TITLE: Fractional Binomial Coefficients
QUESTION [9 upvotes]: I recently examined the binomial coefficient $\binom{\frac{1}{2}}{k}$ and found that the denominator was always a power of two. The same is true of $\binom{\frac{1}{3}}{k}$, where the denominator is always a power of three. While the first proof was simple, the case of $\frac{1}{3}$ was messy and involved counting the powers of 3 in the numerator and denominator. The case of $\frac{1}{p}$ for prime $p$ can be done in the same bashy way. Is it possible there is a more elegant proof?

REPLY [5 votes]: Here’s a purely conceptual proof, more advanced, that has the advantage of involving no computation and no induction:

Let $N$ be a positive number, and let $P_N(t)$ be the polynomial function that tells you the coefficient of $x^N$ in $(1+x)^t$. You know that $P_N$ has rational coefficients, and takes integer values when $t$ is evaluated to a positive integer.
Now consider any prime number $q$. The polynomial $P_N$ has its coefficients in the field ${\mathbb{Q}}_q$ of $q$-adic numbers, is continuous, and takes values in the $q$-adic integers ${\mathbb{Z}}_q$ whenever $t$ is evaluated to a positive integer.
But any rational number without $q$ in its denominator is a $q$-adic integer, so $q$-adically approximable by ordinary integers, and in fact by positive integers. Thus if $\alpha$ is a rational with no $q$ in its denomiator, we get $P_N(\alpha)\in{\mathbb{Z}}_q$, by continuity of $P_N$.
In particular, the coefficients of $(1+x)^{1/p}$ are rational numbers that are in ${\mathbb{Z}}_q$ for each $q\ne p$: the only denominators are powers of $p$.

A more advanced argument allows you to say just how divisible by $p$ the denominators of the coefficients of $(1+x)^{1/p}$ will be, but that’s a story for another day.