Bertrand-Chebyshev Theorem

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Theorem

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


Proof 1

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 $n$.

$\blacksquare$


Proof 2

We will first prove the theorem for the case $n \le 426$.

Consider the following sequence of prime numbers:

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

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

Hence every interval $\set {x: n < x \le 2 n}$, with $n \le 426$, contains one of these prime numbers.


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 $p^k \le 2 n$.

$\Box$


Lemma $4$

For all $n \ge 1$:

$\dbinom {2 n} n \ge \dfrac {2^{2 n}} {2 n}$

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

$\Box$


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

For $n \ge 5$:

$ \sqrt {2 n} < \dfrac {2 n} 3$

Therefore, by Lemma $4$:

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

for $n \ge 3$.

Let $\map \pi n$ denote the prime-counting function.

Then:

$\forall n \ge 1: \map \pi n \le n - 1$

as $1$ is neither prime nor composite.


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

Then we have:

\(\ds \frac {2^{2 n} } {2 n}\) \(\le\) \(\ds \paren {\prod_{p \mathop \le \sqrt {2 n} } 2 n} \paren {\prod_{\sqrt {2 n} \mathop < p \mathop \le \frac {2 n} 3} p}\)
\(\ds \) \(\le\) \(\ds \paren {\prod_{p \mathop \le \sqrt {2 n} } 2 n} \paren {\prod_{p \mathop \le \frac {2 n} 3} p}\)
\(\ds \) \(\le\) \(\ds \paren {2 n}^{\sqrt {2 n} - 1} 2^{4 n / 3}\) Lemma $2$

This is a contradiction for sufficiently large $n$.

Indeed, we have:

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

So:

$2^{2 n} \le \paren {2 n}^{3 \sqrt {2 n} }$

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

Then:

$2^{r^2} \le r^{6 r}$

or equivalently:

$2^r \le r^6$

This fails when $r = 30$.

It fails thereafter, since $2^r$ increases faster than $r^6$.

So our proof works if:

$n \ge 427$

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.


Sources

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