five_fails
plain-language theorem explainer
Multiplier 5 fails to send every Standard Model electric charge into the integers. Anyone citing the face-count integerization scale for the topological Z-map will use this negative result to rule out k=5. The proof is a short rational-arithmetic contradiction: five times two-thirds equals ten-thirds, which is not an integer.
Claim. The multiplier $k=5$ does not integerize all Standard Model charges: it is false that $5Q\in\mathbb{Z}$ for every $Q\in\{-1,\,2/3,\,-1/3\}$.
background
This module derives the charge-to-band polynomial $Z(\tilde Q)$ from recognition boundaries on the 3-cube, without anchors or empirical masses. Stage 1 fixes the integerization scale: a boundary of charge $Q$ couples to the $F$ faces of the cube, and the ledger demands integer entries, so some positive integer $k$ must send every SM charge into $\mathbb{Z}$.
The SM charge set is ${-1,,2/3,,-1/3}$. The face count is $F=2D$ with spatial dimension $D=3$ (forced by the T8/T9 chain), hence $F=6$. The claim under audit is one of the negative cases in the minimality argument: among small positive integers, which $k$ fail to integerize the full charge set.
Upstream dimension constants simply fix $D:=3$; face and count helpers from the simplicial and event layers supply the geometric language for $F$, but the arithmetic here is pure rational number theory on the charge list.
proof idea
Assume toward contradiction that $k=5$ integerizes every SM charge. Specialize to $Q=2/3$ (which sits in the charge list by a trivial membership check). Obtain an integer $n$ with $5\cdot(2/3)=n$. Clear denominators to get $10/3=n$ over $\mathbb{Q}$, then cast to $\mathbb{Z}$ as $10=3n$. The linear Diophantine contradiction is discharged by omega.
why it matters
Stage 1 of the topological Z-map derivation asserts that $F=6$ is the minimal positive even integer $k$ with $kQ\in\mathbb{Z}$ for all three SM charges. This lemma supplies the $k=5$ failure half of that census.
It is packaged immediately into integerization_results, which records the full pattern: $k\in{1,2,4,5}$ fail and $k\in{3,6}$ succeed. That package underwrites the selection of the even scale $F=6$ (rather than the odd integerizer 3) before Stage 2 forces the even quartic $Z=a\tilde Q^2+b\tilde Q^4$ and Stage 3 adds the color offset $2^{D-1}=4$.
Framework landmarks in play: T8 forcing $D=3$, the eight-tick/cube face count $F=2D=6$, and the ledger integrality requirement that $\delta$-units live in $\mathbb{Z}$.
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