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| In [[mathematics]], [[Bertrand's postulate]] (actually a [[theorem]]) states that for each <math>n\ge 1</math> there is a [[prime number|prime]] <math>p</math> such that <math>n<p\le 2n</math>. It was first proven by [[Pafnuty Chebyshev]], and a short but advanced proof was given by [[Srinivasa Ramanujan]].<ref>{{Citation |first=S. |last=Ramanujan |title=A proof of Bertrand's postulate |journal=Journal of the Indian Mathematical Society |volume=11 |year=1919 |pages=181–182 |url=http://www.imsc.res.in/~rao/ramanujan/CamUnivCpapers/Cpaper24/page1.htm }}</ref> The gist of the following elementary proof is due to [[Paul Erdős]]. The basic idea of the proof is to show that a certain [[central binomial coefficient]] needs to have a [[prime factor]] within the desired interval in order to be large enough. This is made possible by a careful analysis of the prime factorization of central binomial coefficients.
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| __TOC__
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| The main steps of the proof are as follows. First, one shows that every [[prime power]] factor <math>p^r</math> that enters into the prime decomposition of
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| the central binomial coefficient <math>\tbinom{2n}{n}:=\frac{(2n)!}{(n!)^2}</math> is at most <math>2n</math>. In particular, every prime larger than <math>\sqrt{2n}</math> can enter at most once into this decomposition; that is, its exponent <math>r</math> is at most one. The next step is to prove that <math>\tbinom{2n}{n}</math> has no prime factors at all in the gap interval <math>\left(\tfrac{2n}{3}, n\right)</math>. As a consequence of these two bounds, the contribution to the size of <math>\tbinom{2n}{n}</math> coming from all the prime factors that are at most <math>n</math> [[asymptotic analysis|grows asymptotically]] as <math>O(\theta^n)</math> for some <math>\theta<4</math>. Since the
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| asymptotic growth of the central binomial coefficient is at least <math>4^n/2n</math>, one concludes that for <math>n</math> large enough the binomial coefficient must have another prime factor, which can only lie between <math>n</math> and <math>2n</math>.
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| Indeed, making these estimates quantitative, one obtains that this argument is valid for all <math>n>468</math>. The remaining smaller values of <math>n</math> are easily settled by direct inspection, completing the proof of the Bertrand's postulate.
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| ==Lemmas and computation==
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| ==={{anchor|Lemma 1}}Lemma 1: A lower bound on the central binomial coefficients===
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| '''Lemma:''' For any integer <math>n>0</math>, we have
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| :<math> \frac{4^n}{2n} \le \binom{2n}{n}.\ </math>
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| '''Proof:''' Applying the [[binomial theorem]],
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| :<math>4^n = (1 + 1)^{2n} = \sum_{k = 0}^{2n} \binom{2n}{k}=2+\sum_{k = 1}^{2n-1} \binom{2n}{k} \le 2n\binom{2n}{n},\ </math>
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| since <math>\tbinom{2n}{n}</math> is the largest term in the sum in the right-hand side, and the sum has <math>2n</math> terms (including the initial two outside the summation).
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| ==={{anchor|Lemma 2}}Lemma 2: An upper bound on prime powers dividing central binomial coefficients===
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| For a fixed prime <math>p</math>, define <math>R(p,n)</math> to be the largest natural number <math>r</math> such that <math>p^r</math> divides <math>\tbinom{2n}{n}</math>.
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| '''Lemma:''' For any prime <math>p</math>, <math>p^{R(p,n)}\le 2n</math>. | |
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| '''Proof:''' The exponent of <math>p</math> in <math>n!</math> is (see [[Factorial#Number theory]]):
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| :<math>\sum_{j = 1}^\infty \left\lfloor \frac{n}{p^j} \right\rfloor,\ </math>
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| so
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| :<math> R(p,n)
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| =\sum_{j = 1}^\infty \left\lfloor \frac{2n}{p^j} \right\rfloor - 2\sum_{j = 1}^\infty \left\lfloor \frac{n}{p^j} \right\rfloor
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| =\sum_{j = 1}^\infty \left(\left\lfloor \frac{2n}{p^j} \right\rfloor - 2\left\lfloor \frac{n}{p^j} \right\rfloor\right).
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| </math>
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| But each term of the last summation can either be zero (if <math>n/p^j \bmod 1< 1/2</math>) or 1 (if <math>n/p^j \bmod 1\ge 1/2</math>) and all terms with <math>j>\log_p(2n)</math> are zero. Therefore
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| :<math>R(p,n) \leq \log_p(2n),\ </math>
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| and
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| :<math>p^{R(p,n)} \leq p^{\log_p{2n}} = 2n.\ </math>
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| This completes the proof of the lemma.
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| ==={{anchor|Lemma 3}}Lemma 3: The exact power of a large prime in a central binomial coefficient===
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| '''Lemma:''' If <math> p </math> is odd and <math> \frac{2n}{3} < p \leq n </math>, then <math>R(p,n) = 0.\ </math>
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| '''Proof:''' The factors of <math>p</math> in the numerator come from the terms <math>p</math> and <math>2p</math>, | |
| and in the denominator from two factors of <math>p</math>. These cancel since <math>p</math> is odd.
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| ==={{anchor|Lemma 4}}Lemma 4: An upper bound on the primorial===
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| We estimate the [[primorial]] function,
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| :<math>x\# = \prod_{p \leq x} p,\ </math> | |
| where the product is taken over all ''prime'' numbers <math>p</math> less than or equal to the real number <math>x</math>.
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| '''Lemma:''' For all real numbers <math>x\ge 1</math>, <math>x\#<4^x</math>
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| '''Proof:'''
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| Considering <math>n=\lfloor x\rfloor </math> it is sufficient to prove the lemma for natural numbers <math>x=n</math>.
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| The proof is by [[mathematical induction]].
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| * <math>n = 1</math>: <math> n\# = 1 < 4 = 4^1.</math>
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| * <math>n = 2</math>: <math> n\# = 2 < 16 = 4^2.</math>
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| * <math>n > 2</math> is even: <math>n\# = (n-1)\# < 4^{n-1} < 4^n.</math>
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| * <math>n > 2</math> is odd. Let <math>n = 2m + 1</math>, then by [[binomial theorem]]:
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| ::<math>
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| \binom{2m + 1}{m} =
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| \frac{1}{2} \left[\binom{2m + 1}{m} + \binom{2m + 1}{m + 1}\right]
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| < \frac{1}{2}\sum_{k = 0}^{2m+1} \binom{2m + 1}{k}
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| = \frac{1}{2}(1 + 1)^{2m + 1}
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| = 4^m.
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| </math>
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| :Each prime ''p'' with <math>m+1<p\le 2m + 1</math> divides <math>\textstyle\binom{2m + 1}{m}</math>, giving us:
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| ::<math>\frac{(2m + 1)\#}{(m + 1)\#} = \prod_{p > m + 1}^{p \leq 2m + 1} p \leq \binom{2m+1}{m} < 4^m.</math>
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| :By induction for <math>m+1<n</math>:
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| ::<math>n\# = (2m + 1)\# < 4^m \cdot (m + 1)\# < 4^m \cdot 4^{m + 1} = 4^{2m + 1} = 4^n.</math>
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| Thus the lemma is proven.
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| ==Proof of Bertrand's Postulate==
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| Assume there is a [[counterexample]]: an integer ''n'' ≥ 2 such that there is no prime ''p'' with ''n'' < ''p'' < 2''n''.
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| If 2 ≤ ''n'' < 468, then ''p'' can be chosen from among the prime numbers 3, 5, 7, 13, 23, 43, 83, 163, 317, 631 (each being less than twice its predecessor) such that ''n'' < ''p'' < 2''n''. Therefore ''n'' ≥ 468.
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| There are no prime factors ''p'' of <math>\textstyle\binom{2n}{n}</math> such that:
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| * 2''n'' < ''p'', because every factor must divide (2''n'')!;
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| * ''p'' = 2''n'', because 2''n'' is not prime;
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| * ''n'' < ''p'' < 2''n'', because we assumed there is no such prime number;
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| * 2''n'' / 3 < ''p'' ≤ ''n'': by [[#Lemma 3|Lemma 3]].
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| Therefore, every prime factor ''p'' satisfies ''p'' ≤ 2''n''/3.
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| When <math> p > \sqrt{2n},</math> the number <math>\textstyle {2n \choose n} </math> has at most one factor of ''p''. By [[#Lemma 2|Lemma 2]], for any prime ''p'' we have ''p''<sup>''R''(''p'',''n'')</sup> ≤ 2''n'', so the product of the ''p''<sup>''R''(''p'',''n'')</sup> over the primes less than or equal to <math>\sqrt{2n}</math> is at most <math>(2n)^{\sqrt{2n}}</math>. Then, starting with [[#Lemma 1|Lemma 1]] and decomposing the right-hand side into its prime factorization, and finally using [[#Lemma 4|Lemma 4]], these bounds give:
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| :<math>\frac{4^n}{2n }
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| \le \binom{2n}{n}
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| = \left(\prod_{p \le \sqrt{2n}} p^{R(p,n)}\right) \left(\prod_{\sqrt{2n} < p \le \frac{2n}{3}} p^{R(p,n)}\right)
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| < (2n)^{\sqrt{2n}} \prod_{1 < p \leq \frac{2n}{3} } p
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| = (2n)^{\sqrt{2n}} \Big( \frac{2n}{3}\Big)\# \le (2n)^{\sqrt{2n}} 4^{2n/3}.\ </math>
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| Taking logarithms yields to
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| :<math>{\frac{\log 4}{3}}n \le (\sqrt{2n}+1)\log 2n\; .</math>
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| By concavity of the right-hand side as a function of ''n'', the last inequality is necessarily verified on an interval. Since it holds true for ''n=467'' and it does not for ''n=468'', we obtain
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| :<math>n < 468.\ </math>
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| But these cases have already been settled, and we conclude that no counterexample to the postulate is possible.
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| ==References==
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| {{Reflist}}
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| * [[Martin Aigner|Aigner, Martin, G.]], [[Günter M. Ziegler]], Karl H. Hofmann, ''[[Proofs from THE BOOK]]'', Fourth edition, Springer, 2009. ISBN 978-3-642-00855-9.
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| {{DEFAULTSORT:Bertrands postulate, proof of}}
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| [[Category:Prime numbers]]
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| [[Category:Factorial and binomial topics]]
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| [[Category:Article proofs]]
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| [[fr:Postulat de Bertrand]]
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