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.