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2 New Transcendental Numbers

For any non-zero algebraic number α, eα, ln α, sin α, cos α, tan α, csc α, sec α, cot α, sinh α, cosh α, tanh α, and coth α are transcendental numbers (see Theorem 9.11 of [5]). In addi- tion ln α is transcendental number where α is a non-zero algebraic number (see Theorem 9.11 of [5]).

Theorem 1 Let f(x) be one of ex, ln x, sin x, cos x, tan x, csc x, sec x, cot x, sinh x, cosh x, tanh x, and coth x. For g Q[f(x)] and h(x) Q[x], if an equation g = h(x) has a solution α such ∈ ∈ that g(f(α)) = 0 and h(α) = 0, then α is a transcendental number. 6 6 Proof. Let α be a solution of the equation g = h(x) in the theorem. If α is transcendental, then there is nothing to prove. Assume that α is algebraic. Then h(α) is algebraic. However g(f(α)) is transcendental. We have a contradiction. Thus we have proven the theorem. ✷

Theorem 2 Let f(x) be one of ex, ln x, sin x, cos x, tan x, csc x, sec x, cot x, sinh x, cosh x, tanh x, and coth x. For a given equation h1(x)f(g(x)) = h2(x), if the equation has a non-zero solution α such that f(g(α)) = 0 and h1(α) = 0 = h2(α) with h1(x) and h2(x) are relatively 6 6 6 prime, then the solution is a transcendental number where g(x), h1(x), h2(x) Q[x]. ∈ Proof. Since every is either an algebraic number or a transcendental num- ber, let us assume that a solution α of the equation of the theorem is algebraic. Then h2(α) is algebraic. Since α is algebraic, h1(α) and g(α) are algebraic. However f(g(α)) is transcen- dental. This gives a contradiction. Thus α is transcendental. ✷

Corollary 1 Let arc sin x = f(x), arc cos x = f(x), arc tan x = f(x), arc cot x = f(x), arc sec x = f(x), arc csc x = f(x) be given well-defined equations where f(x) Q[x]. If one ∈ of those equations has a non-zero solution α such that f(α) = 0, then the solution is a transcendental number. 6

Proof. The proof of the corollary is straightforward by Theorem 1. ✷

Note 1. The real solution of e = x 4 is x = e + 4 which is a transcendental number. The real −

2 number solution of ex + x 12 = 0 is a transcendental number, that is 2.27472787147 . − · · · The equation ex + x 12 = 0 has countably many complex number solutions. The equation − πx + 4x = 49 has a real solution which is the transcendental number x = 3.14097 . The 2 2 · · · following is Thomas’ example: the equation 2(cos x) cos x 1= x x has solutions 0 and − − − 0.4177913944 which is a transcendental number [1]. ✷ · · · j Proposition 1 Let f(x) be one of ex , sin x, cos x, tan x, csc x, sec x, cot x, sinh x, cosh x, tanh x, k and coth x where j is a fixed positive integer. For a given equation (f(x)+ a1) = g(x), if the equation has a real solution α such that f(α) = 0, f(α)+ a1 = 0, and g(α) = 0, then the 6 6 6 solution is a transcendental number where g(x) Q[x], a1 is a given algebraic number, and ∈ k is a given positive integer.

Proof. Let α be the solution of the equation of the proposition. Since √k α is algebraic, the proof of the proposition is straightforward by the proof of Theorem 1. ✷

j Proposition 2 Let f(x) be one of ex , sin x, cos x, tan x, csc x, sec x, cot x, sinh x, cosh x, tanh x, and coth x. For any g Q[f(x)], if 1 deg(g) 4 and h(x) Q[x], if an equation g = h(x) ∈ ≤ ≤ ∈ has a non-zero solution α such that g(f(α)) = 0 and h(α) = 0, then α is a transcendental number. 6 6

Proof. Without loss of generality, suppose f(x) = sin x. Let α be a non-zero solution of the equation g(sin x) = h(x) in the proposition. If α is a transcendental number, then there is nothing to prove. If f((sin x)) is linear, quadratic, cubic, or quartic, then by the solution of a linear, the quadratic, Cardano’s, or Ferarri’s formulas. respectively, we have that the left side sin α is a transcendental number and the right side of the solution is algebraic. This contradiction shows that α is a transcendental number. ✷

Note 2. By Proposition 1, we know that for algebraic number sinh α and cosh α are tran- scendental numbers as well. A solution of xex = x + 12 is between 1.7 and 1.8. − The equation ex x+7 = 0 has no real solution, but it has countably many complex transcen- − dental number solutions. One of the complex number solutions of the equation ex x+7=0 − is x = 1.7701 + 2.669613 i and its complex conjugate x = 1.7701 2.669613 i are · · · · · · ···− · · · transcendental solutions of the equation which has the minimal modulus of all the complex solutions of the equation. The complex solutions of ex x + 7 = 0 are discrete. The equation a1xj k − (e + a2) = f(x) has sometimes countably complex number solutions. ✷

Proposition 3 Let F be a field which is generated by all the algebraic numbers. For any f(x) F [x] and for any transcendental number τ, f(τ) is a transcendental number where ∈ ±1 f(x) / Q[x]. Furthermore for any f(x) F [x ] and for any transcendental number τ, f(τ) is a transcendental∈ number. ∈

Proof. For the proof of the proposition it is enough to prove for the Laurent polynomial ±1 n −1 −m f(x) of F [x ]. Let us put f(x) = anx + + a1x + a0 + a−1x + + a−mx where · · · · · · an, , a−m F , at least one of an, , a1, a−1, , a−m is not zero, and n, ,m are · · · ∈ · · · n· · · −1 · · · −m positive integers or zeros. We have that f(τ)= anτ + +a1τ +a0 +a−1τ + +a−mτ n · · · −1 · ·− ·m and set f(τ) equals to α that is α = anτ + + a1τ + a0 + a−1τ + + a−mτ . If α is a · · · · · ·

3 transcendental number, then there is nothing to prove. Let us assume that α is algebraic. We n+m m+1 m m−1 m have that anτ + + a1τ + a0τ + a−1τ + + a−m = ατ . Since an, , a−m, α · · · · · · · · · are algebraic. This implies that τ is algebraic. We have a contradiction. Thus f(τ) is a transcendental number. ✷

Corollary 2 If τ is a transcendental number and α0, , α4 are algebraic numbers, then 4 3 2 3 · · · 2 −1 f1(τ)= α4τ + α3τ + α2τ + α1τ + α0, f2(τ)= α4τ + α3τ + α2τ + α1τ + α0, f3(τ)= 2 1 −1 −2 −1 −2 −3 −1 α4τ + α3τ + α2τ + α1τ + α0, f4(τ)= α4τ + α3τ + α2τ + α1τ + α0, and α4τ + −2 −3 −4 α3τ + α2τ3 + α1τ + α0 are transcendental numbers where at least one of α1, , α4 is not zero. · · ·

Proof. The proof of the corollary is straightforward by Proposition 3 and the another proof of the corollary is the following. If β is the number of one of the equations of the proposi- tion, then since α0, α1, α2, α3, α4 are algebraic, by Ferrari’s quartic formula, a solution of the equation fi(τ), 1 i 4, is algebraic. But τ is transcendental, thus we have a contradiction. ≤ ≤ So β is a transcendental number. ✷

Corollary 3 If a, b, c, and α are given algebraic numbers, then a(cos α)2+b sin α+c, a sin α2+ b cos α + c, a(tan α)2 + b sec α + c, and a(cot α)2 + b csc α + c are transcendental numbers where at least one of a and b is not zero, and a(cos α)2 + b sin α + c, a sin α2 + b cos α + c, a(tan α)2 + b sec α + c, and a(cot α)2 + b csc α + c are not zeros.

Proof. The proof of the corollary is straightforward by Proposition 3 and the elementary trigonometric identities. ✷

Corollary 4 For any element l of the group [τ]±∞, l is a transcendental number where l / Q ∗ ∈ and τ is a given transcendental number. For any non-zero element l of the group [τ]±∞, l is a transcendental number.

Proof. The proof of the corollary is straightforward by Proposition 3. ✷

Proposition 4 For any two different integers i and j, and a given transcendental number τ, i j k τ τ = φ and k∈Nτ <τ>n,m hold. The cardinality of is uncountable. Furthermore, ∩ ∪ ⊂ T the set τ = τ + α α is an algebraic number is dense in R. { | } Proof. Let τ be the transcendental number of the proposition. Since τ i τ j is a transcendental − number, τ i and τ j are disjoint. For any element l of τ k, l = τ k + r where r is an algebraic number. This implies that l is an element of <τ>n,m. For the second statement of the proposition, for any τ1 τ , we have that τ1 = τ + α for an algebraic number α. Since ∈ ∈ T is uncountable and τ is countable, is uncountable. For the density of τ, for any given T T 1 τ1 of τ and given ǫ> 0, there is a positive integer n such that n <ǫ. This implies that the 1 transcendental number τ1 + of τ is an element (τ ǫ,τ + ǫ). This implies that τ is dense n − in R. This completes the proof of the proposition. ✷

Proposition 5 For any given transcendental numbers τ1 and τ2, τ1 + τ2i and τ1 τ2i are − transcendental numbers. Specifically, e + πi and e πi are transcendental numbers. −

4 Proof. Symmetrically let us assume that τ1 + τ2i is an algebraic number. This implies that τ1 τ2i is algebraic. Assume these algebraic numbers to be α1 and α2 respectively. By − adding them, we have that 2τ1 = α1 + α2. This implies that 2τ1 is a transcendental number but α1 + α2 is an algebraic number. We have a contradiction. This implies that τ1 + τ2i is a transcendental number. Symmetrically τ1 τ2i is a transcendental number. This completes − the proofs of the proposition. ✷

Corollary 5 All most complex numbers of the complex number plane C are transcendental numbers as R.

Proof. The proof of the corollary is straightforward by Proposition 5. ✷

As a result of Proposition 5, for any given real transcendental numbers τ1 and τ2, at least one of τ1 + τ2 or τ1 τ2 is a transcendental number. It is an interesting problem for given − non-zero integers i and j, whether both of πi + ej and πi ej are transcendental numbers or − not.

∗ Proposition 6 For a given transcendental number, the additive groups [τ]n, [τ]n, [τ]n,m, ∗ [τ]n,m, [τ]∞, and [τ]±∞ are Z-modules and they are Q-modules.

We have the following obvious results.

Corollary 6 For a given transcendental number τ, if n1 = n2 or m1 = m2, then the vector 6 6 spaces [τ]n1,m1 and [τ]n1,m1 over Q are not isomorphic. For every element l of [τ]n1,m1 , l is a transcendental number where l / Q. ∈ Proof. The proofs of the corollary is straightforward by Proposition 3. ✷

Proposition 7 For a given transcendental number τ, the rings <τ>∞ (resp. <τ>±∞) and the polynomial ring Q[x] (resp. the Laurent extension Q[x±1] of the polynomial ring Q[x]) are isomorphic. Consequently the polynomial ring Q[x] and the Laurent extension Q[x±1] are integral domains.

Proof. The proof of the proposition is obvious, so let us omit it. ✷

Open question: For given transcendental numbers τi, 1 i 4, is the number τ1 + τ1i + τ2j + τ3k a tran- ≤ ≤ scendental number in the quaternion Q (see [2])?

Acknowledgement: The Authors thank you Thomas Preu for his reviewing in the theorems and corollaries of the manuscript and his interesting examples for improving the manuscript.

5 References

[1] A. Baker, Transcendental Number Theory, Springer-Verlag, 1982.

[2] Samuel Eilenberg and Ivan Niven, The fundamental theorem of algebra for quaternions, Bull. Amer. Math. Soc. 50(4): 246-248 (April 1944).

[3] Jongwoo Lee, Xueqing Chen, Seul Hee Choi, and Ki-Bong Nam, Automorphism groups of some stable Lie algebras, Journal of Lie Theory, Heldermann Verlag, Vol. 21, 2011, 457-468.

[4] William J. LeVeque, Fundamentals of Number Theory, Cambridge University Press, Lon- don, 1975.

[5] Ivan Niven, Irrational Numbers, The Carus Mathematical Monographs, Number 12, 1967.

[6] Thomas Preu, Personal Communications, 2021.

[7] A. B. Shidlovskii, Transcendental Numbers, Translated from Russian by N. Koblitz, Wal- ter de Gruyter, Berlin New York, 1989.

[8] Ian Stewart, Algebraic Number Theory, Chapman and Hall, 1979.

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