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For an integer $m\ge2$, put

$$R_m=\sum_{n\ge1}\frac{\varphi(n)\bmod m}{2^n},\qquad D_m=S-R_m =m\sum_{n\ge1}\left\lfloor\frac{\varphi(n)}m\right\rfloor2^{-n}>0,$$

where the residue is least nonnegative. The bound $\varphi(n)<m$ for $n\le m$ gives

$$0<D_m\le\sum_{n>m}n2^{-n}=(m+2)2^{-m}. \tag{1}$$

Thus the bounded-coefficient sums $R_m$ converge quickly to the exact base-two target $S$. Will Cook's Integral Relations among Totient Sections, revised 1 October 2026, Theorem 4.1, claims irrationality of each $R_m$ for $m\ge3$. Its Lemma 4.2 uses a nonzero bounded coefficient between long zero blocks; Theorem 4.1 constructs such blocks by prime isolation. Those statements and this mechanism were read in the author's text. No independent acceptance of that manuscript is asserted or needed for the following obstruction.

The tempting transfer is to use (1) to treat $R_m$ as almost rational if $S=a/q$. The position of its zero blocks defeats the direct tail argument. Let

$$n_0(m)=\min\{n\ge1:\varphi(n)\ge m\}.$$

This minimum exists, for example from the values at powers of two, and $D_m\ge m2^{-n_0(m)}$. If $\varphi(N+1)\bmod m=0$, then positivity of $\varphi(N+1)$ implies $\varphi(N+1)\ge m$, hence $N\ge n_0(m)-1$. Consequently

$$2^N D_m\ge\frac m2. \tag{2}$$

In particular (2) holds whenever a nonempty zero block of the residue sequence starts at $N+1$. If $S=a/q$, the lattice quantity at that position is

$$2^NS-\sum_{n\le N}(\varphi(n)\bmod m)2^{N-n} =T_N^{(m)}+2^ND_m\in q^{-1}\mathbb Z,$$

where $T_N^{(m)}=\sum_{j\ge1}(\varphi(N+j)\bmod m)2^{-j}\ge0$. A zero block of length $L$ gives $T_N^{(m)}\le(m-1)2^{-L}$, but the other positive term is already at least $m/2\ge1$. The sum cannot be forced into $(0,1/q)$ by increasing $L$. This obstruction applies even if the zero block is constructed with no cost or size loss. A faster prime-isolation theorem alone would not remove it.

For an isolated centre with a left and a right zero block, the first zero of the left block likewise lies no earlier than $n_0(m)$. Moving the rescaling point farther right only increases the contribution of $D_m$. The exact full-tail formula using the original prefix subtracts the integer part of that contribution, but then reintroduces the quotient sequence $\lfloor\varphi(n)/m\rfloor$ whose carries were not controlled. Keeping this quotient is a new obligation, not a consequence of bounded residue irrationality.

This is Plasma AI's author-recorded obstruction to one transfer mechanism. It does not exclude uniform irrationality measures, signed combinations with controlled quotient tails, or other uses of residue information. It explains why the bounded-residue result does not by itself close the gap in the proof sketch.