Ternary scale theorems: Difference between revisions
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Assuming SGA, we have two chains of the aggregate generator '''g''' (going right). In the diagrams below, O represents a note and - represents a generator '''g'''. The two cases are: | Assuming SGA, we have two chains of the aggregate generator '''g''' (going right). In the diagrams below, O represents a note and - represents a generator '''g'''. The two cases are: | ||
<pre> | |||
CASE 1: EVEN LENGTH | CASE 1: EVEN LENGTH | ||
O-O-...-O ( | O-O-...-O (''n''/2 notes) | ||
O-O-...-O ( | O-O-...-O (''n''/2 notes) | ||
and | </pre> | ||
and | |||
<pre> | |||
CASE 2: ODD LENGTH | CASE 2: ODD LENGTH | ||
O-O-O-...-O ( | O-O-O-...-O ((''n'' + 1)/2 notes) | ||
O-O-...-O ( | O-O-...-O ((''n'' − 1)/2 notes). | ||
</pre> | |||
Label the notes (1, ''j'') and (2, ''j''), {{nowrap|1 ≤ ''j'' ≤ ''N''}} where ''N'' is the number of notes in the chain, for notes in the upper and lower chain, respectively. | Label the notes (1, ''j'') and (2, ''j''), {{nowrap|1 ≤ ''j'' ≤ ''N''}} where ''N'' is the number of notes in the chain, for notes in the upper and lower chain, respectively. | ||
==== Statement (1) ==== | ==== Statement (1) ==== | ||
In case 1, let {{nowrap|'''g'''<sub>1</sub> {{=}} (2, 1) − (1, 1)|'''g'''<sub>2</sub> {{=}} (1, 2) − (2, 1)}}, and {{nowrap|'''g'''<sub>3</sub> {{=}} (1, 1) − ({{frac|''n''|2}}, 2)}} {{nowrap|{{=}} (({{frac|−''n''|2}} − 1)*'''g'''<sub>1</sub> − {{frac|''n''|2}}*'''g'''<sub>2</sub>) (mod '''e''')}}. We assume that '''g'''<sub>1</sub>, '''g'''<sub>2</sub> and '''e''' are ℤ-linearly independent. We have the chain '''g'''<sub>1</sub> '''g'''<sub>2</sub> '''g'''<sub>1</sub> '''g'''<sub>2</sub> ... '''g'''<sub>1</sub> '''g'''<sub>3</sub> which visits every note in ''s''. | In case 1, let {{nowrap|'''g'''<sub>1</sub> {{=}} (2, 1) − (1, 1)|'''g'''<sub>2</sub> {{=}} (1, 2) − (2, 1)}}, and {{nowrap|'''g'''<sub>3</sub> {{=}} (1, 1) − ({{frac|''n''|2}}, 2)}} {{nowrap|{{=}} (({{frac|−''n''|2}} − 1)*'''g'''<sub>1</sub> − {{frac|''n''|2}}*'''g'''<sub>2</sub>) (mod '''e''')}}. We assume that '''g'''<sub>1</sub>, '''g'''<sub>2</sub> and '''e''' are ℤ-linearly independent. We have the chain '''g'''<sub>1</sub> '''g'''<sub>2</sub> '''g'''<sub>1</sub> '''g'''<sub>2</sub> ... '''g'''<sub>1</sub> '''g'''<sub>3</sub> which visits every note in ''s''. | ||