Liouville number

In number theory, a Liouville number is a real number x with the property that, for any positive integer n, there exist integers p and q with q > 1 and such that

0 < |x − p/q| < 1/qⁿ.

A Liouville number can thus be approximated "quite closely" by a sequence of rational numbers. In 1844, Joseph Liouville showed that all Liouville numbers are transcendental, and he provided an example of a Liouville number, thus establishing the existence of transcendental numbers for the first time.

Elementary properties

An equivalent definition to the one given above is that for any positive integer n, there exists an infinite number of pairs of integers (p,q) obeying the above inequality.

It is relatively easily proven that if x is a Liouville number, x is irrational. Assume otherwise; then there exists integers c, d with x = c/d. Let n be a positive integer such that 2ⁿ⁻¹ > d. Then if p and q are any integers such that q > 1 and p/q ≠ c/d, then

|x − p/q| = |c/d − p/q| ≥ 1/dq > 1/(2ⁿ⁻¹ q) ≥ 1/qⁿ

which contradicts the definition of Liouville number.

Liouville constant

The number

$c=\sum _{j=1}^{\infty }10^{-j!}=0.110001000000000000000001000....$

is known as Liouville's constant. Liouville's constant is a Liouville number; if we define p_n and q_n as follows:

$p_{n}=\sum _{j=1}^{n}10^{(n!-j!)};\quad q_{n}=10^{n!}$

then we have for all positive integers n

$|c-p_{n}/q_{n}|=\sum _{j=n+1}^{\infty }10^{-j!}=10^{-(n+1)!}+10^{-(n+2)!}+\cdots <10^{-(n!n)}=1/{q_{n}}^{n}$

Liouville numbers and measure

From the point of view of measure theory, the set of all Liouville numbers $L\$ is small. More precisely, its Lebesgue measure $m\$ is zero. The proof given follows some ideas by John Oxtoby.

For positive integers $n>2\$ and $q\geq 2$ set:

$V_{n,q}=\bigcup \limits _{p=-\infty }^{\infty }\left({\frac {p}{q}}-{\frac {1}{q^{n}}},{\frac {p}{q}}+{\frac {1}{q^{n}}}\right)$ – we have $L\subseteq \bigcup \limits _{q=2}^{\infty }V_{n,q}.$

Observe that for each positive integer $n\geq 2,\ m$ we have also:

$L\cap (-m,m)\subseteq \bigcup \limits _{q=2}^{\infty }V_{n,q}\cap (-m,m)\subseteq \bigcup \limits _{q=2}^{\infty }\bigcup \limits _{p=-mq}^{mq}\left({\frac {p}{q}}-{\frac {1}{q^{n}}},{\frac {p}{q}}+{\frac {1}{q^{n}}}\right).$

Since $\left|\left({\frac {p}{q}}+{\frac {1}{q^{n}}}\right)-\left({\frac {p}{q}}-{\frac {1}{q^{n}}}\right)\right|={\frac {2}{q^{n}}}$ and $n>2$ we have

$m(L\cap (-m,\,m))\leq \sum \limits _{q=2}^{\infty }\sum _{p=-mq}^{mq}{\frac {2}{q^{n}}}=\sum \limits _{q=2}^{\infty }{\frac {2(2mq+1)}{q^{n}}}\leq (4m+1)\sum \limits _{q=2}^{\infty }{\frac {1}{q^{n-1}}}\leq (4m+1)\int _{1}^{\infty }{\frac {dq}{q^{n-1}}}\leq {\frac {4m+1}{n-2}}.$

Now $\lim \limits _{n\to \infty }{\frac {4m+1}{n-2}}=0$ and it follows that for each positive integer $m\$, the intersection $L\cap (-m,m)$ has Lebesgue measure zero. Consequently, so has $L\$.

Liouville numbers and topology

For each positive integer set:

$U_{n}=\bigcup \limits _{q=2}^{\infty }\bigcup \limits _{p=-\infty }^{\infty }\left({\frac {p}{q}}-{\frac {1}{q^{n}}},{\frac {p}{q}}+{\frac {1}{q^{n}}}\right)$.

Each of sets $U_{n}\$ is an open dense subset of real line ${\mathbb {R} }$ (note that each $U_{n}\$ contains all rationals). Moreover, $L=\bigcap \limits _{n=1}^{\infty }U_{n}\setminus {\mathbb {Q} }$ and it follows that $L\$ is a dense G-delta set, since its complement is a meagre set. From topological point of view it means that "almost all" real numbers are Liouville numbers.

Liouville numbers and transcendency

All Liouville numbers are transcendental, as will be proven below. Establishing that a given number is a Liouville number provides a useful tool for proving a given number is transcendental. Unfortunately, not every transcendental number is a Liouville number. The terms in the continued fraction expansion of every Liouville number are unbounded; using a counting argument, one can then show that there must be uncountably many transcendental numbers which are not Liouville. Using the explicit continued fraction expansion of e, one can show that e is an example of a transcendental number that is not Liouville. Mahler proved in 1953 that π is another such example.

Irrationality measure

More generally, the irrationality measure (or Liouville–Roth constant) of a real number x is a measure of how "closely" it can be approximated by rationals. Instead of allowing any n in the power of q, we find the least upper bound of the set of real numbers μ such that

0 < |x − p/q| < 1/q^μ

is satisfied by an infinite number of integer pairs (p, q) with q > 0. This least upper bound is defined to be the irrationality measure of x. For any value μ less than this upper bound, the infinite set of all rationals p/q satisfying the above inequality yield an approximation of x; conversely, if μ is greater than the upper bound, then there are no such sequences which get arbitrarily close to x.

The Liouville numbers are precisely those numbers having infinite irrationality measure.

Proof that all Liouville numbers are transcendental

The proof proceeds by first establishing a property of irrational algebraic numbers. This property essentially says that irrational algebraic numbers cannot be well approximated by rational numbers. A Liouville number is irrational but does not have this property, so it can't be algebraic and must be transcendental. The following lemma is usually known as Liouville's theorem (on diophantine approximation), there being several results known as Liouville's theorem.

Lemma: If α is an irrational number which is the root of a polynomial f of degree n > 0 with integer coefficients, then there exists a real number A > 0 such that, for all integers p, q, with q > 0,

|α − p/q| > A/qⁿ.

Proof of Lemma: Let M be the maximum value of |f ′(x)| (the absolute value of the derivative of f) over the interval [α − 1, α + 1]. Let α₁, α₂, ..., α_m be the distinct roots of f which differ from α. Select some value A > 0 satisfying

A < min(1, 1/M, |α − α₁|, |α − α₂|, ..., |α − α_m|)

Now assume that there exists some integers p, q contradicting the lemma. Then

|α − p/q| ≤ A/qⁿ ≤ A < min(1, |α − α₁|, |α − α₂|, ..., |α − α_m|)

Then p/q is in the interval [α − 1, α + 1]; and p/q is not in {α₁, α₂, ..., α_m}, so p/q is not a root of f; and there is no root of f between α and p/q.

By the mean value theorem, there exists an x₀ between p/q and α such that

f(α) − f(p/q) = (α − p/q) · f ′(x₀)

Since α is a root of f but p/q is not, we see that |f ′(x₀)| > 0 and we can rearrange:

|(α − p/q)| = |f(α) − f(p/q)| / |f ′(x₀)| = |f(p/q)| / |f ′(x₀)|

Now, f is of the form ∑_{i = 1 to n} c_i xⁱ where each c_i is an integer; so we can express |f(p/q)| as

|f(p/q)| = |∑_{i = 1 to n} c_i pⁱq⁻ⁱ| = |∑_{i = 1 to n} c_i pⁱqⁿ⁻ⁱ| / qⁿ ≥ 1/qⁿ

the last inequality holding because p/q is not a root of f and the c_i are integers.

Thus we have that |f(p/q)| ≥ 1/qⁿ. Since |f ′(x₀)| ≤ M by the definition of M, and 1/M > A by the definition of A, we have that

|(α − p/q)| = |f(p/q)| / |f ′(x₀)| ≥ 1/(M qⁿ) > A/qⁿ ≥ |(α − p/q)|

which is a contradiction; therefore, no such p, q exist; proving the lemma.

Proof of assertion: As a consequence of this lemma, let x be a Liouville number; as noted in the article text, x is then irrational. If x is algebraic, then by the lemma, there exists some integer n and some positive real A such that for all p, q

|x − p/q| > A/qⁿ

Let r be a positive integer such that 1/(2^r) ≤ A. If we let m = r + n, then, since x is a Liouville number, there exists integers a, b > 1 such that

|x − a/b| < 1/b^m = 1/b^r+n = 1/(b^rbⁿ) ≤ 1/(2^rbⁿ) ≤ A/bⁿ

which contradicts the lemma; therefore x is not algebraic, and is thus transcendental.

External links

• The Beginning of Transcendental Numbers