Liouville number

From Academic Kids

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 < |xp/q| < 1/qn.

A Liouville number can then be approximated "quite closely" by a sequence of rational numbers. An equivalent definition 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 integers such that q>1 and p/qc/d, then

|xp/q| = |c/dp/q| ≥ 1/dq > 1/(2n−1 q) ≥ 1/qn

which contradicts the above definition.

In 1844, Joseph Liouville showed that numbers with this property are not just irrational, but are always transcendental (see proof below). He used this result to provide the first example of a provably transcendental number,

<math>

c = \sum_{j=1}^\infty 10^{-j!} = 0.110001000000000000000001000.... <math>

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

<math>p_n = \sum_{j=1}^n 10^{(n! - j!)}; \quad q_n = 10^{n!}<math>

then for all positive integers n, we have that

<math>|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<math>

This approach provides a useful tool for proving a given number is transcendental. Unfortunately, although every Liouville number is transcendental, not every transcendental number is a Liouville number. It has been proven that π is transcendental, but not a Liouville number.

More generally, the irrationality measure of a real number x is a measure of how "closely" a number 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 < |xp/q| < 1/qμ

is satisfied by an infinite number of integer pairs (p, q) with q > 0. For any value μ less than this upper bound, the set of all rationals p/q satisfying the above inequality is 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 of transcendental property of Liouville numbers

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)| over the interval [α − 1, α + 1]. Let α1, α2, ..., αm 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 − αm|)

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

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

Then p/q is in the interval [α − 1, α + 1]; and p/q is not in {α1, α2, ..., αm}, 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 ci xi where each ci is an integer; so we can express |f(p/q)| as

|f(p/q)| = |∑i = 1 to n ci piqi| = |∑i = 1 to n ci piqni| / qn ≥ 1/qn

the last inequality holding because p/q is not a root of f.

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

|xp/q| > A/qn

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

|xa/b| < 1/bm = 1/br+n = 1/(brbn) ≤ 1/(2rbn) ≤ A/bn

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

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