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

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

(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 $\bold{-3\sqrt{2}}.$