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Source. Y. Yu and K. Chen, Erdős Problem 354(i): Strong Completeness
of Two Dyadic Floor Sequences, manuscript of 13 September 2026, Section
10 "Good rational approximants and long sparse windows" with Subsections
10.1--10.2 and displays (10.1)--(10.2), physical pp. 11--12, in the
seventeen-page PDF held by its library source card,
Yu and Chen (2026).
Standing. This is an author-recorded reconstruction. It is not an
independent review, changes no status and assigns no tier. Dirichlet's
approximation theorem is the one external input; it is imported in the
form stated below and not reproved.
Definitions
The normalized pair, ai,bi, Kn and Pn are as on the
normalization page;
c0, a and Rn as on the
finite-event decay page;
θ=α/β, good rationals, λ, k and En,k as on
the digit-budget page.
log is the natural logarithm. For an integer b≥2 let D(b) be the
least denominator ≥b of a good rational (shown to exist below). Put
b(n)=⌈2nβ⌉,Dn∗=D(b(n)),k(D)=⌈log2(8D)⌉,f(n)=n+k(Dn∗),
m(D)=⌊log2(D/β)⌋ (D≥β),Cβ=⌈log2⌈16β⌉⌉,Cβ′=⌈log2(3⌈β⌉)⌉.
For a reduced p/q its binary height is H=⌈log2(p+q+1)⌉,
and δi=qai−pbi for i≥0.
Imported theorem (Dirichlet). For every real ξ and integer
Q≥1 there are integers j,k with 1≤k≤Q and
∣kξ−j∣≤1/(Q+1). (This is the pigeonhole form; the source cites
Mathlib's Diophantine approximation results as its reference 10.)
Statement
(10.2). Let the normalized pair have irrational θ and suppose
the sequence is incomplete. There is a constant L>0 such that for every
T0≥1 and every ε>0 there are an integer T≥T0 and a
reduced rational p/q with 1<p/q<2 and ∣p/q−θ∣<ε such
that
H2≤4T,KT≤LlogT,∣δi∣<2H (0≤i≤T).
Proof
Step 1: good rationals and the crossing denominator
Let Q≥1. Dirichlet gives 1≤q≤Q and p with
∣θ−p/q∣≤1/(q(Q+1)); reducing p/q can only decrease the
denominator and keeps the bound, and 1/(q(Q+1))<1/q2 since q≤Q.
So a good rational pQ/qQ with ∣θ−pQ/qQ∣≤1/(Q+1) exists for
every Q. None equals θ, which is irrational, so if there were
only finitely many good rationals their distances to θ would have
a positive minimum, contradicting 1/(Q+1)→0. Hence there are
infinitely many good rationals; for a fixed q at most two numerators
satisfy ∣θ−p/q∣<1/q2, so their denominators are unbounded and
D(b) exists for every b≥2.
The pre-crossing rational. Given b≥2, apply Dirichlet with
Q=D(b)−1≥1: there is a reduced p/q with q<D(b) and
∣θ−p/q∣≤1/(D(b)q). It is good, because q<D(b) gives
1/(D(b)q)<1/q2. Minimality of D(b) then forces q<b: a good
denominator in [b,D(b)) would contradict the definition of D(b).
Step 2: matching layers and the cubic advance
For D≥β, 2m(D)β≤D<2m(D)+1β by definition of
m(D), and
k(D)=⌈log2(8D)⌉≤⌈m(D)+1+3+log2β⌉=m(D)+⌈log2(16β)⌉≤m(D)+Cβ.
Apply (DB) from the digit-budget page at depth n≥1 with the good
rational of denominator Dn∗≥b(n)≥2nβ; its k is
k(Dn∗), so under incompleteness
Kf(n)≥Kn+2c0eaKn−3(n≥1).
Claim: f(n)>n3 for arbitrarily large n. Suppose instead
f(n)≤n3 for all n≥n1. Since K is nondecreasing,
Kn3≥Kf(n) for such n. The event set is infinite
(normalization page, item 5), so Kn→∞, and once Kn is large
the exponential term dominates: Kn+2c0eaKn−3≥Kn4.
Hence there is n2≥n1 with Kn3≥Kn4 for all n≥n2.
Choose n0≥n2 with n0>1 and Kn0≥2. By induction on r,
Kn03r≥Kn04r≥24r, because n03r≥n2.
But Km≤m for every m, so 24r≤n03r, that is,
(4/3)r≤log2n0 for every r, which is false. This proves the
claim.
The bound (10.1). Let D be a good denominator with m=m(D)≥1.
Apply the window lemma of the digit-budget page at depth m with this
rational: q=D≥2mβ=λ, so under incompleteness
Rm<3λ/D+Em,k(D)≤3+2k(D), using the crude bound
Em,k≤2k (each of the 2k fractional parts is less than 1).
With (FE-R),
c0eaKm≤Rm+2<2k(D)+5≤2m+2Cβ+5.(10.1)
Step 3: the windows
Fix ε>0 and T0. By the claim, choose n with f(n)>n3
and n as large as needed below. Put D=Dn∗, m=m(D), T=m−1. From
k(D)≤m+Cβ and f(n)=n+k(D)>n3 we get
m>n3−n−Cβ, and for n≥Cβ+4 this gives T≥n2 (as
n3−n2−n≥11n>Cβ+1 for n≥4). Take n large enough that
T≥T0.
Let p/q be the pre-crossing rational for b=b(n) from Step 1: q<b(n)
and ∣θ−p/q∣≤1/(Dq)≤1/D≤1/(2nβ). This tends to 0, so
for n large ∣p/q−θ∣<ε and 1<p/q<2 (as 1<θ<2).
Height. Since p<2q and q<⌈2nβ⌉≤2n⌈β⌉,
p+q+1≤3q<3⋅2n⌈β⌉, so H≤n+Cβ′. For
n≥Cβ′, H≤2n and H2≤4n2≤4T.
Residues. ∣qα−pβ∣=qβ∣θ−p/q∣≤β/D, so for
0≤i≤T=m−1,
2i∣qα−pβ∣≤D2m−1β≤21
because D≥2mβ. Writing ai=2iα−{2iα} and
bi=2iβ−{2iβ},
δi=2i(qα−pβ)−q{2iα}+p{2iβ},∣δi∣<21+max(p,q)<p+q+1≤2H.
Events. D is a good denominator with m(D)=m≥n≥1, so (10.1)
applies; with KT≤Km,
c0eaKT≤2m+2Cβ+5=2T+2Cβ+7,KT≤a1logc02T+2Cβ+7.
For T large the right side is at most LlogT with the fixed constant
L=2/a, since (2T+2Cβ+7)/c0≤T2 eventually. Enlarging n
once more secures this. All four displayed properties of (10.2) now hold
for this T and p/q.
Scope. The good crossing rational of denominator D supplies the
event bound; the pre-crossing rational supplies the small height and the
all-layer residue bound; no continued-fraction indexing is used. The
constants L, Cβ, Cβ′ depend on α,β only. The
construction needs incompleteness (for (DB) and (10.1)) and irrationality
(for the good rationals and the infinitude of events).