This proof shows that the square root of an integer, such as $\sqrt{m}$, must be either an integer (i.e. $m$ is exactly a "perfect" square) or irrational. Another way of saying it is that almost all roots of integers are irrational. Yet another way of saying that is that $\sqrt{m}$ either has

*no decimals at all*, or $\sqrt{m}$ has an

*infinite, non-repeating*decimal expansion!

Consider that Pythagoras and Euclid had proofs that $\sqrt{2}$ and $\sqrt{3}$ are irrational, but these proofs were one-offs. They rely on contradictory inequalities where one side is even and the other is odd, or that irreducible fractions must be reducible. Euclid had a proof that square roots of primes, $\sqrt{p}$, are irrational. However, these are just "special" integers. None of these proofs addressed the vast majority of integer roots.

In a sense, it took two thousand years for the brilliant mathematician, Theodore Estermann, to come along and illuminate the entire landscape. This two-line proof is valid for

*every integer!*Think about that. I learned from an obituary published in a math journal, that Estermann's work formed theoretical bridges for some of the greatest number theorists in history. While few modern theorems are named after him, still there are tons of his students and contemporaries who created "named" theorems, based on his insights.

Since my blog is configured to display equations, and the proof is so simple, I've copied it here. This simple generalization from $\sqrt{2}$ to $\sqrt{m}$ was shown by Harley Flanders:

THEOREM:

*Suppose $m$ is not a perfect square. Then $\sqrt{m}$ is irrational.*

Proof. Let $n$ be the integer with $n < \sqrt{m} < n+1$. It suffices to prove that $\alpha = \sqrt{m}-n$ is irrational. Suppose not. As $0 < \alpha < 1$, we have $\alpha=\frac{p}{q}$ where $0<p<q$. Assume that $q$ is

*as small as possible*(Estermann's key idea). Then we have

\[ \frac{q}{p} = \frac{1}{\sqrt{m}-n} = \frac{\sqrt{m}+n}{m-n^2} = \frac{\alpha+2n}{m-n^2} . \]

We solve for $\alpha$:

\[ \alpha = \frac{(m-n^2)q}{p}-2n = \frac{r}{p} \]

where $r = (m-n^2)q-2np$. Thus $\alpha$ is a fraction with even smaller denominator, a contradiction.