Example: Prove that √3 is irrational.
Solution: Let us assume that √3 is rational.
That is, we can find integers a and b (≠ 0) such that 3 =a/b.
Suppose a and b have a common factor other than 1, then we can divide by the common factor, and assume that a and b are coprime.
Squaring both sides, we get 3b^2= a^2
Therefore, a^2 is divisible by 3, and we know that If p(a prime number) divides a^2 , then p divides a as well.
It means a is also divisible by 3.
So, we can write a = 3c for some integer c.
Substituting for a, we get
that is, b^2= 3c^2
This means that b^2 is divisible by 3, and so b is also divisible by 3.
Therefore, a and b have at least 3 as common factors. But this contradicts the fact that a and b are coprime.
This contradiction has arisen because of our incorrect assumption that 3 is rational.
So, we conclude that √3 is irrational.