Square Root of Prime is Irrational

Theorem: The square Root of Any Prime is Irrational

Proof by contradiction
Let $$p$$ be prime. Suppose the $$\sqrt{p}$$ is rational. Then it can be expressed as $$\frac{m}{n}$$ where $$m$$ and $$n$$ are both integers. Thus we have $$\sqrt{p} = \frac{m}{n}$$ $$p = \frac{m^2}{n^2}$$ $$n^2p = m^2$$ Any prime in the decomposition of $$n^2$$ or $$m^2$$ must occur an even number of times (because it is a square). Thus, p must occur in the decomposition of $$n^2p$$ either once or an odd number of times. This implies that p occurs in $$m^2$$ either once or an odd number of times, a contradiction in either case.  Thus, $$\sqrt{p}$$ must be irrational.  QED