lean_workbook_plus_19672

⚠️ Retired — specification defect

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

Let $n$ be some 100-digit number not composed solely of 2's and 5's, then $\frac{1}{10^{100}} < \frac{1}{n} < \frac{1}{10^{99}}$ .

Why this node was retired

The posted statement is

theorem lean_workbook_plus_19672 (n : ℕ) (hn : n ≠ 0) : (1 / 10^100 : ℝ) < 1 / n ∧ 1 / n < 1 / 10^99   :=  by sorry

The100-digit hypothesis was replaced by only n≠0; the strict upper reciprocal bound additionally excludes n=10^99, which the source permits.

The recorded counterexample refutes the statement as encoded. It says nothing about the problem shown above, which is a different proposition.

Proposed corrected statement

For10^99≤n<10^100 prove1/10^100<1/n≤1/10^99 overℝ. Require n>10^99 if the upper inequality is to be strict; the digits-not-only2/5 condition does not exclude10^99.

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.