Square Root of Prime is Irrational

Theorem: The square Root of Any Prime is Irrational

Proof by contradiction
Let $$p$$ be prime. Suppose that$$\sqrt{p}$$ is rational. Then it can be expressed as $$\frac{m}{n}$$ where $$m$$ and $$n$$ are both integers and are relatively prime. 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.  Q.E.D.

Note: the special case of this proof for 2 is a classic mathematical proof.