Is $e + \pi$ irrational? (open)

Question. Is $e + \pi$ irrational?

Status: open. No proof and no disproof is known. The same is true of $e\pi$, of $e - \pi$, of $e/\pi$, of $\pi^{e}$, of $\pi^{\pi}$ and of $e^{e}$: for none of these is it known whether the number is rational, let alone whether it is transcendental. This is not a gap in this library's prerequisites. It is a gap in the subject.

Not proved in this library, and not provable anywhere at present. Nothing on any page here depends on the value or the arithmetic nature of $e + \pi$.

What is known, and what would settle it. Both constituents are settled individually: $e$ is transcendental (Hermite, 1873, a result that is in scope for this library and will be proved), and $\pi$ is transcendental (Transcendence of $\pi$ (awaiting a scope decision) ‡). A cheap symmetric-function argument already shows that the two candidates cannot both be tame: $e$ and $\pi$ are the roots of

$x^2 - (e + \pi)x + e\pi,$

so if $e + \pi$ and $e\pi$ were both algebraic then $e$ and $\pi$ would be algebraic too. Hence at least one of $e + \pi$ and $e\pi$ is transcendental, and that is essentially the whole of what is known about this pair. Note the asymmetry with $e^{\pi}$, which is known to be transcendental (Gelfond, via the Gelfond-Schneider theorem of 1934 applied to $e^{\pi} = (e^{i\pi})^{-i} = (-1)^{-i}$), and about which much more is known: Nesterenko (1996) proved that $\pi$ and $e^{\pi}$ are algebraically independent over $\mathbb{Q}$. By contrast $\pi^{e}$ is not known even to be irrational, because it is not of the form $\alpha^{\beta}$ with $\alpha, \beta$ algebraic and so Gelfond-Schneider says nothing about it.

Schanuel's conjecture, if proved, would settle all of these at once: it implies that $e$ and $\pi$ are algebraically independent over $\mathbb{Q}$, and hence that $e + \pi$ and $e\pi$ are both transcendental. Schanuel's conjecture is itself open, so this reduces one open problem to a much harder one rather than solving anything.

Why it matters here. The library proves irrationality where irrationality is provable, starting from the refutation of the claim that $\sqrt{2}$ is rational (FALSE: some rational number squares to 2 ↗), and it constructs $\mathbb{R}$ so that $e$ and $\pi$ can be defined at all. This item is the honest boundary marker: the elementary irrationality arguments do not scale, and the first genuinely simple combination of the library's two favourite constants is already past the edge of what anyone can prove.