four_fails

The module derives the charge-to-band map $Z(\tilde{Q})$ from recognition boundaries on the 3-cube, without mass anchors. Stage 1 asks for a positive integer scale $k$ such that the ledger entries $kQ$ are integral for every SM electric charge (T8: $\delta$-units $\simeq\mathbb{Z}$). The face count $F=2D$ of the cube is the candidate scale; at $D=3$ one has $F=6$.

The SM charge list is the three rationals ${-1,,2/3,,-1/3}$. A natural number $k$ integerizes all of them when, for each such $Q$, there exists $n\in\mathbb{Z}$ with $k\cdot Q=n$. The present result is one of the negative cases in that checklist: $k=4$ fails.

Upstream, the predicate integerizes_all and the charge list are the only local ingredients; the proof is pure rational arithmetic and does not invoke Clifford or gravity material.

Assume toward contradiction that $4$ integerizes every SM charge. Specialize to $Q=-1/3$ (membership in the charge list is by simp). Obtain $n\in\mathbb{Z}$ with $4\cdot(-1/3)=n$, i.e. $-4/3=n$. Clearing the denominator yields $-4=3n$ over $\mathbb{Z}$. Then omega discharges the impossibility $3\mid 4$. No external lemmas beyond the local definitions are required.

Stage 1 of the Z-map derivation asserts that $F=6$ is the minimal positive even integerizer of the SM charges. This theorem supplies the $k=4$ failure half of that claim. It is packaged into integerization_results (the six-way conjunction of successes and failures for $k\in{1,2,3,4,5,6}$) and is consumed by six_smallest_positive_even_integerizer, which concludes that every positive even integerizer is at least $6$.

In the broader forcing chain this pins the face-count scale at $D=3$ (T8), so the later polynomial $Z=a\tilde{Q}^2+b\tilde{Q}^4$ and the color offset $2^{D-1}=4$ sit on a topologically forced integerization rung rather than an empirical fit. Note that $k=3$ also integerizes, but is odd and is excluded by the evenness requirement coming from $F=2D$.