Recursive structure of MOS scales: Difference between revisions

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for appropriate integers ''c, d, e, f'', |''c''| < ''n'', and |''e''| < ''n''. We must have that ''c'' = ±''b'' and ''e'' = ∓''a'', as by assumption ''a''L + ''b''s = (''ac'' + ''be'')''g'' + (''ad'' + ''bf'')''p'' = ''p''. In fact, [L, s] is another valid basis for the group generated by ''p'' and ''g'', since by binarity we have ''p, g'' ∈ span(''L, s'').

Assume ''c'' = ''b'' and ''e'' = -''a''. Let χ = L − s > 0; then χ is ''p''-equivalent to ''+ng''. Now by generatedness, binarity, and linear independence of ''p'' and ''g'', the class of ''k''-steps must have two sizes separated by ''ng'', and changing the k-steps outside these two sizes leaves the range −(''n'' − 1)''g'', ..., 0, ..., (''n'' − 1)''g''. Thus the class of ''k''-steps has two sizes for 1 ≤ ''k'' ≤ (''n'' − 1).

Assume ''c'' = ''b'' and ''e'' = −''a''. Let χ = L − s > 0; then χ is ''p''-equivalent to ''+ng''. Now by generatedness, binarity, and linear independence of ''p'' and ''g'', the class of ''k''-steps must have two sizes separated by ''ng'', and changing the k-steps outside these two sizes leaves the range −(''n'' − 1)''g'', ..., 0, ..., (''n'' − 1)''g''. Thus the class of ''k''-steps has two sizes for 1 ≤ ''k'' ≤ (''n'' − 1).

To show binarity for rational step ratios, note that every ''k''-step (which is a specific linear combination of L and s) in the scale with rational step ratio is a limit point of the same linear combination of L and s in versions of the binary scale with irrational step ratios, and thus there must be ''at most'' 2 sizes for each generic interval. Since χ, which separates the two sizes, is ''p''-equivalent to ''ng'', which remains ''p''-inequivalent to 0 since L/s ≠ 1/1, each generic interval has ''exactly'' 2 sizes.

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 for appropriate integers ''c, d, e, f'', |''c''| < ''n'', and |''e''| < ''n''. We must have that ''c'' = ±''b'' and ''e'' = ∓''a'', as by assumption ''a''L + ''b''s = (''ac'' + ''be'')''g'' + (''ad'' + ''bf'')''p'' = ''p''. In fact, [L, s] is another valid basis for the group generated by ''p'' and ''g'', since by binarity we have ''p, g'' ∈ span(''L, s'').
-Assume ''c'' = ''b'' and ''e'' = -''a''. Let χ = L &minus; s > 0; then χ is ''p''-equivalent to ''+ng''. Now by generatedness, binarity, and linear independence of ''p'' and ''g'', the class of ''k''-steps must have two sizes separated by ''ng'', and changing the k-steps outside these two sizes leaves the range &minus;(''n'' &minus; 1)''g'', ..., 0, ..., (''n'' &minus; 1)''g''. Thus the class of ''k''-steps has two sizes for 1 ≤ ''k'' ≤ (''n'' &minus; 1).
+Assume ''c'' = ''b'' and ''e'' = &minus;''a''. Let χ = L &minus; s > 0; then χ is ''p''-equivalent to ''+ng''. Now by generatedness, binarity, and linear independence of ''p'' and ''g'', the class of ''k''-steps must have two sizes separated by ''ng'', and changing the k-steps outside these two sizes leaves the range &minus;(''n'' &minus; 1)''g'', ..., 0, ..., (''n'' &minus; 1)''g''. Thus the class of ''k''-steps has two sizes for 1 ≤ ''k'' ≤ (''n'' &minus; 1).
 To show binarity for rational step ratios, note that every ''k''-step (which is a specific linear combination of L and s) in the scale with rational step ratio is a limit point of the same linear combination of L and s in versions of the binary scale with irrational step ratios, and thus there must be ''at most'' 2 sizes for each generic interval. Since χ, which separates the two sizes, is ''p''-equivalent to ''ng'', which remains ''p''-inequivalent to 0 since L/s ≠ 1/1, each generic interval has ''exactly'' 2 sizes.