math.stackexchange.com/questions/3721179
Hurwitz's Theorem: 设 $\zeta$ 为无理数, 则存在无穷个互素整数 $p$ 与 $q$ 使
$$\lvert \zeta - \frac{p}{q} \rvert < \frac{1}{\sqrt{5} q^2}$$
我们要证明, $\sqrt5$ 是命题为真的最大常数。设 $\varphi = \frac{1 + \sqrt5}{2}$.
先探索一些 $\varphi$ 和 $\frac{p}{q}$ 的性质.
Lemma 1: 令 $\frac{p}{q} \in Q$, 则 $\lvert \varphi - \frac{p}{q} \rvert \lvert \bar{\varphi} - \frac{p}{q} \rvert \geq \frac{1}{q^2}$, 其中 $\bar{\varphi} = \frac{1-\sqrt5}{2}$ 为 $\varphi$ 的共轭.
Lemma 2: 令 $\frac{p}{q} \in Q$, 与 $m \in R$. 若 $\lvert \varphi - \frac{p}{q} \rvert < \frac{1}{mq^2}$, 则 $\lvert \bar{\varphi} - \frac{p}{q} \rvert < \frac{1}{mq^2} + \sqrt5$.
设有无穷个整数 $p_i$ 与 $q_i$ 使 $\lvert \varphi - \frac{p_i}{q_i} \rvert < \frac{1}{mq^2_i}$, 由引理得 $m$ 的一个上界:
$$\frac{1}{q_i^2} \leq\lvert \varphi - \frac{p_i}{q_i} \rvert\lvert \bar{\varphi} - \frac{p_i}{q_i} \rvert < \frac{1 + mq_i^2\sqrt5}{m^2q_i^4}$$
$$\frac{1}{q_i^2} < \frac{1 + mq_i^2\sqrt5}{m^2q_i^4}$$
$$q_i^2m^2 < 1 + mq_i^2\sqrt5$$
$$q_i^2m^2 - \sqrt5q_i^2m -1 < 0$$
$$q_i^2(m^2-\sqrt5m) - 1 < 0$$
回想一下，我们想要找到 $m$ 的与 $q_i$ 无关的上界. 考虑 $\lim \limits_{i \to \infty} q_i^2(m^2-\sqrt5m) - 1$. 若 $q_i \to \infty$ 当 $i \to \infty$, 那么当且仅当 $m^2 - \sqrt5m\le0$ 时，极限才会为负，所以 $m \leq \sqrt5$。这表明 $\sqrt5$ 是Hurwitz定理为真的最大常数。实际上，由于每个分母都有有限多个分子，因此序列 $q_1, q_2, ...$ 必须趋于无穷大。
引理的证明**Proof of Lemma 1:** We consider three cases. First, that $\frac{p}{q} > \varphi > \bar{\varphi}$: we carefully remove the absolute values to get $(\frac{p}{q} - \varphi)(\frac{p}{q} - \bar{\varphi})$, which simplifies to $\frac{p^2}{q^2} -\frac{p}{q}(\varphi + \bar{\varphi}) + \varphi\bar{\varphi}$. Note that $\varphi + \bar{\varphi} = 1$ and $\varphi\bar{\varphi} = -1$, so we end up with $\frac{1}{q^2}(p^2 - pq - q^2)$.
Second, that $\varphi > \frac{p}{q} > \bar{\varphi}$: similarly, we remove the absolute values to get $(\varphi - \frac{p}{q})(\frac{p}{q} - \bar{\varphi})$, which simplifies to $-\frac{p^2}{q^2} +\frac{p}{q}(\varphi + \bar{\varphi}) - \varphi\bar{\varphi} = \frac{1}{q^2}(-p^2 + pq + q^2)$.
And third, that $\varphi > \bar{\varphi} > \frac{p}{q}$: we get $(\varphi - \frac{p}{q})(\bar{\varphi} - \frac{p}{q}) = \frac{p^2}{q^2} -\frac{p}{q}(\varphi + \bar{\varphi}) + \varphi\bar{\varphi} = \frac{1}{q^2}(p^2 - pq - q^2)$.
Observe that, in all three cases, we are left with the product of two things that must be positive. Thus $\lvert \varphi - \frac{p}{q} \rvert \lvert \bar{\varphi} - \frac{p}{q} \rvert = \frac{1}{q^2} \lvert p^2 -pq - q^2 \rvert$. If we consider $p^2 -pq - q^2$ as a polynomial on $p$, we quickly find out that it has no real roots, and thus, that $\lvert p^2 -pq - q^2 \rvert \geq 1$. Hence, $\lvert \varphi - \frac{p}{q} \rvert \lvert \bar{\varphi} - \frac{p}{q} \rvert \geq \frac{1}{q^2}$.
**Proof of Lemma 2:** We consider the same three cases as the first lemma.
$\frac{p}{q} > \varphi > \bar{\varphi}$: $\lvert \varphi - \frac{p}{q} \rvert = \frac{p}{q} - \varphi < \frac{1}{mq^2}$. Adding $\sqrt5$ to the inequality, we get $\frac{p}{q} - \bar{\varphi} < \frac{1}{mq^2} + \sqrt5$. Since $\frac{p}{q} - \bar{\varphi} > 0$, we can put the absolute value back on: $\lvert \frac{p}{q} - \bar{\varphi} \rvert = \lvert \bar{\varphi} - \frac{p}{q} \rvert < \frac{1}{mq^2} + \sqrt5$
$\varphi > \frac{p}{q} > \bar{\varphi}$: we use the following reverse triangle inequalities: $\lvert \varphi - \frac{p}{q} \rvert \geq \lvert \varphi \rvert - \lvert \frac{p}{q} \rvert$ and $\lvert \frac{p}{q} - \varphi \rvert \geq \lvert \frac{p}{q} \rvert - \lvert \varphi \rvert$. To remove the absolute values, we must denote whether $\frac{p}{q}$ is positive or negative.
$\frac{p}{q} \geq 0$: we use the latter inequality to arrive at $\lvert \frac{p}{q} \rvert - \lvert \varphi \rvert = -\varphi + \frac{p}{q} < \frac{1}{mq^2}$. In adding $\sqrt5$, we get $-\bar{\varphi} + \frac{p}{q} = \lvert -\bar{\varphi} + \frac{p}{q} \rvert = \lvert \bar{\varphi} - \frac{p}{q}\rvert < \frac{1}{mq^2} + \sqrt5$
$\frac{p}{q} < 0$: we now use the former inequality to arrive at $\lvert \varphi \rvert - \lvert \frac{p}{q} \rvert = \varphi + \frac{p}{q} < \frac{1}{mq^2}$. Note that $\bar{\varphi} < {\varphi}$, so we have that $\bar{\varphi} + \frac{p}{q} < \frac{1}{mq^2}$. If we add $\sqrt5$, we return to $\varphi + \frac{p}{q} < \frac{1}{mq^2} + \sqrt5$. One simply has to note that $0 < - \bar{\varphi} + \frac{p}{q} < \varphi + \frac{p}{q} < \frac{1}{mq^2} + \sqrt5$. Thus $\lvert -\bar{\varphi} + \frac{p}{q} \rvert = \lvert \bar{\varphi} - \frac{p}{q} \rvert$.
$\varphi > \bar{\varphi} > \frac{p}{q}$: we have that $\bar{\varphi} - \frac{a}{b} < \varphi - \frac{p}{q} < \frac{1}{mq^2}$. Hence, $\bar{\varphi} - \frac{p}{q} < \frac{1}{mq^2}$ becomes $\varphi - \frac{p}{q} < \frac{1}{mq^2} + \sqrt5$. $\bar{\varphi} - \frac{p}{q} < \frac{1}{mq^2} + \sqrt5$. Finally, $\lvert \bar{\varphi} - \frac{p}{q} \rvert < \frac{1}{mq^2} + \sqrt5$. | 
