# Difference between revisions of "Bertrand's Postulate"

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+ | ==Formulation== | ||

'''Bertrand's postulate''' states that for any [[positive integer]] <math>n</math>, there is a [[prime number|prime]] between <math>n</math> and <math>2n</math>. Despite its name, it is in fact a theorem. | '''Bertrand's postulate''' states that for any [[positive integer]] <math>n</math>, there is a [[prime number|prime]] between <math>n</math> and <math>2n</math>. 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|prime number theorem]] article but requires a closer look at the [[combinations|binomial coefficient]] <math>2n\choose n</math>. 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 <math>p</math> satisfying <math>\frac{2n}3<p\le n</math> appears in the prime factorization of <math>2n\choose n</math> is <math>\left\lfloor\frac{2n}{p}\right\rfloor- | ||

+ | 2\left\lfloor\frac{n}{p}\right\rfloor=2-2=0</math>. Thus, | ||

+ | |||

+ | <math>\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)\,. | ||

+ | </math> | ||

+ | |||

+ | The first product does not exceed <math>(2n)^{\sqrt{2n}}</math> and the second one does not exceed <math>4^{\frac {2n}3}</math>. Thus, | ||

+ | |||

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

+ | |||

+ | The right hand side is strictly greater than <math>1</math> for <math>n\ge 500</math>, so it remains to prove the Bertrand postulate for <math>n<500</math>. In order to do it, it suffices to present a sequence of primes starting with <math>2</math> in which each prime does not exceed twice the previous one and the last prime is above <math>500</math>. One such possible sequence is | ||

+ | <math>2,3,5,7,13,23,43,83,163,317,631</math>. | ||

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## Revision as of 13:39, 29 June 2006

## Formulation

**Bertrand's postulate** states that for any positive integer , there is a prime between and . 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 . 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 satisfying appears in the prime factorization of is . Thus,

The first product does not exceed and the second one does not exceed . Thus,

The right hand side is strictly greater than for , so it remains to prove the Bertrand postulate for . In order to do it, it suffices to present a sequence of primes starting with in which each prime does not exceed twice the previous one and the last prime is above . One such possible sequence is .

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