Question

Prove that 3+2√2 is irrational​

Answer 1

$\huge{ \purple{ \bigstar{ \mathfrak{answer♡</p><p>}}}}$

$\huge{ \red{ \ddot{ \smile}}}$

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let us assume that$\large{ \boxed{ \purple{3+2\sqrt{2}}}}$is rational.

then,

=>$3 + 2 \sqrt{2} = \frac{a}{b} \: \:$

where a and b are co prime.

$\large{ \boxed{ \green{\mathbb{EXPLANATION:-}}}}$

$=>3 + 2 \sqrt{2} = \frac{a}{b}\\ \\ \\ =>2 \sqrt{2} = \frac{a}{b} - 3 \\ \\=> 2 \sqrt{2} = \frac{a - 3b}{b} \\ \\ \sqrt{2} = \frac{a - 3b}{2b}$

since,

a and b are intigers, so

$=>\frac{a - 3b}{2b} \: is \: rational$

so,

=>$\large{ \boxed{ \purple{\sqrt{2}}}}$is also rational.

but we know that √2 is irrational.

so ,

this contradiction is arissen because of our wrong assumption.

so,

$\large{ \boxed{ \green{3 + 2 \sqrt{2} \: is \: irrational}}}$

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hops this may help you

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$\huge \blue{ \mathfrak{thanks♡</p><p>}}$

Answer 2

AnswEr :

$\normalsize\sf\ Let \: \red{ 3 + 2\sqrt{2}} \; be \: a \: rational \: number$

$\normalsize\sf\ As; \: we \: know \: Rational \: number \: is \: in \: form \: of \: \green{\frac{a}{b}}$

$\normalsize\dashrightarrow\sf\red{3 + 2 \sqrt{2} } = \green{\frac{a}{b}}$

$\normalsize\quad\sf\ [\because\ a \: and \: b \: are \: co-prime \: and \: b \: \neq\ \: 0]$

$\underline{\bigstar\:\textsf{According \: to \: the \: question \: now:}}$

$\normalsize\ : \implies\sf\ 3 + 2 \sqrt{2} = \frac{a}{b}$

$\normalsize\ : \implies\sf\ 2 \sqrt{2} = \frac{a}{b} - 3$

$\normalsize\ : \implies\sf\ 2 \sqrt{2} = \frac{a - 3b}{b}$

$\normalsize\ : \implies\sf\ \sqrt{2} = \frac{a - 3b}{2b}$

$\boxed{\begin{minipage}{5.3 cm}\underline{\qquad\quad\sf Concept\:Used\qquad\quad}\\\quad\swarrow\qquad\qquad\qquad\searrow\\\sf\frac{a-3b}{2b}(Rational)\qquad\sqrt{2}(Irrational)\\\seaarrow\qquad\qquad\quad(Rational \neq Irrational)\end{minipage}}$

$\normalsize\sf\ This \: contradicts \: the \: fact \: that \: \red{\sqrt{2}} \: is \: \blue{rational}$

$\normalsize\sf\ Hence, \: Our \: assumption \: is \: wrong \: \\ \normalsize\sf\ \red{3 + 2 \sqrt{2}} \: is \: \pink{irrational}$

$\large\maltese \: \: {\boxed{\sf \orange{Hence \: Prove \: !!}}}$