\(\text{Beatty}\) 定理:
若 \(x,y \in \mathbb{R^+},x,y\notin\mathbb{Q}\),且 \(\frac{1}{x}+\frac{1}{y} = 1\),则集合 \(X = \{\lfloor ix\rfloor\ | i \in \mathbb{N_+} \}\) 和集合 \(Y = \{\lfloor iy\rfloor\ | i \in \mathbb{N_+} \}\) 满足:
- \(X\cap Y = \emptyset\)
- \(X\cup Y = \mathbb{N_+}\)
证明:
-
\(X\cap Y = \emptyset\):
反证法。
假设 \(k \in X \cap Y\),则 \(\exists i,j\in\mathbb{N_+}\) 满足:
\(k \le ix,jy < k + 1\)
\(\because k,i,j\in\mathbb{N_+},x,y\notin\mathbb{Q}\)
\(\therefore ix, jy \notin \mathbb{Q}\)
\(\therefore k\not = ix,k \not = jy\)
\(\therefore k < ix, jy < k + 1\)
则有
\(\frac{k}{x} < i < \frac{k + 1}{x}\)
\(\frac{k}{y} < j < \frac{k + 1}{y}\)
两式相加可得:\(k < i + j < k + 1\) 与 \(i,j,k\in\mathbb{N_+}\) 矛盾,则 \(X \cap Y = \emptyset\)
证毕。
-
\(X \cup Y = \mathbb{N_+}\):
由 \(\frac{1}{x}+\frac{1}{y} = 1\) 可得 \(x = \frac{y}{y-1}, y = \frac{x}{x-1}\)
又 \(x, y > 0\) 则 \(x,y > 1\)
则 \(\forall i \in \mathbb{N_+},ix,iy>1 \Rightarrow \lfloor ix\rfloor,\lfloor iy\rfloor\ge 1\)
则 \(X\cup Y \subseteq \mathbb{N_+}\)
反证法,设 \(k \in \mathbb{N_+}\) 且 \(k\notin X, k\notin Y\)
\(\because x,y > 1 \therefore \forall i\in\mathbb{Z}, \lfloor (i-1)x\rfloor < \lfloor ix\rfloor,\lfloor (i-1)y\rfloor < \lfloor iy\rfloor\)
\(k \notin X\),则 \(\exists i\in{\mathbb{N_+}}\) 满足 \(\lfloor(i-1)x\rfloor < k < \lfloor ix\rfloor\)
\(\Rightarrow (i-1)x < k < k + 1 < ix\)
\(\Rightarrow i-1<\frac{k}{x}<\frac{k+1}{x}<i\)
同理可得 \(\exists j\in\mathbb{N_+}, j-1<\frac{k}{y}<\frac{k+1}{y}<j\)
两式相加得 \(i+j-2<k<k+1<i+j\),与 \(i,j,k\in\mathbb{N_+}\) 矛盾,则 \(\mathbb{N_+}\subseteq X\cup Y\)
综上有 \(X \cup Y = \mathbb{N_+}\)
证毕。