Are the Euler–Mascheroni constant, π + e, π − e, πe, π/e, πe, π√2, ππ, eπ2, ln π, 2e, ee, Catalan's constant, or Khinchin's constant rational, algebraic irrational, or transcendental?
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
{\displaystyle 0 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
{\displaystyle {\frac {1}{10^{n}}}\geq \left|x-{\frac {p}{q}}\right|\geq {\frac {1}{q^{\mu +\epsilon }}}}📷
for any ε>0, 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]:246 The Thue–Siegel–Roth theorem states that if α is an algebraic number, real but not rational, then μ(α) = 2.[3]:248
Almost all numbers have an irrationality measure equal to 2.[3]:246
Transcendental numbers have irrationality measure 2 or greater. For example, the transcendental number e has μ(e) = 2.[3]:185 The irrationality measure of π is at most 7.60630853: μ(log 2)
According to a paper of Nelson Carella, the product πe is irrational, but I did not go through the proof yet. Terms like π+e, ππ, ee would be irrational if Schanuels conjecture is proven. The conjecture says: Given any n complex numbers z1, ..., zn which are linearly independent over the rational numbers ℚ, then the field extension ℚ(z1, ..., zn, ez1, ..., ezn) has transcendence degree at least n over ℚ. At least we can say that ab is a transcendental number, if a and b are algebraic numbers with a ≠ 0, 1, and b irrational (the so-called Gelfond-Schneider Theorem). Note that ab is not necessarily transcendental, if b is transcendental!
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