Neutral and interordinal intervals in MOS scales: Difference between revisions
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*** Larger k-step = iL + (k-i)s | *** Larger k-step = iL + (k-i)s | ||
*** Y = Smaller k-step = (i-1)L + (k-i+1)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, | ** 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: | ||
*** <pre> | *** <pre>Y=L^A sL...LsL...LsL...LsL...LsL^B</pre> | ||
*** <pre> | *** <pre>X=L^CsL...LsL...LsL^D</pre> | ||
** 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. | ** 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+C+D+floor(nk/(a+b)) <= |X| <= 1+C+D+ceil(nk/(a+b)) | |||
*** |Y|-|X| >= (A+B)-(C+D) + floor((n+2)k/(a+b)) - ceil(nk/(a+b)) | |||
*** = (A+B)-(C+D)-1 + ceil((n+2)k/(a+b)) - ceil(nk/(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 kX(k+1) would become neutral or semiperfect. | * 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). | * 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). | ||