Question

Prove that √2 is an irrational number.Hence, show that 3-√2 is an irrational number.​

Answer 1

$\bf Let's\ suppose\ \sqrt{2}\ is\ a\ rational\ number.\\\\\\\sf \sqrt{2}\ =\ \dfrac{a}{b},\ where\ a\ and\ b\ are\ co-prime\ integers\ and\ b\ is\ \ne\ 0.\\\\\\\\\sqrt{2}\ =\ \dfrac{a}{b}\\\\\\\bf Squaring\ both\ sides,\\\\\\\sf (\sqrt{2})^2\ =\ \bigg(\dfrac{a}{b}\bigg)^2\\\\\\2\ =\ \dfrac{a^2}{b^2}\\\\\\\implies\ 2\ divides\ a^2\\\\\\\implies\ 2\ divides\ a\\\\\\\bf Now,\ let's\ suppose\ a\ =\ 2c\ for\ some\ integer\ c.\\\\\\\sf (2c)^2\ =\ 2b^2\\\\\\4c^2\ =\ 2b^2$

$\sf 2c^2\ =\ b^2\\\\\\\implies\ 2\ divides\ b^2\\\\\\\implies\ 2\ divides\ b\\\\\\\bf This, 2\ is\ a\ factor\ of\ both\ a\ and\ b.\\\\\\This\ contradicts\ the\ fact\ that\ a\ and\ b\ are\ co-prime\ integers.\\\\\\This\ contradiction\ has\ arisen\ due\ to\ our\ wrong\ assumption.\\\\\\Therefore,\ our\ assumption\ is\ wrong.\\\\\\\sqrt{2}\ is\ an\ irrational\ number.$

$\bf Now,\ to\ prove\ 3\ -\ \sqrt{2}\ is\ irrational.\\\\\\Let's\ suppose\ 3\ -\ \sqrt{2}\ is\ irrational.\\\\\\\sf 3\ -\ \sqrt{2}\ =\ \dfrac{a}{b},\ where\ a\ and\ b\are\ co-primes\ integers\ and\ b\ is\ \ne\ 0.\\\\\\3\ -\ \dfrac{a}{b}\ =\ \sqrt{2}\\\\\\\dfrac{3b\ -\ a}{b}\ =\ \sqrt{2}\\\\\\\bf Since\ a\ and\ b\ are\ integers\ and\ \dfrac{3b\ -\ a}{b}\ is\ rational,\\\\\\\implies\ \sqrt{2}\ is\ also\ rational.\\\\\\This\ contradicts\ the\ fact\ that\ \sqrt{2}\ is\ irrational.$

$\bf This\ contradiction\ has\ arisen\ due\ to\ our\ wrong\ assumption.\\\\\\Therefore,\ our\ assumption\ is\ wrong.\\\\\\3\ -\ \sqrt{2}\ is\ irrational.$