Prove that √2 is irrational. (CBSE 2008, 2009)

Let us assume that √2 is rational. So, we can find integers p and q (≠ 0) such that √2 = .

Suppose p and q have a common factor other than 1.

Then, we divide by the common factor to get √2 = , where a and b are coprime.

So, b√2 = a.

Squaring on both sides, we get

2b² = a²

Therefore, 2 divides a².

Now, by Theorem which states that Let p be a prime number. If p divides a² , then p divides a, where a is a positive integer,

2 divides a².

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

Substituting for a, we get 2b² = 4c² ,i.e. b² = 2c² .

This means that 2 divides b², and so 2 divides b (again using the above 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 2 is rational.

So, we conclude that √2 is irrational.