Square Root of Prime is Irrational

Theorem:
The square root of any prime number is irrational.

Proof by Contradiction
Let $$p$$ be prime.

Suppose that$$\sqrt{p}$$ is rational. Then, there exist coprime integers $$m \,\!$$ and $$n \,\!$$ such that:

$$ $$ $$

Any prime in the prime factorization of $$n^2$$ or $$m^2$$ must occur an even number of times (because they are squares).

Thus, $$p$$ must occur in the prime factorization of $$n^2 p$$ either once or an odd number of times.

Therefore, $$p$$ occurs as a factor of $$m^2$$ either once or an odd number of times, a contradiction in either case.

Thus,$$\sqrt{p}$$ must be irrational.

Note: the special case of p = 2 is a well-known mathematical proof.