Alphabeta Math
Remark‡ sources checked 2026-07-26‡ not proved here
‡ Recorded, not proved here. This statement is included so the library can refer to it honestly, with a citation to the literature. It is not proved anywhere in this library: the track that would prove it has not been developed here yet.

Is e+π irrational? (open)

Statement

Question. Is e+π irrational?

Status: open. No proof and no disproof is known. The same is true of eπ, of e−π, of e/π, of πe, of ππ and of ee: 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.

Remarks

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+π.

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 π is transcendental (Transcendence of π (awaiting a scope decision) ‡). A cheap symmetric-function argument already shows that the two candidates cannot both be tame: e and π are the roots of

x2−(e+π)x+eπ,

so if e+π and eπ were both algebraic then e and π would be algebraic too. Hence at least one of e+π and eπ is transcendental, and that is essentially the whole of what is known about this pair. Note the asymmetry with eπ, which is known to be transcendental (Gelfond, via the Gelfond-Schneider theorem of 1934 applied to eπ=(eiπ)−i=(−1)−i), and about which much more is known: Nesterenko (1996) proved that π and eπ are algebraically independent over Q. By contrast πe is not known even to be irrational, because it is not of the form αβ with α,β 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 π are algebraically independent over Q, and hence that e+π and eπ 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 2 is rational (FALSE: some rational number squares to 2 ↗), and it constructs R so that e and π 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.

Depends on

Used by

Nothing in the library uses this result yet.

Dependency tree · two levels

2 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources