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.
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/q ≠ c/d, then
- |x - p/q| = |c/d - p/q| ≥ 1/dq > 1/(2n-1 q) ≥ 1/qn
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,
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 < |x - p/q| < 1/qμ
The Liouville numbers are precisely those numbers having infinite irrationality measure.
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,
By the mean value theorem, there exists an x0 between p/q and α such that
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
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
Proof of Transcendental Property of Liouville Numbers
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
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 {α1, α2, ..., αm}, so p/q is not a root of f; and there is no root of f between α and p/q.
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 ∑i = 1 to n ci xi where each ci is an integer; so we can express |f(p/q)| as
the last inequality holding because p/q is not a root of f.
which is a contradiction; therefore, no such p, q exist; proving the lemma.
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.External Links