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>Why? because of how I choose to understand the origin of the 7 note major scale: you take any note (the base tone) and multiply the frequency by 3. this creates a fifth (plus one octave). I'll keep in mind that 'the octave' is defined by multiplying the frequency by 2. then, fit the fifth (base tone * 3) into only one octave (3/2). And repeat 'recursively'. This is called 3-limit tuning: https://en.xen.wiki/w/3-limit . 5-limit tuning is what standard western music uses: https://en.xen.wiki/w/5-limit (to include thirds as well as fifths) After reducing the ratios to fit in an octave, you get exactly 8 notes (7 if you subtract the octave itself). Note how https://oeis.org/A054540 shows that 7 notes are a good approximation of the ratios, but so are 12 (which shows why creating a 12-note system was an advantageous move, over 11 or 13). Technically in 12-EDO a fifth is not exactly generated by the ratio 1.5, it's slightly flat at 1.498307... but we choose the note closest to 1.5. > This is the famous circle of fifths, but we all knew that. Finally, after twelve repetitions we're back on the same note, but an octave above. (but why? why stop at twelve? I'm still working through this answer, but it has something to do with convergence maybe? or just the fact that after 12 notes we have now landed within two notes which we 'found' already???) Suppose we already chose a 12-note equal-tempered system. The closest note to the perfect fifth of a fundamental frequency `f` will be `f * 12th-root(2)^7`, (7 notes out just happens to be close to multiplying by 3/2). The next fifth after that would be `f * 12th-root(2)^7 * 12th-root(2)^7 = f * 12th-root(2)^14`. Going out by a fifth 12 times gets you `f * 12th-root(2)^84 = f * 12th-root(2)^(712)`. But we know that `12th-root(2) ^ 12 = 2`, simply from the definition of 12th root. Multiplication is commutative, so we can group the roots-of-twelve by groups of 12 instead of groups of 7, and we get `f 2^7`. Taking that modulo 2, we just get f, i.e. the same (enharmonically equivalent) note. Now suppose we didn't make that choice, instead we chose a 31-note system (I'm a big fan of 31-EDO). In that case, we have the same construction. The fifth in 31-EDO happens to be an interval of 18 notes, and similarly we jump around the scale, but this time an interval of one note is `31st-root(2)`, so we have to do 31 fifths to get back to the same note. This actually tells us something interesting - if we want to form a circle (made out of intervals, that end up hitting the original enharmonically-equivalent-note) to hit all of the notes in our scale the notes we hit must be a permutation of the original scale. It's a little beyond my math to tell you how this works, I think Fermat's little theorem and modular arithmetic has something to do with how it works. Something about how 7 and 12 (or 18 and 31) are relatively prime compared to each other, and it forms a group which generates a permutation. |