Transcendental numbers

TheoremLiouville's theorem on existence of transcendental numbers #

Liouville's theorem on existence of transcendental numbers says that explicit real numbers such as $$\sum_{k=1}^{\infty} 10^{-k!} = 0.110001000000000000000001000\ldots$$ are transcendental: they are not roots of any nonzero polynomial with integer coefficients. The theorem names specific transcendental numbers instead of only proving that some exist. The method behind it is a lower bound on how well an algebraic number can be approximated by fractions: a number that has rational approximations too good for any algebraic number must be transcendental.

Statement

Fix an integer $m \ge 2$ and put $$L_m = \sum_{k=0}^{\infty} \frac{1}{m^{k!}}.$$ The series converges because its terms are dominated by those of a geometric series. The theorem states that $L_m$ is transcendental for every integer $m \ge 2$, meaning that no nonzero polynomial with integer coefficients vanishes at $L_m$. Since $0! = 1! = 1$, the term $1/m$ occurs twice in this sum. Removing one copy gives the more common form $\sum_{k \ge 1} m^{-k!}$, which differs from $L_m$ by the rational number $1/m$ and is therefore transcendental as well; for $m = 10$ it is the decimal number displayed above.

The theorem is a special case of a general criterion. A real number $x$ is a Liouville number if for every natural number $n$ there are integers $a$ and $b$ with $b > 1$ and $$0 < \left| x - \frac{a}{b} \right| < \frac{1}{b^n}$$ (the definition). Liouville's criterion states that every Liouville number is transcendental. The number $L_m$ meets the definition, so $L_m$ is a Liouville number, and the theorem follows.

Proof

$L_m$ is a Liouville number. Fix $n$ and split $L_m$ after the term with index $n$. Every $k!$ with $k \le n$ divides $n!$, so the partial sum is a fraction $a/b$ with denominator $b = m^{n!}$, and $b > 1$. The tail is positive, and comparing it with a geometric series shows that it is less than $2/m^{(n+1)!}$. Since $(n+1)! = n!\,(n+1)$, this bound equals $2/b^{n+1}$, which is at most $1/b^n$ because $b \ge 2$. So the tail is less than $1/b^n$, and $0 < |L_m - a/b| < 1/b^n$.

Liouville numbers are transcendental. Let $x$ be a Liouville number. First, $x$ is irrational: if $x = p/q$ with $q \ge 1$, then every fraction $a/b \neq x$ with $b \ge 2$ satisfies $|x - a/b| \ge 1/(qb)$, and for $n$ with $2^{n-1} \ge q$ this is at least $1/b^n$, against the definition. Now suppose $x$ is a root of a nonzero integer polynomial $f$ of degree $d$. The key estimate is that there is a constant $A > 0$ with $|x - a/q| \ge 1/(A\,q^d)$ for all integers $a$ and $q \ge 1$. For $a/q$ near $x$ this comes from two facts. The number $q^d f(a/q)$ is an integer; since $f$ has only finitely many roots, it is nonzero for $a/q$ close enough to $x$ but different from $x$, so $|f(a/q)| \ge q^{-d}$. On the other hand, the mean value theorem, in its inequality form, gives $|f(a/q)| = |f(x) - f(a/q)| \le M\,|x - a/q|$, where $M$ bounds $|f'|$ near $x$. Fractions far from $x$ are covered by enlarging $A$. Finally choose $n \ge d$ with $2^{\,n-d} > A$. The Liouville property provides $a/b$ with $b \ge 2$ and $|x - a/b| < b^{-n} \le b^{-d}/2^{\,n-d} < 1/(A\,b^d)$, contradicting the estimate.

Abundance of transcendental numbers. The real numbers algebraic over the integers form a countable set; Mathlib proves this for any countable integral domain of coefficients and any integral domain that is a torsion-free algebra over it. Since the real numbers are uncountable (see uncountability of the continuum), all but countably many real numbers are transcendental. This argument shows that transcendental numbers exist but names none; the construction of $L_m$ names them.
Size of the set of Liouville numbers. In the sense of Baire category the Liouville numbers are plentiful: they form a dense subset of the real line, and the set of Liouville numbers is residual, so its complement is meagre. In the sense of measure they are scarce: the set of Liouville numbers has Lebesgue measure zero. As the algebraic numbers are countable, almost every real number is transcendental but not a Liouville number, so Liouville's criterion detects only a small part of the transcendental numbers.
Approximation exponents. A real number $x$ is Liouville with exponent $p$ if there is a constant $C$ such that $0 < |x - a/b| < C/b^p$ for infinitely many denominators $b$. The estimate in the proof shows that an irrational algebraic number of degree $d$ is not Liouville with any exponent above $d$, and a real number is Liouville with every exponent exactly when it is a Liouville number. Roth's theorem, which Mathlib does not contain, sharpens the bound $d$ to $2$ for every irrational algebraic number.

In Mathlib

Mathlib defines the number for every real argument $m$ as the sum of the terms $1/m^{k!}$ over all natural numbers $k$, starting at $k = 0$; when the series diverges, Mathlib's convention for infinite sums assigns it the value $0$. The transcendence theorem is stated only for natural numbers $m \ge 2$, viewed as real numbers. Transcendence is taken over the integers, which for a real number is the same as transcendence over the rationals, since clearing denominators turns a rational polynomial into an integer one. The definition of a Liouville number requires $x \neq a/b$ in place of the strict lower bound $0 < |x - a/b|$, an equivalent condition.

Written for Trough, under CC BY-SA 4.0.

Sources

Mathlibtranscendental_liouvilleNumber

Uses 7

Statement usesTranscendental elements · Polynomial in one indeterminate · Real number · Infinite sum

Proof usesLiouville’s Theorem and the Construction of Transcendental Numbers · The Mean Value Theorem (Freek Wiedijk's 100 theorems) · Geometric series

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