Example 14 Try to convert $-\frac{1}{7}, \frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}$ into decimals.
From the above two lemmas, we know that these six proper fractions can all be converted into pure repeating decimals. Since
$$10^{2} \equiv 2,10^{3} \equiv 20 \equiv-1,10^{6} \equiv 1(\bmod 7)$$
we know that the order of 10 modulo 7 is 6, so the repeating decimals formed by these fractions all have a repeating cycle of exactly 6 digits. The calculations yield
$$\begin{array}{ll}
\frac{1}{7}=0 . \dot{1} 4285 \dot{ }= & \frac{2}{7}=0 . \dot{2} 8571 \dot{4} \\
\frac{3}{7}=0 . \dot{4} 2857 \dot{i}, & \frac{4}{7}=0 . \dot{5} 7142 \dot{8} \\
\frac{5}{7}=0 . \dot{7} 1428 \dot{5}, & \frac{6}{7}=0 . \dot{8} 5714 \dot{2}
\end{array}$$

This set of decimals exhibits an interesting phenomenon: the repeating cycle of each decimal is composed of the same six digits $1,4,2,8,5,7$. If we follow the order of the repeating cycle of $\frac{1}{7}$
Figure 1 shows
$$1 \rightarrow 4 \rightarrow 2 \rightarrow 8 \rightarrow 5 \rightarrow 7$$

and add an arrow from 7 to 1, we get a circle (see left Figure 1). Comparing the order of the digits in the repeating cycles of $\frac{2}{7}, \frac{3}{7}, \frac{4}{7}, \frac{5}{7}, \frac{6}{7}$, we find that they all have the same order as in Figure 1, just starting from different digits among the six, for example, $\frac{2}{7}$ starts from the digit 2, $\frac{3}{7}$ starts from the digit 4, etc.