The following section uses the notation from the chapter on Set Theory. We shall consider the behaviour of inconcrete m, n ∈ ^{ω}ℕ, and we define h, k ∈ ℕ.

Remark: From algebra, we know that the sum, difference, product, and quotient of two algebraic numbers of degree h and k are algebraic of degree at most hk, and that the 1/h-th power of an algebraic number of degree k is algebraic of degree at most hk.

Remark: Transcendental numbers can be viewed as the sum of an algebraic part and a transcendental remainder. When investigating whether a number is transcendental, if the remainder may be expressed as the limit value of a zero sequence (a_{k}) (see Nonstandard Analysis), we cannot simply disregard the values of the sequence for large k: They are important. Transcendental numbers are the numbers that lie between algebraic numbers or on either side of them. By the counting theorem for algebraic numbers, if two distinct transcendental numbers (algebraic of degree h) are sufficiently close, there is no algebraic number (of degree < h) between them.

Bounding theorem for ω-transcendental numbers: Every non-zero complex number whose imaginary or real part has absolute value is ≤ 1/(ώ + 1) or ≥ ώ + 1, is automatically ω-transcendental.

Proof: In a polynomial equation, set a_{m} = 1 and a_{k} = -ώ for k < m, then the claim in the real case follows from the geometric series formula after taking the reciprocal. We can find the exact limit value by replacing ώ + 1 by ω(m) = ώ + 1 - ώ/ω(m)^{m} with ω := ώ + 1 - ώ/ω^{ώ}. In the complex case, substitutions of the form x = (1 + bi)(ώ + 1) with b ∈ ^{ω}ℝ give the desired result.⃞

Coefficient theorem for ω-transcendental numbers: Every normalised irreducible polynomial and series such that |a_{k}| ≥ ώ + 1 for at least one a_{k} only has ω-transcendental zeros.

Proof: The zeros of normalised irreducible polynomials and series are pairwise distinct and uniquely determined. Since they are not ω-algebraic, they must be ω-transcendental.⃞

Product and sum theorem for ω-transcendental numbers: Every unsimplifiable product (or sum) of an infinite complex-rational number a and a complex algebraic or infinite algebraic number b, and every sum and product of an infinite complex algebraic number and a complex ω-algebraic (infinite complex algebraic) number with coprime polynomial or series degrees is ω-transcendental.

Proof: Dividing by an infinite complex-rational number leads always to at least one coefficient with modulus ≥ ώ + 1 in the corresponding minimal polynomial or series. Requiring it to be coprime ensures that there is always at least one infinite algebraic number in the result.⃞

Remark: This simplification is unique if it includes the minimum of the modulus of the coefficients of the minimal polynomial or series of b and every numerator and denominator of the infinite complex-rational number a. Together with the rational numbers, the infinite (complex-)rational numbers already numerically make up the entirety of ℝ (ℂ). Therefore, algebraic and transcendental numbers are (numerically) difficult to distinguish, and approximations are of little use when determining whether a number is algebraic (to a certain degree). The merit of distinguishing between them is therefore questionable. Real continued fractions that do not terminate as a rational number are ω-transcendental, since they are infinite rational. They can only be used as an approximation of ω-algebraic numbers.

Remark: In the case of ω-transcendental numbers we must satisfy ourselves with an arbitrarily precise infinitesimal number, unlike in the case of ω-algebraic numbers, where we can argue using the corresponding minimal polynomial or series. We can therefore represent them in the form of a rational quotient with infinite numerator and denominator. However, when investigating the transcendence of a number using Liouville's approximation theorem, care should be taken to ensure that the approximating rational numbers are in fact conventionally rational and not infinitely rational, such as for example the number 10^{ώ+1}. Similarly, confidence with infinite natural numbers is required when considering prime numbers ≥ ώ + 1 in proofs of transcendence. The following proofs can also be used to show ω-transcendence by replacing the set ^{ω}ℕ by [ώ + 1, n]ℕ for some n that is not excessively large.

Theorem: The sum of all at least conventionally natural powers of a non-zero complex-rational number x that is not a root of unity (geometric series) is already ω-transcendental.

Proof: The modulus of either the numerator or denominator of x^{ώ+1} is ≥ 2^{(ώ+1)/2}. Subtracting 1 and dividing by 1 - x preserves ω-transcendence.⃞

Theorem: Euler's number e is ω-transcendental.

Proof: If we accept the exponential series as a representation for e, it follows that e = (kώ + 1)/ώ! for k ≥ ώ + 1. Therefore, the numerator and the denominator of this fraction must be ≥ ώ + 1, since neither ώ nor a prime divisor of k in the numerator simplifies with ώ!. However, if we accept the representation (1 + 1/ώ)^{ώ} for e, the claim is trivial. Note that these two representations give different numbers.⃞

Theorem: The twin prime constant C_{2} is ω-transcendental as the product of (1 - 1/(p - 1)^{2}) over all primes p ∈ ℙ_{>2}.

Proof: By the prime number theorem (see Number Theory), neither the largest nor the second-largest prime number in ℕ - both of which are = ώ - |O(ln ώ)| - divide the denominator of C_{2}.⃞

Theorem: The Landau-Ramanujan constant K is ω-transcendental as the product of (1 - 1/p^{2})^{-½} over all p ∈ ^{ω}ℙ such that p ≡ 3 mod 4 divided by √2.

Proof: By the (Dirichlet) prime number theorem, neither the largest nor the second-largest prime number ≡ 3 mod 4 of ^{ω}ℙ - both of which are = ώ - |O(ln ώ)| - divide the denominator of K.⃞

Theorem: The Glaisher-Kinkelin constant A = 1^{1} 2^{2} 3^{3} ... ώ^{ώ} (1 + 1/ώ)^{ώ³/4}/ώ^{(ώ²/2+ώ/2+1/12)} is ω-transcendental.

Proof: After simplifying, the largest prime number of ^{ω}ℕ remains in the denominator with exponent > 2.⃞

The greatest-prime criterion for ω-transcendental numbers: If a real number r may be represented as an irreducible fraction a/(bp) ± s/t where a, b, s and t are natural numbers, abst ≠ 0 and b + t > 2, and the (second-)greatest prime number p ∈ ^{ω}ℙ, p ∤ a and p ∤ t, then r is ω-transcendental.

Proof: We have that r = (at ± bps)/(bpt) with denominator ≥ 2p ≥ 2ώ - |O(ln ώ)| > ώ + 1 by the prime number theorem.⃞

Theorem: Pi π is ω-transcendental.

Proof: This follows from its Wallis product representation, or its product representation using the gamma function with value -½, provided that we accept these representations. It should be noted that these two representations yield distinct numbers. Alternatively, we can apply the greatest-prime criterion to the Leibniz series, or the Taylor series of arcsin(x) at x = 1.⃞

Theorem: The trigonometric and hyperbolic functions and their inverse functions, the digamma function ψ, the Lambert-W function, the Ein function, the (hyperbolic) sine integral S(h)i, Euler's Beta function B, and, for positive natural numbers s and u and natural numbers t the generalised error function E_{t}, the hypergeometric functions _{0}F_{t}, the Fresnel integrals C and S and the Bessel function I_{t} and the the Bessel function of the first kind J_{t}, the Legendre function χ_{s}, the polygamma function ψ_{s}, the generalised Mittag-Leffler function E_{s,t}, the Dirichlet series Σ f(u)/u^{s} over u with maximally finite rational |f(u)|, the prime zeta function P(s), the polylogarithm Li_{s}, and the Lerch zeta-function Φ(q, s, r) always yield ω-transcendental values for rational arguments and maximal finite rational |q| and |r| at points where their Taylor series converge.

Proof: The claim follows from the greatest-prime criterion, the Dirichlet prime number theorem, and the Wallis product. For the digamma function, the claim follows from the proof of ω-transcendence of Euler's constant below.⃞

Theorem: The gamma function Γ(z) := m!m^{z}/(z(z + 1) … (z + m)), where m = (ώ + 1)^{(ώ+1)²} and z ∈ ^{ω}ℂ \ -^{ω}ℕ, is ω-transcendental for z ∈ ^{ω}ℚ and for suitable supersets of ^{ω}ℕ resp. ^{ω}ℚ.

Proof: The values of Γ(z) are zeros of minimal polynomials or series with infinite integer coefficients.⃞

Theorem: For x ∈ ^{ω}ℝ, let s(x) be the sum of x^{k}/k over all k ∈ ^{ω}ℕ*. If we define Euler's constant as γ = s(1) - ln ώ ∈ ]0, 1[ (since dx/⌊x⌋ - dx/x ≥ 0 and dx/⌊1 + x⌋ - dx/x ≤ 0 for x ∈ ^{ω}ℝ_{≥1} in the integral representation of γ) and accept m s(½) - s(r/2^{m}) as a representation of ln ώ with ώ = 2^{m} - r, m ∈ ^{ω}ℕ and r ∈ [0, 2^{m-1}[ with a precision of O(m/2^{ώ+1}), then γ is infinite rational and therefore ω-transcendental.

Proof: We obtain -ln(1 - x) = s(x) + O(x^{ώ+1}/(1 - x)) + t(x)dx for x ∈ [-1, 1 - 1/c] and a real function t(x) such that |t(x)| < ώ + 1 by (exact) integration (see Nonstandard Analysis) of the geometric series. The claim follows from the greatest-prime criterion after applying Fermat's little theorem to the denominator of the k-th summand of s, for the largest or second-largest prime in ^{ω}ℕ, whose product is greater than m 2^{m-1} + 2^{m} - r by the prime number theorem.⃞

Remark: At all (higher) precisions, the numerators of the summands contain higher powers of two (after cancelling with the denominator where applicable), so the theorem remains valid. This in particular includes t(x)dx at almost any arbitrary precision, by successively reducing γ to a fraction. If we set dx to be the reciprocal of the maximal power of two in this configuration, with an infinite natural exponent for infinite rational t(x), the claim even holds exactly. In any case, the statement is valid for infinitely many levels of infinity of n, and in particular holds from the conventional perspective.

Definition: When two numbers x, y ∈ ^{ω}ℂ* or their reciprocals do not satisfy any polynomial or series equation p(x, y) = 0, so they are called *ω-algebraically independent*.

Theorem: By the greatest-prime criterion, with e = (1 + 1/p)^{p} for maximal p ∈ ^{ω}ℙ and π as Wallis product, pairwise ω-algebraically independent representations of A, C_{2}, γ, e, K and π do exist.⃞

Theorem: The BBP series Σp(k)/(q(k)b^{k}) over k for b, k ∈ ^{ω}ℕ and polynomials resp. series p and q ∈ ^{ω}ℤ with q(k) ≠ 0 and deg(p) < deg(q) only yield ω-transcendental values.

Proof: We can reduce the sum to a smallest common denominator d ≥ b^{m} > ω with d, m ∈ ℕ*.

Remark: Since all real numbers are (approximately) (infinite) rational numbers, we can compute them in real-time.

Approximation theorem for real algebraic numbers: Every real irrational algebraic number of degree k may be approximated by a real ω-algebraic number of degree h < k with an average asymptotic error of π ζ(h + 1)/(2 |^{ω}ℤ|^{h} ln h).

Proof: On the conventionally real axis, the number of ω-algebraic numbers approximately evenly distributed between fixed limits increases by a factor of approximately |^{ω}ℤ| per degree. The error corresponds to the distance between ω-algebraic numbers. The non-real ω-algebraic numbers are less dense.⃞

Conclusion: Two distinct real ω-algebraic numbers have an average distance of at least π/(|^{ω}ℤ|^{ώ} ln(ώ + 1)). Determining this minimum distance exactly requires an infinite non-linear non-convex optimisation problem to be solved. Therefore, the c-algebraic numbers have an approximate order of O(c), disproving Roth's theorem, which essentially amounts to proving the (trivial) minimum distance between two rational numbers, and thus disproves the abc conjecture, but not Liouville's result.

Theorem: The maximum distance between two neighbouring real ω-algebraic numbers ≠ 0 amounts 1/ώ^{2} - |O(1/ώ^{3})|.

Proof: The distance between two real ω-algebraic numbers ≠ 0 is largest around the points ±1. A ω-rational number r > 1 (0 < r < 1) may be better approximated by the real ω-algebraic x that satisfies the polynomial or series equation ώ/r x^{m} - ώx^{m-1} = 1 (x^{m} - ώx = -ώr). An analogous result holds for negative r. If we wish to approximate 1 by a larger real ω-algebraic x, it must necessarily satisfy a polynomial or series equation such that a_{m} = ώ - 1 and a_{0} = -ώ. By setting a_{m} = ώ - 1 and a_{0} = -ώ the claim follows, since the maximum distance can no longer be reduced.⃞

Gelfond-Schneider theorem: For each α ∈ ^{ω}A_{ℂ}\{0, 1} and β ∈ ^{ω}A_{ℂ}\^{ω}ℚ, α^{β} is ω-transcendental.

Proof: In order for α^{β} to be zero of a minimal polynomial or series, since B := ^{ω}A_{ℂ} is a field, we would have to perform the invalid substitution α := ξη^{q/β} with ξ, η ∈ B* and q ∈ ^{ω}ℚ*.⃞

The above theorem and the property that B is a field immediately imply the following result.

Conclusion: As before, given any α, there does not exist any β such that e = α^{β}, which may be seen by comparing the minimal polynomials or series of e (see above) and α^{β}. If there is precisely one γ ∈ B* for any given arbitrary ω-transcendental τ ∈ ^{ω}ℂ such that τ^{γ } is ω-algebraic, then every τ^{δ} such that δ ∈ B\γ^{ω}ℚ is ω-transcendental.⃞

Definition: A rational number ≠ 0 is said to be *power-free* if it cannot be represented as the power of a rational number with integer exponent ≠ ±1.

Theorem: For any power-free q ∈ Q := ℚ_{>0}, we have that r := q^{x} ∈ Q, if and only if x ∈ ^{ω}ℤ and |x| is not excessively large.

Proof: Since there is no contradiction when ±x ∈ ^{ω}ℕ is not excessively large, we can similarly assume wlog that x ∈ Q\^{ω}ℕ. Since this implies q^{x} ∈ ^{ω}A_{R}\Q with R := ℝ_{>0}, we can similarly assume that x ∈ ^{ω}A_{R}\Q. Then q^{x} is ω-transcendental by the Gelfond-Schneider theorem. We can therefore assume that x := a/b ∈ ^{ω}ℝ\^{ω}A_{R} where a, b ∈ ℤ are ω-transcendental. This implies that q^{a/b} ∈ Q\^{ω}ℕ* and thus a = db with d ∈ ℤ, since q is power-free. We deduce the contradiction that q^{d} = q^{x} which proves the claim, since x must be real. ⃞

Remark: These arguments can be extended to finite transcendental numbers by replacing ^{c}ℕ with ^{ω}ℕ and making the required adjustments. Inconcrete transcendence implies finite transcendence.

Remark: The above theorem proves the Alaoglu and Erdős conjecture, which states that, given distinct p, q ∈ ^{c}ℙ, p^{x} and q^{x} are c-rational if and only if x is a c-integer.

© 06.03.2018 by Boris Haase

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