Question

Prove that following numbers are not rational : 5√3

Answer 1

AnswEr :

$\normalsize\sf\ Let \: 5\sqrt{3} \; be \: a \: rational \: number \\ \\ \sf\ As; \: we \: know \: Rational \: number \: is \: in \: form \: of \: \frac{a}{b} \\ \\ \\ \normalsize\ : \implies\sf\ 5\sqrt{3} = \frac{a}{b} \\ \footnotesize\quad\sf\ [\because\ a \: and \: b \: are \: co-prime \: and \: b \: \neq\ \: 0]$

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$\underline{\bigstar\:\textsf{According \: to \: the \: question \: now:}} \\ \\ \\ \normalsize\ : \implies\sf\ 5\sqrt{2} = \frac{a}{b} \\ \\ \\ \normalsize\ : \implies\sf\ \sqrt{2} = \dfrac{a}{5b} \\ \\ \\ \normalsize\ : \implies\sf\ \sqrt{2} = Irrational \: \& \: \dfrac{a}{5b} = Rational \\ \\ \\ \normalsize\ : \implies\sf\ Rational \neq Irrational$

$\normalsize\sf\ This \: contradicts \: the \: fact \: that \: \sqrt{2} \: is \: rational\\\\ \normalsize\sf\ Hence, \: Our \: assumption \: is \: wrong \:, 5\sqrt{3}\\ \normalsize\sf\ \: is \: irrational\\\\\qquad\begin{aligned}\bf{\dag}\:\:\sf Hence, Verified!! \:\:\quad\end{aligned}$

Answer 2

$\huge\sf\blue{Given}$

➝ 5√3

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$\huge\sf\gray{To\;Prove}$

✭ 5√3 is a irrational number

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$\huge\sf\purple{Steps}$

$\normalsize\sf\ Let \: 5\sqrt{3} \; be \: a \: rational \: number \\ \sf\ We \: know \: Rational \: number \: is \: in \: form \: of \: \frac{a}{b} \\ \\ \normalsize\ \leadsto\sf\ 5\sqrt{3} = \frac{a}{b}$

Where a and b are co primes and b not equal to 0

$\normalsize\ \dashrightarrow{\sf{\orange{5\sqrt{3} = \frac{a}{b}}}} \\ \\ \normalsize\ \dashrightarrow\sf\ \sqrt{3} = \dfrac{a}{5b} \\ \\ \normalsize\ \dashrightarrow\sf\ \sqrt{3} = Irrational \: \& \: \dfrac{a}{5b} = Rational \\ \\ \normalsize\ \dashrightarrow\sf\ Rational ≠ Irrational$

$\normalsize\sf\ But\: it \: contradicts \: the \: fact \: that \: 5\sqrt{3} \: is \: rational\\ \normalsize\sf\ Hence, \: Our \: assumption \: is \: wrong \:, 5\sqrt{3}\\ \normalsize\sf\ \: is \: irrational$

Hence Verified

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