Question

Prove that root p plus root q is an irrational

Answer 1

$\sf Let's\ assume\ that\ \sqrt{p}\ +\ \sqrt{q}\ is\ rational.\\\\\\\sqrt{p}\ +\ \sqrt{q}\ =\ \dfrac{a}{b},\ where\ 'a'\ and\ 'b'\ are\ co\ -\ prime\ integers\ and\ b\ is\ \ne\ 0.$

$\sf \sqrt{p}\ +\ \sqrt{q}\ =\ \dfrac{a}{b}\\\\\\Squaring\ both\ sides.\\\\\\(\sqrt{p}\ +\ \sqrt{q})^2\ =\ \bigg(\dfrac{a}{b}\bigg)\\\\\\p\ +\ 2\sqrt{pq}\ +\ q\ =\ \dfrac{a^2}{b^2}\\\\\\2\sqrt{pq}\ =\ \dfrac{a^2}{b^2}\ -\ p\ -\ q\\\\\\2\sqrt{pq}\ =\ \dfrac{a^2\ -\ pb\ -\ qb}{b}\\\\\\\sqrt{pq}\ =\ \dfrac{a^2\ -\ pb\ -\ qb}{2b}\\\\\\Since\ 'a'\ and\ 'b'\ are\ integers\ and\ \dfrac{a^2\ -\ pb\ -\ qb}{2b}\ is\ rational,\\\\\\\implies\ \sqrt{pq}\ is\ rational.$

$\sf This\ contradicts\ the\ fact\ that\ \sqrt{pq}\ is\ irrational.\\\\\\This\ contradiction\ has\ arisen\ due\ to\ our\ wrong\ assumption.\\\\\\\therefore\ Our\ assumption\ is\ wrong.\\\\\\\sqrt{p}\ +\ \sqrt{q}\ is\ irrational.$