Irrational Number Approximation by Rational Numbers

# Irrational Number Approximation by Rational Numbers

Theorem 1: Let $\alpha \in \mathbb{R} \setminus \mathbb{Q}$. Then there exists infinitely many rational numbers $\frac{a}{b} \in \mathbb{Q}$ such that $\displaystyle{\left | \alpha - \frac{a}{b} \right | < \frac{1}{b^2}}$. |

**Proof:**Let $\alpha \in \mathbb{R} \setminus \mathbb{Q}$ and suppose instead that there exists finitely many such rational numbers, say $\frac{a_1}{b_1}, \frac{a_2}{b_2}, ..., \frac{a_n}{b_n} \in \mathbb{Q}$ are the rationals in lowest terms where:

\begin{align} \quad \left | \alpha - \frac{a_i}{b_i} \right | < \frac{1}{b_i^2} \end{align}

- Let $Q \in \mathbb{R}$ be defined as:

\begin{align} Q = \frac{1}{\min_{1 \leq i \leq n} \left \{ \left | \alpha - \frac{a_i}{b_i} \right | \right \}} \end{align}

- Since $\alpha$ is an irrational, $Q$ is well-defined (we are not dividing by zero), and $Q > 0$. By Dirichlet's Theorem there exists a $\frac{c}{d} \in \mathbb{Q}$ with $1 \leq d \leq Q$ such that:

\begin{align} \left | \alpha - \frac{c}{d} \right | \leq \frac{1}{d(Q + 1)} \end{align}

- Since $d \leq Q$ we have that $d \leq Q + 1$ and hence $\frac{c}{d}$ satisfies:

\begin{align} \quad \left | \alpha - \frac{c}{d} \right | \leq \frac{1}{d^2} \end{align}

- If $d = b_i$ for some $i \in \{ 1, 2, ..., n \}$ then since $d \in \mathbb{Z}$ and $Q < Q + 1$, we have that $Q < d(Q+1)$, so:

\begin{align} \quad \left | \alpha - \frac{c}{d} \right | \leq \frac{1}{d(Q + 1)} < \frac{1}{Q} = \min_{1 \leq i \leq n} \left \{ \left | \alpha - \frac{a_i}{b_i} \right | \right \} \end{align}

- Which is a contradiction since $\frac{c}{d} = \frac{a_i}{b_i}$ for some $i$. Hence $\frac{c}{d}$ is a new rational number such that $\left | \alpha - \frac{c}{d} \right | < \frac{1}{d^2}$ - which contradicts $\frac{a_1}{b_1}, \frac{a_2}{b_2}, ..., \frac{a_k}{b_k}$ being the only ones. Thus the assumption that there are just finitely many such integers existing is false. So there must be infinitely many. $\blacksquare$

Theorem 1 is not true if $\alpha$ is a rational number as we prove in the following theorem.

Theorem 2: If $\beta \in \mathbb{Q}$ then there exists only finite many rational numbers $\frac{a}{b} \in \mathbb{Q}$ such that $\left | \beta - \frac{a}{b} \right | < \frac{1}{b^2}$. |

**Proof:**Let $\beta \in \mathbb{Q}$. Then $\beta = \frac{r}{s}$ for some $r, s \in \mathbb{Z}$ with $s > 0$. If $\frac{a}{b} \neq \frac{r}{s}$ then $|br - as| > 0$ and since $br - as \in \mathbb{Z}$ we further have that $|br - as| \geq 1$. If additionally $b > s$ then:

\begin{align} \quad \left | \frac{r}{s} - \frac{a}{b} \right | = \frac{|br - as|}{sb} \geq \frac{1}{sb} > \frac{1}{b^2} \end{align}

- Therefore, if we want to find all fractions $\frac{a}{b}$ such that $\left | \beta - \frac{a}{b} \right | < \frac{1}{b^2}$ we must have that $0 < b \leq s$. There are only finitely many such $b$. Hence there are only finitely many rational numbers $\frac{a}{b}$ (in lowest terms) such that:

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