Prove that ?6 is an irrational 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 number.

​​​​​​​Regards