If possible Let √2 is rational
∴ √2 = p/q, p, q ∈ z, q ≠ 0
We assume that p and q do not have any common factor
p = √2q ⇒ p = 2q2
p2 is a multiple of 2 ⇒ ∴ p is a multiple of 2
∴ p = 2k, k ∈ z ⇒ p2 = 4k
2q2 = 4k2 ⇒ q2 = 2k2
q2 = 2k2
q2 is a multiple of 2 ⇒ ∴ q is a multiple of z
∴ p and q are both multiple of 2 and hence has a common factor z which is a contradiction
∴ our assumption is wron
∴ √2 is irrational