Question

prove √3 is irrational

Answer 1

this is the one Answer for the given question

Answer 2

Hey Friend ✋✋✋

Let us assume that √3 is a rational number.....

let a and b be any integers where a and b are co-primes...
such that a/b = √3

$\sqrt{3} = \frac{a}{b} \\ \\ squaring \: both \: sides \\ \\ 3 = \frac{ {a}^{2} }{ {b}^{2} } \\ \\ {a}^{2} = 3 {b}^{2} .......(1) \\ \\ {b }^{2} = \frac{ {a}^{2} }{3} ..........therefore \: \: 3 \: divides \: {a}^{2} ...so \: 3 \: will \: divide \: a \\ \\ {a} = 3c \\ \\ 3 {b}^{2} = 9 {c}^{2} \\ \\ {c}^{2} = \frac{ {b}^{2} }{3} ........3 \: divides \: {b}^{2} ...so \: 3 \: divides \: b \\ \\ therefore \: a \: \: and \: b \: has \: common \: factor \: 3 \: so \: they \: are \: not \: coprimes... \\ \\ it \: condricts \: that \sqrt{3} is \: a \: irrational \: number...$


hope it helps .