Some Transcendence Results from a Harmless Irrationality Theorem

J. Ana. Num. Theor. 5, No. 2, 91-96 (2017) 91 Journal of Analysis & Number Theory An International Journal

http://dx.doi.org/10.18576/jant/050201

F. M. S. Lima∗ Institute of Physics, University of Bras´ılia, P.O. Box 04455, 70919-970, Bras´ılia-DF, Brazil

Received: 01 Apr. 2017, Revised: 14 Jun. 2017, Accepted: 16 Jun. 2017 Published online: 1 Jul. 2017

Abstract: The arithmetic nature of values of some functions such as sinz, cosz, sinhz, cosh z, ez, and lnz, is a relevant topic in number theory. For instance, all those functions return transcendental values for non-zero algebraic values of z (z = 1 in the case of lnz). On the other hand, not even an irrationality proof is known for some numbers like lnπ, π +e and π e, though it is6 well-known that at least one of the last two numbers is irrational. In this note, we ﬁrst generalize the last result, showing that at least one of the sum and product of any two transcendental numbers is transcendental. We then use this to show that, given any non-null complex number z = 1/e, at least two of the numbers lnz, z + e and ze are transcendental. It is also shown that coshz, sinhz and tanhz return transcendental6 values for all z = r lnt, r Q, r = 0, t being any transcendental number. The analogue for common trigonometric functions is also proved. ∈ 6 Keywords: Irrationality proofs, Transcendental numbers, Trigonometric functions, Hyperbolic functions

∑∞ k! 1 Introduction k=0 1/(2 ) converges to a transcendental number. From the work of Cantor on set theory in 1874, one knows that As usual, let Q denote the set of all rational numbers, the set A is countable whereas the set R of all real i.e. the numbers which can be written as p/q, p and q numbers is uncountable, so ‘almost all’ real numbers are being integers, q = 0. Also, let A denote the set of all transcendental. However, it remained an important 6 algebraic numbers (over Q), i.e. the complex numbers z unsolved problem to prove the transcendence of naturally which are roots of some polynomial equation in Z[z]. All occurring numbers, such as e (the natural logarithm base) other complex numbers — i.e. z A — are called and π (the Archimedes’ constant). Then, in 1873 Hermite 1 6∈ transcendental numbers. Though the existence of proved that er is transcendental for all rational r = 0 (in irrational numbers such as √2 remounts to the ancient particular, e is transcendental) [5]. In 1882, Lindemann6 Greeks, no example of a transcendental number was proved the following extension of Hermite’s result [9]. known at the beginning of the 19th century, which reflects the difficulty of showing that a given number is Lemma 1 (Hermite-Lindemann) The number eα is transcendental. The existence of ‘non-algebraic’ numbers transcendental for all algebraic α = 0. was conjectured by Euler in 1744, in his Introductio in 6 analysin infinitorum, where he claims, without a proof, This implies the transcendence of π, as follows from π that “the logarithms of (rational) numbers which are not Euler’s identity ei = 1, which is equivalent to the − powers of the base are neither rational nor (algebraic) impossibility of squaring the circle with only ruler and irrational, so they should be called transcendental.” A compass, a problem that remained open by more than two such proof appeared only in 1844, when Liouville showed thousand years. Lemma 1 is equivalent to the that any number that has a rapidly converging sequence of transcendence of lnα for all α A , α = 0,1.2 Based ∈ 6 distinct rational approximations must be upon these first results, in 1885 Weierstrass succeeded in transcendental [10]. In particular, he used his proving a much more general result. approximation theorem to show that the series

1 All rational numbers are roots of qz p = 0, thus algebraic. 2 We are interpreting the complex function lnz as its principal Hence all transcendental numbers are irrational.− value, with the argument lying in the interval ( π,π]. −

∗ Corresponding author e-mail: fabio@ﬁs.unb.br c 2017 NSP Natural Sciences Publishing Cor. 92 F. M. S. Lima: Some transcendence results

Lemma 2 (Lindemann-Weierstrass) Given an integer 2 Further transcendence results n > 0, whenever α0,...,αn are distinct algebraic α α numbers, the numbers e 0 ,...,e n are linearly Lemma 3 has a logarithmic version, namely independent over A . That is, for any β ,...,β A not 0 n Lemma 4 (Log version) log α = lnα/lnβ is a all zero, ∈ β n transcendental number whenever α and β are non-zero α ∑ β e k = 0. algebraic numbers, β = 1, and logβ α Q. k 6 6∈ k 0 6 = This form appears, e.g., in Theorem 10.2 of Ref. [13]. For a proof, see, e.g., Theorem 1.4 of Ref. [1] or It has a consequence for tangent arcs, as noted by Theorem 1.8 of Ref. [2]. As an immediate consequence, Margolius in Ref. [11]. when one takes α0 = 0 and β0 = 0, one concludes that 6 Corollary 3 (Margolius) If x is rationaland x = 0, 1, arctan x 6 ± then the number is transcendental. Corollary 1 Given an integer n > 0, being α1,...,αn π distinct non-zero algebraic numbers and β1,...,βn A α θ ∑n β k ∈ Proof. Write x = tan , x Q, x = 0, 1. Then not all zero, k=1 k e is a transcendental number. ∈ 6 ± arctanx θ 1/i ln(z/ z ) iθ θ θ = = · | | From Euler’s formula e± = cos i sin , it follows π π 1/i ln( 1) iθ iθ ± iθ iθ that cosθ = (e + e− )/2 and sinθ = (e e− )/(2i). · − − √ 2 √ 2 Analogously, the basic hyperbolic functions are deﬁned as ln 1/ 1 + x + xi/ 1 + x θ θ θ θ θ θ = ± , (1) cosh := (e +e− )/2 and sinh := (e e− )/2. From ln( 1) Corollary 1, it follows that − − which follows by taking z = 1 + xi in ln(z/ z )= iθ, ± | | Corollary 2 For any algebraic α 0, all numbers which in turn comes from the exponential representation = θ cosα, sinα, coshα, and sinhα are transcendental.6 z = z ei . Clearly, the last expression in Eq. (1) is a ratio of two| | logs with algebraic arguments,6 so Lemma 4 The relevance of investigating the transcendence of applies and θ/π has to be either rational or powers and logarithms of algebraic numbers was transcendental. However, it is irrational because, being acknowledged by Hilbert in his famous lecture r Q, x = tanθ = tan(r π) is rational only when ∈ 7 “Mathematical Problems” in 1900, at the 2nd x = 0, 1, as proved in Corollary 3.12 of Ref. [13]. ± International Congress of Mathematicians [6], being the content of his 7th problem, in which he questioned the π arithmetic nature of ei z for z A .3 Of course, for z = r In particular, it follows that the Plouffe’s constant a rational it was already known∈ that both cos(rπ) and arctan 1 /π is transcendental [11]. Let us extend π 2 sin(rπ) are algebraic, so ei r is algebraic,4 but the case Margolius’ result to all basic trigonometric arcs. of irrational algebraic values of z remained unsolved Hereafter, the word ‘trig’ will stand for any of until 1934, when Gelfond and Schneider, working cos,sin,tan,cot,sec,csc . independently, showed that [4,15] { } Theorem 1 (Extension of Margolius’ result) If x is a arctrig(x) Lemma 3 (Gelfond-Schneider) If α = 0,1 and β Q real algebraic number, then is either a rational 6 α β6∈ π are algebraic numbers, then any value of is or transcendental number. transcendental. Proof. The proof is similar to the previous one, being For a proof, see e.g. Theorem 2.1 of Ref. [2] or enough to take x = trig(θ), x A R, and write Theorem 10.1 (and Sec. 4 of Chap. 10) of Ref. [13]. This ∈ T √2 arctrig(x) θ ln(z/ z ) lemma promptly implies the transcendence of 2 and = = | | , (2) π 2i π π e = i − , two real numbers mentioned by Hilbert [6]. It ln( 1) πβ π β β − also follows that ei = ei = ( 1) is a 6 A α β A − Since is a ﬁeld, then, given any , , all the transcendental number for every β A Q, which solves numbers α β, α β, and α/β (β = 0) are also algebraic∈ (see, 5 ∈ \ Hilbert’s 7th problem. e.g., Sec. 6.6± and Theorem 6.12 of6 Ref. [3]). More generally, given r Q and α A , α = 0, if z is any complex algebraic 3 Hilbert himself remarked that he expected this problem to be (respectively,∈ transcendental)∈ 6 number then all numbers z α, αz, harder than showing the Riemann hypothesis! z/α, and zr are also algebraic (respectively, transcendental),± the 4 It was proved by Lehmer in 1933 that, for rational r = k/n, only exception being z0 = 1 for z A . 6∈ n > 2, the numbers 2cos (2πr) and 2sin(2πr) are algebraic 7 The irrationality of θ/π is also nicely proved by Margolius integers (i.e., roots of monic polynomial equations in Z[x])[7]. in Theorem 3 of Ref. [11] by exploring the properties of 5 Indeed, given a,b A R, if either a = 0 and b Q or sequences of primitive Pythagorean triples formed on writing π T π a = 0, then e(a+bi) = ∈e(b ia)i =( 1)b ia is transcendental.6∈ x = a/b, a and b being distinct non-zero integers. 6 − − −

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z = 0. The choice of z now changes accordingly to the For a proof, see Corollary 2 of Ref. [8]. Note that 6 α πβ function represented by ‘trig’. For arccosx, choose e + is transcendental even if α = 0, as long as iβ Q z = x √1 x2 i. For arcsecx, choose (see Footnote 5). Note also that Corollary 7 implies6∈ the z = 1 ± √x−2 1 i. For arcsinx, choose transcendence of (α + lnβ)/π for any non-zero z = ± √±1 x2 +− xi. For arccscx, choose α,β A . z = √±x2 1− i. For arctanx, choose z = 1 + xi,as in ∈ the previous± − proof.± For arccotx, choose z =± x i. In all All this said, it is embarrassing that the numbers lnπ, these cases, the ratio z/ z is an algebraic function± of x, π + e and π e are still not known to be transcendental. In so it is an algebraic number| | for all x A R. The last fact, not even an irrationality proof is known, though it is expression in Eq. (2) is then a ratio∈ of twoT logs with easy to show that at least one of π + e and π e must be algebraic arguments, so Lemma 4 applies. irrational. This is proved, e.g., in a nice survey on irrational numbers by Ross in Ref. [14], but let us present Conversely, if x R then it will be transcendental a short proof for completeness. Let us call quadratic any ∈ whenever arctrig(x)/π A Q. This implies, e.g., the algebraic number which is a root of a 2nd-order ∈ \ 9 transcendence of any trig √2 π . polynomial equation with rational coefficients. From the fact that π is not a quadratic number (since it is not even The following extension of Lemma 3 was conjectured an algebraic number), it follows that by Gelfond and proved by Baker in 1966, becoming the definitive result in this area. Lemma 6 (Harmless irrationality) At least one of the numbers π + e and π e is irrational. Lemma 5 (Baker) Given non-zero algebraic numbers α1,...,αn such that lnα1,...,lnαn are linearly Proof. Consider the quadratic equation π π independent over Q, then the numbers 1,lnα1,...,lnαn (x ) (x e) = 0, whose roots are and e. By − · − 2 are linearly independent over A . That is, for any expanding the product, one has x (π + e)x + πe = 0. − β0,...,βn A not all zero, we have Assume, towards a contradiction, that both coefficients ∈ π + e and π e are rational numbers. Then, our quadratic n equation would have rational coefficients and both roots β0 + ∑ βk lnαk = 0. π 6 would be quadratic numbers. However, is not a k=1 quadratic number. For a proof, see Theorem 2.1 of Ref. [1].8 This lemma has several interesting consequences. Since this proof does not make use of any property of Corollary 4 Given non-zero algebraic numbers e, it is clear that Lemma 6 can be generalized. α1,...,αn, for any β1,...,βn A the number β α β α ∈ 1 ln 1 + ... + n ln n is either null or transcendental. It Lemma 7 (General irrationality) Given any irrational α α is transcendental when ln 1,...,ln n are linearly number u which is not quadratic and any complex number β β independent over Q and 1,..., n are not all zero. v, at least one of the numbers u + v and uv is irrational. For a proof, see Theorem 2.2 of Ref. [1]. Proof. The proof is identical to the previous one, being enough to substitute π by u and e by v. Corollary 5 Let α1,...,αn,β0,β1,...,βn be non-zero β β βn algebraic numbers. Then the product e 0 α1 1 ... αn is a transcendental number. In particular, this lemma applies when u t is a For a proof, see Theorem 2.3 of Ref. [1]. = transcendental number, so at least one of the numbers t v and tv is irrational. Of course, for any algebraic Corollary 6 For any algebraic numbers α1,...,αn + β β v = 0 both t + v and tv are transcendental numbers,10 so other than 0 or 1, let 1, 1,..., n be algebraic numbers 6 linearly independent over Q. Then the number the interesting case is when v is also a transcendental β1 βn number. This leads us to the following result. α1 ... αn is transcendental. For a proof, see Theorem 2.4 of Ref. [1]. Theorem 2 (Transcendence of sums and products) Given two transcendental numbers t1 and t2, at least one α+πβ Corollary 7 The number e is transcendental for of the numbers t1 +t2 and t1 t2 is transcendental. all algebraic values of α and β, α = 0. 6 9 All rational numbers are quadratic because they are a root of 8 Baker also gave a quantitative lower bound for these linear x(x p/q) = 0. − forms in logs, which had profound consequences for diophantine 10 This basic transcendence rule is easily proved by equations. This work won him a Fields medal in 1970. contradiction.

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r Proof. Given t1,t2 A , consider the quadratic equation Given any transcendental number t, t is 6∈ (x t1)(x t2)= 0, whose roots are t1 and t2. As it is transcendental for all rational r = 0, which can be readily − − 2 6 equivalent to x (t1 +t2)x +t1 t2 = 0, assume, towards a proved by contradiction, writing r = p/q, p and q being − α contradiction, that both s = t1 + t2 and p = t1 t2 are non-zero integers. What about t , α being an irrational algebraic numbers. The equation then reads algebraic? On taking t = eπ , we know that t i A 2 ∈ x sx + p = 0, so, by completing the square, one ﬁnds whereas t√2 A (see Footnote 5). The next theorem − sheds some light6∈ on this question. s2 s2 x2 sx + = p − 4 4 − Theorem 4 (Existence of an irrational algebraic exponent) s 2 s2 Given any transcendental number t, there is an = x = p. (3) α α ⇒ − 2 4 − irrational algebraic such that t is also transcendental. This implies that (x s/2)2 is algebraic (see Footnote 6), − which is impossible because, being x one of the roots t1 Proof. Given any transcendental number t, assume, α β and t2, the number x s/2 must be transcendental. towards a contradiction, that t = is algebraic for all − α A Q. From Corollary 8, tr (β tr) is This theorem implies, in particular, that at least one transcendental∈ \ for all r Q, r = 0, which means− that of π + e and π e is transcendental. However, we are in a tr+α t2r = tr+α (1 tr∈α ) is6 transcendental. Clearly, − − − position to prove a stronger result. α1,2 := r α is an irrational algebraic, so α α ± t 1 (1 t 2)= β1 (1 β2) should also be transcendental, Theorem 3 (Transcendence of two numbers) Given which− is false because− it is the product of two algebraic any non-zero complex number z = 1/e, at least two of the numbers. numbers z + e, ze, and lnz are transcendental.6 Proof. If z = 0 is an algebraic number, then both z + e 6 Note that the similar proposition “for all transcendental and ze are transcendental numbers, so let us restrict our α attention to z A . If lnz is transcendental then we are number t, there is an irrational algebraic α such that t is done because we6∈ know, from Theorem 2, that at least one algebraic” is false, as follows from Lemma 1, taking t = e. of z + e and ze is transcendental. All that remains is to For t = π, however, it remains open the question if there α 12 check whether lnz A implies that both z + e and ze is some algebraic α = 0 for which π is algebraic. α 6 are transcendental numbers.∈ Since lnz = α = z = e , Another consequence of Corollary 8 is as follows. then z + e = eα + e is a transcendental number⇒ for all Theorem 5 (Linear independence of hyperbolic functions) α A , α = 1, according to Corollary 1.11 Also, since For any transcendental number t and any rational lnz∈ A , then6 1 + lnz = ln(ez) is also algebraic, and r = 0, the numbers 1, cosh(r lnt), and sinh(r lnt) are then,∈ according to Lemma 1, the number ze has to be linearly6 independent over A . In particular, both transcendental for all z such that ln(ze) = 0, i.e. z = 1/e. 6 6 cosh(r lnt) and sinh(r lnt) are transcendental numbers. Proof. Since t r is transcendental for any t A and any 6∈ In particular, this theorem implies that at least two of r Q, r = 0, then it follows from Corollary 8 that, for any π π π α∈,β A6 not both zero, + e, e,and ln are transcendental numbers. ∈ β r r r Indeed, we can make suitable choices of t and t in α t + = α t + β t− A 1 2 tr 6∈ Theorem 2 in order to get further transcendence results. r lnt r lnt = α e + β e− A ⇒ 6∈ Corollary 8 For any transcendental number t and = (α + β) cosh(r lnt) + (α β) sinh(r lnt) A . (4) algebraic numbers α and β not both zero, the numbers ⇒ − 6∈ α t + β/t and t (α t) are both transcendental. Since α and β are arbitrary algebraic numbers, then − Proof. For any t A and α,β A , not both zero, α cosh(r lnt)+ β sinh(r lnt) A , (5) α 6∈ β ∈ 6∈ take t1 = t and t2 = /t in Theorem 2. If exactly one e e of α,β is null, then the proof is immediate. Otherwise, where α = α + β and β = α β are also algebraic αβ A − since t1 t2 = (see Footnote 6), then numbers (not both zero), soe ∈ e t1 + t2 = α t + β/t has to be a transcendental number. Finally, for any α A , take t1 = t and t2 = α t in α cosh(r lnt)+ β sinh(r lnt) = γ , γ A . (6) Theorem 2. Since ∈t t α A , then the number− 6 ∀ ∈ 1 + 2 = e e t t = t (α t) has to be transcendental.∈ 12 Note that this question is relevant for the transcendence of 1 2 α − lnπ, because, given non-null α,β A , β = 1, π = β 11 Note that α = 1 implies z = e, a case in which our theorem α lnπ = lnβ is transcendental, according∈ to the6 log-version⇒ of also holds since both z + e = 2e and ze = e2 are transcendental. Lemma 1.

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Therefore, γ + α cosh(r lnt) + β sinh(r lnt) = 0, Theorem 8 (Transcendence of tan(β lnα)) For any − 6 α β α β which shows thate 1, cosh(r lnt) ande sinh(r lnt) are algebraic numbers , , = 0,1 and i Q, the linearly independent over A . number tan(β lnα) is transcendental.6 6∈ The transcendence of cosh(r lnt) follows on taking α = 0 and β = 0 in Eq. (6), whereas that of sinh(r lnt) Proof. Given non-zero algebraic numbers α,β, α = 1, 6 assume, towards a contradiction, that tan(β lnα)= γ6 for followse on takinge α = 0 and β = 0. 6 some γ A . Then e e ∈ iβ lnα iβ lnα sin(β lnα) 1 e e− In addition, it is easy to prove the transcendence of γ = = − cos(β lnα) i eiβ lnα + eiβ lnα tanh(r lnt). αiβ α iβ α2iβ γ − 1 Theorem 6 (Transcendence of tanh r lnt ) For any = i = β − β = β − . (9) ( ) ⇒ αi + α i α2i + 1 transcendental number t and any r Q, r = 0, the − number tanh(r lnt) is transcendental. ∈ 6 The last equality implies that iγ = 1. From Lemma 3, we β 6 know that t = αi is transcendental for all algebraic Proof. For any transcendental number t and any r Q, values of β such that iβ Q, therefore r = 0, we have tanh(r lnt) := sinh(r lnt)/cosh(r ln∈t)= 6∈ r6 r r r 2r 2r (t t− )/(t + t− ) = (t 1)/(t + 1) = 1. Now, t2 1 assume,− towards a contradiction,− that tanh r ln6 t α, for iγ = γ = − ( )= t2 + 1 some α A , α = 0,1. Then e ∈ 6 = γ t2 + γ = t2 1 t2r 1 ⇒ − α = (e1 γ )et2 = 1 + γ 2r − = t + 1 ⇒ − γ γ 2 e1 + 1 +e i = t2r 1 = α t2r + 1 = α t2r + α = t = = , (10) ⇒ − ⇒ 1 γe 1 iγ = (1 α)t2r = α + 1 − − ⇒ − which should bee algebraic since it is a quotient of two 1 + α = t2r = , (7) algebraic numbers. However, this is impossible because ⇒ 1 α 2 A − t is transcendental for all t . which is impossible since the quotient of two algebraic 6∈ numbers is also algebraic, whereas t 2r A . 6∈ Theorems 7 and 8 imply, for instance, that all numbers √ π It follows, in particular, that cosh(lnπ), sinh(lnπ), trig(ln2) and trig 2 are transcendental. and tanh(lnπ) are transcendental numbers. Similar results can be derived for the basic trigonometric functions. 3 Conclusion Theorem 7 (Linear independence of trig values) For any algebraic numbers α,β, α = 0,1 and iβ Q, the Summarizing, we reviewed in this note the main 6 6∈ numbers 1, cos(β lnα) and sin(β lnα) are linearly transcendence results presently known involving the basic independent over A . In particular, both cos(β lnα) and trigonometric and hyperbolic functions, as well as ez and sin(β lnα) are transcendental numbers. lnz. Since not even an irrationality proof is known for some numbers like lnπ, π + e and π e,13 we decided to Proof. Since iβ A Q, then, from Lemma 3, t = αiβ ∈ \ explore a generalization of a well-known ‘harmless’ is a transcendental number. From Corollary 8, the sum irrationality theorem, our Lemma 6, towards the at + b/t is also transcendental for all algebraic a and b derivation of conditional transcendence results for those not both zero, so numbers. Hopefully, the results put forward here in this iβ iβ iβ lnα iβ lnα paper should be useful for those researchers who are aα + bα− = ae + be− A 6∈ investigating the irrationality and/or transcendence of = (a + b) cos(β lnα)+ i(a b) sin(β lnα) A , (8) ⇒ − 6∈ such numbers. which is equivalent to a cos(β lnα)+ b sin(β lnα) = c, 6 13 There is a recent work on modular functions by Nesterenko for all c A . Therefore,e for all a,b eA not both zero, π √n π ∈ ∈ (1996) [12], in which he shows that and e are a cos(β lnα)+ b sin(β lnα) c =e 0,e for all c A . The algebraically independent over Q for all integer n > 0. This − 6 ∈ π π transcendencee ofe cos(β lnα) follows by taking a = 0, b = implies that, for any rational r = 0, q ln = √n p lnr for 6 all non-negative integers p and q6 (not both zero).6 For r−= 1, e.g., 0 and that of sin(β lnα) follows by taking a =e0, b =e0. 6 one concludes that lnπ and √n π are linearly independent over e e Q. In particular, lnπ is not a rational multiple of π.

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Acknowledgement Fabio´ M. S. Lima received the PhD degree The author thanks Mr. Bruno S. S. Lima for all helpful in Physics at University of discussions on classical transcendence results. Bras´ılia in 2003. His research interests in mathematics are in analytic number theory, References particularly the properties of the transcendental functions [1] A. Baker, Transcendental number theory, 2nd ed., related to the Gamma and Cambridge Univ. Press, Cambridge, 1990. Riemann zeta functions, as [2] A. Baker and G. Wüstholz, Logarithmic Forms and well as irrationality and transcendence proofs for Diophantine Geometry, Cambridge Univ. Press, New York, numbers related to e, π, odd zeta values, and Catalan’s 2007. constant G. He has published research articles in reputed [3] P. J. Cameron, Introduction to Algebra, 2nd ed., Oxford international journals of both theoretical physics and Univ. Press, New York, 2008. mathematical sciences. He is referee of a number of [4] A. O. Gelfond, Sur le septième problème de Hilbert, Izv. physics and mathematics journals. Akad. Nauk SSSR 7 (1934), 623–630. [5] Ch. Hermite, Sur la fonction exponentielle, C. R. Math. Acad. Sci. Paris (Sér. A) 77 (1873), 18– 24. [6] D. Hilbert, Mathematical problems, Göttinger Nachrichten (1900), 253–297. See also: Bull. Amer. Math. Soc. 8 (1902), 437–479. [7] D. H. Lehmer, A note on trigonometric algebraic numbers, Amer. Math. Monthly 40 (1933), 165–166. [8] F. M. S. Lima and D. Marques, Some transcendental functions with an empty exceptional set. Available at arXiv: 1004.1668v2 (2010). [9] F. Lindemann, Uber¨ die zahl π, Math. Ann. 20 (1882), 213– 225. [10] J. Liouville, Sur des classes très-étendue de quantités dont la valeur n’est ni algébrique, ni mêne reductibles àdes irrationnelles algébriques, C. R. Math. Acad. Sci. Paris 18 (1844), 883–885. [11] B. H. Margolius, Plouffe’s constant is transcendental. Pre-print (2003). Available at: http://www.plouffe.fr/simon/articles/plouffe.pdf [12] Yu. V. Nesterenko, Modular functions and transcendence questions (Russian), Mat. Sb. 187 (1996), no. 9, 65–96; English transl.: Sb. Math. 187 (1996), 1319–1348. [13] I. Niven, Irrational numbers, 2nd print, The Carus mathematical monographs, MAA, New York, 1963. [14] M. Ross, Irrational thoughts, Math. Gaz. 88 (2000), 68–78. [15] T. Schneider, Transzendenzuntersuchungen periodischer Funktionen: I. Transzendenz von Potenzen; II. Transzendenzeigenschaften elliptischer Funktionen, J. Reine Angew. Math. 172 (1934), 65–74.

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