I just started learning about equivalence classes in the context of modular arithmetic, and learned that the union of the $n$ equivalence classes modulo $n$ is $\mathbb{Z}$.
My professor then applied this to the fact that we've seen equivalence classes before when talking about anti-derivatives - how $\int f(x) dx + c$ is the equivalence class of anti-derivatives of $f$.
I apologize if this is not quite correct, and I'd welcome a correction, because I'm not too sure of the precise language used when speaking of relations.
I'm wondering if because the union of equivalence classes is the whole set, that any differentiable function must have a class of anti-derivatives $g(x)$ such that $\frac{d}{dx} g(x) = f(x)$? I was reading the response to this question, which is similar to mine, but I'm wondering if due to the fact that we can form an equivalence relation, there must be an anti-derivative for any differentiable function?
No. The fact that the union of the equivalence classes is the whole set just means that any differentiable function $f:\mathbb{R}\to\mathbb{R}$ is in some equivalence class. This is obvious, because the set $S$ of all $g$ which are equivalent to $f$ is one of the equivalence classes, and $f\in S$ because $f$ is equivalent to itself. This in no way says anything about the existence of $g$ whose derivative is $f$.
(In fact, it is true that any such $f$ has an antiderivative, but this has nothing to do with this equivalence relation. Instead, it is true by the fundamental theorem of calculus, since if $g(x)=\int_0^xf(t)dt$ then $f$ is the derivative of $g$.)