ĀRU INTELLIGENCE / INWARD MATHEMATICS™

Inward Discovery / research archive v1

Find the flaw.
Keep the proof honest.

Five mathematical statements meet an exact, reproducible search. Three fail with rational counterexamples. Two survive this finite grid; neither is labeled a proven theorem by the archive.

Explore five candidates ↓Download exact archive ↓

01 / Scientific status

A counterexample can settle a universal claim.

One valid counterexample disproves a universal statement under the specified model. An exhaustive check of a chosen finite grid says only that no counterexample was found there. To prove a statement for all real readings and targets, a separate mathematical argument is required.

Exact inputs

Readings and targets are rational values from the displayed grid. The mean, median, and error comparisons use fractions.

Reproducible search

Each card records the number of cases checked and its first counterexample when one exists.

Independent replay

A separate checker reconstructs the finite domain, statuses, counts, and witnesses without running the generator.

Generator source SHA-256: 5aa654b506dcb020201540a9716ed08e5f1f778edc3e1420531096db1dba107eRead the method and limitations · Inspect the independent verifier

02 / Established identities

Past the finite grid.

Two familiar identities admit general arguments over real numbers. An exact symbolic checker validates their recorded algebra and every index reordering. A separate Lean module proves the midpoint identity and its conditional error comparisons for all real estimates and targets. Reordering invariance has a symbolic check here, but is not formalized in Lean. These established results are separate from the finite search statuses below.

A midpoint always ties.

For any two real estimates a and b, compare squared absolute errors at target t:

(a−t)² − (b−t)² = (a−b)(a+b−2t)

At t = (a+b)/2 the second factor vanishes, so the absolute errors are equal.

Order cannot change these estimates.

Reordering five real readings preserves their sum and sorted multiset. The mean and middle sorted value are unchanged, and so are their absolute errors at every target.

mean = (r₁+r₂+r₃+r₄+r₅)/5

The checker covers all 120 permutations of five formal positions, including reorderings of repeated values.

Proof record SHA-256: 05c51ba6c7df0e3cc7c9176c4b1be69bdcd42a75d8b1aba58290f9123ea89357Read the derivation and trust boundary · Lean proof and limits · Download proof record · Inspect the checker

03 / Evidence ledger

Inspect each claim.

The diagrams locate values on a 0–100 scale for orientation. The fraction labels and archived JSON carry the exact mathematics.

Showing 5 of 5 candidates

Candidate 01 · 7 cases checkedMedian is at least as close as mean for every chosen target.Counterexample found

One exact witness disproves this universal claim in the five-reading model.

Exact counterexample

  1. 0
  2. 0
  3. 0
  4. 0
  5. 25
ReadingsMeanMedianTarget
Target25Mean5Median0

Mean is strictly closer at target 25.

Candidate 02 · 6 cases checkedMean is at least as close as median for every chosen target.Counterexample found

One exact witness disproves this universal claim in the five-reading model.

Exact counterexample

  1. 0
  2. 0
  3. 0
  4. 0
  5. 25
ReadingsMeanMedianTarget
Target0Mean5Median0

Median is strictly closer at target 0.

Candidate 03 · 2 cases checkedChanging the target cannot reverse a strict winner for fixed readings.Counterexample found

The same five readings reverse the strict winner when only the target changes.

First target

  1. 0
  2. 0
  3. 0
  4. 0
  5. 25
ReadingsMeanMedianTarget
Target0Mean5Median0

Median is strictly closer at target 0.

Second target

  1. 0
  2. 0
  3. 0
  4. 0
  5. 25
ReadingsMeanMedianTarget
Target25Mean5Median0

Mean is strictly closer at target 25.

Candidate 04 · 108 cases checkedFor distinct estimates, their exact midpoint is an equal-error target.Unrefuted on finite grid
No counterexample on this grid.

For each distinct-estimate multiset, the exact midpoint was checked. The separate squared-error factorization supplies a general mathematical argument. This status alone is not a proof over all real values.

Candidate 05 · 15,625 cases checkedPermuting the five readings cannot change the winner at a fixed target.Unrefuted on finite grid
No counterexample on this grid.

Every ordered five-reading tuple on this grid was compared with its sorted readings at each target. This status alone is not a proof over all real values.

The print view expands every candidate and includes its exact evidence. It records the current archive snapshot, not a claim of new general proof.