Multiplication distributes over any other operator? We know that multiplication distributes over addition $(a + b) * c = a * c + b * c$
Is there any other associative operator/function $f$ such that this holds for an “interesting” subset of real numbers?
i.e. does there exist S, f:
$$
S \subseteq \mathbb{R}, f : S^2 \rightarrow S \\
s.t. \forall a, b, c \in S \\
f(a, b) * c = f(a * c, b * c)
$$
 A: Yes, let $S = \{x \in \mathbb{R} | x \ge 0\}$ and $f(a,b)=\sqrt{ab}$. Then $f(a,b) \cdot c = \sqrt{ab} \cdot c = \sqrt{ab} \cdot \sqrt{c^2} = \sqrt{abc^2} = \sqrt{(ac)(bc)} = f(ac, bc)$.
Edit: I found a few more functions which satisfy $c\cdot f(a,b) = f(ac,bc)$.
Credit to the post you linked in the comments: any linear function will work. I.e. for $S = \mathbb{R}$ with any real $\alpha, \beta$, we can define $f_{\alpha, \beta}(x,y)=\alpha x + \beta y$, and then we have $c \cdot (\alpha x + \beta y) = \alpha(cx) + \beta(cy)$. This gives us a plethora of corollaries (e.g. subtraction, which I didn't mention earlier due to triviality).
Also inspired by that post, the max and min functions satisfy the criterion over the nonnegative reals. This introduces us to a family of statistical functions that also satisfy the criterion over $n$-ary operations on the reals (proof left as an exercise):

*

*average / (arithmetic) mean

*median

*mode if we define the mode to be $0$ if there is no most common element

*range
This line of thinking helped me to realize that both the arithmetic and geometric means satisfy the distributive criterion, so I verified that the harmonic mean also satisfies for $S = \mathbb{R} \setminus \{0\}$, i.e.:
\begin{equation}
f(a,b) = \frac{1}{\frac{1}{a} + \frac{1}{b}} \\
f(ac,bc) = \frac{1}{\frac{1}{ac} + \frac{1}{bc}} = \frac{1}{\frac{1}{c} \left(\frac{1}{a} + \frac{1}{b}\right)} = \frac{c}{\frac{1}{a} + \frac{1}{b}} = c\cdot f(a,b)
\end{equation}
We can similarly define functions $f'_{\alpha,\beta}(x,y) = f(\alpha x, \beta y)$ for any of the means, and multiplication is still distributive in the resulting functions. I suspect that this will hold for some additional cases of generalized means, and perhaps in general, but I haven't looked into this too deeply.
Aside: If we allow $0\in S$, then we get $c\cdot f(0,0) = f(0,0)$, so $f(0,0) = 0$ if $c$ is arbitrary. This was an interesting observation, but it did not yield any fruitful examples for me.
