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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 Prime constant ： ウィキペディア英語版 Prime constant The prime constant is the real number $\backslash rho$ whose $n$th binary digit is 1 if $n$ is prime and 0 if ''n'' is composite or 1. In other words, $\backslash rho$ is simply the number whose binary expansion corresponds to the indicator function of the set of prime numbers. That is, :$\backslash rho\; =\; \backslash sum\_\; \backslash frac\; =\; \backslash sum\_^\backslash infty\; \backslash frac$ where $p$ indicates a prime and $\backslash chi\_\{\backslash mathbb\{P\}\}$ is the characteristic function of the primes. The beginning of the decimal expansion of ''ρ'' is: $\backslash rho\; =\; 0.414682509851111660248109622\backslash ldots$ The beginning of the binary expansion is: $\backslash rho\; =\; 0.011010100010100010100010000\backslash ldots\_2$ ==Irrationality== The number $\backslash rho$ is easily shown to be irrational. To see why, suppose it were rational. Denote the $k$th digit of the binary expansion of $\backslash rho$ by $r\_k$. Then, since $\backslash rho$ is assumed rational, there must exist $N$, $k$ positive integers such that $r\_n=r\_$ for all $n\; >\; N$ and all $i\; \backslash in\; \backslash mathbb$. Since there are an infinite number of primes, we may choose a prime $p\; >\; N$. By definition we see that $r\_p=1$. As noted, we have $r\_p=r\_$ for all $i\; \backslash in\; \backslash mathbb$. Now consider the case $i=p$. We have $r\_=r\_=r\_=0$, since $p(k+1)$ is composite because $k+1\; \backslash geq\; 2$. Since $r\_p\; \backslash neq\; r\_$ we see that 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「Prime constant」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.