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 2^{n1} > d. Then if p and q are integers such that q>1 and p/q ≠ c/d, then
which contradicts the above definition.
In 1844, Joseph Liouville[?] showed that numbers of 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,
known as Liouville's constant. Liouville's constant is a Liouville number; if we define p_{n} and q_{n} as follows:
then for all positive integers n, we have that
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
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.
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) over the interval [α1, α+1]. Let α_{1}, α_{2}, ..., 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 [α  1, α + 1]; and p/q is not in {α_{1}, α_{2}, ..., 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 x_{0} between p/q and α such that
Since α is a root of f but p/q is not, we see that f ′(x_{0}) > 0 and we can rearrange:
Now, f is of the form ∑_{i = 1 to n} c_{i} x^{i} 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.
Thus we have that f(p/q) ≥ 1/q^{n}. Since f ′(x_{0}) ≤ 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/(2^{r}) ≤ 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.
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