Generator-offset property: Difference between revisions
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'''Claim 2''': If a binary scale ''U'' has ''b'' Y's and ''b'' Z's, gcd(''j'', 2''b'') = 1, and consecutively stacked ''j''-steps in ''U'' occur in 2 alternating sizes, then ''U'' = (YZ)<sup>''b''</sup>. | '''Claim 2''': If a binary scale ''U'' has ''b'' Y's and ''b'' Z's, gcd(''j'', 2''b'') = 1, and consecutively stacked ''j''-steps in ''U'' occur in 2 alternating sizes, then ''U'' = (YZ)<sup>''b''</sup>. | ||
Proof: Write u and v for the two sizes of ''j''-steps. Since gcd(''j'', 2''b'') = 1 there exists ''m'' such that stacking ''m''-many ''j''-steps yields scale steps of ''U'', and ''m'' is odd because gcd(''m'', 2''b'') = 1. Hence the scale steps of ''U'' are (uv)<sup>(''m''−1)/2</sup>u mod e and (vu)<sup>(''m''−1)/2</sup>v mod e, and the step sizes clearly alternate. | Proof: Write u and v for the two sizes of ''j''-steps. Since gcd(''j'', 2''b'') = 1, there exists ''m'' such that stacking ''m''-many ''j''-steps yields scale steps of ''U'', and ''m'' is odd because gcd(''m'', 2''b'') = 1. Hence the scale steps of ''U'' are (uv)<sup>(''m''−1)/2</sup>u mod e and (vu)<sup>(''m''−1)/2</sup>v mod e, and the step sizes clearly alternate. | ||
These two claims prove that ''E''<sub>X</sub>(S) = (YZ)<sup>''b''</sup> and that the two alternants' sizes differ by replacing one Y for a Z. | These two claims prove that ''E''<sub>X</sub>(S) = (YZ)<sup>''b''</sup> and that the two alternants' sizes differ by replacing one Y for a Z. | ||