√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.