Ternary parallelogram scales are MOS substitution: Difference between revisions
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Proof: The order cannot be less than ''n'', lest we have {{nowrap|φ((0, ''uk''<sub>'''w'''</sub>)) {{=}} 0}} for some {{nowrap|0 < ''u'' < ''n'',}} contradicting injectivity of φ within a fundamental domain (following from Step 2). If the order is ''N'' > ''n'', we have two cases. | Proof: The order cannot be less than ''n'', lest we have {{nowrap|φ((0, ''uk''<sub>'''w'''</sub>)) {{=}} 0}} for some {{nowrap|0 < ''u'' < ''n'',}} contradicting injectivity of φ within a fundamental domain (following from Step 2). If the order is ''N'' > ''n'', we have two cases. | ||
* If ''n'' = 2, then by disjointness the image of the (0, 1) translation of [0 : ''m''] must be [''m'' : 2''m''], forcing ''k''<sub>'''w'''</sub> = ''m'' which is of order 2. | * If ''n'' = 2, then by disjointness the image of the (0, 1) translation of [0 : ''m''] must be [''m'' : 2''m''], forcing ''k''<sub>'''w'''</sub> = ''m'' which is of order 2. | ||
* If ''n'' ≥ 3, assume by way of contradiction that {{nowrap|[0 : ''m''], [0 : ''m''] + ''a'', ..., [0 : ''m''] + (''n'' - 1)''a''}}, where ''a'' = ψ(''k''<sub>'''w'''</sub>), are disjoint. Then {{nowrap|[0 : ''m''], [0 : ''m''] + ''a'', ..., [0 : ''m''] + (''n'' - 2)''a''}} are disjoint | * If ''n'' ≥ 3, assume by way of contradiction that {{nowrap|[0 : ''m''], [0 : ''m''] + ''a'', ..., [0 : ''m''] + (''n'' - 1)''a''}}, where ''a'' = ψ(''k''<sub>'''w'''</sub>), are disjoint. Then {{nowrap|[0 : ''m''], [0 : ''m''] + ''a'', ..., [0 : ''m''] + (''n'' - 2)''a''}} are disjoint. We ask: where do we place {{nowrap|[0 : ''m''] + (''n'' - 1)''a''?}} | ||
** The number of remaining slots on all of {{nowrap|ℤ/''mn''ℤ}} is ''m''. | ** The number of remaining slots on all of {{nowrap|ℤ/''mn''ℤ}} is ''m''. | ||
** There is no contiguous region of ''m'' unoccupied slots. If there were, then the (''n'' - 1) windows that have already been placed must have no gap between them. This implies that ''a'' generates {{angbr|''m''}} and has order ''n''. | ** There is no contiguous region of ''m'' unoccupied slots. If there were, then the (''n'' - 1) windows that have already been placed must have no gap between them. This implies that ''a'' generates {{angbr|''m''}} and has order ''n''. Hence there is no way to place {{nowrap|[0 : ''m''] + (''n'' - 1)''a''}} without it overlapping with one of the other windows. | ||
The claim establishes that | The claim establishes that | ||