Liouville Number - Liouville Numbers and Transcendence

Liouville Numbers and Transcendence

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.

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,

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

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

Then p/q is in the interval ; 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

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

Now, f is of the form c_i xi where each c_i is an integer; so we can express |f(p/q)| as

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

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

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

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