Assume that √3is rational
We  took 2 integers  i.e, r& s
Therefore, √3= r/s ,[ p/q form]
√3= r/x/s/x [xis  a  common  factor] ,  =a/b
squaring on both sides 
√3 b2  = a2
3b2 a2
=3divides a 
Similarly  ,  we  have to proove   3dvides b also
From  above  situation
 It i clear that a& b are integers & √3 is  rational
But bythe fat √3 Is irrtional
So our assumption is wrong.

√3 is irrational, to prove that, assume  that √3 is rational. 
Then, √3 = a/b
when we square both sides,
3 = a²/b², which implies,
3b² = a²
Since 3 is a factor of 3b², a² is divisible by 3. Now, if a² is divisible by 3 then 'a' is also divisible by 3.(1)
Now, since 'a' is divisible by 3, 3 is a factor of 'a'. That is, a = 3k, where 'k' is some integer. Now, 3b² = a² ⇒ 3b² = (3k)² ⇒ 3b² = 9k². Dividing both sides by 3, we have b² = 3k² which means that b² is divisible by 3. This implies that 'b' is also divisible by 3.(2) Now, in (1) 'a' is divisible by 3 and in (2) 'b' is also divisible by 3. This contradicts the fact that 'a' and 'b' have no common factor except 1.Therefore, our assumption is false, and hence √3 is irrational.