Hilbert's 7th Problem: Transcendence of Certain Numbers

Definitions and mathematical statement

Shared notation.

$\mathbb N=\{1,2,\ldots\}$, and $\mathbb Z,\mathbb Q,\mathbb R,\mathbb C$ denote the integers, rationals, reals and complex numbers. Any other conventions are specified locally.

A complex number is algebraic if it is a zero of a nonzero polynomial in $\mathbb Q[X]$; otherwise it is transcendental. Let $\overline{\mathbb Q}\subset\mathbb C$ be the field of algebraic numbers. Here irrational means not in $\mathbb Q$. For $\alpha\ne0$, the values of the possibly multivalued power $\alpha^\beta$ are $\exp(\beta L)$ as $L\in\mathbb C$ ranges over the logarithms satisfying $\exp(L)=\alpha$.

Statement recorded in the supplied source.

For every $\alpha,\beta\in\overline{\mathbb Q}\setminus\mathbb Q$ and every $L\in\mathbb C$ such that $\exp(L)=\alpha$, is $\exp(\beta L)$ transcendental? Equivalently, is every value of $\alpha^\beta$ transcendental when both $\alpha$ and $\beta$ are algebraic and irrational? [S1][S2] This historical question has an affirmative answer by the Gelfond--Schneider theorem (1934), which also permits rational algebraic bases other than $0$ and $1$. The theorem applies to every fixed choice of logarithm. [S2]

Short English statement

Must raising an algebraic irrational number to an algebraic irrational power always give a transcendental number, for every complex value of the power? This historical question is answered affirmatively by the Gelfond--Schneider theorem.

Sources

• S1 ULAM.AI and the UnsolvedMath Contributors, UnsolvedMath: A Curated Collection of Open Mathematics Problems, record 85 (HIL-007). CC BY 4.0. https://huggingface.co/datasets/ulamai/UnsolvedMath.
• S2 Michail Karatarakis and Freek Wiedijk, A formalization of the Gelfond--Schneider theorem, arXiv:2603.24823v1, 25 March 2026, section 1, pp. 1--2, Theorem 3.1 and section 3.1, pp. 5--6 (including arbitrary logarithm values). https://arxiv.org/pdf/2603.24823v1.