Question

prove 3 by 5 is an irrational number​

Answer 1

Proof

$\sf Let \ 3\sqrt{5} \ be \ a \ rational \ number.$

$\sf Rational \ numbers \ are \ in \ the \ form \ \dfrac{a}{b} \ where \ a \ and \ b \ are \ coprime \ and \ b \neq 0$

$\implies 3\sqrt{5}=\dfrac{a}{b}$

$\implies \sqrt{5}=\dfrac{a}{3b}$

$\sf \bold{Now, \dfrac{a}{3b} \ is \ a \ rational \ number}$

$\implies \bold{\sqrt{5} \ is \ a \ rational \ number}$

$\sf \bold{But \ this \ contradict \ to\ the\ fact \ that \ \sqrt{5} \ is \ an \ irrational \ number.}$

$\bold{Hence \ our \ assumption \ is \ wrong}$

$\boxed{\boxed{\bold{Therefore, \ 3\sqrt{5} \ is \ an \ irrational \ number}}}}$

                                                                           

Answer 2

Answer:

answer = 5

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