Question

Prove that root3 is an irrational number​

Answer 1

root 3 = 1.7329508076...

it is not terminating not reapting

so it is irrational number

Answer 2

$\huge\mathfrak\color{red}AnSwEr$

$\sf\orange {\underline {\sf ☆Let's}}$

Assume, to contrary, that $\sqrt {3}$ is a rational number

$\sf\green{\underline{\sf ☆So,}}$

We can find integers a and b such that, $\sqrt {5}=\frac {a}{b}$

$\sqrt{5}=\frac{a}{b}$where a and b are

co-primes.

$\sf\pink {\underline{\sf ☆So,}}$

$(b{ \sqrt{2}) }^{2} = {a}^{2}$

${2b}^{2} = {a}^{2}$

$b = \frac {\sqrt {a}^{2}}{2}$

$a = 2c$

$\sf\purple{\underbrace {\sf ☆Now,}$

Substituting for a :-

$\sqrt {2b}^{2} = \sqrt {a}^{2}$

$\sqrt {2b}^{2} = \sqrt ({2c}^{2})$

$\sqrt {2b}^{2} = \sqrt{4c}^{2}$

$\sqrt {b}^{2} = \frac {\sqrt {4c}^{2}}{2}$

$\sqrt {b}^{2} = \sqrt {2c}^{2}$

$\huge\mathcal\color {grey}Might\:Help\:You$