回復 6# anyway13 的帖子
分別去假設\(x\)的範圍
if \(\displaystyle n \leq x <n+\frac{1}{4}, n\in \mathbb{N}\)，\(\displaystyle n=\frac{102}{7}\)不合
以下同理去看
\(\displaystyle n+\frac{1}{4}\leq x <n+\frac{1}{2}, n\in \mathbb{N}\)
\(\displaystyle n+\frac{1}{2}\leq x <n+\frac{3}{4}, n\in \mathbb{N}\)
\(\displaystyle n+\frac{3}{4}\leq x <n+1, n\in \mathbb{N}\)
其中只有\(\displaystyle n+\frac{3}{4}\leq x <n+1, n\in \mathbb{N}\)，求出\(n=14 \in \mathbb{N}\)
所以\(\displaystyle \frac{59}{4}\leq x <15\)
