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
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
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