Let us consider to be rational

where a & b are integers (b≠0)

Rearranging we get

The R.H.S of the above expression is a rational number since it can be expressed as a numerator by a denominator

Let L.H.S = where p and q are integers (q≠0)

⇒

⇒

Squaring both sides we get

2q² = p²…Equation 1

Since 2 divides p² so it must also divide p

so p is a multiple of 2

let p = 2k where k is an integer

Putting in Equation 1 the value of p we get

2q² = 4k²

⇒ q² = 2k²

Since 2 divides q² so it must also divide q

so q is a multiple of 2

But this contradicts our previously assumed data since we had considered p & q has been resolved in their simplest form and they shouldn't have any common factors.

So √2 is irrational and hence

is also irrational

Hence Proved