"小於" 的定義可以當作 "大於" 的衍生 : 若且唯若 $\beta > \alpha$ 時, 有 $\alpha < \beta$

$\alpha = \beta, \ \ \ \alpha < \beta, \ \ \ \alpha > \beta$

$證明$ :

$\alpha > r \geq \beta$

$\alpha > r \geq \beta, \ \ \ r = \beta$

$引理證畢$

${s}' \geq \alpha \geq s,\ \ \ {s}' \geq \beta \geq s$,

${s}' - s < \varepsilon$,

$證明$ :

$\alpha > {r}' > r > \beta$

${s}' \geq \alpha \geq s, \ \ \ {s}' \geq \beta \geq s$

${s}' \geq \alpha > {r}' > r > \beta \geq s$

${s}' \geq {r}' > r > s$, 亦即

${s}' - s > {r}' - r > 0$

$\varepsilon = {r}' - r$, 於是 ${s}' - s > \varepsilon$. 這與結論相矛盾, 因此假設不成立

$定理證畢$

$m < \alpha < M$

$m$ 逐次加 $1$, 我們一定可以得到這樣一對相鄰的整數 $C$$C + 1$, 使得

$C < \alpha < C + 1$

$C.1, C.2, ..., C.9$

$C.c_{1} < \alpha < C.c_{1} + \frac{1}{10}$

$C.c_{1}c_{2}...c_{n} < \alpha < C.c_{1}c_{2}...c_{n} + \frac{1}{10^{n}} \ \ \ \ \ \ \ \ \ \ \ \ \ (I)$

$\alpha$ 為非負無理數, 則稱有理數

$x_{n}^{-} = C.c_{1}c_{2}...c_{n}$

$\alpha$$n$ 位不足近似, 稱有理數

$x_{n}^{+} = C.c_{1}c_{2}...c_{n} + \frac{1}{10^{n}}$

$\alpha$$n$ 位過剩近似. 其中, $x_{n}^{-} < \alpha < x_{n}^{+}$

$\alpha$ 為非正無理數, 則稱有理數

$x_{n}^{-} = C.c_{1}c_{2}...c_{n} - \frac{1}{10^{n}}$

$\alpha$$n$ 位不足近似, 稱有理數

$x_{n}^{+} = C.c_{1}c_{2}...c_{n}$

$\alpha$$n$ 位過剩近似. 其中, $x_{n}^{-} < \alpha < x_{n}^{+}$

$x_{0}^{-} < x_{1}^{-} < x_{2}^{-} < ... < x_{n}^{-} < ... < \alpha < ... < x_{n}^{+} < ... < x_{2}^{+} < x_{1}^{+} < x_{0}^{+}$

$C.c_{1}c_{2}...c_{n} \leq \alpha \leq C.c_{1}c_{2}...c_{n} + \frac{1}{10^{n}} \ \ \ \ \ \ \ \ \ \ \ \ \ (II)$

$x_{0}^{-} \leq x_{1}^{-} \leq x_{2}^{-} \leq ... \leq x_{n}^{-} \leq ... \leq \alpha \leq ... \leq x_{n}^{+} \leq ... \leq x_{2}^{+} \leq x_{1}^{+} \leq x_{0}^{+}$

$C.c_{1}c_{2}...c_{n - 1}k99...9... \leq C.c_{1}c_{2}...c_{n} \leq C.c_{1}c_{2}...c_{n}00...0...$

1. 零循環表示法 : $C.c_{1}c_{2}...c_{n}00...0...$
2. $9$ 循環表示法 : $C.c_{1}c_{2}...c_{n - 1}k99...9...$, 其中 $k = c_{n} - 1$

$\alpha = C.c_{1}c_{2}...c_{n} = C.c_{1}c_{2}...c_{n}00...0... = C.c_{1}c_{2}...c_{n - 1}k99...9...$

$\alpha < 0$, 還有另外一種表示法. 設數 $\alpha = -C.c_{1}c_{2}...c_{n}$, 此處 $C > 0$, 於是有

$\alpha = -C.c_{1}c_{2}...c_{n} = \overline{B}.b_{1}b_{2}...b_{n} = \overline{B}.b_{1}b_{2}...b_{n} = \overline{B}.b_{1}b_{2}...b_{n - 1}k99...9...$

$N = N.00...0... = M.99...9...$

$-N = -N.00...0... = -M.99...9.. = \overline{K}.00...0... = \overline{N}.99...9...$

$C.c_{1}c_{2}...c_{n}$$C.c_{1}c_{2}...c_{n} + \frac{1}{10^{n}}$

$\frac{1}{10^{n}} \geq \varepsilon \ \ \ \ \ \Rightarrow \ \ \ \ \ \frac{1}{\varepsilon} \geq 10^{n}$

$\frac{1}{10^{n}} < \varepsilon$

$C_{n} = C.c_{1}c_{2}...c_{n}$${C_{n}}' = C.c_{1}c_{2}...c_{n} + \frac{1}{10^{n}}$

$C_{n} \leq \alpha \leq {C_{n}}'$

$a < \alpha < {a}'$