The transcendence of \( e \) (Hermite, 1873) and \( \pi \) (Lindemann, 1882) were landmark achievements, but they were only the beginning. The 20th century brought two major breakthroughs: the Gelfond-Schneider theorem (1934), which resolved Hilbert’s seventh problem, and Baker’s theorem (1966), which generalized it into a tool with far-reaching applications. Yet despite these advances, some of the most natural questions about transcendental numbers remain stubbornly open.
This lesson surveys the modern landscape of transcendental number theory: what we know, what we can prove, and what remains tantalizingly out of reach.
Hilbert’s Seventh Problem
At the International Congress of Mathematicians in 1900, David Hilbert posed 23 problems that would shape the course of 20th-century mathematics. His seventh problem asked:
Is \( \alpha^\beta \) transcendental when \( \alpha \) is algebraic (\( \neq 0, 1 \)) and \( \beta \) is algebraic irrational?
Hilbert considered this problem so difficult that he expected it to be solved after the Riemann Hypothesis. In fact, it was resolved independently in 1934 by Aleksandr Gelfond and Theodor Schneider, well before the Riemann Hypothesis (which remains open to this day).
The Gelfond-Schneider Theorem
Statement
Theorem (Gelfond-Schneider, 1934). If \( \alpha \) is an algebraic number with \( \alpha \neq 0 \) and \( \alpha \neq 1 \), and \( \beta \) is an algebraic irrational number, then \( \alpha^\beta \) is transcendental.
Here \( \alpha^\beta = e^{\beta \ln \alpha} \) for any branch of the logarithm; the theorem holds regardless of the branch chosen.
Spectacular Consequences
Important: The Gelfond-Schneider theorem immediately implies the transcendence of:
- \( 2^{\sqrt{2}} \), the **Gelfond-Schneider constant** (\( \alpha = 2 \), \( \beta = \sqrt{2} \)).
- \( e^{\pi} = (-1)^{-i} \), **Gelfond’s constant** (\( \alpha = -1 \), \( \beta = -i \), both algebraic, \( -i \) is irrational).
- \( i^i = e^{-\pi/2} \), since \( \alpha = i \) (algebraic, \( \neq 0,1 \)) and \( \beta = i \) (algebraic irrational).
- \( \sqrt{2}^{\sqrt{2}} \) (\( \alpha = \sqrt{2} \), \( \beta = \sqrt{2} \)).
- \( 2^{\sqrt[3]{3}} \) (\( \alpha = 2 \), \( \beta = \sqrt[3]{3} \)).
Example (Gelfond’s constant). Gelfond’s constant \( e^\pi \approx 23.14069584\ldots \) is transcendental. This is remarkable because \( e \) and \( \pi \) are both transcendental, yet neither \( e + \pi \) nor \( e\pi \) is known to be transcendental. The specific combination \( e^\pi \), however, is provably transcendental via the Gelfond-Schneider theorem applied to \( (-1)^{-i} \).
Proof Outline
The proof of the Gelfond-Schneider theorem uses a fundamentally different technique from Hermite-Lindemann. We outline the main ideas.
Stage 1: Constructing an auxiliary function.
Consider the function
$$ \Phi(z) = \sum_{j=0}^{J}\sum_{k=0}^{K} a_{jk}\, e^{(j + k\beta)z\ln\alpha}, $$
where the \( a_{jk} \) are integers not all zero, to be determined. The parameters \( J \) and \( K \) are chosen so that \( (J+1)(K+1) \) exceeds the number of conditions we impose.
Stage 2: Forcing many zeros.
We require \( \Phi(z) \) to vanish to high order at \( z = 0, 1, 2, \ldots, L \). The values \( \Phi(m) \) for \( m \in \mathbb{N} \) are algebraic (since \( \alpha^m \) and \( (\alpha^\beta)^m \) are algebraic by hypothesis). Siegel’s lemma guarantees an integer solution \( a_{jk} \) with controlled size.
Stage 3: Extrapolation.
An analytic argument (using the maximum modulus principle) shows that if \( \Phi \) vanishes to high order at enough integer points, it must vanish to even higher order. Iterating, one shows \( \Phi \) vanishes identically.
Stage 4: Contradiction.
But \( \Phi \equiv 0 \) implies a nontrivial linear relation among the exponentials \( e^{(j+k\beta)z\ln\alpha} \). Since \( \beta \) is irrational, the exponents \( j + k\beta \) are distinct, and the exponential functions are linearly independent, giving a contradiction. \( \blacksquare \)
This proof introduced the auxiliary function method (sometimes called the Gelfond-Schneider method), which became a central tool in transcendental number theory.
Baker’s Theorem
The Breakthrough
Alan Baker dramatically generalized the Gelfond-Schneider theorem in 1966, earning the Fields Medal in 1970.
Theorem (Baker, 1966). Let \( \alpha_1, \alpha_2, \ldots, \alpha_n \) be nonzero algebraic numbers. If \( \ln\alpha_1, \ln\alpha_2, \ldots, \ln\alpha_n \) are linearly independent over \( \mathbb{Q} \), then \( 1, \ln\alpha_1, \ln\alpha_2, \ldots, \ln\alpha_n \) are linearly independent over the algebraic numbers.
In other words, if \( \beta_0, \beta_1, \ldots, \beta_n \) are algebraic numbers with
$$ \beta_0 + \beta_1\ln\alpha_1 + \beta_2\ln\alpha_2 + \cdots + \beta_n\ln\alpha_n = 0, $$
then \( \beta_0 = \beta_1 = \cdots = \beta_n = 0 \).
Consequences of Baker’s Theorem
Baker’s theorem implies:
(i) The Gelfond-Schneider theorem (take \( n = 2 \)).
(ii) Any nonzero linear combination \( \beta_1\ln\alpha_1 + \cdots + \beta_n\ln\alpha_n \) with algebraic \( \alpha_i, \beta_j \) is either zero or transcendental.
(iii) If \( \alpha_1^{\beta_1}\alpha_2^{\beta_2}\cdots\alpha_n^{\beta_n} = 1 \) with algebraic \( \alpha_i \) and \( \beta_i \), then either all exponents are zero or the \( \ln\alpha_i \) are \( \mathbb{Q} \)-linearly dependent.
Effective Bounds
A crucial feature of Baker’s theorem, distinguishing it from earlier results, is that it is effective: Baker provided explicit lower bounds for \( |\beta_1\ln\alpha_1 + \cdots + \beta_n\ln\alpha_n| \) in terms of the heights of the \( \alpha_i \) and \( \beta_j \):
$$ |\beta_1\ln\alpha_1 + \cdots + \beta_n\ln\alpha_n| > B^{-C} $$
for a computable constant \( C \) depending on \( n \) and the \( \alpha_i \), where \( B = \max(|\beta_1|, \ldots, |\beta_n|) \).
This effectiveness has profound applications to Diophantine equations, because it allows one to bound the size of solutions rather than merely proving their finiteness.
Application: The Catalan Conjecture
Theorem (Mihailescu, 2002). The only solution in integers \( x, y \geq 1 \) and \( p, q \geq 2 \) to
$$ x^p – y^q = 1 $$
is \( 3^2 – 2^3 = 1 \).
This was conjectured by Catalan in 1844 and proved by Preda Mihailescu using techniques from the theory of cyclotomic fields, building on ideas related to Baker’s bounds on linear forms in logarithms. Baker’s theorem was used in earlier partial results that limited the possible size of solutions.
What We Don’t Know
Despite over 150 years of progress, some of the most natural questions remain wide open.
The \( e + \pi \) Problem
Proposition. At least one of \( e + \pi \) and \( e\pi \) is transcendental.
Proof. Consider the quadratic \( t^2 – (e+\pi)t + e\pi = (t-e)(t-\pi) = 0 \). If both \( e + \pi \) and \( e\pi \) were algebraic, this would be a polynomial with algebraic coefficients, and its roots \( e \) and \( \pi \) would be algebraic, contradicting the known transcendence of \( e \) (or \( \pi \)). \( \blacksquare \)
Important: Despite the elegant argument above, we cannot determine which of \( e + \pi \) and \( e\pi \) is transcendental, or whether both are. Most experts believe both are transcendental, but no proof is known for either individually. This is one of the most embarrassing gaps in our understanding of basic constants.
Euler’s Constant
Definition. The Euler-Mascheroni constant is
$$ \gamma = \lim_{n\to\infty}\left(\sum_{k=1}^n \frac{1}{k} – \ln n\right) \approx 0.5772156649\ldots $$
Despite being one of the most important constants in analysis and number theory, we do not even know whether \( \gamma \) is rational or irrational, let alone transcendental. This is widely regarded as one of the most important open problems in number theory.
A Catalog of Unknown Cases
The following are all open:
- Is \( e^e \) transcendental? Almost certainly, but unproven.
- Is \( \pi^e \) transcendental? Unknown.
- Is \( \pi^\pi \) transcendental? Unknown.
- Is \( 2^e \) transcendental? Unknown (the Gelfond-Schneider theorem requires the exponent to be algebraic irrational, and \( e \) is transcendental).
- Is \( \ln\pi \) transcendental? Almost certainly, but unproven.
- Is \( \pi + e^\pi \) transcendental? Unknown.
In contrast, the following are known to be transcendental: \( e \), \( \pi \), \( e^\pi \), \( \sqrt{2}^{\sqrt{2}} \), \( \ln 2 \), \( \sin(1) \), \( \cos(1) \), \( \Gamma(1/4) \), and \( e^{\pi\sqrt{163}} \) (though the last is very close to an integer: \( e^{\pi\sqrt{163}} \approx 640320^3 + 744 – 7.5\times 10^{-13} \)).
Schanuel’s Conjecture
The “master conjecture” of transcendental number theory, if true, would resolve most of the open questions above.
Statement
Conjecture (Schanuel). If \( z_1, z_2, \ldots, z_n \) are complex numbers linearly independent over \( \mathbb{Q} \), then the transcendence degree of
$$ \mathbb{Q}(z_1, z_2, \ldots, z_n, e^{z_1}, e^{z_2}, \ldots, e^{z_n}) $$
over \( \mathbb{Q} \) is at least \( n \).
In less technical language: given \( n \) numbers that are “independent” over the rationals, the \( 2n \) quantities \( z_1, \ldots, z_n, e^{z_1}, \ldots, e^{z_n} \) together have at least \( n \) “algebraically independent” members. They cannot all be bound together by algebraic relations.
What Schanuel Would Imply
If Schanuel’s conjecture is true:
(i) \( e \) and \( \pi \) are algebraically independent (hence both \( e + \pi \) and \( e\pi \) are transcendental).
(ii) \( e^e \) is transcendental.
(iii) \( \pi^e \), \( \pi^\pi \), \( e^{e^e} \) are all transcendental.
(iv) The Lindemann-Weierstrass theorem (already known unconditionally).
(v) The Gelfond-Schneider theorem (already known unconditionally).
Proof of (i). Take \( n=2 \), \( z_1 = 1 \), \( z_2 = i\pi \). These are \( \mathbb{Q} \)-linearly independent (if \( a + bi\pi = 0 \) for rationals \( a,b \) not both zero, then \( \pi \) would be rational, contradiction). By Schanuel’s conjecture, the transcendence degree of \( \mathbb{Q}(1, i\pi, e, e^{i\pi}) = \mathbb{Q}(i\pi, e, -1) = \mathbb{Q}(\pi, e) \) over \( \mathbb{Q} \) is at least 2. Therefore \( e \) and \( \pi \) are algebraically independent. \( \blacksquare \)
Status
Only very special cases of Schanuel’s conjecture have been proved: the Lindemann-Weierstrass theorem corresponds to the case where the \( z_i \) are algebraic, and Baker’s theorem handles certain linear combinations. The full conjecture remains far out of reach of current techniques.
Irrationality Measures Revisited
The Connection to Transcendence
Recall that the irrationality measure of \( \alpha \) is
$$ \mu(\alpha) = \inf\left\{\mu : \left|\alpha – \frac{p}{q}\right| > \frac{1}{q^\mu} \text{ for all but finitely many } \frac{p}{q}\right\}. $$
By Roth’s theorem (1955), every algebraic irrational has \( \mu(\alpha) = 2 \). Liouville numbers have \( \mu = \infty \). Most transcendental numbers we care about sit at or near \( \mu = 2 \).
Known Values and Bounds
- Number: Any rational | Irrationality measure: \( \mu = 1 \) | Notes: Exact
- Number: Any algebraic irrational | Irrationality measure: \( \mu = 2 \) | Notes: Roth’s theorem (exact)
- Number: \( e \) | Irrationality measure: \( \mu = 2 \) | Notes: Exact, despite being transcendental
- Number: \( \pi \) | Irrationality measure: \( \mu \leq 7.103 \) | Notes: Best known bound (Zeilberger-Zudilin, 2020)
- Number: \( \ln 2 \) | Irrationality measure: \( \mu \leq 3.574 \) | Notes: Marcovecchio, 2009
- Number: Liouville numbers | Irrationality measure: \( \mu = \infty \) | Notes: By definition
The fact that \( \mu(e) = 2 \) exactly, the same as algebraic irrationals, is remarkable. It shows that irrationality measure alone cannot distinguish \( e \) from algebraic numbers. The deeper structure of transcendence requires the machinery of Hermite, Lindemann, and their successors.
The Evolution of Transcendence Theory
The progression of major results traces a beautiful arc:
- Year: 1844 | Result: First transcendental number | Mathematician: Liouville | Method: Rational approximation
- Year: 1873 | Result: \( e \) is transcendental | Mathematician: Hermite | Method: Auxiliary polynomial + integration
- Year: 1882 | Result: \( \pi \) is transcendental | Mathematician: Lindemann | Method: Generalized Hermite’s method
- Year: 1885 | Result: Lindemann-Weierstrass theorem | Mathematician: Weierstrass | Method: Full generalization
- Year: 1934 | Result: \( \alpha^\beta \) transcendental | Mathematician: Gelfond, Schneider | Method: Auxiliary function method
- Year: 1955 | Result: Optimal approximation bound | Mathematician: Roth | Method: Irrationality measure = 2
- Year: 1966 | Result: Linear independence of logarithms | Mathematician: Baker | Method: Effective bounds
- Year: 2002 | Result: Catalan conjecture (\( x^p – y^q = 1 \)) | Mathematician: Mihailescu | Method: Cyclotomic fields + Baker
The Afterword: Numbers That Escape Algebra
The story of transcendental numbers is a story about the limits of algebra. Algebraic numbers, roots of polynomials with integer coefficients, are the numbers that algebra can “name.” Transcendental numbers escape this naming; they live outside every polynomial equation, no matter how complicated.
The proofs reveal a beautiful pattern. To show a number is transcendental, we assume it is algebraic and then construct an integer that is simultaneously nonzero and arbitrarily small, a contradiction. The art lies in the construction: Liouville used rational approximation, Hermite and Lindemann used auxiliary polynomials integrated against exponentials, Gelfond and Schneider used auxiliary functions with many zeros.
What remains unknown is humbling. We cannot tell whether \( e + \pi \) is rational, much less transcendental. Euler’s constant \( \gamma \) resists even the question of irrationality. Schanuel’s conjecture, if proved, would resolve these questions at a stroke, but it seems far beyond current techniques.
Perhaps the deepest lesson is this: almost all numbers are transcendental, yet we can barely recognize them.
Exercises
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Use the Gelfond-Schneider theorem to prove that \( 2^{\sqrt{3}} \) is transcendental.
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Prove that \( e^\pi \) is transcendental by expressing it as \( (-1)^{-i} \) and applying Gelfond-Schneider. Carefully verify that all hypotheses of the theorem are satisfied.
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Is \( (\sqrt{2})^{\sqrt{2} \cdot \sqrt{3}} = (\sqrt{2})^{\sqrt{6}} \) transcendental? Prove your answer.
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Explain why the Gelfond-Schneider theorem does not apply to \( 2^e \). (What condition fails?)
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Verify the proof that at least one of \( e + \pi \) and \( e\pi \) is transcendental. Then explain why we cannot determine which one (or both) it is.
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Baker’s theorem says that if \( \ln 2 \) and \( \ln 3 \) are \( \mathbb{Q} \)-linearly independent, then \( 1, \ln 2, \ln 3 \) are linearly independent over the algebraic numbers. Prove that \( \ln 2 \) and \( \ln 3 \) are indeed \( \mathbb{Q} \)-linearly independent. (Hint: suppose \( a \ln 2 + b \ln 3 = 0 \) with integers \( a, b \).)
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Assuming Schanuel’s conjecture, prove that \( e^e \) is transcendental. (Hint: take \( z_1 = 1, z_2 = e \). What can you conclude about the transcendence degree of \( \mathbb{Q}(1, e, e, e^e) = \mathbb{Q}(e, e^e) \)?)
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The number \( e^{\pi\sqrt{163}} \) is known to be transcendental, yet it is approximately \( 640320^3 + 744 – 7.5 \times 10^{-13} \), which is astonishingly close to the integer 262537412640768744. Does this near-integer property conflict with Roth’s theorem? Explain carefully.