Hurwitz's Theorem

# Hurwitz's Theorem

Recall that if $\alpha \in \mathbb{R} \setminus \mathbb{Q}$ then there exists infinitely many rational numbers $\frac{a}{b} \in \mathbb{Q}$ such that:

(1)\begin{align} \quad \left | \alpha - \frac{a}{b} \right | < \frac{1}{b^2} \end{align}

We also saw that if $\beta \in \mathbb{Q}$ then there exists only finitely many rational numbers which satisfy this property.

We now present Hurwitz's theorem which gives us an improved and optimal bound for the same sort of conclusion.

Theorem 1 (Hurwitz's Theorem): Let $\alpha \in \mathbb{R} \setminus \mathbb{Q}$. Then there exists infinitely many $\frac{a}{b} \in \mathbb{Q}$ such that $\displaystyle{\left | \alpha - \frac{a}{b} \right | < \frac{1}{\sqrt{5} b^2}}$. |

Hurwitz's theorem gives as an optimal sort of bounded for irrational number approximation in the following sense. If $c$ is a constant such that $0 < c < \frac{1}{\sqrt{5}}$ and if $\alpha \in \mathbb{R} \setminus \mathbb{Q}$ then there are only finitely many rational numbers $\frac{a}{b} \in \mathbb{Q}$ for which:

(2)\begin{align} \quad \left | \alpha - \frac{a}{b} \right | < \frac{c}{b^2} \end{align}

We prove this in the following corollary.

Corollary 1: Let $\alpha \in \mathbb{R} \setminus \mathbb{Q}$. If $0 < c < \frac{1}{\sqrt{5}}$ then there exists only finitely many $\frac{a}{b} \in \mathbb{Q}$ such that $\displaystyle{\left | \alpha - \frac{a}{b} \right | < \frac{c}{b^2}}$. |

**Proof:**Let $p(x) =x^2 - x + 1$. This polynomial has two real roots, they are:

\begin{align} \quad x = \frac{1 + \sqrt{5}}{2} \quad , \quad \frac{1 - \sqrt{5}}{2} \end{align}

- Let $\alpha = \frac{1 + \sqrt{5}}{2}$ and let $\frac{a}{b} \in \mathbb{Q}$ be such that $\left | \frac{a}{b} - \alpha \right | < \frac{c}{b^2}$. Then:

\begin{align} \quad \left | P \left ( \frac{a}{b} \right ) \right | &= \left | \frac{a}{b} - \alpha \right | \left | \frac{a}{b} - \frac{1 - \sqrt{5}}{2} \right | \\ & \leq \left | \frac{a}{b} - \alpha \right | \left | \frac{a}{b} - \alpha + \sqrt{5} \right | \\ & \leq \left | \frac{a}{b} - \alpha \right |^2 + \sqrt{5} \left | \frac{a}{b} - \alpha \right | \\ & \leq \frac{c^2}{b^4} + \sqrt{5} \frac{c}{b^2} \quad (*) \end{align}

- Now observe that since $p$ only has irrational roots, that then:

\begin{align} \quad \left | P \left ( \frac{a}{b} \right ) \right | = \frac{|a^2 - ab - b^2|}{b^2} \geq \frac{1}{b^2} \quad (**) \end{align}

- Combining the inequalities at $(*)$ and $(**)$ give us that:

\begin{align} \quad \frac{1}{b^2} & \leq \frac{c^2}{b^4} + \sqrt{5} \frac{c}{b^2} \\ 1 & \leq \frac{c^2}{b^2} + \sqrt{5}c \\ 1 - \sqrt{5}{c} & \leq \frac{c^2}{b^2} \\ \frac{1}{1 - \sqrt{5}c} & \geq \frac{b^2}{c^2} \\ \frac{c^2}{1 - \sqrt{5}c} & \geq b^2 \\ \sqrt {\frac{c^2}{1 - \sqrt{5}c}} & \geq b \end{align}

- So if $b$ is bounded. Hence there are only finitely many such $\frac{a}{b}$ for which $\left | \alpha - \frac{a}{b} \right | < \frac{c}{b^2}$. $\blacksquare$