Ringer scale: Difference between revisions

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=== Sketch of the proof ===
=== Sketch of the proof ===
Consider an ''N''-note [[periodic scale]] with period ''P'' as being defined by a function <math>f: \mathbb{Z} \to \mathbb{Q}_0</math> with <math>f(Nk) = P^k.</math>
Consider an ''N''-note [[periodic scale]] with period ''P'' as being defined by a function <math>f: \mathbb{Z} \to \mathbb{Q}_{>0}</math> with <math>f(Nk) = P^k.</math>


By the construction of a ringer scale, we are given some [[val]] [[map]] <math>m : \mathbb{Q}_0 \to \mathbb{Z}</math> that satisfies <math>m(f(k+1)/f(k)) = 1</math> for all ''k'' in '''Z'''. (This can be checked by hand or by computer as we only need to check one period <i>P</i>s worth of 1-scalestep intervals.)
By the construction of a ringer scale, we are given some [[val]] [[map]] <math>m : \mathbb{Q}_{>0} \to \mathbb{Z}</math> that satisfies <math>m(f(k+1)/f(k)) = 1</math> for all ''k'' in '''Z'''. (This can be checked by hand or by computer as we only need to check one period <i>P</i>s worth of 1-scalestep intervals.)


By induction this implies <math>m(f(k+s)/f(k)) = s</math> because the intervals from ''k'' to ''k''+1, from ''k''+1 to ''k''+2, ..., from ''k''+''s''-1 to ''k''+''s'' all multiply together. This also implies <math>m(f(k))=k,</math> proving ''f'' to be [[epimorphic]], therefore CS. {{qed}}
By induction this implies <math>m(f(k+s)/f(k)) = s</math> because the intervals from ''k'' to ''k''+1, from ''k''+1 to ''k''+2, ..., from ''k''+''s''-1 to ''k''+''s'' all multiply together. This also implies <math>m(f(k))=k,</math> proving ''f'' to be [[epimorphic]], therefore CS. {{qed}}