Jul 042012

In 1949, Leo Moser gave a proof of  Bertrand’s postulate in the paper “A theorem of the distribution of primes”(American Mathematical Monthly,624-624,1949,vol56). Maybe this proof  is the simplest one among all proofs of  Bertrand’s postulate.

  1. If \(n<p\leqslant 2n\), then \(p\) occurs exactly once in \({2n\choose n}\);
  2. If \(n\geqslant 3, \frac{2n}3<p<n\), then \(p\) doesn’t occurs in \({2n\choose n}\);
  3. If \(p^2>2n\), then \(p\) occurs at most once in \({2n\choose n}\);
  4. If \(2^\alpha\leqslant 2n<2^{\alpha+1}\), then \(p\) occurs at most \(\alpha\) times in \({2n\choose n}\).

If  \(2^\alpha<2n\leqslant  2^{\alpha+1}\), suppose there is no prime \(p\) with



\begin{equation}{2n\choose n}\leqslant{2a_1\choose a_1}{2a_2\choose a_2}\dotsm{2\choose 1}\bigg({2a_k\choose a_k}{2a_{k+1}\choose a_{k+1}}\dotsm{2\choose 1}\bigg)^\alpha,\end{equation}

where \(a_1=\big[\frac{n+1}3\big], k=\big[\frac{\alpha+1}2\big]\), and \(a_i=\big[\frac{a_{i-1}+1}3\big]\), for \(i=2,\,3,\,\dotsc\,.\)

By \(1,\,2,\,3\) and \((1)\), every prime which appears on the left-hand side of \((2)\) appears also on the right; and those primes which appear with multiplicity greater than \(1\) on the left appear on the right with multiplicity at least \(2\alpha+1\), which by \(4\) is at least equal to the multiplicity with which they appear on the left.

On the other hand, It is obvious that

\begin{equation}2n>2a_1+2a_2+\dotsb+ 2+\alpha(2a_k+2a_{k+1}+\dotsb+2)\end{equation}

for \(n>2^{11}.\) Hence the inequality in \((2)\) should be reversed. So for \(n>2^{11}\), we have a contradiction which proves Bertrand’s postulate for these values of \(n\).

Jun 262012

Sylvester’s theorem  the binomial coefficient

\begin{equation}{n\choose k}=\frac{n(n-1)(n-2)\cdots(n-k+1)}{k!}\end{equation}

has a prime divisor \(p\) greater than \(k\). In other words, if \(n\geq 2k\), then the product of  \(k\) consecutive integers greater than \(k\) is divisible by a prime greater than \(k\). Note that for \(n=2k\) we obtain precisely Bertrand’s postulate.


  1. if \(n>6\) is an integer, then there always exists at least two prime numbers \(p\), one  is the form \(4k+1\) and the other \(4k+3\), such that \(n<p<2n\).
  2. if \(k\) is a positive  integer, then there always exists a positive  integer \(N\) such that  for every  integer \(n>N\),  there exist at least \(k\) prime numbers \(p\) such that \(n<p<2n\).
  • 对任意正整数 \(n>6\) ,  至少存在一个 \(4k+1\) 型和一个 \(4k+3\)  型素数 \(p\)  使得  \(n<p<2n\).
  • 对任意正整数 \(k\),  存在正整数 \(N\),  使得只要正整数 \(n>N\),  就存在 \(k\) 个素数 \(p\) 使得 \(n<p<2n\).

注意, 定理 \(2\) 已经隐含在 Ramanujan 给出的 Bertrand’s postulate 的证明中 Ramanujan’s proof of Bertrand’s postulate.