lean_workbook_plus_35332
Disproved⚠️ Retired — mistyped fractional exponent
The exponent
^(1/5)in the Lean statement below is natural-number division, not a real exponent, so it collapses to an integer power and the displayed radical is not what is being asserted. Do not import this node or use it as a dependency.
Let , and be roots of the equation . Prove that:\n\n
Why this node was retired
The posted statement is
theorem lean_workbook_plus_35332 (a b c : ℂ) (ha : a^3 + 15*a^2 - 198*a + 1 = 0) (hb : b^3 + 15*b^2 - 198*b + 1 = 0) (hc : c^3 + 15*c^2 - 198*c + 1 = 0) : a^(1/5) + b^(1/5) + c^(1/5) = 0 := by sorry
A numeric-literal exponent carrying no type ascription is elaborated at type ℕ (via HPow ℝ ℕ ℝ, Monoid.npow), and natural-number division truncates. Each occurrence therefore collapses:
^(1/5)→^0, so the radical is1for every base
The recorded counterexample refutes the collapsed statement. It says nothing about the exercise displayed on this page, which is a different proposition.
What a faithful statement would require
Every fractional exponent must be given a real type, e.g. x ^ ((1 : ℝ) / 3) using Real.rpow, or be written with the intended root operation (Real.sqrt, or the signed real cube root where the radicand may be negative). Real powers also need their own domain hypotheses: rpow is only the intended root for a nonnegative base.
Proposed corrected statement
Take a,b,c to enumerate the three distinct real roots (or use the full root multiset), and interpret fifth roots as real odd roots, not natural powers on arbitrary repeated complex roots.
From the public Prove2Me statement audit (wamlat/prove2me-errors). That proposal is natural-language mathematics and is not Lean-verified; it is a specification for a corrected node, not a drop-in replacement.
import Mathlib.Analysis.Complex.Basic
theorem lean_workbook_plus_35332 (a b c : ℂ) (ha : a^3 + 15*a^2 - 198*a + 1 = 0) (hb : b^3 + 15*b^2 - 198*b + 1 = 0) (hc : c^3 + 15*c^2 - 198*c + 1 = 0) : a^(1/5) + b^(1/5) + c^(1/5) = 0 := by sorry