1.3: Algebraic and Transcendental Numbers

The set of polynomials with coefficients in $\mathbb{Z}, \mathbb{Q}, \mathbb{R}$, or $\mathbb{C}$ is denoted by $\mathbb{Z} [x], \mathbb{Q}[x], \mathbb{R}[x]$, and $\mathbb{C}[x]$, respectively.

Definition 1.12

An element $x \in \mathbb{R}$ is called an algebraic number if it satisfies $p(x) = 0$, where $p$ is a non-zero polynomial in $\mathbb{Z}[x]$. Otherwise it is called a transcendental number.

The transcendental numbers are even harder to pin down than the general irrational numbers. We do know that $e$ and $π$ are transcendental, but the proofs are considerably more difficult (see [12]). We’ll see below that the transcendental numbers are far more abundant than the rationals or the algebraic numbers. In spite of this, they are harder to analyze and, in fact, even hard to find. This paradoxical situation where the most prevalent numbers are hardest to find, is actually pretty common in number theory.

The most accessible tool to construct transcendental numbers is Liouville’s Theorem. The setting is the following. Given an algebraic number $y$, it is the root of a polynomial with integer coefficients $f(x) = \sum^{d}_{i=0} a_{i} x^{i}$, where we always assume that the coefficient $a_{d}$ of the highest power is non-zero. That highest power is called the degree of the polynomial. Note that we can always find a polynomial of higher degree that has $y$ as a root. Namely, multiply $f$ by any other polynomial $g$.

Definition 1.13

We say that $f(x) = \sum^{d}_{i=0} a_{i} x^{i}$ in $\mathbb{Z}[x]$ is a minimal polynomial for $\rho$ if f is a non-zero polynomial of minimal degree such that $f(\rho) = 0$.

Theorem 1.14 Liouville's Theorem

Let $f$ be a minimal polynomial of degree $d \ge 2$ for $\rho \in \mathbb{R}$. Then

$$\exists c(\rho) > 0 \mbox{ such that } \forall \frac{p}{q} \in \mathbb{Q} : |\rho-\frac{p}{q}| > \frac{c(\rho)}{q^d} \nonumber$$

Proof

Clearly, if $|\rho-\frac{p}{q}| \ge 1$, the inequality is satisfied. So assume that $|\rho-\frac{p}{q}| < 1$.

Now let $f$ be a minimal polynomial for $\rho$, and set

$$K = \mbox{max}_{t \in [\rho-1, \rho+1]} |f'(t)| \nonumber$$

We know that $f(\frac{p}{q})$ is not zero, because otherwise $f$ would have a factor $(x-\frac{p}{q})$. In that case, the quotient $g$ of $f$ and $(x-\frac{p}{q})$ would not necessarily have integer coefficients, but some integral multiple $mg$ of $g$ would. However, $mg$ would be of lower degree, thus contradicting the minimality of $f$. This gives us that

$$|q^{d} f(\frac{p}{q})| = |\sum_{i=0}^{d} a_{i}p^{i}q^{d-1}| \ge 1 \Rightarrow |f(\frac{p}{q})| \ge q^{-d} \nonumber$$

because it is a non-zero integer. Finally, we use the mean value theorem which tells us that there is a $t$ between $\rho$ and $\frac{p}{q}$ such that

$$K \ge |f'(t)| = |\frac{f(\frac{p}{q})-f(\rho)}{\frac{p}{q}-\rho}| \ge \frac{q^{-d}}{|\frac{p}{q}-\rho} \nonumber$$

since $f(\rho) = 0$. For $K$ as above, this gives us the desired inequality.

Definition 1.15

A real number $\rho$ is called a Liouville number if for all $n \in \mathbb{N}$, there is a rational number $\frac{p}{q}$ such that

$$|\rho-\frac{p}{q}| <\frac{1}{q^n} \nonumber$$

It follows directly from Liouville's theorem that sauch numbers must be transcendental. Liouville numbers can be constructed fairly easily. The number

$$\rho = \sum_{k=1}^{\infty} 10^{-k!} \nonumber$$

is an example. If we set $\frac{p}{q}$ equal to $\sum_{k=1}^{\infty} 10^{-k!}$, then $q = 10^{n!}$. Then

$$\left|\rho-\frac{p}{q}\right| = \sum_{k=n+1}^{\infty} 10^{-k!} \nonumber$$

It is east to show that this is less than $q^{-n}$ (exercise 1.17).

It is worth nothing that there is an optimal version of Louville's Theorem. We record it here without proof.

Theorem 1.16 Roth's Theorem

Let $\rho \in \mathbb{R}$ be algebraic. Then for all $\epsilon > 0$

$$\exists c(\rho, \epsilon) > 0 \mbox{ such that } \forall \frac{p}{q} \in \mathbb{Q} : |\rho-\frac{p}{q}| > \frac{c(\alpha, \epsilon)}{q^{2+\epsilon}} \nonumber$$

where $c(\rho, \epsilon)$ depends only on $\rho$ and $\epsilon$.

This result is all the more remarkable if we consider it in the context of the following more general result (which we will have occasion to prove in Chapter 6).

Theorem 1.17

Let $\rho \in \mathbb{R}$ be irrational. Then there are infinitely many $\frac{p}{q} \in \mathbb{Q}$ such that $|\rho-\frac{p}{q} <\frac{1}{q^2}$.