Question

prove that √3 is an irrational number.​

Answer 1

_____________________________________________

$\huge\underline \mathfrak\blue{solution:-}$

Let us assume on the contrary that $\sqrt{3}$ is rational. Then there exists coprime positive integers a and b such that:-

$\green\longrightarrow\tt\sqrt{3} = \frac{a}{b}$

$\green\longrightarrow\tt a = \sqrt{3} \: b$

$\green\longrightarrow\tt {a}^{2} = 3 {b}^{2}$

$\green\longrightarrow\tt 3 \: divides \: {a}^{2}$

$\green\longrightarrow\tt 3 \: divides \: a.........(i)$

Again,

$\red\longrightarrow\tt a = 3c\:(For\:some\:integer\: c)$

$\red\longrightarrow\tt {a}^{2} = 9 {c}^{2}$

$\red\longrightarrow\tt 3 {b}^{2} = 9 {c}^{2}$

$\red\longrightarrow\tt {b}^{2} = 3 {c}^{2}$

$\red\longrightarrow\tt 3 \: divides \: {b}^{2}$

$\red\longrightarrow\tt 3 \: divides \: b...........(ii)$

From (i) and (ii) we observed that a and b have at least 3 as a common factor. But this contradicts the fact that a and b are coprime. So our assumption is not correct.

Thus, $\sqrt{3}$ is irrational.

I HOPE YOU WILL MARK ME AS BRAINLIEST❤

_____________________________________________