FormalConjectures/ErdosProblems/1050.lean
Explanation legend Lean4 Domain Explanation Language (L4DEL)
L4DEL explanations are a separate layer. They do not change the Lean source. Intuition is explanatory prose; numbered items are a reading path; sources link the wording back to evidence.
- L4DEL
- Lean4 Domain Explanation Language
Credit: The upstream authors and contributors retain credit for the project and mathematics. lean4-explainer only renders their pinned source and adds the separately disclosed explanation layer.
Project sources: Pinned project documentation, Formal Conjectures paper, arXiv v1
4 declarations, 4 contain sorry, 0 declared as axioms
Erdős Problem 1050
References:
- erdosproblems.com/1050
- [Bo91] Borwein, Peter B., On the irrationality of {$\sum(1/(q^n+r))$}. J. Number Theory (1991), 253--259.
- [Bo92] Borwein, Peter B., On the irrationality of certain series. Math. Proc. Camb. Phil. Soc. (1992), 141--146.
- [Er48] Erdős, P., On arithmetical properties of Lambert series. J. Indian Math. Soc. (N.S.) (1948), 63-66.
- [Er88c] Erdős, P., On the irrationality of certain series: problems and results. New advances in transcendence theory (Durham, 1986) (1988), 102-109.
Is $\sum_{n=1}^\infty \frac{1}{2^n - 3}$ irrational?
The answer is yes, proved by P. B. Borwein [Bo91] (with a cleaner self-contained proof in [Bo92]), specialized to $q = 2$, $r = -3$.
A formal Lean proof is given in an external repository,
gotrevor/lean-gallery, formalized by Trevor Morris with
Claude Code and Harmonic's Aristotle.
Imports (1)
- theorem
Erdos1050.erdos_1050sorry line 51 sourcedepends on sorry what this means
Statement uses:
Irrational,tsum,Real,Nat,HDiv.hDiv,OfNat.ofNat,HSub.hSub,HPow.hPow,HAdd.hAdd,SummationFilter.unconditionalProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous44 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
- theorem
Erdos1050.erdos_1050.variants.two_pow_sub_onesorry line 61 sourcedepends on sorry what this means
Statement uses:
Irrational,tsum,Real,Nat,HDiv.hDiv,OfNat.ofNat,HSub.hSub,HPow.hPow,HAdd.hAdd,SummationFilter.unconditionalProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous44 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
- theorem
Erdos1050.erdos_1050.variants.borweinsorry line 71 sourcedepends on sorry what this means
Statement uses:
Int,LE.le,OfNat.ofNat,Rat,Ne,Nat,Neg.neg,HPow.hPow,Irrational,tsum,Real,HDiv.hDiv,HAdd.hAdd,SummationFilter.unconditionalProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous56 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
- theorem
Erdos1050.erdos_1050.variants.transcendentalErdős Problem 1050, variant [Er88c] (open) sorry line 84 sourceSource documentation
Erdős Problem 1050, variant [Er88c] (open). Erdős conjectured that $\sum_{n=1}^\infty \frac{1}{2^n + t}$ is transcendental for every integer $t \ne 0$ — strictly stronger than the irrationality established by Borwein [Bo91]. This remains open.
Two exclusions make the series well-posed and the claim non-vacuous, exactly as in Borwein's theorem: $t \ne -2^n$ for all $n \ge 1$ (so no denominator vanishes), and $t \ne 0$ (at $t = 0$ the series is the rational $\sum_{n \ge 1} 2^{-n} = 1$, hence not transcendental).
depends on sorry what this means
Statement uses:
Iff,True,Int,Ne,OfNat.ofNat,Nat,LE.le,Neg.neg,HPow.hPow,Transcendental,Rat,Real,tsum,HDiv.hDiv,HAdd.hAdd,SummationFilter.unconditionalProof uses:
sorryAx,Lean.Name,Bool.false,Lean.Name.num,Lean.Name.str,Lean.Name.anonymous76 instances, coercions and auxiliary constants left out.
Tactic trace: 1 steps
Open to load the checked before-and-after goal states.
theorem erdos_1050.variants.transcendental : answer(sorry) ↔ ∀ t : ℤ, t ≠ 0 → (∀ n : ℕ, 1 ≤ n → t ≠ -(2 : ℤ) ^ n) → Transcendental ℚ (∑' n : ℕ, (1 : ℝ) / ((2 : ℝ) ^ (n + 1) + (t : ℝ))) := by …