Let the selectivity be α(P) = N(P)/D(P), where each side is a sum of Langmuir-type channels
channel(P) = K + A · bP/(1 + bP) K ≥ 0, A ≥ 0, b > 0Every channel is monotone increasing and bounded, and so is every sum. The two limits are immediate:
α(0) = K_N / K_D
α(∞) = (K_N + A_N) / (K_D + A_D)Both are finite and positive. They are generally different, because the amplitudes A enter one limit and not the other. This is the fact the earlier conjecture missed.
If α is continuous on (0, ∞) and α(P) = 1 at finitely many points, then the number of crossings is odd when α(0) and α(∞) lie on opposite sides of 1, and even when they lie on the same side.
Proof. α is continuous and never zero or infinite (both K terms are positive in the cases of interest). Let g = α − 1, also continuous. Each crossing is a sign change of g. A continuous function whose endpoint signs agree must change sign an even number of times; if the endpoints disagree, an odd number. Since crossings are exactly the sign changes, the count has the parity of the endpoints. ∎
The theorem is elementary — it is the intermediate value argument applied to a ratio of monotone sums. Its value is not depth but use: it converts a numerical conjecture into a constraint that any reported reversal must satisfy, and it tells you which crossing counts are even possible.
3,000 random channel sets, 1–4 channels per side, coefficients drawn over 1.5–2.5 decades, crossings counted on a 4,000-point log grid from P = 10⁻³ to 10³.
| Crossings | 0 | 1 | 2 | 3 | 4+ |
|---|---|---|---|---|---|
| Count | 2,310 | 593 | 95 | 2 | 0 |
Parity violations: 0. Every case's crossing count was consistent with its own two end signs. That is the theorem being exercised 3,000 times without a counterexample.
Two cases produced three crossings, and both have the signature the theorem requires — opposite limits:
| case | channels (N, D) | α(0) | α(∞) | crossings |
|---|---|---|---|---|
| 1 | 3, 1 | 0.535 | 1.049 | 3 |
| 2 | 2, 1 | 0.403 | 1.159 | 3 |
Both start below 1 and finish above it. The curve must therefore cross at least once; it happens to cross three times, which requires two returns — a shape the earlier "at most two" claim excluded.
Rate: 2 in 3,000, or 0.07%. Rare, but real, and reproducible from the seed.
| Old claim | New statement | |
|---|---|---|
| Form | "At most two crossings" | Parity fixed by the two limits; no upper bound established |
| Status | Conjecture, no counterexample found | Theorem for parity; open for the upper bound |
| Screening use | "A third crossing indicates a model outside the class" | "A crossing count inconsistent with the end signs indicates an error or a model outside the class" |
The new rule is weaker but usable, and it cannot be wrong in the way the old one was. A reported reversal with three crossings and both limits on the same side of 1 is now a definite error — either in the model or in the analysis.
| # | Hypothesis | Disposition |
|---|---|---|
| N-23 | An upper bound of 4 crossings exists for 4-channel-per-side Langmuir sums, and the bound grows more slowly than the channel count. | test now — extend the sweep to S = 6 with more trials |
| N-24 | Three-crossing cases require at least one side to have more channels than the other; equal-channel sums cannot produce three. | test now — stratify by channel count |
| N-25 | Published reversals with an even number of crossings and equal limits are parameter artefacts; odd-crossing reversals with unequal limits are the physically possible class. | note — needs a corpus of reversals |
| N-26 | The parity constraint applies to any ratio of monotone bounded sums, so it transfers to thermal rectification and to diode rectification without modification. | test now — restate the proof for the general class |
If one sentence survives: the number of selectivity crossings has the parity of the two limits of α, this is provable and held in 3,000 trials, and it kills my own earlier claim of a maximum of two.