lean_workbook_plus_25372

⚠️ Retired — specification defect

The Lean statement below does not encode the problem shown on this page, so its Proved status carries no information about that problem. Do not import this node or use it as a dependency.

The function $f$ has the property that for each real number $x$ in its domain, $1/x$ is also in its domain and $f(x) + f\left(\frac {1}{x}\right) = x.$ What is the largest set of real numbers that can be in the domain of $f$ ?

Why this node was retired

The posted statement is

theorem lean_workbook_plus_25372 (f : ℝ → ℝ) (hf: ∀ x, x ≠ 0 → f x + f (1/x) = x) : {x | x ≠ 0} = {x | ∃ y, y ≠ 0 ∧ y = x ∨ y = 1/x}   :=  by sorry

The unknown domain was replaced by all nonzero reals, which is inconsistent by swapping2 and1/2. The existential set on the RHS is itself all reals under the posted parsing.

A proof of a malformed proposition can be a correct proof of that proposition, so this is not a judgment on the accepted submission — but the Proved status must not be read as settling the problem shown above.

Proposed corrected statement

Let D⊆ℝ∖{0} be closed under reciprocals and f:D→ℝ satisfy f(x)+f(1/x)=x on D. Prove D⊆{−1,1}, and that D={−1,1} is realizable by f(1)=1/2 and f(−1)=−1/2; hence this is the largest possible domain.

Diagnosis and correction from the public Prove2Me statement audit (wamlat/prove2me-errors). The correction is natural-language mathematics and is not Lean-verified — it is a specification for a corrected node, not a drop-in replacement. No corrected replacement node exists yet.