# Neutral and interordinal intervals in MOS scales

*This page assumes that the reader is familiar with TAMNAMS mos interval and step ratio names.*

Given a tuning of a primitive (i.e. single-period) mos pattern aLbs〈E〉 with arbitrary equave E in a specific tuning (i.e. with a specific hardness value for L/s), we may define two types of notes "in the cracks of" interval categories defined by aLbs〈E〉:

- Given 1 ≤
*k*≤ a + b − 1, the**neutral***k*-mosstep or*k*-step (abbrev. n*k*ms, n*k*s) is the interval exactly halfway between the smaller*k*-step and the larger*k*-step of the mos. When the mos is generated by a (perfect)*k*-step, this may instead be called the**semiperfect***k*-step (abbrev. sP*k*s), since it is halfway between the perfect and imperfect (either diminished or augmented, depending on whether the generator is bright or dark)*k*-step. The following*always*holds for a given interval class of*k*-steps between 1-steps and (a + b − 1)-steps, inclusive:- neutral
*k*-step = smaller*k*-step + c/2 = larger*k*-step − c/2, where c = L − s is the chroma of the mos.

- neutral
- Given 1 ≤
*k*≤ a + b − 2, and assuming that the larger*k*-step < the smaller (*k*+ 1)-step, the**interordinal**between*k*-steps and (*k*+ 1)-steps, denoted*k*×(*k*+ 1)(m)s (read "*k*cross (*k*+ 1) (mos)step" or "*k*inter (*k*+ 1) (mos)step"), is the interval exactly halfway between the larger*k*-step and the smaller (*k*+ 1)-step.- If the smaller (
*k*+ 1)-step is*strictly larger*than the larger*k*-step in*basic*aLbs,*k*×(*k*+ 1) is called a**proper interordinal**. If a > b, then aLbs〈E〉 has b − 1 proper interordinals. The following holds for any proper interordinal interval*k*-inter-(*k*+ 1)-step: *k*-inter-(*k*+ 1)-step = larger*k*-step + s/2 = smaller (*k*+ 1)-step − s/2.

- If the smaller (

Neutral *k*-steps generalize neutral interval categories based on the diatonic mos, which are:

- neutral 1-diastep = neutral 2nd (A-Bd)
- neutral 2-diastep = neutral 3rd (A-Ct)
- semiperfect 3-diastep = semiperfect 4th (A-Dt)
- semiperfect 4-diastep = semiperfect 5th (A-Ed)
- neutral 5-diastep = neutral 6th (A-Ft)
- neutral 6-diastep = neutral 7th (A-Gt)

Though the term *interordinal* is intended to be JI-agnostic and generalizable to non-diatonic mosses, the term comes from the fact that *k*-steps in the diatonic mos are conventionally called "(*k* + 1)ths". Proper interordinals in other mosses generalize diatonic interordinals (aka "interseptimals"), which are:

- 1-inter-2-diastep = "2nd-inter-3rd" = semifourth = chthonic = ultramajor 2nd
- 2-inter-3-diastep = "3rd-inter-4th" = semisixth = naiadic = ultramajor 3rd
- 4-inter-5-diastep = "5th-inter-6th" = semitenth = cocytic = inframinor 6th
- 5-inter-6-diastep = "6th-inter-7th" = semitwelfth = ouranic = inframinor 7th

Given a primitive mos aLbs with a > b, it's easy to observe the following properties of the simplest equal tunings for the mos, due to the way they divide the small step (s) and the chroma (c = L − s). Note that s separates adjacent ordinal categories (i.e. interval classes) while c separates larger and smaller intervals in the same ordinal category.

- The basic equal tuning (2a + b)-edE contains neither neutrals nor interordinals, since both s and c are one edo step. (For the diatonic mos 5L2s, this tuning is 12edo.)
- The monohard equal tuning (3a + b)-edE contains neutrals of that mos but not interordinals, since c is two edo steps but s is one edo step. (For diatonic, this tuning is 17edo.)
- The monosoft equal tuning (3a + 2b)-edE contains interordinals but not neutrals, since c is one edo step and s is two edo steps. (For diatonic, this tuning is 19edo.)
- 2(2a + b)-edE, twice the basic equal tuning, contains both types of intervals, since both c and s are two edo steps. (For diatonic, this tuning is 24edo.)

## Examples

These charts show where neutrals and interordinals corresponding to example 2/1-equivalent mosses are located in 2(2a+b)-edo (twice the basic edo tuning). Intervals larger than 600c have been omitted, as the structure is symmetrical about 600c.

### Diatonic (5L2s)

Basic 5L2s (diatonic, dia-): 12edo Parent mos: soft 2L5s (pentic, pt-) 1\24 2\24 m1dias (m2nd) 3\24 n1dias (n2nd) 4\24 M1dias (M2nd) == m1pts 5\24 1×2dias (2nd×3rd) == n1pts 6\24 m2dias (m3rd) == M1pts 7\24 n2dias (n3rd) == 1×2pts 8\24 M2dias (M3rd) == d2pts 9\24 2×3dias (3rd×4th) == sP2pts 10\24 P3dias (P4th) == P2pts 11\24 sP3dias (sP4th) 12\24 A3/d4dias (A4th/d5th) == 2×3pts

### m-chromatic (7L5s)

Basic 7L5s (m-chromatic, mchr-): 19edo Parent mos: soft 5L2s (diatonic, dia-) 1\38 2\38 m1s 3\38 n1s 4\38 M1s == m1dias (m2nd) 5\38 1×2s == n1dias (n2nd) 6\38 m2s == M1dias (M2nd) 7\38 n2s 8\38 M2s/m3s == 1×2dias (2nd×3rd) 9\38 n3s 10\38 M3s == m2dias (m3rd) 11\38 3×4s == n2dias (n3rd) 12\38 m4s == M2dias (M3rd) 13\38 n4s 14\38 M4s/d5s == 2×3dias (3rd×4th) 15\38 sPs 16\38 P5s == P3dias (P4th) 17\38 5×6s == sP3dias (sP4th) 18\38 m6s == A3dias (A4th) 19\38 n6s == 3×4dias (4th×5th)

### Manual (4L1s)

Basic 2/1-equivalent 4L1s (manual, man-): 9edo Parent mos: soft 1L3s (antetric, att-) 1\18 2\18 d1mans 3\18 sP1mans 4\18 P1mans == P1atts 5\18 1×2mans == sP1atts 6\18 m2mans == A1atts 7\18 n2mans == 1×2atts 8\18 M2mans == m2atts 9\18 2×3mans == n2atts

### Oneirotonic (5L3s)

Basic 5L3s (oneirotonic, onei-): 13edo Parent mos: soft 3L2s (anpentic, apt-) 1\26 2\26 m1oneis 3\26 n1oneis 4\26 M1oneis == m1apts 5\26 1×2oneis == n1apts 6\26 m2oneis == M1apts 7\26 n2oneis 8\26 M2/d3oneis == 1×2apts 9\26 sP3oneis 10\26 P3oneis == P2apts 11\26 3×4oneis == sP2apts 12\26 m4oneis == A2apts 13\26 n4oneis == 2×3apts

## Interordinal Theorem

The Interordinal Theorem relates the neutral (resp. interordinal) intervals of aLbs (with a > b) with the interordinal (resp. neutral) intervals of its parent mos, bL(a − b)s.

### Statement

Suppose a > b and gcd(a, b) = 1.

- Every interordinal of basic aLbs〈E〉 is a neutral or semiperfect interval of the parent mos bL(a-b)s〈E〉.
- Every interordinal interval of the parent mos bL(a-b)s〈E〉 of basic aLbs〈E〉 is a neutral or semiperfect interval of basic aLbs〈E〉.
- Except the neutral/semiperfect 1-step and the neutral/semiperfect (a+b-1)-step, every neutral or semiperfect interval of basic aLbs〈E〉 is a proper interordinal of bL(a-b)s〈E〉. The number (b − 1) counts the places in 2(2a+b)edE (twice the basic mos tuning for aLbs〈E〉) where the parent's interordinal is improper, being two steps away, instead of one step away, from each of the adjacent ordinal categories.

### Preliminaries for the proof

Below we assume that the equave is 2/1, but the proof generalizes to any equave.

Consider a primitive mos aLbs. Recall that (b − 1) satisfies:

(b − 1) = |(brightest mode of basic aLbs, ignoring equaves) ∩ (darkest mode of basic aLbs, ignoring equaves)| = #{k : 0 < k < a+b and larger k-step of basic aLbs = smaller (k+1)-step of basic aLbs} = # of "improprieties".

Also recall that the following are equivalent for a mos aLbs:

- a > b.
- The parent mos, which is bL(a-b)s, has steps L + s and L. In this context, since aLbs is assumed to have hardness 2/1, bL(a-b)s has hardness 3/2 thus is strictly proper.

To prove the theorem, we need a couple lemmas.

### Lemma 1

Let n and k be integers, and let x be a real number such that kx is not an integer. Then floor((n+k)x) − floor(nx) ≥ floor(kx).

#### Proof

floor((n+k)x) − floor(nx) = -1 + ceil((n+k)x) + ceil(-nx) ≥ ceil((n+k)x − nx) − 1 = ceil(kx) − 1 = floor(kx).

### Discretizing Lemma

Consider an m-note maximally even mos of an n-equal division, and let 1 ≤ k ≤ m-1. Then a k-step of that mos is either floor(nk/m)\n or ceil(nk/m)\n.

#### Proof

The circular word formed by stacking k-steps of the mos is itself a maximally even mos, considered as a subset of a kn-note equal tuning. One step of an m'-note maximally even mos of an n'-note equal tuning is either floor(n'/m')\n'<equave> or ceil(n'/m')\n'<equave>, implying the lemma.

### Proof of Theorem

Let n = 2a + b (the basic edo tuning of aLbs) and suppose that m\(2n) is an interordinal (where m must be odd) between k-steps and (k+1)-steps, denoted k×(k+1)ms. Recall that:

- In basic aLbs, s = 1\n = 2\2n.
- A concrete mos tuning is improper iff its hardness is > 2/1 and the number of s steps it has is > 1.

Part (1) and (2) take some step size arithmetic:

- Smaller k+1-step of aLbs minus larger k-step of aLbs ≥ 0, with equality at improprieties (essentially by definition). At the values of k and k+1 that are proper, this equals s.
- To see why, suppose the difference is L (here k ≥ i ≥ 1, 0 < k < k+1 < a+b):
- X = Larger (k+1)-step = (i+2)L + (k-i-1)s
- Smaller (k+1)-step = (i+1)L + (k-i)s
- Larger k-step = iL + (k-i)s
- Y = Smaller k-step = (i-1)L + (k-i+1)s

- Since the smaller k-step has two more s steps than the larger (k+1)-step, Y has two more complete chunks of L's than X:
Y=L^A sL...LsL...LsL...LsL...LsL^B

X=L^CsL...LsL...LsL^D

- Let r be the number of complete chunks (maximal contiguous strings of L's) in X, and let μ = n/(a+b). Note that the spacings between the s steps of a mos form a maximally even mos, and that since a > b, 3/2 < μ < 2. By the Discretizing Lemma, we have (|·| denotes length):
- 1+A+B+floor((r+2)μ) ≤ |Y| ≤ 1+A+B+ceil((r+2)μ)
- 1+C+D+floor(rμ) ≤ |X| ≤ 1+C+D+ceil(rμ)
- -1 = |Y|-|X| ≥ (A+B)-(C+D) + floor((r+2)μ) − ceil(rμ)
- = (A+B)-(C+D)-1 + floor((r+2)μ) − floor(rμ)
- (Lemma 1) ≥ (A+B)-(C+D)-1 + floor(2μ)
- Hence, (C+D)-(A+B) ≥ floor(2μ).
- Also, (C+D)-(A+B) ≤ 2*ceil(μ)-2 = 2(ceil(μ)-1) = 2*floor(μ). This is a contradiction: as 3/2 < μ < 2, we have floor(2μ) − 2*floor(μ) = 1.

- To see why, suppose the difference is L (here k ≥ i ≥ 1, 0 < k < k+1 < a+b):
- As s is the chroma of bL(a-b)s, it
*would*be the difference between major and minor intervals in the parent mos, assuming these interval sizes (smaller k+1-step, larger k-step) occur in the parent; so k×(k+1) would become neutral or semiperfect. - To show that these actually occur in bL(a-b)s, consider smaller and larger j-steps ( 1 ≤ j ≤ a-1) in the parent mos. These intervals also occur in the mos aLbs separated by s, and the number of j’s (“junctures”) that correspond to these places in aLbs is exactly a-1. These j's correspond to values of k such that larger k-step < smaller k+1-step. Note that we are considering “junctures” between k-steps and k+1-steps in aLbs, excluding k = 0 and k = a+b-1, so the total number of “junctures” to consider is finite, namely a+b-2. This proves parts (1) and (2).

Part (3) is also immediate now: when larger k-step = smaller k+1-step, larger k+1-step − smaller k-step = 2(L-s) = 2s = L. The step L is 4 steps in 2n-edo.

### Corollary

If a > b, then aLbs〈E〉 has b − 1 proper interordinals.