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

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 $m \in \N$:

$\displaystyle \prod_{p \mathop \le m} p \le 2^{2 m}$

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

$\Box$


Lemma 3

If $p$ is a prime number and $p^k \mathrel \backslash \dbinom {2 n} n$, then $p^k \le 2 n$.

$\Box$


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:

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

for $n \ge 3$.

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

Then we have:

\(\displaystyle \frac {2^{2 n} } {2 n + 1}\) \(\le\) \(\displaystyle \prod_{p \mathop \le \sqrt {2 n} } 2 n \prod_{\sqrt {2 n} \mathop < p \mathop \le \frac 2 3 n} p\)
\(\displaystyle \) \(\le\) \(\displaystyle \paren {2 n}^{\sqrt {2 n} } \prod_{p \mathop \le \frac 2 3 n} p\)
\(\displaystyle \) \(\le\) \(\displaystyle \paren {2 n}^{\sqrt {2 n} } 2^{\frac 4 3 n}\) by Lemma 2

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 $n$.

$\blacksquare$


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