Neutral and interordinal intervals in MOS scales: Difference between revisions
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==== Proof ==== | ==== 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: | 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. | * 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. | * 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 (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. | ||
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*** <pre>X=L^CsL...LsL...LsL^D</pre> | *** <pre>X=L^CsL...LsL...LsL^D</pre> | ||
** Let r be the number of complete chunks in X, and let μ = n/(a+b). We'll obtain a contradiction. By the Discretizing Lemma, we have: | ** Let r be the number of complete chunks in X, and let μ = n/(a+b). We'll obtain a contradiction. By the Discretizing Lemma, we have: | ||
*** 1+A+B+floor( | *** 1+A+B+floor((r+2)μ) <= |Y| <= 1+A+B+ceil((r+2)μ) | ||
*** 1+C+D+floor(μr) <= |X| <= 1+C+D+ceil( | *** 1+C+D+floor(μr) <= |X| <= 1+C+D+ceil(rμ) | ||
*** -1 = |Y|-|X| >= (A+B)-(C+D) + floor( | *** -1 = |Y|-|X| >= (A+B)-(C+D) + floor((r+2)μ) - ceil(rμ) | ||
*** = (A+B)-(C+D)-1 + ceil( | *** = (A+B)-(C+D)-1 + ceil((r+2)μ) - ceil(rμ) | ||
*** >= (A+B)-(C+D)-1 + floor(2μ) | *** >= (A+B)-(C+D)-1 + floor(2μ) | ||
*** Hence, (C+D)-(A+B) >= floor(2μ). | *** Hence, (C+D)-(A+B) >= floor(2μ). | ||