Question

Prove that 1 upon 2 minus under root 5 is irrational number ​

Answer 1

$\sf Rationalizing\ it,\\\\\frac{1}{2\ -\ \sqrt{5}} = \frac{1 (2\ +\ \sqrt{5})}{(2\ -\ \sqrt{5})(2\ +\ \sqrt{5)}} = \frac{2\ +\ \sqrt{5}}{2^{2}\ -\ \sqrt{5} ^{2}} = \frac{2\ +\ \sqrt{5}}{4\ -\ 5} = \frac{2\ +\ \sqrt{5}}{-1} = -2\ -\ \sqrt{5}$$\sf Let\ us\ assume\ that -2\ -\ \sqrt{5}\ is\ rational.\\\\-2\ -\ \sqrt{5} = \frac{a}{b}, where\ a\ and\ b\ are\ co-prime\ integers\ and\ b\neq  0.\\\\-2\ -\ \sqrt{5} = \frac{a}{b}\\ \\-2\ - \frac{a}{b} = \sqrt{5} \\\\\frac{-2b\ -\ a}{b} = \sqrt{5} \\\\We\ know\ that \sqrt{5}\ is\ irrational.\\\\Since\ a\ and\ b\ are\ integers,\ \frac{-2b\ -\ a}{b}\ is\ rational.\\\\This\ implies\ that\ \sqrt{5}\ is\ rational\ too.\\\\This\ contradicts\ the\ fact\ that\ \sqrt{5}\ is irrational.$ $\sf Therefore,\ our\ assumption\ is\ wrong.\\\\\frac{1}{2\ -\ \sqrt{5}}\ is\ irrational.$