Question

50 points ♥️♥️♥️

Find 3 irrational between 4 and 5.​

Answer 1

$\fbox{\fbox{\mathcal{\large{Answer:-}}}}$

$\sf{\red{4 \times 5 = 20 \: isn't \: a \: square}}$

$\sf{\pink{Hence \: 1\: number\: is \sqrt{20} = 2 \sqrt{5} }}$

$\sf{\green{Similarly, 8 \sqrt{5} \: isn't a\: square \:}}$

$\sf{\blue{so\: \sqrt{8 \sqrt{5} }= \sqrt{300}\:is \: the\: second\: number}}$

$\sf{\purple{so\: {4}^{2} = 46 \: and \: {5}^{2} = 25}}$

$\sf{\orange{Hence \sqrt{19} is\: the\: third}}$

Answer 2

$\huge\mathfrak{\red{\bigstar\color{midnightblue}{\underline{\underline{Solution:-}}}}}$

$\sf{4\times 5=20\:is\:not\:a\: square}$

$\sf{Hence,\:1\: number\:is\: \sqrt{20} = \sqrt[2]{5} }$

$\sf{Similarly,\sqrt[8]{5} \:is\: not\:a\: square}$

$\sf{So\:\sqrt{8 \sqrt{2} } = \sqrt{300} \:is\: the\: second\: number}$

$\sf{\large{So\: {4}^{2} = 16 \: and \: {5}^{2} = 25}}$

$\bf{\red{ Hence \: \sqrt{19} \: is \: the \: third}}$