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 of these prime number is smaller than twice the previous one.

Hence every interval $\set {x: n < x \le 2 n}$, 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 $\paren {2 n}!$ exactly twice.

Therefore, by Lemma 1:


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

for $n \ge 3$.

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:


 * $2^{\frac 2 3 n} \le \paren {2 n + 1} \paren {2 n}^{\sqrt {2 n} }$

Now:
 * $2 n + 1 \le \paren {2 n}^2 \le \paren {2 n}^{\frac 1 3 \sqrt {2 n} }$

for $n \ge 18$.

So:
 * $2^{2 n} \le \paren {2 n}^{4 \sqrt {2 n} }$

Put $r = \sqrt {2 n}$.

Then:
 * $2^{r^2} \le r^{8 r}$

or equivalently:
 * $2^r \le r^8$

This fails when $r = 2^6 = 64$.

It fails 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 $n$.