### Shortest Proof of Irrationality of sqrt(N), where N is not a perfect square

This proof is really the shortest proof I have ever seen for this theorem, and it’s really witty! I stumbled this proof while browsing the American Mathematical Monthly periodicals, to be specific on page 524 in the June-July 2008 issue. I reproduce the proof here.

Theorem  Let ${N}$ be a positive integer. If ${N}$ is a not perfect square, then ${\sqrt{N}}$ is irrational.

Proof: Before embarking on the proof, recall that the standard proof uses the method of contradiction. Here we shall prove the contrapositive statement to prove the theorem. That is, we shall prove that if ${\sqrt{N}}$ is rational then ${N}$ is a perfect square.

Thus assume that ${\sqrt{N}}$ is rational. This means we can write

$\displaystyle \sqrt{N}=\frac{p}{q},$

where ${p,q}$ are integers, ${q\neq 0}$, and that it is in the lowest form. This implies ${p^2=Nq^2}$, which we can rearrange so as to write

$\displaystyle \frac{p}{q}=\frac{Nq}{p}.$

Since we assumed that ${p/q}$ is in lowest form, we see that both ${Nq}$ and ${p}$ must be a integral multiple of ${p}$ and ${q}$ respectively. Taking that constant multiple to be ${c}$, we may write ${Nq=cp}$ and ${p=qc}$. The latter equation says

$\displaystyle \frac{p}{q}=c,$

but since ${p/q}$ is nothing but ${\sqrt{N}}$, this implies that ${\sqrt{N}=c}$, which simply means that ${N}$ is a perfect square. This proves the contrapositive statement, and therefore the theorem follows. $\Box$

Isn’t it beautiful!