If an operator preserves divisibility, does that imply that it preserves multiplicability?

Question

Specifically, if it were true that $$\int_a^b \frac{f(x)}{g(x)} \, dx = \frac{\int_a^b f(x) \, dx}{\int_a^b g(x) \, dx}\,,$$ then would that imply $$ \int_a^b f(x) \cdot g(x) \, dx = \int_a^b f(x) \, dx \cdot \int_a^b g(x) \, dx\,. $$ Or more succinctly, is the statement $$O\left(\frac{a}{b}\right) = \frac{O\left(a\right)}{O\left(b\right)} \implies O\left(a\cdot b\right) = O\left(a\right) \cdot O\left(b\right) $$ true for some operator $O$? How can this be proved or disproved?

Answer 1

Are you sure about integration? It looks to me like you are assuming $\int{1/f}$ = $1/\int{f}$, which I don't think it true (for integrals).

As for your more general question, "preserves division" and "preserves inverses" would imply "preserves multiplication", which you should be able to prove easily, but "definite integral" is not such an operator.