Neutral and interordinal intervals in MOS scales: Difference between revisions
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*** = (A + B) − (C + D) − 1 + floor((r + 2)μ) − floor(rμ) | *** = (A + B) − (C + D) − 1 + floor((r + 2)μ) − floor(rμ) | ||
*** (by Lemma 1) ≥ (A + B) − (C + D) − 1 + floor(2μ) | *** (by Lemma 1) ≥ (A + B) − (C + D) − 1 + floor(2μ) | ||
*** Hence, floor(2μ) ≤ (C + D) − (A + B) ≤ 2*ceil(μ) − 2 = 2(ceil(μ) − 1) = 2*floor(μ). This is a contradiction | *** Hence, floor(2μ) ≤ (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. | ||
* 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. | * 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). | * 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). | ||