Generator-offset property: Difference between revisions

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For (4), assume S is aX bY bZ, a odd. If b = 1, there’s nothing to prove. So assume b > 1. Suppose for the sake of contradiction that Y’s and Z’s don't alternate perfectly. Assume that the perfect generator of aX 2bW is iX + jW with j ≥ 2. (If j = 1, we can invert the generator to make j ≥ 2, since b > 1.)

In ''S'', consider (i+j)-steps (representing the generator) with: (1) the maximum possible number of Y’s (at least 2 more than the # of Z's), (2) the maximum number of Z’s (at least 2 more than the # of Y's), or (3) an intermediate number of Y’s and Z’s between the two or (4) (the preimage of) the imperfect generator, having a different number of X’s than (1), (2), and (3). Since a + 2b ≥ 5, there are at least 4 perfect generators, so there must be at least one of each of (1), (2), and (3), giving a contradiction to SV3. [Whenever the root of the (i+j)-step are moved within S, the numbers of Y's and Z's change one at a time and reach a maximum at some choice of the root and a minimum with another choice, guaranteeing that intermediate values are reached. We can use this "intermediate value theorem" argument because the preimage of the imperfect generator occurs at only one ~~root~~.]

In ''S'', consider (i+j)-steps (representing the generator) with: (1) the maximum possible number of Y’s (at least 2 more than the # of Z's), (2) the maximum number of Z’s (at least 2 more than the # of Y's), or (3) an intermediate number of Y’s and Z’s between the two or (4) (the preimage of) the imperfect generator, having a different number of X’s than (1), (2), and (3). Since a + 2b ≥ 5, there are at least 4 perfect generators, so there must be at least one of each of (1), (2), and (3), giving a contradiction to SV3. [Whenever the root of the (i+j)-step are moved within S, the numbers of Y's and Z's change one at a time and reach a maximum at some choice of the root and a minimum with another choice, guaranteeing that intermediate values are reached. We can use this "intermediate value theorem" argument because the preimage of the imperfect generator occurs at only one note and the rest of the notes only have (1), (2) and (3).]

Any generator of aX 2bW must have an odd number of W steps (Otherwise, intervals with an odd number of W steps can't be generated.). This with (4) immediately gives (5).

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 For (4), assume S is aX bY bZ, a odd. If b = 1, there’s nothing to prove. So assume b > 1. Suppose for the sake of contradiction that Y’s and Z’s don't alternate perfectly. Assume that the perfect generator of aX 2bW is iX + jW with j ≥ 2. (If j = 1, we can invert the generator to make j ≥ 2, since b > 1.)
-In ''S'', consider (i+j)-steps (representing the generator) with: (1) the maximum possible number of Y’s (at least 2 more than the # of Z's), (2) the maximum number of Z’s (at least 2 more than the # of Y's), or (3) an intermediate number of Y’s and Z’s between the two or (4) (the preimage of) the imperfect generator, having a different number of X’s than (1), (2), and (3). Since a + 2b ≥ 5, there are at least 4 perfect generators, so there must be at least one of each of (1), (2), and (3), giving a contradiction to SV3. [Whenever the root of the (i+j)-step are moved within S, the numbers of Y's and Z's change one at a time and reach a maximum at some choice of the root and a minimum with another choice, guaranteeing that intermediate values are reached. We can use this "intermediate value theorem" argument because the preimage of the imperfect generator occurs at only one root.]
+In ''S'', consider (i+j)-steps (representing the generator) with: (1) the maximum possible number of Y’s (at least 2 more than the # of Z's), (2) the maximum number of Z’s (at least 2 more than the # of Y's), or (3) an intermediate number of Y’s and Z’s between the two or (4) (the preimage of) the imperfect generator, having a different number of X’s than (1), (2), and (3). Since a + 2b ≥ 5, there are at least 4 perfect generators, so there must be at least one of each of (1), (2), and (3), giving a contradiction to SV3. [Whenever the root of the (i+j)-step are moved within S, the numbers of Y's and Z's change one at a time and reach a maximum at some choice of the root and a minimum with another choice, guaranteeing that intermediate values are reached. We can use this "intermediate value theorem" argument because the preimage of the imperfect generator occurs at only one note and the rest of the notes only have (1), (2) and (3).]
 Any generator of aX 2bW must have an odd number of W steps (Otherwise, intervals with an odd number of W steps can't be generated.). This with (4) immediately gives (5).