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翻訳と辞書　辞書検索 [ 開発暫定版 ] スポンサード リンク Proof of Bertrand's postulate ： ウィキペディア英語版 Proof of Bertrand's postulate In mathematics, Bertrand's postulate (actually a theorem) states that for each $n\backslash ge\; 1$ there is a prime $p$ such that The main steps of the proof are as follows. First, one shows that every prime power factor $p^r$ that enters into the prime decomposition of the central binomial coefficient $\backslash tbinom:=\backslash frac$ is at most $2n$. In particular, every prime larger than $\backslash sqrt$ can enter at most once into this decomposition; that is, its exponent $r$ is at most one. The next step is to prove that $\backslash tbinom$ has no prime factors at all in the gap interval $\backslash left(\backslash tfrac,\; n\backslash right)$. As a consequence of these two bounds, the contribution to the size of $\backslash tbinom$ coming from all the prime factors that are at most $n$ grows asymptotically as $O(\backslash theta^n)$ for some . Since the asymptotic growth of the central binomial coefficient is at least $4^n/2n$, one concludes that for $n$ large enough the binomial coefficient must have another prime factor, which can only lie between $n$ and $2n$. Indeed, making these estimates quantitative, one obtains that this argument is valid for all $n>468$. The remaining smaller values of $n$ are easily settled by direct inspection, completing the proof of Bertrand's postulate. ==Lemmas and computation== 抄文引用元・出典: フリー百科事典『 ウィキペディア（Wikipedia）』 ■ウィキペディアで「Proof of Bertrand's postulate」の詳細全文を読む スポンサード リンク 翻訳と辞書 : 翻訳のためのインターネットリソース Copyright(C) kotoba.ne.jp 1997-2016. All Rights Reserved.