Rational and Irrational Numbers — Exercise 1.4

Simplify the following:

$\begin{matrix} \text{(i)} & \sqrt{45} - 3\sqrt{20} + 4\sqrt{5} \\[1.5em] \text{(ii)} & 3\sqrt{3} + 2\sqrt{27} + \dfrac{7}{\sqrt{3}} \\[1.5em] \text{(iii)} & 6\sqrt{5} × 2\sqrt{5} \\[1.5em] \text{(iv)} & 8\sqrt{15} ÷ 2\sqrt{3} \\[1.5em] \text{(v)} & \dfrac{\sqrt{24}}{8} + \dfrac{\sqrt{54}}{9} \\[1.5em] \text{(vi)} & \dfrac{3}{\sqrt{8}} + \dfrac{1}{\sqrt{2}} \\[1.5em] \end{matrix}$

Answer

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

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

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

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

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

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

Simplify the following:

$\begin{matrix} \text{(i)} & (5 + \sqrt{7})(2 + \sqrt{5}) \\[1.5em] \text{(ii)} & (5 + \sqrt{5})(5 - \sqrt{5}) \\[1.5em] \text{(iii)} & (\sqrt{5} + \sqrt{2})^2 \\[1.5em] \text{(iv)} & (\sqrt{3} - \sqrt{7})^2 \\[1.5em] \text{(v)} & (\sqrt{2} + \sqrt{3})(\sqrt{5} + \sqrt{7}) \\[1.5em] \text{(vi)} & (4 + \sqrt{5})(\sqrt{3} - \sqrt{7}) \\[1.5em] \end{matrix}$

Answer

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

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

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

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

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

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

If $\sqrt{2}$=1.414, then find the value of :

$\begin{matrix} \text{(i)} & \sqrt{8} + \sqrt{50} + \sqrt{72} + \sqrt{98} \\[1.5em] \text{(ii)} & 3\sqrt{32} - 2\sqrt{50} + 4\sqrt{128} - 20\sqrt{18} \\[1.5em] \end{matrix}$

Answer

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

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

If $\sqrt{3}$ = 1.732, then find the value of :

$\begin{matrix} \text{(i)} & \sqrt{27} + \sqrt{75} + \sqrt{108} - \sqrt{243} \\[1.5em] \text{(ii)} & 5\sqrt{12} - 3\sqrt{48} + 6\sqrt{75} + 7\sqrt{108} \\[1.5em] \end{matrix}$

Answer

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

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

State which of the following numbers are irrational :

$\begin{matrix} \text{(i)} & \sqrt{\dfrac{4}{9}}, -{\dfrac{3}{70}},\sqrt{\dfrac{7}{25}},\sqrt{\dfrac{16}{5}} \\[1.5em] \text{(ii)} & -{\sqrt{\dfrac{2}{49}}}, {\dfrac{3}{200}},\sqrt{\dfrac{25}{3}},-{\sqrt{\dfrac{49}{16}}} \\[1.5em] \end{matrix}$

Answer

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

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

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

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

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

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

State which of the following numbers will change into non-terminating, non-recurring decimals :

$\begin{matrix} \text{(i)} & - 3\sqrt{2} \\[1.5em] \text{(ii)} & \sqrt{\dfrac{256}{81}} \\[1.5em] \text{(iii)} & \sqrt{27 × 16} \\[1.5em] \text{(iv)} & \sqrt{\dfrac{5}{36}} \\[1.5em] \end{matrix}$

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

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 .

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

Here, $\sqrt{\dfrac{256}{81}} = \dfrac{16}{9}$ 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 .

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

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

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

State which of the following numbers are irrational:

$\begin{matrix} \text{(i)} & 3-\sqrt{\dfrac{7}{25}}\\[1.5em] \text{(ii)} & -\dfrac{2}{3}+\sqrt[3]{2} \\[1.5em] \text{(iii)} & \dfrac{3}{\sqrt{3}} \\[1.5em] \text{(iv)} & -\dfrac{2}{7}\sqrt[3]{5} \\[1.5em] \text{(v)} & (2-\sqrt{3})(2+\sqrt{3}) \\[1.5em] \text{(vi)} & (3+\sqrt{5})^2 \\[1.5em] \text{(vii)} &(\dfrac{2}{5}\sqrt{7})^2 \\[1.5em] \text{(viii)} & (3-\sqrt{6})^2 \\[1.5em] \end{matrix}$

Answer

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

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

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

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

Here, 2 is not perfect cube

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

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

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

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

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

Here, 5 is not perfect cube

∴ $-\dfrac{2}{7}\sqrt[3]{5}$ 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

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

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

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

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

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

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

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

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

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

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

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

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

Prove that the following numbers are irrational:

$\begin{matrix} \text{(i)} & \sqrt[3]{2} \\[1.5em] \text{(ii)} & \sqrt[3]{3} \\[1.5em] \text{(iii)} & \sqrt[4]{5} \\[1.5em] \end{matrix}$

Answer

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

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

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

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

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

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

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 $\dfrac{p}{q}$, 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 .

Find the greatest and the smallest real numbers among the following real numbers :

$\begin{matrix} \text{(i)} & 2\sqrt{3},\dfrac{3}{\sqrt{2}},-\sqrt{7},\sqrt{15} \\[1.5em] \text{(ii)} & -3\sqrt{2},\dfrac{9}{\sqrt{5}},-4,{\dfrac{4}{3}}{\sqrt{5}},{\dfrac{3}{2}}{\sqrt{3}} \\[1.5em] \end{matrix}$

Answer

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

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

∴ 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.

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

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

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

Write the following numbers in ascending order:

$\begin{matrix} \text{(i)} & 3\sqrt{2} , 2\sqrt{3} , \sqrt{15} , 4 \\[1.5em] \text{(ii)} & 3\sqrt{2} , 2\sqrt{8} , 4, \sqrt{50} ,4\sqrt{3} \\[1.5em] \end{matrix}$

Answer

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

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

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 :

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

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

Write the following real numbers in descending order:

$\begin{matrix} \text{(i)} & \dfrac{9}{\sqrt{2}} , {\dfrac{3}{2}}\sqrt{5} , 4\sqrt{3} , 3\sqrt{\dfrac{6}{5}} \\[1.5em] \text{(ii)} & \dfrac{5}{\sqrt{3}} , {\dfrac{7}{3}}{\sqrt{2}} , -\sqrt{3} , 3\sqrt{5} , 2\sqrt{7}. \\[1.5em] \end{matrix}$

Answer

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

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

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

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

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

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

Arrange the following numbers in ascending order : $\sqrt[3]{2} , \sqrt3 , \sqrt[6]{5}$.

Answer

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

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

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