# 刘维尔数

$0 < \left| x - \frac{p}{q} \right| < \frac{1}{q^{n}}$

## 基本性质

$\left| x - \frac{p}{q} \right| = \left| \frac{c}{d} - \frac{p}{q} \right| = \left| \frac{cq-dp}{dq} \right| \ge \frac{1}{dq} > \frac{1}{2^{n-1}q} \ge \frac{1}{q^n}$

## 刘维尔常数

$c = \sum_{j=1}^\infty 10^{-j!} = 0.110001000000000000000001000\ldots$

$p_n = \sum_{j=1}^n 10^{n! - j!}, \quad q_n = 10^{n!}$

$\left|c - \frac{p_n}{q_n}\right| = \sum_{j=n+1}^\infty 10^{-j!} = 10^{-(n+1)!} + 10^{-(n+2)!} + {} \cdots < 10\cdot10^{-(n+1)!} \le \Big(10^{-n!}\Big)^n = \frac{1}{{q_n}^n}.$

## 超越性

### 证明

$\left\vert \alpha - \frac{p}{q} \right\vert > \frac{A}{q^n}$

$0 < A < \min \left(1, \frac{1}{M}, \left\vert \alpha - \alpha_1 \right\vert, \left\vert \alpha - \alpha_2 \right\vert, \ldots , \left\vert \alpha-\alpha_m \right\vert \right)$

$\left\vert \alpha - \frac{p}{q} \right\vert \le \frac{A}{q^n} \le A< \min\left(1, \frac{1}{M}, \left\vert \alpha - \alpha_1 \right\vert, \left\vert \alpha - \alpha_2 \right\vert, \ldots , \left\vert \alpha-\alpha_m \right\vert \right)$

$f(\alpha)-f(p/q) = (\alpha - p/q) \cdot f'(x_0)$

$\left\vert (\alpha -p/q)\right\vert = \left\vert f(\alpha)- f(p/q)\right\vert / \left\vert f'(x_0) \right\vert = \left\vert f(p/q) / f'(x_0) \right\vert \,$

$f$是多项式，所以

$\left\vert f(p/q) \right\vert = \left\vert \sum_{i=0}^n c_i p^i q^{-i} \right\vert = \frac{\left\vert \sum_{i=0}^n c_i p^i q^{n-i} \right\vert}{q^n} \ge \frac {1}{q^n}$

$\left\vert \alpha - p/q \right\vert = \left\vert f(p/q) / f'(x_0) \right\vert \ge 1/(Mq^n) > A/q^n \ge \left\vert \alpha - p/q \right\vert$

$\exists n \in \mathbb Z, A > 0 \forall p, q \left( \left\vert x - \frac{p}{q} \right\vert > \frac{A}{q^{n}} \right)$

$\left|x-\frac ab\right|<\frac1{b^m}=\frac1{b^{r+n}}=\frac1{b^rb^n} \le \frac1{2^r}\frac1{b^n} \le \frac A{b^n}$