Question

Prove that following numbers are irrational : 2 + √2

Answer 1

AnswEr :

$\normalsize\sf\ Let \: 2 + \sqrt{2} \; be \: a \: rational \: number \\ \\ \sf\ As; \: we \: know \: Rational \: number \: is \: in \: form \: of \: \frac{a}{b} \\ \\ \\ \normalsize\dashrightarrow\sf\ 2 + \sqrt{2} = \frac{a}{b} \\ \footnotesize\quad\sf\ [\because\ a \: and \: b \: are \: co-prime \: and \: b \: \neq\ \: 0]$

$\rule{120}2$

$\underline{\bigstar\:\textsf{According \: to \: the \: question \: now:}} \\ \\ \\ \normalsize\ : \implies\sf\ 2 + \sqrt{2} = \frac{a}{b} \\ \\ \\\normalsize\ : \implies\sf\ \sqrt{2} = \dfrac{a}{b} - 2 \\ \\ \\ \normalsize\ : \implies\sf\ \sqrt{2} = \dfrac{a - 2b}{b} \\ \\ \\ \normalsize\ : \implies\sf\ \sqrt{2} = Irrational \: \& \: \dfrac{a - 2b}{b} = Rational \\ \\ \\ \normalsize\ : \implies\sf\ Rational \neq Irrational$

$\normalsize\sf\ This \: contradicts \: the \: fact \: that \: \sqrt{2} \: is \: rational\\\\ \normalsize\sf\ Hence, \: Our \: assumption \: is \: wrong \:, 2 + 2\sqrt{2}\\ \normalsize\sf\ \: is \: irrational\\\\\qquad\begin{aligned}\bf{\dag}\:\:\sf Hence, Verified!! \:\:\quad\end{aligned}$

Answer 2

$\sf Let\ us\ assume\ that\ 2\ +\ \sqrt{2}\ is\ rational.\\\\\\2\ +\ \sqrt{2}\ =\ \dfrac{a}{b},\ where\ 'a'\ and\ 'b'\ are\ co\ -\ prime\ integers\ and\ b\ is\ \ne\ 0.\\\\\\2\ +\ \sqrt{2}\ =\ \dfrac{a}{b}\\\\\\\sqrt{2}\ =\ \dfrac{a}{b}\ -\ 2\\\\\\\sqrt{2}\ =\ \dfrac{a\ -\ 2b}{b}\\\\\\Since\ 'a'\ and\ 'b'\ are\ integers\ and\ \dfrac{a\ -\ 2b}{b}\ is\ rational,\\\\\\\implies\ \sqrt{2}\ is\ rational\ as\ well.\\\\\\This\ contradicts\ the\ fact\ that\ \sqrt{2}\ is\ irrational.$

$\sf This\ contradiction\ has\ arisen\ due\ to\ our\ wrong\ assumption.\\\\\\\therefore\ Our\ assumption\ is\ wrong.\\\\\\2\ +\ \sqrt{2}\ is\ irrational.$