Question

prove that 3 minus root 2 is an irrational number​

Answer 1

Solution:

$\sf Let\;assume\;that\;3-\sqrt{2}\;is\;rational\;number.\\ \\ \\ \implies \sf 3-\sqrt{2}=\dfrac{a}{b}\;\;\;\;[Where\;a\;and\;b\;are\;integers]\\ \\ \\ \implies \sf \sqrt{2} = 3-\dfrac{a}{b}\\ \\ \\ \sf Since\;a\;and\;b\;are\;integers.\;So,\;3-\dfrac{a}{b}\;is\;rational\;number.\\ \\ \sf But\;as\;we\;know\;that\;\sqrt{2}\;is\;irrational\;number.\\ \\ \sf And,\;Irrational\neq Rational.\\ \\ \sf So,\;our\;assumption\;is\;wrong.\\ \\ \sf 3-\sqrt{2}\;is\;Irrational\;number.\\ \\ \underline{\bf Hence\;Proved!!!}$

Answer 2

$\huge{\underline{\underline{\bf{\purple{Solution:-}}}}}$

$\sf Let\;3-\sqrt{2}\;be\;the\;rational\;number.$

$\bold{Its\: simplest\: form}$$\large\frac{p}{q}$$\bold{where\: p\: and\: q\: have\: no\: common\: factor.}$$\bold{other\: than\: one}$

➠${3}{-}\sqrt{2}\:{=}\:\large\frac{p}{q}$

➠${3}{-}\large\frac{p}{q}\:{=}\:\sqrt{2}$

➠$\sqrt{2}\:{=}\:\large\frac{3q-p}{q}$

$\bold{Since,\: p\: and\: q\: are\: integers\: so}$$\large\frac{3q-p}{q}$$\bold{is\: rational}$

.•.$\sqrt{2}$$\bold{is\: rational}$

$\bold{But\: the\: contradiction\: the\: fact\: is\: that}$

$\sqrt{2}$$\bold{is\: irrational.}$

$\bold{ So\: our\: assumption\:is\: wrong.}$

$\sf Hence\;3-\sqrt{2}\;is\;irrational.$