证明：对无理数 ξ，集合 {nξ-[nξ]：n∈N+} 不含有理数且在 [0,1] 稠密

elim

题：试证对无理数\(a,\{na-\lfloor na\rfloor:n\in\mathbb{N}^+\}\)不含有理数且在[0,1]稠密.

elim

题：对无理数\(\small\,\xi,\,\{n\xi-\lfloor n\xi\rfloor: n\in\mathbb{N}^+\}\)不含有理数且在\(\small\,[0,1]\,\)稠密.
证：记\(\,\langle x\rangle=x-\lfloor x\rfloor,\;\Lambda=\{\langle n\xi\rangle: n\in\mathbb{N}^+\}.\)
\((1)\quad\langle x\xi\rangle=n\xi-\lfloor n\xi\rfloor\not\in\mathbb{Q},\)
\((2)\quad(m,n\in\mathbb{N}^+)\wedge(m\ne n)\implies\langle m\xi\rangle\ne\langle n\xi\rangle,\)
\((3)\quad\forall k\in\mathbb{N}^+\exists n\in\mathbb{N}^+\,(\langle nx\rangle < \frac{1}{k}).\)
\(\qquad\)现证\((3).\)取\(\,k\in\mathbb{N},k>1.\)据\((2),\)及鸽笼原理,存在
\(\qquad\,m,n+m\in\mathbb{N}\,\)使\(\,|\langle(n+m)\xi\rangle-\langle m\xi\rangle|< \frac{1}{k}.\;\)即有
\(\qquad-\frac{1}{k}<\langle n\xi\rangle+\lfloor n\xi\rfloor+\lfloor m\xi\rfloor-\lfloor (n+m)\xi\rfloor< \frac{1}{k}\)
\(\qquad\)易见\(\;\tau=\lfloor n\xi\rfloor+\lfloor m\xi\rfloor-\lfloor(n+m)\xi\rfloor\in\{-1,0\}.\)
\(\qquad\)若\(\,\tau=0,\)则\(\,0< \langle n\xi\rangle<\frac{1}{k}.\,\)否则\(\,\tau{\small=-1,\;}0< 1-\langle n\xi\rangle<\frac{1}{k}\)
\(\qquad\)令\(\,\beta=1-\langle n\xi\rangle,\,\)取\(\,p\in\mathbb{N}\,\)使\(\,1< p\beta< 1+\frac{1}{k},\,\)则有
\(\qquad\,p\langle n\xi\rangle=p(1-\beta)=p-p\beta\in(p-1,p-1+\frac{1}{k})\,\)即
\(\qquad\,0<(pn)\xi-p\lfloor n\xi\rfloor-p+1<\frac{1}{k}.\;\)可见\(\, 0<\langle(pn)\xi\rangle<\frac{1}{k}.\)
据\((1),\;\Lambda\cap\mathbb{Q}=\varnothing.\,\)据(3),对\(\,0\le\alpha<\beta\le 1,\,\)有\(\,k,m,n\in\mathbb{N}^+\)使
\(\langle n\xi\rangle<\frac{1}{k}<\beta-\alpha,\;0\le(m-1)\langle n\xi\rangle\le\alpha< m\langle n\xi\rangle<\beta.\;\)
可见\(\,\langle mn\xi\rangle\in(\alpha,\beta)\cap\Lambda.\quad\small\square\)

elim

注记
(1) 若\(,\,\langle n\xi\rangle = n\xi-\lfloor n\xi\rfloor\,\)是有理数, 则\(\,\xi=\large\frac{\langle n\xi\rangle+\lfloor n\xi\rfloor}{n}\,\)亦然.
(2) 当\(\,m,n\in\mathbb{Z},\,n\ne m\)时无理数\(\xi=\large\frac{\langle n\xi\rangle-\langle m\xi\rangle+\lfloor n\xi\rfloor-\lfloor m\xi\rfloor}{n-m}\)
\(\therefore\quad\langle m\xi\rangle\ne\langle n\xi\rangle.\quad\therefore\quad\varphi: n\mapsto \langle n\xi\rangle\,\)是单射.

luyuanhong

楼上 elim 的帖子很好！已收藏。

elim

luyuanhong

楼上 elim 的帖子很好！已收藏。