# Bertrand-Chebyshev Theorem/Proof 1

## 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 $1$

For all $n \in \N$:

- $\dbinom {2 n} n \ge \dfrac {2^{2 n}} {2 n + 1}$

where $\dbinom {2 n} n$ denotes a binomial coefficient.

$\Box$

### Lemma $2$

For all $m \in \N$:

- $\ds \prod_{p \mathop \le m} p \le 2^{2 m}$

where the continued product is taken over all prime numbers $p \le m$.

$\Box$

### Lemma $3$

Let $p$ be a prime number.

Let $p^k \divides \dbinom {2 n} n$.

Then $p^k \le 2 n$.

$\Box$

From Lemma $3$:

- 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 $\dfrac {2 n} 3 < p \le n$, because such a prime number divides $n!$ exactly once and $\paren {2 n}!$ exactly twice.

Therefore, by Lemma $1$:

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

for $n \ge 3$.

Aiming for a contradiction, suppose there is no prime number $p$ with $n < p \le 2 n$.

Then we have:

\(\ds \frac {2^{2 n} } {2 n + 1}\) | \(\le\) | \(\ds \prod_{p \mathop \le \sqrt {2 n} } 2 n \prod_{\sqrt {2 n} \mathop < p \mathop \le \frac {2 n} 3} p\) | ||||||||||||

\(\ds \) | \(\le\) | \(\ds \paren {2 n}^{\sqrt {2 n} } \prod_{p \mathop \le \frac {2 n} 3} p\) | ||||||||||||

\(\ds \) | \(\le\) | \(\ds \paren {2 n}^{\sqrt {2 n} } 2^{\frac {4 n} 3}\) | Lemma $2$ |

This is a contradiction for sufficiently large $n$.

Indeed, we have:

- $2^{2 n / 3} \le \paren {2 n + 1} \paren {2 n}^{\sqrt {2 n} }$

Now:

- $2 n + 1 \le \paren {2 n}^2 \le \paren {2 n}^{\sqrt {2 n} / 3}$

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 $n$.

$\blacksquare$

## Also known as

The **Bertrand-Chebyshev Theorem** is also known as **Bertrand's Postulate** or **Bertrand's Conjecture**.

Some sources give this as **Chebyshev's theorem (in number theory)** to distinguish it from a theorem in statistics.

## Source of Name

This entry was named for Joseph Louis François Bertrand and Pafnuty Lvovich Chebyshev.

## Historical Note

The Bertrand-Chebyshev Theorem was first postulated by Bertrand in $1845$. He verified it for $n < 3 \, 000 \, 000$.

It became known as **Bertrand's Postulate**.

The first proof was given by Chebyshev in $1850$ as a by-product of his work attempting to prove the Prime Number Theorem.

After this point, it no longer being a postulate, **Bertrand's Postulate** was referred to as the Bertrand-Chebyshev Theorem.

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

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