Algebraic Number Approximation

Lemma 1: Let $p \in \mathbb{Z}[x]$ be a polynomial of degree $d$. If $\frac{a}{b} \in \mathbb{Q}$ and $P \left ( \frac{a}{b} \right ) \neq 0$ then $\displaystyle{\left | P \left ( \frac{a}{b} \right ) \right | \geq \frac{1}{b^d}}$.

• Proof: Let $p(x) = a_0 + a_1x + ... + a_dx^d$ where $a_d \neq 0$. Then:

(1)

\begin{align} \quad \left | P \left ( \frac{a}{b} \right ) \right | = \left | a_0 + a_1 \frac{a}{b} + ... + a_d \frac{a^d}{b^d} \right | = \frac{|a_0b^d + a_1ab^{d-1} + ... + a_da^d|}{|b^d|} \overset{(*)} \geq \frac{1}{|b^d|} \geq \frac{1}{b^d} \end{align}

• The inequality at $(*)$ comes from the assumption that $P \left ( \frac{a}{b} \right ) \neq 0$ and so the absolute value of a sum of integers not summing to zero is at least $1$. $\blacksquare$

Theorem 2: If $\alpha$ is an algebraic number of degree $d$ then there exists a number $C(\alpha) > 0$ such that for all $\frac{a}{b} \in \mathbb{Q}$ with $\alpha \neq \frac{a}{b}$ we have that $\displaystyle{\left | \alpha - \frac{a}{b} \right | \geq \frac{C(\alpha)}{b^d}}$.

• Proof: Since $\alpha$ is algebraic with degree $d$ there exists a polynomial $P \in \mathbb{Z}[x]$ of minimal degree for which $P(\alpha) = 0$. Let $\frac{a}{b} \in \mathbb{Q}$. Denote that if $\alpha \neq \frac{a}{b}$ then $\frac{a}{b}$ is not a root of $P$, otherwise the polynomial $\frac{P(x)}{\left ( x - \frac{a}{b} \right )}$ is a polynomial of lesser degree with $\alpha$ as a root, contradicting $P$ being such a polynomial with minimum degree. So $\frac{a}{b}$ cannot be a root of $P$.

• Case 1: Suppose that $\left | \alpha - \frac{a}{b} \right | > 1$. Since $b^d \geq 1$ we have that:

(2)

\begin{align} \quad \left | \alpha - \frac{a}{b} \right | \geq 1 \geq \frac{1}{b^d} \end{align}

• Case 2: Suppose that $\left | \alpha - \frac{a}{b} \right | \leq 1$. By the previous Lemma, since $\frac{a}{b}$ is not a root of $P$ we have that:

(3)

\begin{align} \left | P \left ( \frac{a}{b} \right ) \right | \geq 1 \end{align}

• So by the Mean-Value theorem we have that:

(4)

\begin{align} \quad P \left ( \frac{a}{b} \right ) &= P \left ( \frac{a}{b} \right ) - P(\alpha) \\ &= \left ( \frac{a}{b} - \alpha \right ) P'(t) \end{align}

• For some $t$ between $\frac{a}{b}$ and $\alpha$. (Here, $P'(t)$ denotes the derivative of $P$, which exists since $P$ is just a polynomial). Since $\left | \alpha - \frac{a}{b} \right | \leq 1$ by assumption, we must have that $t \in [\alpha - 1, \alpha + 1]$. Since $P$ is a polynomial it is continuous on this closed bounded interval there exists a maximum value by the Extreme-Value theorem, call it $M > 0$ for which:

(5)

\begin{align} \quad |P'(t)| \leq \max_{x \in [\alpha - 1, \alpha +1]} |P'(x)| = M \end{align}

• Therefore:

(6)

\begin{align} \quad \frac{1}{b^d} \leq \left | P \left ( \frac{a}{b} \right ) \right | = \left | \alpha - \frac{a}{b} \right | |P'(t)| \leq \left |\alpha - \frac{a}{b} \right | M \end{align}

• Hence:

(7)

\begin{align} \quad \left | \alpha - \frac{a}{b} \right | \geq \frac{1}{Mb^d} \end{align}

• So let $C(\alpha) = \frac{1}{M}$. Then:

(8)

\begin{align} \quad \left | \alpha - \frac{a}{b} \right | \geq \frac{C(\alpha)}{b^d} \quad \blacksquare \end{align}