Three and the Golden Ratio
Abstract
In decimal arithmetic, let n=3m, where every prime factor of m is 2 or 5. At one common long-division depth, the fractions k/n split into three residue classes. One terminates, one carries the same one-digit repetend as 1/n, and one carries its nines complement. Counting terminating fractions as aligned gives \alpha_{10}(3m) \;=\; \frac{2m-1}{3m-1}. This alignment reaches the reciprocal golden ratio precisely when m\ge \varphi^2=\varphi+1; among decimal-supported integers, the first case is m=4.
The same one-digit argument works in every base b when p\nmid b, b\equiv1\pmod p, and m is b-supported, giving \alpha_b(pm)=(2m-1)/(pm-1). Among odd primes, only p=3 can reach 1/\varphi; the prime 2 is degenerate, with alignment identically one.
The equality at p=3 suggests the reciprocal-square condition m^{*}(\tau,p)=1/\tau^{2}. It yields a cubic with a unique root in (0,1) for every prime p \ge 3. Viewed formally for all primes, the same cubic factors over \mathbb{Q} exactly for p \in \{2, 3, 5\}. The two non-trivial factorizations produce thresholds 1/\varphi and 1/\varphi^{2}, both in the golden field \mathbb{Q}(\sqrt{5}). Of these, only 1/\varphi lies in the separating interval (2/5,\, 2/3). The minimal polynomial of the reciprocal golden ratio \tau = 1/\varphi, namely \tau^{2} + \tau - 1, divides the cubic if and only if p = 3, with coefficient 3 produced by polynomial division and yielding the fourth-power identity \tau^{4} = 2 - 3\tau. For p \ge 7, the cubic is irreducible over \mathbb{Q} and its threshold lies outside every quadratic field.
The Selection Problem
The golden ratio usually enters arithmetic through the Fibonacci recurrence. Here it appears in long division and distinguishes the prime 3. For the family n=3m, with m built from powers of 2 and 5, the synchronized digit count crosses 1/\varphi precisely when m\geq\varphi^2=\varphi+1.
Is this meeting of 3 and \varphi a decimal accident, or does it survive a change of base?
The base-b expansion of a rational number k/n either terminates or eventually repeats. When it repeats, the shortest repeating block is called the repetend. For \gcd(n, b) = 1, the length of the repetend of 1/n equals the multiplicative order \mathop{\mathrm{ord}}_n(b), and the repetends of the reduced fractions \{k/n : \gcd(k,n)=1\} are organized by the orbits of \langle b \rangle acting on (\mathbb{Z}/n\mathbb{Z})^\times. Fractions within a single orbit are cyclic permutations of one another [3].
Throughout, base-b expansions are taken in their canonical form. Terminating expansions are written with trailing zeros, not with an eventual string of digits equal to b-1.
For a fixed denominator n, write \mathcal{F}(n)=\{k/n:1\le k\le n-1\}. The pattern occurs in the decimal family n=3m, where m has no prime factors other than 2 and 5. After the terminating part of the denominator is cleared, the fractions k/(3m) fall into three classes according to k\pmod3. Multiples of 3 terminate. The class k\equiv1\pmod3 carries the same one-digit repetend as 1/(3m). The class k\equiv2\pmod3 carries its nines complement.
This visible split suggests a simple question. What proportion of \mathcal F(3m) belongs to the terminating or matching class? We call that proportion the repetend alignment. The three classes give 2m-1 aligned fractions among the 3m-1 fractions in \mathcal F(3m), hence \alpha_{10}(3m)=(2m-1)/(3m-1). The value increases toward 2/3 and crosses 1/\varphi at the exact real threshold m=\varphi^2=\varphi+1.
The decimal count is the starting point. We then show that the same argument works for every prime p\nmid b in the one-digit regime b\equiv1\pmod p, with every prime factor of m dividing b. The formula becomes \alpha_b(pm)=(2m-1)/(pm-1). Among odd primes, only p=3 reaches the reciprocal golden ratio.
Threshold separation alone does not distinguish 1/\varphi from the rest of (2/5,2/3). The equality \varphi^2=(1/\varphi)^{-2} suggests a second question. Which prime thresholds have a critical resolution equal to their reciprocal square? The resulting cubic places the non-trivial quadratic cases p=3 and p=5 in \mathbb Q(\sqrt5) and places every p\ge7 in a cubic extension. At p=3, polynomial division gives \tau^4=2-3\tau for \tau=1/\varphi.
The alignment formula and the three-class decomposition rest on classical arithmetic of repetends; the underlying orbit structure and complement pairing appear in various forms in [3] and the Midy literature. The reciprocal-square classification and its consequences in \mathbb Q(\sqrt5) are the new algebraic step.
Definitions
Fix a base b \ge 2. For positive integers k and n with 1 \le k \le n-1, the base-b expansion of k/n takes the form \frac{k}{n} \;=\; 0.\, a_{1} a_{2} \cdots a_{s}\, \overline{d_{1} d_{2} \cdots d_{\ell}}, where a_{1} \cdots a_{s} is the non-repeating prefix and in the nonterminating case, d_{1} \cdots d_{\ell} is the repetend of length \ell \ge 1. In the terminating case we set \ell = 0 and regard the repetend as empty.
Definition 1. A positive integer m is b-supported if every prime factor of m divides b. Equivalently, m is b-supported if and only if 1/m has a terminating base-b expansion. We write \mathcal{S}_b for the set of b-supported positive integers.
In base 10, \mathcal{S}_{10} is the set of integers of the form 2^{a} 5^{c}, the denominators with terminating decimal expansions.
Definition 2 (Synchronized repetend alignment). Let p\nmid b be prime, let m\in\mathcal S_b, and put n=pm\ge2. Choose a common clearing depth D\ge0 with m\mid b^D, and set L=\mathop{\mathrm{ord}}_p(b). Beginning immediately after the first D base-b digits, compare the next L digits of every k/n with the corresponding digits of 1/n. If k/n terminates, set a_b(k/n,1/n)=1. Otherwise, let a_b(k/n,1/n) be the proportion of these L synchronized positions at which the digits agree. The repetend alignment is \alpha_b(n) \;=\; \frac{1}{n-1} \sum_{k=1}^{n-1} a_b(k/n,\,1/n).
The definition does not depend on the clearing depth. Increasing D advances every nonterminating tail by the same power of b, so it rotates the two length-L words together and preserves their position-wise match proportion. Individual repetends are never restarted at separately chosen phases. The common depth belongs to the definition and is the clock used by every comparison below.
All alignment theorems below impose b\equiv1\pmod p, and hence L=1. Synchronization still fixes the digit supplied by division, even though there is no nontrivial cyclic phase in a one-digit word.
For the family n=3m with m\in\mathcal S_{10}, the synchronized window has length one because \mathop{\mathrm{ord}}_3(10)=1.
The Three Classes
The decimal case supplies the three-class count. The same argument then extends to the stated one-digit periodic regime.
Lemma 3. Let n = 3m with m \in \mathcal{S}_{10} and m \ge 1. For each k \in \{1, \ldots, n-1\}, exactly one of the alternatives (A)–(C) holds.
k \equiv 0 \pmod 3. Then k/n terminates.
k \equiv 1 \pmod 3. Then k/n has a one-digit repetend equal to that of 1/n.
k \equiv 2 \pmod 3. Then k/n has a one-digit repetend equal to the nines complement of that of 1/n.
Proof. Choose a common clearing depth D with m\mid10^D, and put u=10^D/m. Since 3\nmid10, we have 3\nmid u. At the synchronized position after depth D, the reference fraction has remainder u\pmod3, while k/(3m) has remainder ku\pmod3.
Write k=3q+r with r\in\{0,1,2\}. If r=0, then k/(3m)=q/m terminates. If r=1, then ku\equiv u\pmod3, so the synchronized digit agrees with the reference digit. If r=2, then ku\equiv-u\pmod3. The two nonzero remainders modulo 3 produce the digits \left\lfloor\frac{10}{3}\right\rfloor=3, \qquad \left\lfloor\frac{20}{3}\right\rfloor=6, in one order or the other. They are nines complements. Finally, \mathop{\mathrm{ord}}_3(10)=1, so this synchronized digit is the entire repetend. ◻
Remark 4. The nines complement between classes (B) and (C) is the decimal manifestation of the universal identity k/n + (n-k)/n = 1, which pairs every element of \mathcal{F}(n) with its complement. The alignment classes correspond to the digit pair \{3, 6\}, the pair associated with the prime 3 in base ten.
The Alignment Formula and the Golden Threshold
Theorem 5. Let n = 3m with m \in \mathcal{S}_{10} and m \ge 1. Then \alpha_{10}(n) \;=\; \frac{2m - 1}{3m - 1}.
Proof. By Lemma 3, aligned elements of \mathcal{F}(n) are those in classes (A) and (B). Class (A) contributes m - 1 multiples of 3 in \{1, \ldots, 3m-1\}. Class (B) contributes m integers congruent to 1 \pmod 3. Class (C), the remaining m elements, contributes nothing. The total is (m-1) + m = 2m-1 out of 3m-1. ◻
The normalization in Definition 2 requires n\ge2; n=1 is not assigned an alignment. The present theorem starts at n=3. The edge case n=2 occurs in Theorem 10 with p=2 and m=1, where the formula gives \alpha_b(2)=1.
Corollary 6. As m \to \infty through \mathcal{S}_{10}, \alpha_{10}(3m) \;\longrightarrow\; \frac{2}{3}. The convergence is monotone from below, since \frac{d}{dm}\frac{2m-1}{3m-1} \;=\; \frac{1}{(3m-1)^2} \;>\; 0.
| m | n = 3m | \alpha_{10}(n) | Decimal |
|---|---|---|---|
| 1 | 3 | 1/2 | 0.5000 |
| 2 | 6 | 3/5 | 0.6000 |
| 4 | 12 | 7/11 | 0.6364 |
| 5 | 15 | 9/14 | 0.6429 |
| 8 | 24 | 15/23 | 0.6522 |
| 10 | 30 | 19/29 | 0.6552 |
| 16 | 48 | 31/47 | 0.6596 |
| 32 | 96 | 63/95 | 0.6632 |
| 100 | 300 | 199/299 | 0.6656 |
| 256 | 768 | 511/767 | 0.6662 |
We now ask for which m the alignment exceeds 1/\varphi, where \varphi = (1 + \sqrt{5})/2 is the golden ratio?
Theorem 7. Let m \ge 1 be a positive integer. Then \frac{2m - 1}{3m - 1} \;\ge\; \frac{1}{\varphi} \quad\Longleftrightarrow\quad m \;\ge\; \varphi^{2}. Since 2<\varphi^{2}<3, the integer threshold is m\ge3.
Proof. The inequality is equivalent to \varphi(2m - 1) \ge 3m - 1, or m(2\varphi - 3) \;\ge\; \varphi - 1. Using \varphi - 1 = 1/\varphi and 2\varphi - 3 = \sqrt{5} - 2 > 0, we may divide. m \;\ge\; \frac{1/\varphi}{\sqrt{5} - 2}. Rationalizing the denominator and applying \varphi(\sqrt{5} - 2) = (3 - \sqrt{5})/2 yields m \;\ge\; \frac{2}{3 - \sqrt{5}} \;=\; \frac{3 + \sqrt{5}}{2} \;=\; \varphi^{2}. Finally, 1<\sqrt5<3 gives 2<\varphi^2<3, so an integer m satisfies this inequality exactly when m\ge3. ◻
The identity \varphi^{2} = \varphi + 1 is the defining algebraic property of the golden ratio [4]. Its appearance here is not assumed but derived. It emerges from the interaction between the threshold 1/\varphi and the ternary residue structure modulo three.
Remark 8. Under the prime and one-digit hypotheses of Theorem 10, any threshold t\in(2/5,2/3) separates the same primes, since the alignment limit 2/p exceeds t only for p\in\{2,3\}. The golden ratio is one such threshold. Under the reciprocal-square criterion of Section 6, 1/\varphi is the distinguished value in this interval.
Corollary 9. Among denominators n = 3m with m \in \mathcal{S}_{10}, the repetend alignment exceeds 1/\varphi if and only if m \ge 4. The first admissible denominator is n = 12, with \alpha_{10}(12) = 7/11 \approx 0.6364.
Proof. By Theorem 7 the integer threshold is m \ge 3. Since 3 \notin \mathcal{S}_{10}, the smallest admissible value is m = 4 = 2^{2}. ◻
Why the Prime Three
Theorem 7 shows that among denominators n=3m with m\in\mathcal S_{10}, the golden threshold produces a clean classification. We now show this is no coincidence of base ten. The alignment formula extends to every base b and prime p\nmid b with b\equiv1\pmod p, and within that class the prime 3 is selected by the reciprocal-golden criterion.
Theorem 10 (General prime threshold). Let b \ge 2, let p \nmid b be a prime with b \equiv 1 \pmod p. Then, for every m\in\mathcal S_b, \alpha_b(pm) \;=\; \frac{2m - 1}{pm - 1}. There exists an m\in\mathcal S_b satisfying \alpha_b(pm)\ge1/\varphi if and only if p\in\{2,3\}. Among odd primes this holds only for p=3, with threshold m \ge \varphi^{2} = \varphi + 1. For p \ge 5, no positive m\in\mathcal S_b satisfies \alpha_b(pm) \ge 1/\varphi.
Proof. Choose s \ge 0 with m \mid b^{s} and set u = b^{s}/m. Since every prime factor of m divides b and p \nmid b, we have p \nmid m, and hence p \nmid u. Then \frac{k}{pm} \;=\; \frac{ku}{p b^{s}}, so the base-b expansion of k/(pm) is obtained from that of ku/p by shifting the radix point s places. The repetend of k/(pm) is therefore determined by the residue of ku modulo p.
Since b \equiv 1 \pmod p, we have \mathop{\mathrm{ord}}_p(b) = 1, so every nonzero residue class modulo p yields a one-digit repetend. Write b = 1 + cp with c \ge 1. For 1 \le j \le p-1, the one-digit repetend of j/p is \left\lfloor \frac{bj}{p} \right\rfloor \;=\; \left\lfloor cj + \frac{j}{p} \right\rfloor \;=\; cj, since 1 \le j \le p-1 gives 0 < j/p < 1. Thus distinct nonzero residues modulo p give distinct one-digit repetends. Since p \nmid u, multiplication by u permutes the nonzero residues.
Write k = pq + r with r \in \{0, 1, \ldots, p-1\}. If r = 0, then k/(pm) = q/m terminates. If r \ne 0, then ku \equiv ru \pmod p, and the map r \mapsto ru is a bijection on \{1, \ldots, p-1\}. Hence the p-1 nonzero residues produce p-1 distinct one-digit repetends, exactly one of which (from r = 1) matches the repetend of 1/(pm).
So the aligned elements are the m-1 multiples of p and the m integers congruent to 1 \pmod p, giving (m-1) + m = 2m-1 out of pm-1.
For the threshold, the inequality (2m - 1)/(pm - 1) \ge 1/\varphi reduces to m(2\varphi - p) \;\ge\; \varphi - 1. Since 2<\sqrt5<3, we have 3<2\varphi<4. The primes below 2\varphi are exactly 2 and 3. For p = 2 the alignment formula gives \alpha_b(2m) = (2m-1)/(2m-1) = 1 for all m, so the threshold is exceeded trivially. For p = 3 the threshold is m \ge \varphi^{2}, as in Theorem 7. Such an admissible value always exists. The congruence b\equiv1\pmod3 gives b\ge4, and m=b belongs to \mathcal S_b. If p \ge 5, then 2\varphi - p < 0, so dividing reverses the inequality and gives m \;\le\; \frac{\varphi - 1}{2\varphi - p} \;<\; 0, which has no positive integer solutions. ◻
Among odd primes, p = 3 is therefore the unique prime for which the golden threshold produces a non-trivial classification. The prime 2 is degenerate, since the alignment formula gives \alpha_b(2m) = 1 for all admissible m. The critical resolution \varphi^{2} = \varphi + 1 is the algebraic signature of the selection.
Corollary 11. Let t \in (2/5,\,2/3). Under the hypotheses of Theorem 10, the inequality \alpha_b(pm) \ge t admits an admissible positive integer solution m \in \mathcal{S}_b if and only if p \in \{2, 3\}. Among odd primes, this occurs only for p = 3.
Proof. For p = 2, the alignment formula gives \alpha_b(2m) = 1, so the inequality holds for all admissible m. For p = 3, since t < 2/3 and the alignment increases monotonically toward 2/3, sufficiently large admissible m \in \mathcal{S}_b satisfy the inequality. Such values are unbounded because b^r\in\mathcal S_b for every r\ge0. For p \ge 5, \alpha_b(pm) < 2/p \le 2/5 < t, so no positive integer m satisfies the inequality. ◻
Remark 12. The p=3 specialization holds in any base b \ge 2 satisfying b \equiv 1 \pmod 3, namely b \in \{4, 7, 10, 13, 16, 19, \ldots\}. If 3 \mid b, then the p = 3 case is absorbed into the terminating part, and this one-digit periodic mechanism does not apply. Other primes may satisfy the theorem’s hypotheses depending on the base. Within the stated hypotheses, the algebraic calculation is independent of the base; its applicable range is not.
The Self-Referential Cubic
Remark 8 shows that threshold separation alone does not distinguish 1/\varphi. At p=3, however, the critical resolution is \varphi^2=(1/\varphi)^{-2}. This equality suggests the closure condition m^{*}(\tau,p)=\tau^{-2}. We now classify its prime solutions.
Definition 13. For a prime p \ge 3 and a threshold \tau with 0 < \tau < 2/p, the critical resolution is the positive real value m^{*}(\tau, p) at which (2m - 1)/(pm - 1) = \tau. Solving for m gives m^{*}(\tau, p) \;=\; \frac{1 - \tau}{2 - p\tau}. The classification at (\tau, p) is self-referential if the critical resolution equals the reciprocal square of the threshold. m^{*}(\tau, p) \;=\; \frac{1}{\tau^{2}}.
Remark 14. For p = 2 the alignment formula gives \alpha_b(2m) = 1 for all m, so the critical resolution is not defined in the same sense. We treat p = 2 separately.
The self-referential condition requires \tau to satisfy \tau^{3} - \tau^{2} - p\tau + 2 \;=\; 0. Write f_p(\tau)=\tau^{3}-\tau^{2}-p\tau+2. Although the self-referential threshold was defined only for primes p\ge3, the same polynomial can be evaluated at the degenerate prime p=2 for the factorization comparison below. For p\ge3, it satisfies f_p(0)=2>0 and f_p(1)=2-p<0, so (1) admits a root in (0, 1) for every such p.
Lemma 15. For p \ge 3, the root of (1) in (0,1) is unique, and the corresponding critical resolution m^* is positive.
Proof. On [0,1] we have f_p'(\tau) = 3\tau^2 - 2\tau - p. Since 3\tau^2 - 2\tau \le 1 on this interval and p \ge 3, we get f_p'(\tau) \le 1 - 3 < 0. Hence f_p is strictly decreasing on [0,1], and the root is unique. Moreover, f_p(2/p) = 4(2-p)/p^3 < 0, so the root lies in (0, 2/p). Since \tau < 2/p implies 2 - p\tau > 0, the critical resolution m^* = (1-\tau)/(2-p\tau) is positive. ◻
What distinguishes p = 3 is not the existence of a solution but its algebraic nature.
Theorem 16. For prime p, the polynomial f_p factors over \mathbb{Q} if and only if p \in \{2, 3, 5\}, with factorizations \begin{aligned} p = 2: &\quad (\tau - 1)(\tau^{2} - 2), \\ p = 3: &\quad (\tau - 2)(\tau^{2} + \tau - 1), \\ p = 5: &\quad (\tau + 2)(\tau^{2} - 3\tau + 1). \end{aligned} For p \ge 7, the cubic is irreducible over \mathbb{Q}.
Proof. Direct expansion gives \begin{aligned} (\tau-1)(\tau^2-2) &=\tau^3-\tau^2-2\tau+2,\\ (\tau-2)(\tau^2+\tau-1) &=\tau^3-\tau^2-3\tau+2,\\ (\tau+2)(\tau^2-3\tau+1) &=\tau^3-\tau^2-5\tau+2. \end{aligned} By the rational root theorem, the possible rational roots of f_p are \pm 1, \pm 2. Direct evaluation gives f_p(1) = 2 - p, f_p(-1) = p, f_p(2) = 6 - 2p, f_p(-2) = 2p - 10. These vanish at p = 2, 3, 5 respectively; f_p(-1)=p never vanishes at a prime. For p \ge 7 no rational root exists. Since a reducible cubic over \mathbb{Q} has a rational root, this proves irreducibility for p \ge 7. ◻
The case p = 2 yields \tau = 1 (the interval boundary) and \tau = \pm\sqrt{2}, all outside (0,1), so no self-referential threshold arises. The two non-trivial factorizations are at p = 3 and p = 5.
Theorem 17. The self-referential thresholds at p = 3 and p = 5 are \tau(3) \;=\; \frac{\sqrt{5} - 1}{2} \;=\; \frac{1}{\varphi}, \qquad \tau(5) \;=\; \frac{3 - \sqrt{5}}{2} \;=\; \frac{1}{\varphi^{2}}. Both lie in the golden field \mathbb{Q}(\sqrt{5}). They are consecutive powers of 1/\varphi.
For p \ge 7, the self-referential threshold is a cubic irrational and lies in no quadratic field.
Proof. The quadratic factors at p = 3 and p = 5 have discriminants 1 + 4 = 5 and 9 - 4 = 5 respectively, so their roots lie in \mathbb{Q}(\sqrt{5}). Put \rho=\frac{\sqrt5-1}{2}. Then \rho^2+\rho-1=0, so substitution into the p=3 factor shows f_3(\rho)=0. Since 0<\rho<1 and \rho=1/\varphi, uniqueness gives \tau(3)=1/\varphi.
Now put \sigma=\rho^2=(3-\sqrt5)/2. Direct substitution gives \sigma^2-3\sigma+1=0, so f_5(\sigma)=0. Since 0<\sigma<1, uniqueness gives \tau(5)=\sigma=1/\varphi^2.
For p \ge 7, irreducibility by Theorem 16 forces the roots into a cubic extension of \mathbb{Q}, which cannot embed in any quadratic field. ◻
Corollary 18. Among self-referential thresholds associated to primes, the only one lying in the separating interval (2/5,\,2/3) is \tau = 1/\varphi, occurring at p = 3.
Proof. The case p = 2 produces no threshold in (0,1). For p = 3, Theorem 17 gives 2/5<\tau=1/\varphi<2/3. Both inequalities are exact. 5\sqrt5>9 and 3\sqrt5<7, whose squares are 125>81 and 45<49. For p=5, the threshold satisfies 1/\varphi^2<2/5 because 5\sqrt5>11, whose square is 125>121. For p \ge 7, Lemma 15 gives \tau < 2/p \le 2/7 < 2/5. Thus no other prime contributes a self-referential threshold in (2/5,\, 2/3). ◻
Corollary 19. The minimal polynomial of the reciprocal golden ratio \tau = 1/\varphi, namely \tau^{2} + \tau - 1, divides the self-referential cubic if and only if p = 3. The resulting identity is \tau^{4} \;=\; 2 - 3\tau, where 3 is the unique prime annihilating the remainder of the polynomial division.
Proof. Dividing \tau^{3} - \tau^{2} - p\tau + 2 by \tau^{2} + \tau - 1 gives quotient \tau - 2 and remainder (3 - p)\tau. The remainder vanishes iff p = 3. The identity \tau^{4} = 2 - 3\tau then follows from \tau^{2} = 1 - \tau (valid for \tau = 1/\varphi, the positive root of \tau^2 + \tau - 1). ◻
The self-referential structure has three regimes.
When p = 3, the threshold is 1/\varphi and is governed by the minimal polynomial of 1/\varphi. The alignment limit 2/3 exceeds the threshold, producing the classification of Theorem 7. This is the unique self-referential threshold in the separating interval (Corollary 18).
When p = 5, the threshold is 1/\varphi^{2} and is governed by the companion quadratic \tau^{2} - 3\tau + 1. The alignment limit 2/5 exceeds this threshold but falls below 1/\varphi, so no classification at the golden threshold arises.
For p \ge 7, the thresholds are cubic irrationals outside \mathbb{Q}(\sqrt{5}). The golden ratio has no further role.
Thus exactly two prime cases produce self-referential thresholds in \mathbb{Q}(\sqrt5), namely p=3 and p=5. For every prime p\ge7, the threshold has degree three over \mathbb Q.
The Scope of the Selection
Terminating fractions as aligned
The alignment statistic counts terminating fractions as aligned. If they are excluded instead, the hypotheses of Theorem 10 leave m(p-1) nonterminating fractions, of which m align, and the resulting proportion is 1/(p-1). It is therefore 1/2 throughout the p=3 family and the golden threshold does not enter. Every theorem above uses the stated terminating-aligned convention.
Connection to Midy’s theorem
Midy’s theorem [1, 2] and its generalizations study digit-sum relations within a single repetend. The present result addresses the collective organization of an entire set of fractions rather than the internal structure of one repetend. The nines complement between classes (B) and (C) is a manifestation of the same underlying arithmetic.
References
[1]E. Midy, De quelques propriétés des nombres et des fractions décimales périodiques, Nantes, 1836.
[2]W. G. Leavitt, A theorem on repeating decimals, Amer. Math. Monthly 74 (1967), no. 6, 669–673.
[3]G. H. Hardy and E. M. Wright, An Introduction to the Theory of Numbers, 6th ed., Oxford University Press, 2008.
[4]T. Koshy, Fibonacci and Lucas Numbers with Applications, vol. 1, 2nd ed., Wiley, 2017. https://doi.org/10.1002/9781118742327.