Let us assume, to the contrary, that √3 is rational.
So, we can find coprime integers a and b(b ≠ 0)
such that √3 = a/b
=> √3b = a
=> 3b2 = a2 ….(i) [Squaring both the sides]
=> a2 is divisible by 3
=> a is divisible by 3
So, we can write a = 3c for some integer c.
=> a2 = 9c2 ….[Squaring both the sides]
=> 3b2 = 9c2 ….[From (i)]
=> b2 = 3c2
=> b2 is divisible by 3
=> b is divisible by 3
=> 3 divides both a and b.
=> a and b have at least 3 as a common factor.
But this contradicts the fact that a and b are coprime.
This contradiction arises because we have assumed that √3 is rational.
Hence, √3 is irrational.