Neutral and interordinal intervals in MOS scales

The Interordinal Theorem

Recall that the “impropriety number” (b - 1) of a primitive (i.e. single-period) mos aLbs 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}.

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

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

Discretizing Lemma

Consider an m-note maximally even mos of n-edo. 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.

Statemwnt of the theorem

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

The impropriety number (b - 1) of the mos aLbs counts the places in 2(2a+b)edo (twice the basic mos tuning for aLbs) where assigning interordinals to the parent mos of aLbs fails.
Every interordinal in basic aLbs (an interval that is exactly halfway between the larger k-step and the smaller (k+1)-step) is a neutral or semiperfect interval in the parent mos bL(a-b)s, but a neutral interval in basic aLbs is not always an interordinal interval in the parent mos.

Proof

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 kX(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) is easier to see: where basic aLbs is improper, larger k-step = smaller k+1-step, and larger k+1-step - smaller k-step = 2(L-s) = 2s = L. But the step L is not two steps in 2n-edo.

Part (2) takes some step size arithmetic:

Larger k+1-step of aLbs minus larger k-step of aLbs = Smaller k+1-step of aLbs minus smaller k-step of aLbs must be L, if 0 < k < k+1 < a+b. The reason is that larger 1-step = L, larger 2-step = LL, because there are more L’s than s’s.
Smaller k+1-step of aLbs minus larger k-step of aLbs >= 0, with =0 at improprieties. 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, X has two more complete chunks of L's than Y:
  - ```
  X=L^A sL...LsL...LsL...LsL...LsL^B
```
- ```
Y=L^CsL...LsL...LsL^D
```
- Chunk sizes in a mos differ by at most 1; call the two chunk sizes in aLbs m-1 >= 1 and m >= 2. We'll obtain a contradiction.
- First assume that m-1 is the more common chunk size.
- Now assume that m is the more common chunk size.
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 kX(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. 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 a+b-2. This proves part (2).