The reason for this is simply because you include the base note when you start counting, so if you stop midway and stack another interval on top, you have to account for that stop because it decreases the total travel distance by one.
I’m arguing that ‘counting the base note’ is necessary because there is no zero. If you travel 1 unit of your favorite unit of length, stop, and move one unit more, you have moved 2 units, not 3 as it is in music. Stops don’t add to or reduce travel distance.
Redjard@lemmy.dbzer0.com 6 days ago
Name the prime a “none”, the octave a sept.
Now, 2 “septaves”, c1 to c3, are a 14th. 2*7 = 14.
You can make off-by-one intervals work, but you have to constantly juggle some +1s or -1s compared to what we usually use.
If you counted distance in steps, then moving from your front door to your front door would be 0 steps, not one, and moving by 6 steps is twice the distance of 3 steps.
A piano with 5 septaves has 5*7 = 60 keys, wait.
So anyway mathematically one dodecave, one 12th, c1 to c2, has 12 segments, the frequency diffefence is 2. So a second, 2 notes, has 2/12 of that interval, the ratio is 2^2/12^.
A first, a halftone, has 2^1/12^ as its frequency ratio, and a none has 2^0/12^ = 1, the same frequency.
No matter if you count physical keys, distance on a keyboard to change a note by, or mathematical frequency in the air, starting at 1 goes against our intuition, and when you try to add or multiply it is easy to get completely wrong results.
PS: You might want to go C to C on your 5 dodecave keyboard, in which case the concept of “started hour” etc. is familiar, you know to add one arriving at 61 keys, and you know that means an assymetry where one C doesn’t have 11 other keys to itself.
The other way around you’d have to subtract 4, so probably subtract 5 and add 1 since you were dealing with 5 tredecaves in your head not 1 base tredecave followed by 4 extension tredecaves.