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In the wiki section on prime limit tuning, one reads:

p-Limit Tuning. Given a prime number p, the subset of $Q^+$ consisting of those rational numbers x whose prime factorization has the form $ x=p_1^{\alpha_1}\ldots p_r^{\alpha_r}$ with $p_1, \ldots, p_r\leq p$ forms a subgroup of $(Q^+, *)$ We say that a scale or system of tuning uses p-limit tuning if all interval ratios between pitches lie in this subgroup

Notice that for $p=3$ one gets the just tuning version of the usual cycle of fifths (except that its is an infinite scale, there is no cycle here. To get the usual chromatic 12-notes scale people have fudged it via equal temperament)

Because of the fundamental law of the octaves, one does not actually work with the above, but with the quotient spaces given by identifying two notes $q_1$ and $q_2$ differing by a multiple of $2^n$.

Modding out by this equivalence one can work with the "infinite musical scale" $[1, 2]\cap Q$: let us think of $1$ as the DO ( C, if you like ) of the infinite keyboard and 2 as the next DO, and all rationals in-between as the "notes" available.

Now my question (which maybe trivial, maybe not so trivial, I have no clue): if you take a p-prime scale, is it dense in the infinite musical scale? In other words, can I approximate arbitrarily close any rationals in $[0, 1]$ by a suitable element in the p-scale for a fixed p? If not, what is the distribution of the infinite subscale in the infinite keyboard ?

Clearly, the infinite keyboard is the colimit of the intermediate scales, as p goes to infinity. But how does each one "sit" in the higher ones?

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up vote 3 down vote accepted

I assume that you are everywhere working up to multiplication by a power of $2$ (or else the result is trivially false if $p = 2$ for example) and assuming that your scale has some nontrivial interval ratio $n$ in it which is not a power of $2$. Then this is straightforward.

Lemma: A subgroup of $\mathbb{R}$ is dense if and only if it contains arbitrarily small elements.

Proof. The integer multiples of a small element $r$ are at most $|r|$ apart from each other.

We apply the lemma as follows. The subgroup generated by $1$ and $\log_2 n$ contains arbitrarily small elements because $\log_2 n$ is irrational by prime factorization, so it is dense in $\mathbb{R}$. Given $\alpha \in (0, 1]$, it then follows that we can find a sequence $a_k, b_k$ of integers such that

$$a_k + b_k \log_2 n \to \log_2 \alpha$$

hence such that

$$2^{a_k} n^{b_k} \to \alpha$$

and the conclusion follows.

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The question was trivial after all. But anyway, your proof is still beautiful: very simple and very elegant: Kudos Qiaochu! PS Do you know if those subgroups play any role whatever aside musical tuning? – Mirco A. Mannucci Jul 15 '12 at 15:55

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