Pitch Classes as Equivalence Classes
Published:
When two pitches are a number of octaves apart, we say that they belong to the same pitch class. This is due to octave equivalence.
If we represent each pitch by a frequency \(p \in (0, \infty)\), we can define this relationship in mathematical terms:
\[p \sim p' \iff p = 2^k p' \text{, for some } k \in \mathbb{Z}\]
This is an equivalence relation on \((0, \infty)\). To prove this, we simply need to prove the relation is reflexive, symmetric and transitive. I will not explain here what these terms mean. See Wikipedia for an article on equivalence relations. For brevity's sake, I'll define \(P = (0, \infty)\).
Reflexivity
Let \(p \in P\). Then: \(p = 1 p = 2^0 p\). So \(p \sim p\).
Symmetry
Let \(p, p' \in P\) such that \(p \sim p'\). This means there is a \(k \in \mathbb{Z}\) such that \(p = 2^k p'\). Multiplying both sides of the equation by \(2^{-k}\) and flipping it gives us: \(p' = 2^{-k} p\). And so \(p' \sim p\).
Transitivity
Let \(p, p', p'' \in P\) such that \(p \sim p'\) and \(p' \sim p''\). This gives us two equations, where \(k, l \in \mathbb{Z}\):
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\(p = 2^k p'\)
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\(p' = 2^l p''\)
Multiplying equation 2 by \(2^k\) gives us:
\[2^k p' = 2^{kl} p''\]
We can now use equation 1 to substitute here the left-hand side:
\[p = 2^{kl} p''\]
And so \(p \sim p''\).
Conclusion
By representing each pitch by a frequency, we have defined an equivalence relation on pitches based on the principle of octave equivalence alone. This relation induces a quotient set \(P / \sim\). I believe it's fitting to define pitch classes as the members of \(P / \sim\).
I'm guessing this is what people have in mind when they talk about pitch classes, but I haven't found anyone explictly defining it this way, so I thought I'd post it here. I'm sure this has been done before and I simply haven't searched thoroughly enough.