I hope this question is appropriate for MO:

Let $n$ be a natural number, $U_n := \{ d | d \text{ divides } n, \gcd(d,n/d)=1\}$ be the set of unitary divisors.

We can make $U_n$ to a boolean ring:

$$a \oplus b := \frac{ab}{\gcd(a,b)^2} = \frac{\operatorname{lcm}(a,b)}{\gcd(a,b)}$$ and $$a \otimes b := \gcd(a,b)$$

Let $\Pi(n) := \{ p | p \text{ is prime}, p| n\}$ be the set of prime divisors of $n$. We can define a topology on this set where the open sets are

$$ \{ \Pi(d) | d \text{ divides } n \}$$

then $$\Pi(\operatorname{rad}(ab)) = \Pi(a) \cup \Pi(b)$$ and $$\Pi(\gcd(a,b)) = \Pi(a) \cap \Pi(b)$$

where $\operatorname{rad}(x) = \prod_{p|x}p$ is the radical of $x$.

To each open set $U$ we define a number

$$\operatorname{rad}(U):= \prod_{p \in U}p$$

The open sets build also a boolean ring with:

$$U \oplus V := U \Delta V$$ where $\Delta$ denotes the symmetric difference, and $$U \otimes V := U \cap V$$

Then $\operatorname{rad}$ is a isomorphism of boolean rings:

$$\operatorname{rad}(U \oplus V) = \operatorname{rad}(U) \oplus \operatorname{rad}(V)$$ $$\operatorname{rad}(U \otimes V) = \operatorname{rad}(U) \otimes \operatorname{rad}(V)$$ Also $\operatorname{rad}(\emptyset) = 1$, where $1$ is the zero in $U_{\operatorname{rad}(n)}$ and $\operatorname{rad}(\Pi(n)) = \operatorname{rad}(n)$, where $\operatorname{rad}(n)$ is the one in $U_{\operatorname{rad}(n)}$.

Furthermore, since $k(a,b) = \frac{\gcd(a,b)^2}{ab} = \frac{1}{a\oplus b}$ is a positive definite function on the natural numbers and a simililarity, we can embedd this boolean ring $U_n$ isometrically in Euclidean space $\mathbb{R}^{2^{\omega(n)}}$ (on the sphere of radius one with center $0$) where $\omega(n)$ counts the distinct prime divisors of $n$ and we can define a distance between two unitary divisors:

$$ d(a,b) = \sqrt{k(a,a)+k(b,b)-2k(a,b)} = \sqrt{2(1-\frac{1}{a\oplus b})}$$

Also for all $a,b,c \in U_n$ we have:

$$k(c\oplus a , c \oplus b ) = k(a,b)$$

My (soft) question is this:

Is this of any use for anything, maybe in number theory? :) Thanks for your help.