Notation
We define compound intervals as those intervals which can be 'factored' into a non-empty set of intervallic atoms. The Unison (1:1) serves as an identity element and is excluded from this definition because it can be trivially represented as an empty set of atomic factors.
A convenient way of representing a compound interval in terms of its constituent atoms is to use the exponent (^) notation to indicate the number of times each atom occurs in the factoring of that interval. For instance, the Just Fourth {T, S, s, S} may be written {T^1, S^2, s^1}.
An alternative method is to represent the intervals as points in a 3-dimensional space
Representation Lemma
An arbitrary repeated sequence of a given intervallic atom cannot be represented in terms of the other atoms.
{T^p} ≠ {S^q , s^r } ∀ p, q, r ≥ 0 provided [p, q, r]
Proof
Proceeding by the method of Proof by Contradiction, we assume that the above inequality is actually false. In that case, after substituting the values for T, S, and s, we can write:
(9/8)^p = (16/15)^q ⋅ (25/24)^r
or
3^(2p) / 2^(3p) = 2^(4q) / (5^q ⋅ 3^q) ⋅ 5^(2r) / (2^3r ⋅ 3^r)
which gives
5^q ⋅ 3^(2p+q+r) ⋅ 2^(3r) = 5^(2r) ⋅ 3^0 ⋅ 2^(4q+3p)
By the Fundamental Theorem of Arithmetic [And1971], the exponents on both sides must be equal. So,
q = 2r
2p+q+r = 0
3r = 4q+3p
Putting q = 2r into the other two equations gives,
2p +3r = 0
3p + 5r = 0
which implies,
p = -2r
However, since p and r are both positive, the only solution then is the trivial
p = r = 0 ⇒ q = 0
This solution, however, represents the Unison {T^0, S^0, s^0} and also contradicts the starting condition that not all p, q, r are 0 simultaneously. Hence, the inequality {T^p} ≠ {S^q, s^r} must be true.
Similar proofs apply for {S^p} ≠ {T^q, s^r} and {s^p} ≠ {T^q, S^r}.
QED.
The Fundamental Theorem of Intervallic Atoms
There is a unique way of factoring a compound interval into constituent intervallic atoms.
{T^a, S^b, s^c} = {T^p, S^q, s^r} ⇔ a = p, b = q, c = r, where a, b, c, p, q, r ≥ 0.
Proof
We proceed by the method of Proof by Contradiction. Let us assume that there exist compound intervals which can be factored in more than one way. Let Д be the smallest such compound interval. Then, we can write:
Д = {T^a, S^b, s^c} = {T^p, S^q, s^r}, where a, b, c, p, q, r ≥ 0.
Trivially, at least one of the following identities must not hold
a = p
b = q
c = r
Otherwise both the representations of Д would be identical.
Also,
Now, let min(x, y) be the minimum of the two values x & y. Define
e = min(a, p)
f = min(b, q)
g = min(c, r)
Then we have two cases.
Case 1: [e, f, g]
This implies that the interval {T^e, S^f, s^g} is common to both the representations of Д and can be factored out:
Д ∖ {T^e, S^f, s^g } = {T^(a-e), S^(b-f), s^(c-g)} = {T^(p-e), S^(q-f), s^(r-g)}
Now, by the conditions of the assumption, Д is the smallest compound interval representable in more than one way. Clearly, both {T^(a-e), S^(b-f), s^(c-g)} and {T^(p-e), S^(q-f), s^(r-g)} are smaller than Д and both are compound intervals since they obviously have a representation in terms of the intervallic atoms.
Thus, by the conditions of the assumption, both necessarily have a unique representation. Since the intervals are identical, it implies that corresponding indices are identical:
a = p
b = q
c = r
which contradicts the conditions of the assumption. Hence, Д cannot exist in this case.
Case 2:
Since not all the three indices (a, b, c or (p, q, r can be zero simultaneously, this case happens only if one index is 0 in one group and the other two indices are 0 in the other group. Without loss of generality, assume that a = q = r = 0. This means that Д can be written,
Д = {S^b, s^c}= {T^p}, where b, c, p 0.
However, by the Representation Lemma, this is impossible. Hence, Д cannot exist in this case.
Thus, there is no such Д representable in more than one way.
QED.
This theorem is termed the Fundamental Theorem of Intervallic Atoms.
Intervals as Mathematical Groups
Consider L = {m:n | m, n ∈ N, gcd(m, n = 1, n <= m <= 2n} which represents the set of all intervals in the range of an octave. Then ƛ = {L, ⋅} is a multiplicative Abelian Group if the multiplication operation ⋅ is defined consistently to be modulo 2:1 and closed under L. The neutral (or identity) element is the Unison 1:1.
References
[And1971] George E. Andrews, "Number Theory". Dover Publications Inc., 1971.
