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For denominators n=3m with m=2^a5^c and a,c\ge0 integers, the proper fractions split into three residue classes. Counting terminating fractions together with those whose repetend matches 1/n gives the alignment \alpha_{10}(3m)=\frac{2m-1}{3m-1}. This exceeds 1/\varphi, where \varphi=(1+\sqrt5)/2, exactly when m\ge4 in the supported family. The real crossing occurs at m=\varphi^2. In a base b\equiv1\pmod p with p prime, the same count for b-supported m gives (2m-1)/(pm-1). Among odd primes only p=3 can cross the golden threshold; prime two has alignment one.
The reciprocal-square condition on the real critical resolution yields the cubic \tau^3-\tau^2-p\tau+2=0. For prime parameters it factors over \mathbb Q exactly at 2,3,5. Its thresholds at 3 and 5 are 1/\varphi and 1/\varphi^2; at primes p\ge7 they have degree three and lie in no quadratic field. Among prime-associated thresholds satisfying this condition, 1/\varphi is the unique value in (2/5,2/3).
Of the eleven proper fractions with denominator twelve, three terminate, four repeat the digit 3, and four repeat the digit 6. The reference fraction 1/12=0.08\overline3 belongs to the first repeating class. Counting the three terminating fractions together with the four matching fractions gives an alignment of 7/11. This is above the reciprocal golden ratio 1/\varphi\approx0.618. At denominator six the corresponding alignment is 3/5, below that threshold.
The count comes from residues modulo three. For every denominator n=3m whose multiplier m is built from powers of 2 and 5, the same three classes have sizes m-1, m, and m. Counting the terminating and matching classes gives (2m-1)/(3m-1). The terminating class is assigned full alignment by convention; the statistic measures membership in these two classes.
As m increases, this proportion rises toward 2/3. Solving for its crossing of 1/\varphi gives the real value \varphi^2, the reciprocal square of the threshold itself. The first admissible integer multiplier is m=4, giving denominator twelve. The reciprocal-square relation leads to the question of which prime families admit the same equality.
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 denominator n\ge2, write \mathcal F(n)=\{k/n:1\le k\le n-1\}. Comparing these fractions at a common long-division depth fixes the phase of every periodic tail. In the decimal family above, each nonterminating tail has only one repeating digit.
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 characterization of Section 6 classifies the additional relation between threshold and critical scale.
The sets \mathcal F(n) were first explored computationally using nfield [5]. The threshold separation was found computationally first; the algebraic proof came later.
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.
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 decimal family has just two nonzero remainder states.
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.
Midy’s theorem [1, 2] and its generalizations study digit-sum relations within a single repetend. Here the same complement arithmetic pairs classes across the complete set \mathcal F(3m).
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 count includes the m-1 terminating fractions. Excluding them from both the count and the normalization leaves m matching fractions among 2m nonterminating fractions, so that proportion is always 1/2. More generally, under the hypotheses of Theorem 10, the nonterminating proportion is m/[m(p-1)]=1/(p-1). Including the terminating class is what makes the alignment vary with m.
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]. At threshold 1/\varphi, the real crossing scale is \varphi^2, the reciprocal square of that threshold.
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\}.
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}. ◻
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 independent of decimal notation. 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 golden crossing at p=3 occurs at the real scale m=\varphi^2=1/\tau^2, where \tau=1/\varphi. This equality suggests the reciprocal-square condition m^*(\tau,p)=1/\tau^2. It is an additional condition on the real alignment profile, chosen from the observed equality. The classification below determines exactly which prime thresholds satisfy it. The real critical scale is distinct from the first admissible integer multiplier.
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). ◻
At p=3, the threshold lies below its alignment limit 2/3 and inside the separating interval (2/5,2/3). At p=5, the threshold 1/\varphi^2 lies below its alignment limit 2/5 and outside that interval. Exactly these two prime-associated reciprocal-square thresholds lie in \mathbb Q(\sqrt5). For every prime p\ge7, the threshold has degree three over \mathbb Q and belongs to no quadratic field.
[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.
[5]A. S. Petty, nfield, software repository. https://github.com/alexspetty/nfield
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