and s ( ¹ ) such that = r s . Suppose r and s have a common factor other than . Then, we divide by the common factor to get where a and b are coprime. So, b = a .
Squaring on both sides and rearranging, we get b = a . Therefore, divides a . Now, by Theorem . , it follows that divides a .
So, we can write a = c for some integer c . Substituting for a , we get b = c , that is, b = c . This means that divides b , and so divides b (again using Theorem . with p = ).
Therefore, a and b have at least as a common factor. But this contradicts the fact that a and b have no common factors other than . This contradiction has arisen because of our incorrect assumption that is rational. So, we conclude that Example : Prove that is irrational.
Solution : Let us assume, to the contrary, that is rational. That is, we can find integers a and b ( ¹ ) such that = a b Suppose a and b have a common factor other than , then we can divide by the common factor, and assume that a and b are coprime. Squaring on both sides, and rearranging, we get b = a . Therefore, a is divisible by , and by Theorem .
, it follows that a is also divisible by . So, we can write a = c for some integer c . Substituting for a , we get b = c , that is, b = c . This means that b is divisible by , and so b is also divisible by (using Theorem .
with p = ). Therefore, a and b have at least as a