Question

prove that √3 is irrational number

Answer 1

if possible let √3 is rational and let it's simplest form be a/b.
then , a and b are integer having no common factor other than 1 , and b is not equal to 0.
now,√3= a/b
3= a^2/b^2 (on squaring both side)
3b^2=a^2--------(1)
3 divides a^2 (3 divides 3b^2)
3 divides a (3is a prime and divides a^2= 3divides a)
let a= 3cfor some integer c .
putting a= 3c in (1) , we get
3b^2= 9c^2
b^2=3c^2
= 3 divides b^2(3 divides c^2)
= 3 divides b ( 3 is prime and 3 divides b^2 = 3 divides b)
thus , 3 is common factor of a and b .
but this contradict the fact that a bad b have no common factor other than 1 .
the contradiction arises by assuming that √3 is rational.
hence, √3 is irrational.