Generator-offset property: Difference between revisions

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For (4), assume S is ''a''X ''b''Y ''b''Z, ''a'' odd. If ''b'' = 1, there's nothing to prove, so assume ''b'' > 1.
For (4), assume S is ''a''X ''b''Y ''b''Z, ''a'' odd. If ''b'' = 1, there's nothing to prove, so assume ''b'' > 1.


Consider the two alternants, detemperings of the generator iX + jW of the ''a''X 2''b''W mos ''T''(X, W) = ''S''(X, W, W) where gcd(''j'', 2''k'') = 1.
Consider the two alternants, detemperings of the generator ''i''X + ''j''W of the ''a''X 2''b''W mos ''T''(X, W) = ''S''(X, W, W) where gcd(''j'', 2''k'') = 1.


'''Claim 1''': Deleting X's from the alternants of ''S'' gives every ''j''-step subword in the scale ''E''<sub>X</sub>(''S'')(Y, Z), the scale word obtained by deleting all X's from X.  
'''Claim 1''': Deleting X's from the alternants of ''S'' gives every ''j''-step subword in the scale ''E''<sub>X</sub>(''S'')(Y, Z), the scale word obtained by deleting all X's from X.