Neutral and interordinal intervals in MOS scales: Difference between revisions

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 ** Chunk sizes in a mos differ by at most 1; call the two chunk sizes in aLbs m-1 >= 1 and m >= 2, and let r be the number of complete chunks in X. We'll obtain a contradiction.
 ** By the Discretizing Lemma, we have:
-*** 1+A+B+floor(n(k+2)/(a+b)) <= |Y| <= 1+A+B+ceil(n(k+2)/(a+b))
+*** 1+A+B+floor(n(r+2)/(a+b)) <= |Y| <= 1+A+B+ceil(n(r+2)/(a+b))
-*** 1+C+D+floor(nk/(a+b)) <= |X| <= 1+C+D+ceil(nk/(a+b))
+*** 1+C+D+floor(nr/(a+b)) <= |X| <= 1+C+D+ceil(nr/(a+b))
-*** -1 = |Y|-|X| >= (A+B)-(C+D) + floor(n(k+2)/(a+b)) - ceil(nk/(a+b))
+*** -1 = |Y|-|X| >= (A+B)-(C+D) + floor(n(r+2)/(a+b)) - ceil(nr/(a+b))
-*** = (A+B)-(C+D)-1 + ceil((2a+b)(k+2)/(a+b)) - ceil((2a+b)k/(a+b))
+*** = (A+B)-(C+D)-1 + ceil((2a+b)(r+2)/(a+b)) - ceil((2a+b)r/(a+b))
 *** > (A+B)-(C+D)-2 + (4a+2b)/(a+b)
 *** Hence, (C+D)-(A+B) > (4a+2b)/(a+b)-1.
 * 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).