How does counting E twice explain the discrepancy between the third between C and E, third between E and G v. fifth between C and G?
I still don't grasp the "source of the discrepancy". "the E got counted twice when we went C,D,E and then E,F,G, but only got counted once when we went C,D,E,F,G." — So what? How does this expound the discrepancy?
The music theorists of the Middle Ages committed a fencepost error that’s too entrenched to dig up now. Consider the chord made of the notes C, E, and G (a C major triad). If music theory nomenclature for intervals made sense, the distance from C up to E plus the distance from E up to G would equal the distance from C up to G. And that’s true if you measure the intervals by counting upward steps. The problem comes when you describe intervals with what I suppose might be termed “ordinal nomenclature”: going from C to E is called going up by a third (because you count 1,2,3 when you play C,D,E) and going from E to G is called going up by a third for the same reason, but going from C to G is called going up by a fifth (because you count 1,2,3,4,5 as you play C,D,E,F,G).
The source of the discrepancy should be clear: the E got counted twice when we went C,D,E and then E,F,G, but only got counted once when we went C,D,E,F,G. So in music theory, when you stack a third on top of a third, you get a fifth. We musicians are stuck with nomenclature that essentially makes us say “3+3=5” so many times that we eventually stop noticing we’re saying it. (Of course, saying “a third plus a third is a fifth” is confusing on a different level, since it sounds like “1/3 + 1/3 = 1/5”. But I digress.)
1 answer
The discrepancy comes from the way musical intervals are named.
The third+third=fifth equation is thus not $$3+3=5$$ but $$(3-1) + (3-1) = (5-1)$$
The E is counted double in the naïve addition $3+3=5$ because it is counted as the end of one interval and the start of the next interval.
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