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Difference between revisions of "Bertrand's Postulate"

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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>\binom{2n}{n}</math>. Assuming that the reader is familiar with that proof, the Bertrand postulate can be proved as follows.
 
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>\binom{2n}{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-
+
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>\binom{2n}{n}</math> is <math>\left\lfloor\frac{2n}{p}\right\rfloor-
 
2\left\lfloor\frac{n}{p}\right\rfloor=2-2=0</math>. Thus,
 
2\left\lfloor\frac{n}{p}\right\rfloor=2-2=0</math>. Thus,
  

Revision as of 20:43, 11 January 2010

Formulation

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

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 $\binom{2n}{n}$. 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 $\binom{2n}{n}$ is $\left\lfloor\frac{2n}{p}\right\rfloor- 2\left\lfloor\frac{n}{p}\right\rfloor=2-2=0$. Thus,

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

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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