Unit II — Five Species of Inequality & the Equality Unfolding Tree — Critical Commentary

Algebraic Proof of the Three Commands, SL(2, Z) Projective Action & Canon Reception · Book I, Chs. 17–23

Nicomachus I.17–22 — Equality & the Five Species of Relative Inequality Every greater ratio m : n = q + r/n classifies uniquely by quotient q and remainder r 0 · Equality (ἰσότης) 1 : 1 m = 1, n = 1 1 · Multiplex (πολλαπλάσιος) 2 : 1 (duple) m = 2, n = 1 2 · Superparticularis (ἐπιμόριος) 3 : 2 (hemiolios) m = 3, n = 2 3 · Superpartiens (ἐπιμερής) 5 : 3 (epidimoiros) m = 5, n = 3 4 · Mult. superparticularis 5 : 2 (2 + 1/2) m = 5, n = 2 5 · Mult. superpartiens 8 : 3 (2 + 2/3) m = 8, n = 3
Exact bar-length comparison of Equality (1:1) and the five Nicomachean species of greater inequality, computed in geometor.nicomachus.classify_nicomachean_ratio.

I. Why Every Node Has Square Extremes & Bijects with Coprime Rationals (I.23)

Nicomachus observes with wonder at the close of Book I (I.23.16, Hoche p. 71; D’Ooge p. 229) that in every triple generated from Equality (1, 1, 1), the first and third terms are squares while the middle term is their geometric mean. Exact algebra reveals why Nicomachus’s three prostagmata (attributed by Iamblichus [iamblichus:pistelli:1894] and Theon of Smyrna [theon:hiller:1878, p. 107] to Eratosthenes and Adrastus) are nothing less than the symmetric square representation \(\mathrm{Sym}^2(\mathbb{Z}^2)\) of the Euclidean unimodular group \(\mathrm{SL}_2(\mathbb{Z}_{\ge 0})\)!

Let a coprime pair \((p, q)\) with \(0 < p \le q\) represent the three-term geometric progression:

\[(a, b, c) = (p^2,\; p q,\; q^2).\]
  1. Direct Command \(D(a, b, c) = (a,\; a + b,\; a + 2b + c)\):

    \[D(p^2, pq, q^2) = \bigl(p^2,\; p(p + q),\; (p + q)^2\bigr),\]

    which acts on the underlying root ratio \(\frac{q}{p}\) by \((p, q) \mapsto (p,\; p + q)\), i.e. \(\frac{q}{p} \mapsto \frac{q}{p} + 1\)! Hence if \(\frac{q}{p} = m : 1\) is a multiple, \(D\) increments it to \((m+1) : 1\); if \(\frac{q}{p} = \frac{n+1}{n}\) is superparticular ((4,6,9)), \(D\) adds \(1\) to produce \(\frac{2n+1}{n}\), a multiple superparticular ((4,10,25)); and if \(\frac{q}{p} = \frac{n+r}{n}\) is superpartient ((9,15,25)), \(D\) adds \(1\) to produce \(\frac{2n+r}{n}\), a multiple superpartient ((9,24,64)).

  2. Reversed (*Anastrophe*) Command \(R(a, b, c) = D(c, b, a)\):

    \[R(p^2, pq, q^2) = D(q^2, pq, p^2) = \bigl(q^2,\; q(q + p),\; (q + p)^2\bigr),\]

    which acts on the root pair by \((p, q) \mapsto (q,\; p + q)\), mapping the ratio \(\frac{q}{p} > 1\) to \(\frac{p + q}{q} = 1 + \frac{p}{q} \in (1, 2)\)! When \(p = 1\) (a multiple \(q : 1\)), \(1 + \frac{1}{q} = \frac{q + 1}{q}\) is strictly superparticular ((4,6,9), (9,12,16)); when \(p \ge 2\) (any non-multiple \(\frac{q}{p}\) with \(\gcd(p, q) = 1\)), \(1 + \frac{p}{q} = \frac{q + p}{q}\) has remainder \(1 < p < q\) and is strictly superpartient ((9,15,25))!

Because \((p, q) \mapsto (p, p+q)\) and \((p, q) \mapsto (q, p+q)\) are the exact generators of the Calkin–Wilf / Stern–Brocot binary tree restricted to ratios \(\ge 1\) (since at the root \((1, 1)\) both branches coincide at (1, 2, 4)), every coprime rational inequality \(q : p > 1\) appears once and only once in Nicomachus’s tree (geometor.nicomachus.generate_equality_tree)!

II. Archytas’s Theorem on Superparticular Ratios & Timaeus Harmonics (I.19)

Among the five species, the superparticular (*epimorios*) \((n + 1) : n\) holds supreme status in Pythagorean harmonics (plato:timaeus:27d-36d; archytas:dk47b1):

  • The first three superparticulars are the musical fifth (\(3:2\), hēmiolios), fourth (\(4:3\), epitritos), and major third (\(5:4\), epitetartos), while \(9:8\) (epogdoos) is the whole tone.

  • As Archytas proved (preserved in Boethius, De Institutione Musica III.11, and Euclid’s Sectio Canonis Prop. 3), no geometric mean in rational integers exists between two terms in a superparticular ratio—because in Nicomachus’s tree every superparticular node \((n^2, n(n+1), (n+1)^2)\) has adjacent integer roots \(n\) and \(n+1\) with no integer between them, proving immediately that neither the octave’s fifth/fourth nor the whole tone \(9:8\) can be halved rationally!