Ternary scale theorems: Difference between revisions

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In case 1, let '''g'''1 = (2, 1) − (1, 1), '''g'''2 = (1, 2) − (2, 1), and '''g'''3 = (1, 1) − (''n''/2, 2) = ((−''n''/2 − 1)*'''g'''1 − ''n''/2*'''g'''2) mod '''e'''. We assume that '''g'''1, '''g'''2 and '''e''' are ℤ-linearly independent. We have the chain '''g'''1 '''g'''2 '''g'''1 '''g'''2 ... '''g'''1 '''g'''3 which visits every note in ''S''.

Since ''S'' is generator-offset it is well-formed with respect to '''g''' = ('''g'''2 + '''g'''1). Since '''g'''1 and '''g'''2 subtend the same number of steps, all multiples of the generator '''g''' must be even-steps, and those intervals that are "offset" by '''g'''1 must be odd-steps. Letting ''M'' be the subset consisting of all even-numbered notes (which are generated by '''g''') and considering ''M'' as a scale by dividing degree indices in ''M'' by two, ''M'' is well-formed with respect to '''g''', thus ''M'' (and its offset) must be a mos subset. Hence ('''g'''3 + '''g'''1), the imperfect generator of the mos generated by g, subtends the same number of steps as g. Thus '''g'''2 and '''g'''3 subtend the same number of steps, a fact we need in order to be able to substitute one instance of '''g'''2 with '''g'''3 in the next part.

Since ''S'' is generator-offset it is well-formed with respect to the aggregate generator '''g''' = ('''g'''2 + '''g'''1). Since '''g'''1 and '''g'''2 subtend the same number of steps, all multiples of the generator '''g''' must be even-steps, and those intervals that are "offset" by '''g'''1 must be odd-steps. Letting ''M'' be the subset consisting of all even-numbered notes (which are generated by '''g''') and considering ''M'' as a scale by dividing degree indices in ''M'' by two, ''M'' is well-formed with respect to '''g''', thus ''M'' (and its offset) must be a mos subset. Hence ('''g'''3 + '''g'''1), the imperfect generator of the mos generated by g, subtends the same number of steps as g. Thus '''g'''2 and '''g'''3 subtend the same number of steps, a fact we need in order to be able to substitute one instance of '''g'''2 with '''g'''3 in the next part.

Let ''r'' be odd and ''r'' ≥ 3. Consider the following abstract sizes for the interval class of ''k''-steps reached by stacking ''r'' generators:

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 In case 1, let '''g'''<sub>1</sub> = (2, 1) &minus; (1, 1), '''g'''<sub>2</sub> = (1, 2) &minus; (2, 1), and '''g'''<sub>3</sub> = (1, 1) &minus; (''n''/2, 2) = ((&minus;''n''/2 &minus; 1)*'''g'''<sub>1</sub> &minus; ''n''/2*'''g'''<sub>2</sub>) mod '''e'''. We assume that '''g'''<sub>1</sub>, '''g'''<sub>2</sub> and '''e''' are ℤ-linearly independent. We have the chain '''g'''<sub>1</sub> '''g'''<sub>2</sub> '''g'''<sub>1</sub> '''g'''<sub>2</sub> ... '''g'''<sub>1</sub> '''g'''<sub>3</sub> which visits every note in ''S''.
-Since ''S'' is generator-offset it is well-formed with respect to '''g''' = ('''g'''<sub>2</sub> + '''g'''<sub>1</sub>). Since '''g'''<sub>1</sub> and '''g'''<sub>2</sub> subtend the same number of steps, all multiples of the generator '''g''' must be even-steps, and those intervals that are "offset" by '''g'''<sub>1</sub> must be odd-steps. Letting ''M'' be the subset consisting of all even-numbered notes (which are generated by '''g''') and considering ''M'' as a scale by dividing degree indices in ''M'' by two, ''M'' is well-formed with respect to '''g''', thus ''M'' (and its offset) must be a mos subset. Hence ('''g'''<sub>3</sub> + '''g'''<sub>1</sub>), the imperfect generator of the mos generated by g, subtends the same number of steps as g. Thus '''g'''<sub>2</sub> and '''g'''<sub>3</sub> subtend the same number of steps, a fact we need in order to be able to substitute one instance of '''g'''<sub>2</sub> with '''g'''<sub>3</sub> in the next part.
+Since ''S'' is generator-offset it is well-formed with respect to the aggregate generator '''g''' = ('''g'''<sub>2</sub> + '''g'''<sub>1</sub>). Since '''g'''<sub>1</sub> and '''g'''<sub>2</sub> subtend the same number of steps, all multiples of the generator '''g''' must be even-steps, and those intervals that are "offset" by '''g'''<sub>1</sub> must be odd-steps. Letting ''M'' be the subset consisting of all even-numbered notes (which are generated by '''g''') and considering ''M'' as a scale by dividing degree indices in ''M'' by two, ''M'' is well-formed with respect to '''g''', thus ''M'' (and its offset) must be a mos subset. Hence ('''g'''<sub>3</sub> + '''g'''<sub>1</sub>), the imperfect generator of the mos generated by g, subtends the same number of steps as g. Thus '''g'''<sub>2</sub> and '''g'''<sub>3</sub> subtend the same number of steps, a fact we need in order to be able to substitute one instance of '''g'''<sub>2</sub> with '''g'''<sub>3</sub> in the next part.
 Let ''r'' be odd and ''r'' ≥ 3. Consider the following abstract sizes for the interval class of ''k''-steps reached by stacking ''r'' generators: