Let us assume, to the contrary, 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.
So, b√3 =a
Squaring on both sides, and rearranging, we get 3b2 = a2.
Therefore, a2 is divisible by 3, and by Theorem 1.3, it follows that a is also divisible by 3.
So, we can write a = 3c for some integer c.
Substituting for a, we get 3b2 = 9c2, that is, b2 = 3c2.
This means that b2 is divisible by 3, and so b is also divisible by 3 (using Theorem 1.3 with p = 3).
Therefore, a and b have at least 3 as a common factor.
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.