Neutral and interordinal intervals in MOS scales: Difference between revisions
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Let n and k be integers, and let x be a real number such that kx is not an integer. Then floor((n + k)x) − floor(nx) ≥ floor(kx). | Let n and k be integers, and let x be a real number such that kx is not an integer. Then floor((n + k)x) − floor(nx) ≥ floor(kx). | ||
==== Proof ==== | ==== Proof ==== | ||
floor((n + k)x) − floor(nx) = −1 + ceil((n + k)x) + ceil(−nx) ≥ ceil((n + k)x − nx) − 1 = ceil(kx) − 1 = floor(kx). | floor((n + k)x) − floor(nx) = −1 + ceil((n + k)x) + ceil(−nx) ≥ ceil((n + k)x − nx) − 1 = ceil(kx) − 1 = floor(kx). {{qed}} | ||
=== Discretizing Lemma === | === Discretizing Lemma === | ||
Consider an m-note [[maximal evenness|maximally even]] mos of an n-equal division, and let 1 ≤ k ≤ m − 1. Then a k-step of that mos is either floor(nk/m)\n or ceil(nk/m)\n. | Consider an m-note [[maximal evenness|maximally even]] mos of an n-equal division, and let 1 ≤ k ≤ m − 1. Then a k-step of that mos is either floor(nk/m)\n or ceil(nk/m)\n. | ||
==== Proof ==== | ==== Proof ==== | ||
The circular word formed by stacking k-steps of the mos is itself a maximally even mos, considered as a subset of a kn-note equal tuning. One step of an m'-note maximally even mos of an n'-note equal tuning is either floor(n'/m')\n'{{angbr|E}} or ceil(n'/m')\n'{{angbr|E}}, implying the lemma. | The circular word formed by stacking k-steps of the mos is itself a maximally even mos, considered as a subset of a kn-note equal tuning. One step of an m'-note maximally even mos of an n'-note equal tuning is either floor(n'/m')\n'{{angbr|E}} or ceil(n'/m')\n'{{angbr|E}}, implying the lemma. {{qed}} | ||
=== Proof of Theorem === | === Proof of Theorem === | ||