Question

prove that
$\sqrt{3 } \: \: \: is \: an \: irrational \: no$

Answer 1

suppose root 3 is a rational no.

such that root 3 = a/b , where a and b both r integers and b = nonzero integer 'a ' and 'b ' have no common factor other than 1

therefore root 3 = a/b .......where a nad b r coprime nos.

therefore , b root 3 = a

therefore, 3b2 = a2 (squaring both sides)..... (1)

therefore , b2 = a2/3

therefore, 3 divides a2 ,so 3 divides a

so we write, a = 3c..........where 'c ' is an integer.....(2)

a2 = (3c)2 .....squaring both sides

therefore, 3b2 = 9c2 .........substuting (2) in (1)

therefore, b2 = 3 c2

therefore, c2 = b2 / 3

3 divides b2 , means 3 divides b

therefore 'a ' and 'b ' have at least 3 as common factor but it was stated before that it was stated that a and b had no common factors other than 1.

this contradiction arises because we have assumed that root 3 is rational. therefore, root 3 is irritional no