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Neutral and interordinal intervals in MOS scales: Difference between revisions

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Part (1) and (2) take some step size arithmetic:
Part (1) and (2) take 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 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.  
* 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):
** To see why, suppose the difference is L (here k ≥ i ≥ 1, 0 < k < k+1 < a+b):
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