User:Contribution/Successive superparticular complementary pair: Difference between revisions

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{{todo|Finish the article and move it|inline=1|comment=When the article is finished, move it to the main root}}
For each pair of superparticular ratios <math>{s1}/{s2}</math>​ and <math>{s2}/{s3}</math>, there exists a ratio <math>{a}/{b}</math> such that <math>{s1}/{s2}</math>​ and <math>{s2}/{s3}</math>​ are <math>{a}/{b}</math> complementary; it is observed that <math>a−b=1</math> or <math>a−b=2</math>.
For each pair of superparticular ratios <math>{s1}/{s2}</math>​ and <math>{s2}/{s3}</math>, there exists a ratio <math>{a}/{b}</math> such that <math>{s1}/{s2}</math>​ and <math>{s2}/{s3}</math>​ are <math>{a}/{b}</math> complementary; it is observed that <math>a−b=1</math> or <math>a−b=2</math>.
In other words, for each ratio <math>a/b</math> where <math>a−b=1</math> or <math>a−b=2</math>, there exists a pair of superparticular ratios <math>{s1}/{s2}</math>​ and <math>{s2}/{s3}</math> that are <math>{a}/{b}</math> complementary.
In other words, for each ratio <math>a/b</math> where <math>a−b=1</math> or <math>a−b=2</math>, there exists a pair of superparticular ratios <math>{s1}/{s2}</math>​ and <math>{s2}/{s3}</math> that are <math>{a}/{b}</math> complementary.
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{{todo|Table|inline=1|comment=Explain the table.}}
{{todo|Table|inline=1|comment=Explain the table.}}
{{todo|Finish the article and move it|inline=1|comment=When the article is finished, move it to the main root}}


{| class="wikitable sortable right-1 left-2 right-3 left-4 right-5 left-6 right-7 left-8 right-9 left-10 right-11 left-12 right-13 left-14 right-15 left-16 right-17 left-18 right-19 left-20"
{| class="wikitable sortable right-1 left-2 right-3 left-4 right-5 left-6 right-7 left-8 right-9 left-10 right-11 left-12 right-13 left-14 right-15 left-16 right-17 left-18 right-19 left-20"