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The irrationality measure (or irrationality exponent or approximation exponent or Liouville–Roth constant) of a real number x is a measure of how "closely" it can be approximated by rationals. Generalizing the definition of Liouville numbers, 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< \left| x- \frac{p}{q} \right| < \frac{1}{q^{\mu}} 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.[3] 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 at most finitely many (p, q) with q > 0 that satisfy the inequality; thus, the opposite inequality holds for all larger values of q. In other words, given the irrationality measure μ of a real number x, whenever a rational approximation x ≅ p/q, p,q ∈ N yields n + 1 exact decimal digits, we have \frac{1}{10^n} \ge \left| x- \frac{p}{q} \right| \ge \frac{1}{q^{\mu}} except for at most a finite number of "lucky" pairs (p, q). For a rational number α the irrationality measure is μ(α) = 1.[3] The Thue–Siegel–Roth theorem states that if α is an algebraic number, real but not rational, then μ(α) = 2.[4] Almost all numbers have an irrationality measure equal to 2.[3] Transcendental numbers have irrationality measure 2 or greater. For example, the transcendental number e has μ(e) = 2.[5] The irrationality measure of π is at most 7.60630853: μ(log 2)<3.57455391 and μ(log 3)<5.125.[6] The Liouville numbers are precisely those numbers having infinite irrationality measure.[4] 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.[7] 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, \left| \alpha - \frac{p}{q} \right | > \frac{A}{q^n} 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, ..., αm be the distinct roots of f which differ from α. Select some value A > 0 satisfying A< \min \left(1, \frac{1}{M}, \left| \alpha - \alpha_1 \right|, \left| \alpha - \alpha_2 \right|, \ldots , \left| \alpha-\alpha_m \right| \right) Now assume that there exist some integers p, q contradicting the lemma. Then \left| \alpha - \frac{p}{q}\right| \le \frac{A}{q^n} \le A< \min\left(1, \frac{1}{M}, \left| \alpha - \alpha_1 \right|, \left|\alpha - \alpha_2 \right|, \ldots , \left| \alpha-\alpha_m \right| \right) 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. By the mean value theorem, there exists an x0 between p/q and α such that f(\alpha)-f(\tfrac{p}{q}) = (\alpha - \frac{p}{q}) \cdot f'(x_0) Since α is a root of f but p/q is not, we see that |f ′(x0)| > 0 and we can rearrange: \left|\alpha -\frac{p}{q}\right |= \frac{\left | f(\alpha)- f(\tfrac{p}{q})\right |}{|f'(x_0)|} = \left | \frac{f(\tfrac{p}{q})}{f'(x_0)} \right | Now, f is of the form \sum_{i=0}^n ci xi where each ci is an integer; so we can express |f(p/q)| as \left|f \left (\frac{p}{q} \right) \right| = \left| \sum_{i=0}^n c_i p^i q^{-i} \right| = \frac{1}{q^n} \left| \sum_{i=0}^n c_i p^i q^{n-i} \right | \ge \frac {1}{q^n} the last inequality holding because p/q is not a root of f and the ci 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 \left | \alpha - \frac{p}{q} \right | = \left|\frac{f(\tfrac{p}{q})}{f'(x_0)}\right| \ge \frac{1}{Mq^n} > \frac{A}{q^n} \ge \left| \alpha - \frac{p}{q} \right| 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 \left| x - \frac{p}{q} \right|> \frac{A}{q^{n}} 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 \left|x-\frac ab\right|<\frac1{b^m}=\frac1{b^{r+n}}=\frac1{b^rb^n} \le \frac1{2^r}\frac1{b^n} \le \frac A{b^n} which contradicts the lemma; therefore x is not algebraic, and is thus transcendental. |
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