Generator-offset property: Difference between revisions

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==== 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 ~~ceil~~(''n''/2) notes and on ''p′'' = ''p'' + offset we'll have ~~floor~~(''n''/2) notes.

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.

We must have gcd(''k'', ''n'') = 1. If not, 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.

By modular arithmetic we have ''rk'' ≡ ''k''/2 mod ''n'' iff ''r'' ≡ ~~ceil~~(''n''/2) mod ''n''. (~~Since gcd(~~2, ''n''~~) = 1, 2 is~~ multiplicatively invertible mod ''n''~~, and we can multiply both sides by 2 to check this~~.) This proves that the offset, which must be reached after ~~ceil~~(''n''/2) k-steps, is a ''k''/2-step, as desired. (If the offset wasn't reached in ~~ceil~~(''n''/2) steps, the two generator chains either wouldn't be disjoint or wouldn't have the assumed lengths.)

By modular arithmetic we have ''rk'' ≡ ''k''/2 mod ''n'' iff ''r'' ≡ (''n'' + 1)/2 mod ''n''. (Note that both 2 and ''k'' are 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. (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 GO scales) ===

@@ Line 120: / Line 120: @@
 ==== 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 ceil(''n''/2) notes and on ''p′'' = ''p'' + offset we'll have floor(''n''/2) notes.
+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.
 We must have gcd(''k'', ''n'') = 1. If not, 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.
-By modular arithmetic we have ''rk'' ≡ ''k''/2 mod ''n'' iff ''r'' ≡ ceil(''n''/2) mod ''n''. (Since gcd(2, ''n'') = 1, 2 is multiplicatively invertible mod ''n'', and we can multiply both sides by 2 to check this.) This proves that the offset, which must be reached after ceil(''n''/2) k-steps, is a ''k''/2-step, as desired. (If the offset wasn't reached in ceil(''n''/2) steps, the two generator chains either wouldn't be disjoint or wouldn't have the assumed lengths.)
+By modular arithmetic we have ''rk'' ≡ ''k''/2 mod ''n'' iff ''r'' ≡ (''n'' + 1)/2 mod ''n''. (Note that both 2 and ''k'' are 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. (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 GO scales) ===