Proof of Bertrand's postulate

In mathematics, Bertrand's postulate (now a theorem) states that, for each ${\displaystyle n\geq 2}$, there is a prime ${\displaystyle p}$ such that ${\displaystyle n<p<2n}$. First conjectured in 1845 by Joseph Bertrand,^[1] it was first proven by Chebyshev, and a shorter but also advanced proof was given by Ramanujan.^[2]

The following elementary proof was published by Paul Erdős in 1932, as one of his earliest mathematical publications.^[3] The basic idea is to show that the central binomial coefficients must have a prime factor within the interval ${\displaystyle (n,2n)}$ in order to be large enough. This is achieved through analysis of their factorizations.

The main steps of the proof are as follows. First, one shows that the contribution of every prime power factor ${\displaystyle p^{r}}$ in the prime decomposition of the central binomial coefficient ${\displaystyle \textstyle {\binom {2n}{n}}=(2n)!/(n!)^{2}}$ is at most ${\displaystyle 2n}$; then, one shows that every prime larger than ${\displaystyle {\sqrt {2n}}}$ appears at most once.

The next step is to prove that ${\displaystyle {\tbinom {2n}{n}}}$ has no prime factors in the interval ${\displaystyle ({\tfrac {2n}{3}},n)}$. As a consequence of these bounds, the contribution to the size of ${\displaystyle {\tbinom {2n}{n}}}$ coming from the prime factors that are at most ${\displaystyle n}$ grows asymptotically as ${\displaystyle \theta ^{\!\;n}}$ for some ${\displaystyle \theta <4}$. Since the asymptotic growth of the central binomial coefficient is at least ${\displaystyle 4^{n}\!/2n}$, the conclusion is that, by contradiction and for large enough ${\displaystyle n}$, the binomial coefficient must have another prime factor, which can only lie between ${\displaystyle n}$ and ${\displaystyle 2n}$.

The argument given is valid for all ${\displaystyle n\geq 427}$. The remaining values of ${\displaystyle n}$ are by direct inspection, which completes the proof.