Ternary scale theorems: Difference between revisions

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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.

If gcd(''k'', ''n'') > 1. since ''n'' is odd, gcd(''k'', ''n'') is an odd number at least 3, and by well-formedness with respect to the generator, the generators must form more than 2 parallel chains. Thus, gcd(''k'', ''n'') = 1~~, and by~~ modular arithmetic ~~we have~~ ''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.)

If gcd(''k'', ''n'') > 1. since ''n'' is odd, gcd(''k'', ''n'') is an odd number at least 3, and by well-formedness with respect to the generator, the generators must form more than 2 parallel chains. 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.)

== Proposition 3 (Properties of even generator-offset ternary scales) ==

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 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'' &minus; 1)/2) notes.
-If gcd(''k'', ''n'') > 1. since ''n'' is odd, gcd(''k'', ''n'') is an odd number at least 3, and by well-formedness with respect to the generator, the generators must form more than 2 parallel chains. Thus, gcd(''k'', ''n'') = 1, and by modular arithmetic we have ''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'' &minus; 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.)
+If gcd(''k'', ''n'') > 1. since ''n'' is odd, gcd(''k'', ''n'') is an odd number at least 3, and by well-formedness with respect to the generator, the generators must form more than 2 parallel chains. Thus, gcd(''k'', ''n'') = 1. By modular arithmetic, if 0 &le; 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'' &minus; 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.)
 == Proposition 3 (Properties of even generator-offset ternary scales) ==