Difference between revisions of "Bertrand's Postulate"

Revision as of 14:42, 13 June 2007

Formulation

Bertrand's postulate states that for any positive integer $n$ , there is a prime between $n$ and $2n$ . Despite its name, it is, in fact, a theorem.

Proof

It is similar to the proof of Chebyshev's estimates in the prime number theorem article but requires a closer look at the binomial coefficient $2n\choose n$ . Assuming that the reader is familiar with that proof, the Bertrand postulate can be proved as follows.

Note that the power with which a prime $p$ satisfying $\frac{2n}3<p\le n$ appears in the prime factorization of $2n\choose n$ is $\left\lfloor\frac{2n}{p}\right\rfloor- 2\left\lfloor\frac{n}{p}\right\rfloor=2-2=0$ . Thus,

$\frac{2^{2n}}{(2n+1)}\le{2n\choose n}\le \left(\prod_{p\le\sqrt{2n}}p^{\lfloor\log_p (2n)\rfloor}\right)\cdot \left(\prod_{\sqrt{2n}<p\le\frac{2n}3}p\right)\cdot \left(\prod_{n<p\le {2n}}p\right)\,.$ .

The first product does not exceed $(2n)^{\sqrt{2n}}$ and the second one does not exceed $4^{\frac {2n}3}$ . Thus,

$\left(\prod_{n<p\le{2n}}p\right)\ge \frac{4^{\frac n3}}{(2n+1)(2n)^{\sqrt {2n}}}$

The right hand side is strictly greater than $1$ for $n\ge 500$ , so it remains to prove the Bertrand postulate for $n<500$ . In order to do it, it suffices to present a sequence of primes starting with $2$ in which each prime does not exceed twice the previous one, and the last prime is above $500$ . One such possible sequence is $2,3,5,7,13,23,43,83,163,317,631$ .

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Difference between revisions of "Bertrand's Postulate"

Revision as of 14:42, 13 June 2007

Formulation

Proof

Revision as of 13:47, 6 July 2006 (view source) Inscrutableroot (talk \| contribs) m (proofreading) ← Older edit	Revision as of 14:42, 13 June 2007 (view source) Boy Soprano II (talk \| contribs) m (Bertrand's postulate moved to Bertrand's Postulate: Convention of capitalizing named theorems) Newer edit →
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