Let us assume that √2 is rational.

If it is rational, then there must exist two integers r and s (s ≠ 0) such that √2=

r

s

.

If r and s have a common factor other than 1, then,

we divide r and s by their highest common factor to get √2=

a

b

,

where a and b are co-prime. So,b√2 = a.

we get 2b² = a² . Therefore, 2 divides a².

On squaring both sides and rearranging

Now, by Theorem 1.6, it follows that since 2 is dividing a², it also divides a.

So, we can write a= 2c for some integer c.

Substituting for a, we get 2b²=4c², that is , b²=2c².

This means that 2 divides b², and so 2 divides b

(again using Theorem 1.6 with p-2)

Therefore, both a and b have 2 as a common factor.

But this contradicts the fact that a and b are .

This contradiction has arisen because of our assumption that √2 is rational.

Thus our assumption is false. So, we conclude that √2 is irrational.

So it is √2 is