We are now ready to give a proof that √2 is irrational. We will use a technique called proof by contradiction.

Example 7. Show that √2 is irrational.

Solution : Let us assume that √2 is rational.

If it is rational, then there must exist two integers r and s (s ¹ 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.

On squaring both sides and rearranging, we get 2b² = a² .

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

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

Substituting for a, we get 2b ² = 4c² 2 , that is, b ² = 2c ²

. This means that 2 divides b 2 , and so 2 divides b² (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 co-prime.

This contradiction has arisen because of our assumption that √2 is rational. Thus our assumption is false. So, we conclude that √2 is irrational.

In general, it can be shown that d is irrational whenever √d is a positive integer which is not the square of another integer. As such, it follows that √6, √8, √15 √24 etc. are all irrational numbers.

In class IX, we mentioned that :

• the sum or difference of a rational and an irrational number is irrational

• the product or quotient of a non-zero rational and irrational number is irrational.