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 To meet Wikipedias quality standards, this article or section may require cleanup. ... In mathematics, the real numbers are intuitively defined as numbers that are in one-to-one correspondence with the points on an infinite line—the number line. ... The integers consist of the positive natural numbers (1, 2, 3, …), their negatives (−1, −2, −3, ...) and the number zero. ...

0 < |x − p/q| < 1/qⁿ.

A Liouville number can thus be approximated "quite closely" by a sequence of rational numbers. In 1844, Joseph Liouville showed that all Liouville numbers are transcendental, and he provided an example of a Liouville number, thus establishing the existence of transcendental numbers for the first time. In mathematics, a sequence is a list of objects (or events) arranged in a linear fashion, such that the order of the members is well defined and significant. ... 1844 was a leap year starting on Monday (see link for calendar). ... Joseph Liouville (born March 24, 1809, died September 8, 1882) was a French mathematician. ... In mathematics, a transcendental number is any real number that is not algebraic, that is, not the solution of a non-zero polynomial equation with integer (or, equivalently, rational) coefficients. ...

Elementary properties

An equivalent definition to the one given above 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 any integers such that q > 1 and p/q ≠ c/d, then In mathematics, an irrational number is any real number that is not a rational number, i. ...

|x − p/q| = |c/d − p/q| ≥ 1/dq > 1/(2ⁿ⁻¹ q) ≥ 1/qⁿ

which contradicts the definition of Liouville number.

Liouville constant

The number

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

then we have for all positive integers n

Liouville numbers and transcendentality

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. In mathematics, a continued fraction is an expression such as where a0 is some integer and all the other numbers an are positive integers. ... e is the unique number such that the derivative (slope) of f(x)=ex at any point is equal to the height of the function at that point. ... Lower-case Ï€ (the lower case letter is usually used for the constant) The mathematical constant Ï€ is an irrational number, approximately equal to 3. ...

Irrationality measure

More generally, the irrationality measure of a real number x is a measure of how "closely" it 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 In mathematics, the supremum of an ordered set S is the least element (not necessarily in S) which is greater than or equal to each element of S. Consequently, it is also referred to as the least upper bound. ...

0 < |x − p/q| < 1/q^μ

is satisfied by an infinite number of integer pairs (p, q) with q > 0. This least upper bound is defined to be the irrationality measure of x. For any value μ less than this upper bound, the infinite set of all rationals p/q satisfying the above inequality yield 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 that all Liouville numbers are transcendental

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. In mathematics, an irrational number is any real number that is not a rational number, i. ... In mathematics, an algebraic number is any number that is a root of a non-zero polynomial with integer (or equivalently, rational) coefficients. ... In mathematics, a lemma is a proven proposition which is used as a stepping stone to a larger result rather than an independent statement, in and of itself. ... Liouvilles theorem has various meanings: In complex analysis, see Liouvilles theorem (complex analysis). ...

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, In mathematics, a polynomial is an expression in which constants and variables are combined using only addition, subtraction, multiplication, and positive whole number exponents (raising to a power). ... This article is about the term degree as used in mathematics. ...

|α − p/q| > A/qⁿ.

Proof of Lemma: Let M be the maximum value of |f ′(x)| (the absolute value of the derivative of f) over the interval [α − 1, α + 1]. Let α₁, α₂, ..., α_m be the distinct roots of f which differ from α. Select some value A > 0 satisfying In mathematics, the absolute value (or modulus1) of a real number is its numerical value without regard to its sign. ... In mathematics, the derivative is defined as the instantaneous rate of change of a function. ... In mathematics, interval is a concept relating to the sequence and set-membership of one or more numbers. ...

A < min(1, 1/M, |α − α₁|, |α − α₂|, ..., |α − α_m|)

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

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

Then p/q is in the interval [α − 1, α + 1]; 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 For any function that is continuous on [a, b] and differentiable on (a, b) there exists some c in the interval (a, b) such that the secant joining the endpoints of the interval [a, b] is parallel to the tangent at c. ...

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} c_i xⁱ where each c_i is an integer; so we can express |f(p/q)| as

|f(p/q)| = |∑_{i = 1 to n} c_i pⁱq⁻ⁱ| = |∑_{i = 1 to n} c_i pⁱqⁿ⁻ⁱ| / qⁿ ≥ 1/qⁿ

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

|(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

|x − p/q| > A/qⁿ

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

|x − a/b| < 1/b^m = 1/b^r+n = 1/(b^rbⁿ) ≤ 1/(2^rbⁿ) ≤ A/bⁿ

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

External links

• The Beginning of Transcendental Numbers