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A strict lower bound for the fractional part of a nonintegral cube root

Proved
WorkbookCorrected.plus_70623

by wamlart · Sep 13, 2026 · Mathlib 0df444a (Lean v4.33.1)

corrected-formalizationlean-workbooksource-checked

Prove: {n3}>13n23\{\sqrt[3]{n}\}>\frac{1}{3\sqrt[3]{n^2}}{3n​}>33n2​1​ where nnn is a positive integer not equal to a cube of any integer

Formalization Note: The source formalization omitted the fractional-part operation and used a square root with a natural-number fractional exponent. This correction represents the cube root by a nonnegative real r with r³=n and restores r−floor(r). Its denominator3r² equals3 times the cube root of n². The noncube condition is preserved for positive integers.

Source: InternLM Lean-Workbook, record lean_workbook_plus_70623 (Apache-2.0).

Preamble
import Mathlib
Formal statement
theorem WorkbookCorrected.plus_70623 (n : ℕ) (hn : 0<n) (hnc : ¬ ∃ k : ℕ, k^3=n)
    (r : ℝ) (hr : 0 ≤ r) (he : r^3=(n:ℝ)) : r-(⌊r⌋₊:ℝ) > 1/(3*r^2) := by sorry
Source
https://huggingface.co/datasets/internlm/Lean-Workbook/blob/main/lean_workbook.json, record lean_workbook_plus_70623; Apache-2.0

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