Question

Prove that 3+2√2 is irrational​

Answer 1

AnswEr :

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

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

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

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

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

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

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

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

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

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

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

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

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

Answer 2

Question :

Prove that 3+2√2 is irrational

Solution :

we have to prove that 3+2√2 is irrational.

Let us assume that 3+2√2 is irrational.

we know that if any no is rational then it can be written in the form $\sf\dfrac{p}{q}$

Hence,$\sf3+2\sqrt{2}$ can be writing in the form $\sf\dfrac{a}{b}$ where a,b(b≠0 ) are co- prime .

Thus ,

$\sf3+2\sqrt{2}=\dfrac{a}{b}$

$\sf2\sqrt{2}=\dfrac{a}{b}-3$

$\sf2\sqrt{2}=\dfrac{a-3b}{b}$

$\sf\sqrt{2}=\dfrac{a-3b}{2b}$

Here ,

$\sf\dfrac{a-3b}{2b}\:is\:rational$

But √3 is irrational.

Since, Rational≠iraational

This is a contradiction

∴ Our assumption is wrong

Hence ,3+2√2 is irrational.

Hence , Proved