Prove that is `sqrt2` irrational number.

#### Solution

Let us assume, to the contrary, that `sqrt2` is rational. So, we can find integers r and s (≠0) such that `sqrt2=r/s`. Suppose r and s not having a common factor other than 1. Then, we divide by the common factor to get `sqrt2=a/b` where a and b are coprime.

So, b`sqrt2` = a.

Squaring on both sides and rearranging, we get 2b^{2} = a^{2}.

Therefore, 2 divides a^{2}. Now, by Theorem it following that 2 divides a.

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

Substituting for a, we get 2b^{2} = 4c^{2}, that is, b^{2} = 2c^{2}.

This means that 2 divides b^{2}, and so 2 divides b (again using Theorem with p = 2).

Therefore, a and b have at least 2 as a common factor.

But this contradicts the fact that a and b have no common factors other than 1.

This contradiction has arisen because of our incorrect assumption that `sqrt2` is rational.

So, we conclude that `sqrt2` is irrational.