Liouville's Inequality and Liouville Numbers

Liouville was the first to produce transcendental numbers.^[1] These "Liouville numbers" are of the form:
\begin{align*}
\frac{1}{a} + \frac{1}{a^{2!}} + \frac{1}{a^{3!}} + \frac{1}{a^{4!}} + \cdots,
\end{align*}
where \(a\) is a positive integer. The key is Liouville's inequality:^[2]

For every real irrational algebraic real number \(\alpha\) of degree \(n\), there exists a positive number \(C\) such that for arbitrary integers \(p\) and \( q \; (q > 0) \):
\begin{align*}
\left|\alpha - \frac{p}{q}\right| > \frac{C}{q^n}.\tag{1}
\end{align*}

Proof. An algebraic number \(\alpha\) of degree \(n\) satisfies an irreducible polynomial \(f(x)\) of degree \(n\) with integral coefficients, but does not satisfy any such polynomial of lesser degree. Put:
\begin{align*}
f(x) = a_0 + a_1x + a_2x^2 + \cdots + a_nx^n, \hspace{8pt} a_i \in \mathbb{Z}.
\end{align*}
\(\alpha\) can't be a double root of \(f(x)\). To see this, factor \(f(x)\) in \(\mathbb{R}[x]\):
\begin{align*}
f(x) = (x - \alpha) \cdot g(x), \hspace{5pt} g(x) \in \mathbb{R}[x].\tag{2}
\end{align*}
If \(\alpha\) is a double root of \(f(x)\), then \((x-\alpha) | g(x)\) — that is, \(g(x) = (x-\alpha) \cdot h(x)\) for some \(h(x)\) in \(\mathbb{R}[x]\), so:
\begin{align*}
f(x) &= (x - \alpha)^2 \cdot h(x)\\
\therefore f'(x) &= 2(x-\alpha) \cdot h(x) + (x-\alpha)^2 \cdot h'(x)\\
&= (x- \alpha) \cdot k(x).
\end{align*}
Therefore \(\alpha\) is a root of \(f'(x)\), a polynomial of degree \(n-1\) with integral coefficients, contradicting the minimality of \(n\). It follows that \(g(\alpha) \not = 0\), so there is a \(\delta > 0\) such that \(g(x) \not = 0\) for \(\alpha - \delta \leq x \leq \alpha + \delta\). For \(x\) in this range, there is a positive \(M\) such that \(|g(x)| < M\).

Let \(p\) and \(q\) be given integers with \(q > 0\). First suppose \(|\alpha - p/q| > \delta\). Since \(q > 1, \; q^n > 1\) and:
\begin{align*}
\left|\alpha - \frac{p}{q}\right| > \frac{\delta}{q^n}.
\end{align*}
That is, for \(p/q\) further away from \(\alpha\) than \(\delta, \; C = \delta\) will do.

Secondly, suppose \(|\alpha - p/q| \leq \delta\). Then \(g(p/q) \not = 0\), so substituting \(x = p/q\) into (2):
\begin{align*}
\frac{p}{q} - \alpha &= \frac{f(p/q)}{g(p/q)}\\[0.7em]
&= \frac{a_0 + a_1(p/q) + \cdots + a_n(p/q)^n}{g(p/q)}\\[0.7em]
&= \frac{a_0q^n + a_1pq^{n-1} + \cdots + a_np^n}{q^n \cdot g(p/q)}.\tag{3}
\end{align*}
The left side of (3) is not zero, so neither is the numerator on the right, but that numerator is an integer, so it is at least one in absolute value. Taking the absolute value of (3) and bringing in this fact plus the bound \(M\) on \(g(x)\):
\begin{align*}
\left|\alpha - \frac{p}{q}\right| > \frac{1}{q^n \cdot M},
\end{align*}
so in the interval \((\alpha-\delta, \alpha-\delta), \; C = 1/M\) is good. Both situations are covered when \(C = \min(\delta, 1/M)\), proving the theorem. QED.

Pause for a moment to appreciate this proof, which brings in the Fundamental Theorem of Algebra, the derivative, and the boundedness of continuous functions to prove a fact about integers.

To make this concrete, work through the case with \(\alpha = \sqrt{2}\), where \(f(x)=x^2-2\) and \(n=2\). Then:
\begin{align*}
f(x) &= x^2 - 2 = (x-\sqrt{2})\underbrace{(x+\sqrt{2})}_{\large g(x)}.
\end{align*}
In order to stay away from the other root of \(f(x), \; \alpha' = -\sqrt{2}\), it suffices to take \(\delta=2\sqrt{2}\). First, for \(p/q\) such that \(|\sqrt{2}-p/q| > 2\sqrt{2}\):
\begin{align*}
\left|\sqrt{2}-\frac{p}{q}\right| > 2\sqrt{2} \geq \frac{2\sqrt{2}}{q^2},
\end{align*}
because \(|q^2| \geq 1\). So in this range, \(C = \delta = 2\sqrt{2}\) will do. Secondly, let \(p/q\) be such that \(|\sqrt{2}-p/q| \leq 2\sqrt{2}\). In this range:
\begin{align*}
-2\sqrt{2} &\leq \frac{p}{q} - \sqrt{2} \leq 2\sqrt{2}\\[0.7em]
0 &\leq \frac{p}{q} + \sqrt{2} \leq 4\sqrt{2}\\[0.7em]
\therefore \; &\left|\frac{p}{q} + \sqrt{2}\right| \leq 4\sqrt{2} = M.
\end{align*}
So for \(p/q\) in this interval:
\begin{align*}
\frac{p}{q} - \sqrt{2} &= \frac{f(p/q)}{g(p/q)}\\[0.7em]
&= \frac{p^2/q^2-2}{p/q+\sqrt{2}}\\[0.7em]
&= \frac{p^2-2q^2}{q^2 \cdot (p/q+\sqrt{2})}.\tag{4}
\end{align*}
The parenthesized expression in the denominator of (4) has absolute value less than or equal to \(M = 4\sqrt{2}\). Furthermore, the numerator is an integer not equal to zero, since otherwise \(\alpha = \sqrt{2}\) would be rational. Therefore the numerator is greater than or equal to one in absolute value. Consequently:
\begin{align*}
\left|\frac{p}{q} - \sqrt{2}\right| \geq \frac{1}{q^2 \cdot M}.
\end{align*}
So in this interval, \(C = 1/M = \sqrt{2}/8\) will do and the theorem holds true for \(\alpha=\sqrt{2}\) for this value of \(C\):
\begin{align*}
C &= \min{(\delta, 1/M)}\\
&= \min{(2\sqrt{2}, \sqrt{2}/8)}\\
&= \sqrt{2}/8 = 0.1767 \; 7670
\end{align*}
It is possible to do better by choosing a lower value of \(\delta\) (better = a higher value of \(C\)). In the same vein as above, \(M = 2\sqrt{2} + \delta\), so \(1/M\) increases as \(\delta\) decreases, the happy medium being when \(\delta = 1/M\):
\begin{equation}
\begin{gathered}
\delta = 1/M\\
\delta^2 + 2\sqrt{2} \cdot \delta - 1 = 0\\
\delta = \sqrt{3} - \sqrt{2}\\
C = \min{(\delta, 1/M)} = \delta = 0.3178 \; 3724
\end{gathered}
\end{equation}
In fact, anything under \(C = 6 - 4\sqrt{2} = 0.3431 \; 4575\) works and nothing higher is possible, as is shown by considering \(p/q = 3/2\).

Consider the following number:
\begin{align*}
\alpha = \frac{1}{2^{1!}} + \frac{1}{2^{2!}} + \frac{1}{2^{3!}} + \frac{1}{2^{4!}} + \cdots.
\end{align*}
\(\alpha\) is called a Liouville number and is irrational because it is a non-repeating base two decimal. The goal is to prove that \(\alpha\) is transcendental by showing that it does not satisfy Liouville's inequality — that is, for all \(C > 0\), there are natural numbers \(p, \; q, \; n\) such that (1) does not hold.^[3]

Let \(p_n/q_n\) be the \(n^{\text{th}}\) partial sum of \(\alpha\):
\begin{align*}
\frac{p_n}{q_n} &= \frac{1}{2^{1!}} + \frac{1}{2^{2!}} + \cdots + \frac{1}{2^{n!}}\\[0.7em]
&= \sum_{k=1}^n \frac{1}{2^{k!}}\\[0.7em]
&= \sum_{k=1}^n \frac{2^{n!}}{2^{n!} \cdot 2^{k!}}\\[0.7em]
&= \frac{1}{2^{n!}} \sum_{k+1}^n 2^{n!-k!}.\tag{5}
\end{align*}
\(p_n\) is the sum in (5) and \(q_n = 2^{n!}\). Also, \(p_n/q_n\) is an increasing sequence converging to \(\alpha\), so the absolute value is not needed in the following:
\begin{align*}
\alpha - \frac{p_n}{q_n} &= \frac{1}{2^{{(n+1)}!}} + \frac{1}{2^{{(n+2)}!}} + \frac{1}{2^{{(n+3)}!}} + \cdots\\[0.7em]
&< \frac{1}{2^{{(n+1)}!}} + \frac{1}{2^{{(n+1)!+1}}} + \frac{1}{2^{{(n+1)!+2}}} + \cdots\\[0.7em]
&= \frac{1}{2^{{(n+1)}!-1}}.
\end{align*}
Let's take a moment to justify these steps. The first line subtracts the partial sum from \(\alpha\), leaving the tail of the sum. The middle line follows because each denominator after the first one is replaced by a smaller value. Note that the exponents in the middle line are sequential, resulting in an infinite geometric sum whose value is given on the third line. Continuing:
\begin{align*}
\alpha - \frac{p_n}{q_n} & < \frac{1}{2^{{(n+1)}!-1}}\\[0.7em]
&= \frac{2}{2^{(n+1)!}} = \frac{2}{\left(2^{n!}\right)^{n+1}} = \frac{2}{q_n^{n+1}}.\tag{6}
\end{align*}
In order to violate Liouville's inequality, it is necessary, given \(C > 0\), to find a \(q_n\) such that the right side of (6) is relatively small, namely:
\begin{align*}
\frac{2}{q_n^{n+1}} & < \frac{C}{q_n^n}\\[0.7em]
\frac{2}{q_n} & < C\\[0.7em]
\frac{2}{2^{n!}} & < C.\\[0.7em]
\end{align*}
These three inequalities are equivalent, and for any \(C > 0, \; n\) can be chosen large enough so the third one is true. For that \(n\), Liouville's inequality does not hold for \(\alpha, \; p_n, \; q_n\) and therefore \(\alpha\) is transcendental. There is nothing special about the value 2 whose powers appear in \(\alpha\); any integer \(a \geq 2\) can be used, leading to an entire family of transcendental numbers, the Liouville numbers.

It seems that Liouville was the first to prove that transcendental numbers actually exist. The great paradox is that it's harder to produce a transcendental number than to prove that almost all real numbers are transcendental, as Cantor famously did using a cardinality argument: the real numbers are uncountable and the algebraic numbers are countable, so the set difference, the transcendental numbers, are also uncountable. But that was in the future (Liouville: 1844, Cantor: 1874).^[4]

Mike Bertrand

September 21, 2024

^ 1. Liouville's results first appeared in two short articles in Comptes Rendus in 1844. The first is "Mémoires et communications", verbally transmitted by Joseph Liouville at Comptes rendus de l'Académie des Sciences 18, (1844), pp. 883–885. The second a few pages later is "Nouvelle démonstration d'un théorème sur les irrationnelles algébriques" at pp. 910–911.

^ 2. The proof is from Continued Fractions by A. Ya. Khinchin, Dover Publications (1977), ISBN‎ 978-0486696300, pp. 45-47.

^ 3. The proof that Liouville numbers are trancendental is from An Introduction to Number Theory, by Harold M. Stark, The MIT Press (1970, 1987), ISBN 0-262-69060-8, pp. 174-175. Stark also proves Liouville's inequality.

^ 4. Robert Gray showed that Cantor used a diagonal argument to produce actual transcendental numbers just as surely as Liouville did. Cantor's idea was to enumerate the algebraic numbers in a concrete way, then construct a decimal number disagreeing with the \(k^\text{th}\) algebraic number in the \(k^\text{th}\) place to the right of the decimal in a concrete way (each digit of the constructed number being one more than that of the corresponding algebraic number, 0 in case 9 in encountered, for example). The constructed decimal number disagrees with every algebraic number and therefore must be transcendental. Changing the way the constructed decimal number varies from each of the listed algebraic numbers produces an infinite number of concrete transcendental numbers. See "Georg Cantor and Transcendental Numbers", by Robert Gray, The American Mathematical Monthly, Vol. 101, No. 9 (Nov. 1994), pp. 819-832.