Ternary scale theorems: Difference between revisions
m →Theorem 2 (Odd generator-offset scales are SGA): Commenting this out until further review. |
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and use the vectors (-1, 2) and (ceil(n/2), 1) as the Fokker block chromas. A rank-3 Fokker block has the property that tempering out by each of the chromas gives two mosses. These correspond to two of the temperings '''X''' = '''Y''', '''Y''' = '''Z''' and '''X''' = '''Z'''. The third tempering follows by symmetry (by taking the other chirality). {{qed}} | and use the vectors (-1, 2) and (ceil(n/2), 1) as the Fokker block chromas. A rank-3 Fokker block has the property that tempering out by each of the chromas gives two mosses. These correspond to two of the temperings '''X''' = '''Y''', '''Y''' = '''Z''' and '''X''' = '''Z'''. The third tempering follows by symmetry (by taking the other chirality). {{qed}} | ||
== Theorem 2 (Odd generator-offset scales are SGA) == | |||
Suppose that a periodic scale satisfies the following: | Suppose that a periodic scale satisfies the following: | ||
* is generator-offset | * is generator-offset | ||
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=== Proof === | === Proof === | ||
Assume that the generator is a ''k''-step and ''k'' is even. (If ''k'' is not even, invert the generator.) On some note ''p'' we have a chain of (''n'' + 1)/2 notes and on ''p′'' = ''p'' + offset we'll have (''n'' − 1)/2) notes. | Assume that the generator '''g''' is a ''k''-step and ''k'' is even. (If ''k'' is not even, invert the generator.) On some note ''p'' we have a chain of (''n'' + 1)/2 notes and on ''p′'' = ''p'' + offset we'll have (''n'' − 1)/2) notes. | ||
Assume gcd(''k'', ''n'') > 1, and by eliminating 3, 5, and 7, assume ''n'' ≥ 9. Since ''n'' is odd, ''d'' = gcd(''k'', ''n'') is an odd number at least 3, and by well-formedness with respect to the generator, there must be a circle of ''n''/''d'' < floor(''n''/2) notes formed by '''g''', contrary to the assumption of GO. Thus, gcd(''k'', ''n'') = 1. | |||
By modular arithmetic, if 0 ≤ ''r'' < ''n'', ''rk'' ≡ ''k''/2 mod ''n'' iff ''r'' ≡ (''n'' + 1)/2 mod ''n''. (Note that both 2 and ''k'' are coprime with ''n'', hence multiplicatively invertible mod ''n''.) This proves that the offset, which must be reached after (''n'' + 1)/2 ''k''-steps, is a ''k''/2-step, as desired. (As [''k''] is a generator of ℤ/''n'', stacking (''n'' − 1)-many ''k''-steps must visit every note exactly once. Thus if the offset wasn't reached in (''n'' + 1)/2 steps, the two generator chains either wouldn't be disjoint or wouldn't have the assumed lengths.) {{qed}} | By modular arithmetic, if 0 ≤ ''r'' < ''n'', ''rk'' ≡ ''k''/2 mod ''n'' iff ''r'' ≡ (''n'' + 1)/2 mod ''n''. (Note that both 2 and ''k'' are coprime with ''n'', hence multiplicatively invertible mod ''n''.) This proves that the offset, which must be reached after (''n'' + 1)/2 ''k''-steps, is a ''k''/2-step, as desired. (As [''k''] is a generator of ℤ/''n'', stacking (''n'' − 1)-many ''k''-steps must visit every note exactly once. Thus if the offset wasn't reached in (''n'' + 1)/2 steps, the two generator chains either wouldn't be disjoint or wouldn't have the assumed lengths.) {{qed}} | ||