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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.
| | Your treadmill provides you absolutely desirable weight loss workouts. Currently many fitness treadmills prepared with a working computer provide fat loss or fat reduction programs.<br><br>Ditch the Soda. Colas are loaded with glucose that leads to excess fat gain. One cup of soda has about 180 calories while contributing no health value to your diet, so imagine how numerous clear calories are in a Big Gulp. Try switching a soda for water if you would like to lose weight fast.<br><br>The ideal time to do any cardio exercise is initially thing each morning before you eat anything. What happens is that because you haven't eaten anything during the night, as you're asleep, a body has to utilize calories that are stored in the body and burning calories is what we require if you need to lose weight.<br><br>I am not a fan of celery, however, this soup has a entire bunch. I then recalled reading once which takes more calories to burn than to eat. I looked it up online and according to the Urban Legends Reference Pages, it certain is true. Celery is a negative-calorie food. The website says which an eight-inch stalk of celery takes six calories to consume and six to burn. But yet, it could still fill we up, which makes it ideal to consume to fight cravings. Realizing this leads me to find how this soup diet can really work.<br><br>1 Practice controlled part by eating little balanced meals every time daily. Ideally [http://safedietplansforwomen.com/how-to-lose-weight-fast lose weight fast] go for just lean protein and those foods which are low in fat. You cannot turn to food for psychological comfort.<br><br>The biggest idea to remember is to not go to extremes but do include inside a diet lean protein plus also small amounts of nuts, seeds plus oils.<br><br>Fiber gives we the bulk and slows down a food digestion so we won't feel hungry easily and won't probably eat over the body requirements. Fiber assists to better a bowel movement for daily waste removal. That explains why you need to include adequate fiber inside your vegetarian diet to get rid of weight effectively and healthily. |
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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>,
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| 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>
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| 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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Your treadmill provides you absolutely desirable weight loss workouts. Currently many fitness treadmills prepared with a working computer provide fat loss or fat reduction programs.
Ditch the Soda. Colas are loaded with glucose that leads to excess fat gain. One cup of soda has about 180 calories while contributing no health value to your diet, so imagine how numerous clear calories are in a Big Gulp. Try switching a soda for water if you would like to lose weight fast.
The ideal time to do any cardio exercise is initially thing each morning before you eat anything. What happens is that because you haven't eaten anything during the night, as you're asleep, a body has to utilize calories that are stored in the body and burning calories is what we require if you need to lose weight.
I am not a fan of celery, however, this soup has a entire bunch. I then recalled reading once which takes more calories to burn than to eat. I looked it up online and according to the Urban Legends Reference Pages, it certain is true. Celery is a negative-calorie food. The website says which an eight-inch stalk of celery takes six calories to consume and six to burn. But yet, it could still fill we up, which makes it ideal to consume to fight cravings. Realizing this leads me to find how this soup diet can really work.
1 Practice controlled part by eating little balanced meals every time daily. Ideally lose weight fast go for just lean protein and those foods which are low in fat. You cannot turn to food for psychological comfort.
The biggest idea to remember is to not go to extremes but do include inside a diet lean protein plus also small amounts of nuts, seeds plus oils.
Fiber gives we the bulk and slows down a food digestion so we won't feel hungry easily and won't probably eat over the body requirements. Fiber assists to better a bowel movement for daily waste removal. That explains why you need to include adequate fiber inside your vegetarian diet to get rid of weight effectively and healthily.