \\[1.5em] = 1000 - 30 \\[1.5em] = 970 \\[1.5em] \therefore \bold{x^3+ y^3 = 970}\end{aligned}$$

Write all the numbers as square root under one radical :

$$\begin{aligned}\sqrt{2} = \sqrt{2} \\[1.5em] 3.5 = \sqrt{12.25} \\[1.5em] \sqrt{10} = \sqrt{10} \\[1.5em] {-\frac{5}{\sqrt{2}}} = - \sqrt{\frac{25}{2}} = -\sqrt{12.5} \\[1.5em] {\frac{5}{2}}{\sqrt{3}} = \sqrt{\frac{25 × 3}{4}} = \sqrt{\frac{75}{4}} = \sqrt{18.75} \\[1.5em] \text{Since} , 18.75 \gt 12.25 \gt 10 \gt 2 \gt - 12.5 \\[1.5em] \Rightarrow \sqrt{18.75} \gt \sqrt{12.25} \gt \sqrt{10} \gt \sqrt{2} \gt -\sqrt{12.5} \\[1.5em] \Rightarrow {\frac{5}{2}}{\sqrt{3}}\gt 3.5 \gt \sqrt{10} \gt \sqrt{2} \gt -\frac{5}{\sqrt{2}}\end{aligned}$$

Hence, the given numbers in descending order are $\bold{\frac{5}{2}{\sqrt{3}}} , \bold{3.5} , \bold{\sqrt{10}} ,\bold{\sqrt{2}} , \bold{-\frac{5}{\sqrt{2}}}$.

Consider the squares of $\sqrt{3}$ and $\sqrt{5}$

$(\sqrt{3})^2$ = 3 and $(\sqrt{5})^2$ = 5

Here, 4 is rational number between 3 and 5

$\sqrt{4}$ = 2 is rational number between $\sqrt{3}$ and $\sqrt{5}$.

Irrational number between $\sqrt{3}$ and $\sqrt{5}$ = $\frac{\sqrt{3} + \sqrt{5}}{2}$

Consider the squares of $2\sqrt{3}$ and $2\sqrt{5}$.

$(2\sqrt{3})^2$ = 4 × 3 = 12 and $(2\sqrt{5})^2$ = 4 × 5 = 20

As , $18 \gt 17 \gt 15$ it follows that

$\sqrt{18} \gt \sqrt{17} \gt \sqrt{15}$ , therefore

$\sqrt{18}$ , $\sqrt{17}$ , $\sqrt{15}$ lie between $\sqrt{12}$ and $\sqrt{20}$ i.e. $2\sqrt{3}$ and $2\sqrt{5}$.

Hence, three irrational number between $2\sqrt{3}$ and $2\sqrt{5}$ in descending order are $\sqrt{18}$ , $\sqrt{17}$ , $\sqrt{15}$.

Let a = $\sqrt{2}$ and b = $\sqrt{3}$ are two different irrational numbers :

(i) a + b = $\sqrt{2}$ + $\sqrt{3}$ is also an irrational number.

(ii) a × b = $\sqrt{2}$ × $\sqrt{3}$ = $\sqrt{6}$ is also an irrational number.

Let a = $2\sqrt{3}$ and b = $5\sqrt{3}$ be two different irrational numbers

Here, $\frac{a}{b}$ = $\frac{2\sqrt{3}}{5\sqrt{3}}$ = $\frac{2}{5}$

$\therefore \bold{\frac{2}{5}}$ is a rational number.

34.0356 can be expressed in the form $\frac{p}{q}$

This can be written as 34.0356 = $\frac{340356}{10000}$ = $\frac{85089}{2500}$

Here, 85089 and 2500 are coprime integers .

Since , it is terminating decimal

It is Rational number and the prime factors of its denominator q will be 2 or 5 or both .

(i) 279.034

This can be written as 279.034 = $\frac{279034}{1000}$

Since, it is terminating decimal

It is Rational number and the prime factors of its denominator q will be 2 or 5 or both .

(ii) $76.\overline{17893}$

Since it is non-terminating recurring decimal,

$76.\overline{17893}$ = 76.1789317893...

It is a rational number which is non-terminating and repeating. Its denominator q will have prime factors other than 2 or 5.

(iii) 3.010010001...

Since, it is non-terminating non-repeating decimal number

∴ It is an Irrational number.

(iv) 39.546782

This can be written as 39.546782 = $\frac{39546782}{1000000}$

Since , it is terminating decimal

It is Rational number and the prime factors of its denominator q will be 2 or 5 or both .

(v) 2.3476817681... = $2.34\overline{7681}$

Since, it is a non-terminating repeating decimal number,

∴ It is a Rational number and its denominator q will have prime factors other than 2 or 5.

(vi) 59.120120012000...

Since, it is non-terminating non-repeating decimal number

∴ It is an Irrational number.