Bertrand-Chebyshev Theorem

Theorem
For all $n \in \N_{>0}$, there exists a prime number $p$ with $n < p \le 2 n$.

Proof
We will first prove the theorem for the case $n \le 2047$.

Consider the following sequence of prime numbers:


 * $2, 3, 5, 7, 13, 23, 43, 83, 163, 317, 631, 1259, 2503$

Each prime number is smaller than twice the previous one.

Hence every interval $\{ x: n < x \le 2 n \}$, with $n \le 2047$, contains one of these prime numbers.

Lemma 3
In particular, if $p > \sqrt{2 n}$, then $p$ appears at most once in $\dbinom {2 n} n$.

For $n \ge 3$, there is no prime factor $p$ with $\dfrac 2 3 n < p \le n$, for such a prime number divides $n!$ exactly once and $\left({2 n}\right)!$ exactly twice.

Therefore, by Lemma 1:


 * $\displaystyle \frac {2^{2 n} } {2 n + 1} \le \binom {2 n} n \le \prod_{p \mathop \le \sqrt{2 n}} 2 n \prod_{\sqrt{2 n} \mathop < p \mathop \le \frac 2 3 n} p \prod_{n \mathop < p \mathop \le 2 n} p$

for $n \ge 3$.

Now assume there is no prime number $p$ with $n < p \le 2 n$.

Then we have:

This is a contradiction if $n$ is large enough. Indeed, we have:


 * $2^{\frac 2 3 n} \le \left({2 n + 1}\right) \left({2 n}\right)^{\sqrt {2 n}}$

Now $2 n + 1 \le \left({2 n}\right)^2 \le \left({2 n}\right)^{\frac 1 3 \sqrt{2 n}}$ for $n \ge 18$.

So $2^{2 n} \le \left({2 n}\right)^{4 \sqrt{2 n}}$.

Put $r = \sqrt{2 n}$; then $2^{r^2} \le r^{8 r}$, or equivalently $2^r \le r^8$.

This fails when $r = 2^6 = 64$, and thereafter, since $2^r$ increases faster than $r^8$.

So our proof works if $n \ge 2^{11} = 2048$, and the examples show it is true for smaller $n$.

Also known as
The Bertrand-Chebyshev theorem is also known as Bertrand's postulate or Bertrand's conjecture.

Historical note
This result was first postulated by Bertrand in 1845. He verified it for $n < 3 \, 000 \, 000$.

The first proof was given by Chebyshev in 1850.

In 1919, Ramanujan gave a simpler proof based on the Gamma function.

In 1932, Erdős gave an even simpler proof based on basic properties of binomial coefficients. That proof is the one which is presented here.