Đặt \(x=\left[x\right]+\left\{x\right\}\)
\(\Rightarrow\left[x\right]+\left[x+\frac{1}{2}\right]=\left[x\right]+\left[\left[x\right]+\left\{x\right\}+\frac{1}{2}\right]=2\left[x\right]+\left[\left\{x\right\}+\frac{1}{2}\right]\)
\(\left[2x\right]=\left[2\left[x\right]+2\left\{x\right\}\right]=2\left[x\right]+\left[2\left\{x\right\}\right]\)
Ta cần chứng minh \(\left[\left\{x\right\}+\frac{1}{2}\right]=\left[2\left\{x\right\}\right]\)
Thật vậy:
- Với \(0\le\left\{x\right\}< \frac{1}{2}\Rightarrow\left\{{}\begin{matrix}\left\{x\right\}+\frac{1}{2}< 1\\2\left\{x\right\}< 1\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[\left\{x\right\}+\frac{1}{2}\right]=0\\\left[2\left\{x\right\}\right]=0\end{matrix}\right.\)
- Với \(\frac{1}{2}\le\left\{x\right\}< 1\Rightarrow\left\{{}\begin{matrix}1\le\left\{x\right\}+\frac{1}{2}< 2\\1\le2\left\{x\right\}< 2\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}\left[\left\{x\right\}+\frac{1}{2}\right]=1\\\left[2\left\{x\right\}\right]=1\end{matrix}\right.\)
