1. Introduction
In 1900, in the International Congress of Mathematicians (in Paris), Hilbert provided a list of 23 problems and his seventh problem was about the arithmetic nature of the power
of two algebraic numbers
and
. In 1934–1935, Gelfond and Schneider [
1] (p. 9) (independently) completely solved this problem: if
and
are algebraic numbers, with
or 1, and
irrational. Then
is transcendental. This outstanding result is called the Gelfond–Schneider theorem.
As immediate consequences, we have the transcendence of the numbers , and .
The Gelfond–Schneider theorem classifies the (arithmetic) nature of
, for algebraic numbers
x and
y (because
is algebraic if
y is rational). However, when
, one may has all possibilities (see
Table 1).
The case
may be of more interest: is it possible
to be algebraic, for some transcendental number
T? We point out that a negative answer for this question is expected (but still unproved), since the numbers
, and
are expected to be transcendental. However, related to the previous question, Marques and Sondow [
2] showed that its answer is Yes. In fact, they proved that the set of algebraic numbers of the form
(where
T runs over all transcendental numbers) is dense in the interval
.
In this direction, it may be interesting to consider the following generalization of the previous question: given arbitrary non-constant polynomials
, is there always a transcendental number
T, such that
is algebraic? The answer for this question is also Yes and it was given by Marques [
3]: the set of algebraic numbers of the form
, with
T transcendental, is dense in some connected subset of
or
. See [
4,
5] for generalizations of this result and some new results in [
6,
7].
We still point out that, in 2020, Trojovský [
8] proved some results related to the transcendence of numbers of the form
for positive integers
n and complex numbers
.
In this paper, we continue this program by studying the existence of algebraic numbers of the form , where is algebraic and T is transcendental. More precisely, our main result is the following:
Theorem 1. Let be a positive algebraic number and let be a non-constant rational function. Then the set of algebraic numbers of the form , with T transcendental, is dense in some open interval of .
Example 1. In particular, for and , we have that the previous result implies in the existence of an open interval such that the set of algebraic numbers of the form with T transcendental, is dense in I. As we shall see in the next section, it is possible to make this interval explicit (if α and are previously given). In the case of and , we infer that .
Example 2. We remark that the transcendence of still remains as an open problem. However, as an application of the previous example, we have the existence of a sequence of transcendental numbers, such that , for all , and tends to , as .
Remark 1. We recall that a rational function is the quotient of two polynomials, say (with no common zeros). This function is defined in the whole complex plane but the set of zeroes of (which are poles of ). The coefficients of are defined as the coefficients of and .
A well-known generalization of the notion of an algebraic number is a “like-algebraic” tuple: the complex numbers are said to be algebraically dependent if there exists a nonzero polynomial such that . Otherwise, are called algebraically independent (in particular, they are all transcendental numbers).
A major open problem in transcendental number theory is a conjecture raised by Schanuel (in the 1960s) during a course at Yale given by Lang [
9] (pp. 30–31).
Conjecture 1 (Schanuel’s conjecture (SC)).
Let be complex numbers which are -linearly independent. Then there are at least n algebraically independent numbers among We remark that Schanuel’s conjecture is proved only for : Lindemann’s theorem asserts that is a transcendental number, for any non-zero algebraic number . As an immediate consequence, we deduce that is transcendental for any (since is algebraic).
There are many research topics on the deep consequences of the veracity of the Schanuel’s conjecture (for some of them, we refer the reader to [
10,
11] and references therein). In the opposite direction of Theorem 1, here, we still prove the following conditional result:
Theorem 2. Assume Schanuel’s conjecture (SC) and let be a non-constant rational function. If is an algebraic number, then is transcendental.
Example 3. We point out that even the transcendence of 2e is still unproved. Thus, the previous result gives (conditionally) the transcendence of this number (for the choice of and ).
We point out that the above fact is an open problem if (SC) is not assumed to be true (even for a particular rational function ). The number e can be replaced by some other known transcendental numbers, as etc.
2. Proof of the Theorem 1
Before proceeding further, we shall provide a field-theoretical result which will be an essential ingredient in the proof.
We know that the set of the real algebraic numbers is dense in
(because
). Indeed, we know that the set
of the
m-degree real algebraic numbers is dense in
, for all
, see [
12] (Theorem 7.4). Before the proof of the theorem, we shall need a slightly stronger fact. More precisely,
Lemma 1. For any , the setis dense in . Proof. Let
p be a prime number. It suffices to prove that the set
because clearly
is dense in
. First, note that
is an
m-degree algebraic number, for any non-zero rational number
Q. In fact,
is a root of
which is irreducible over
(since
is irreducible by the Eisenstein’s irreducibility criterion).
Towards a contradiction, we suppose the existence of an integer number
such that
is a rational number. Thus, so is
. We know that
is a basis of the field extension
. By the Binomial Theorem,
. If
, then
, by the
-linear independence of
. When
, we get
We then rewrite the second summatory as
If
, the result follows again by the
-linear independence of
. Otherwise, we have
and so
Repeating this procedure
times, we arrive at
where
and we used that
. Thus
, again because
is linearly independent over
. □
Now, we are ready to deal with the proof of theorem.
Proof. Set and let us consider an open interval such that , for all . Now, the function given by is well-defined and analytical in . We claim that is a non-constant function. Indeed, on the contrary, , for all . However, this would imply that is constant (contradicting the hypothesis) since . Thus, there exists with . Therefore, by the Inverse Function Theorem, there are open connected intervals and (with ) such that is a diffeomorphism (i.e., a differentiable function which has a differentiable inverse). In particular, is an open map (i.e., it maps open sets to open sets). In particular, is an open interval of .
Set
as the union of all sets
, where
m runs over all positive integers coprime to
(the degree of
over
). That is,
By Lemma 1, the set is dense in and therefore is dense in the interval . Now, it suffices to prove that every element of this intersection has the desired form. In fact, if , then , for some (since ). We claim that T is a transcendental number. Indeed, assume that T is algebraic. Since , we use the Gelfond–Schneider theorem to conclude that is a rational number, say with . Thus, we can write . Hence divides (since ). Since and , we conclude that yielding that is a rational number which contradicts the choice of . This contradiction implies that T is transcendental and the proof is complete. □