Is $2\pi$ not in the same equivalence class as $\pi$? This is my first encounter with equivalence classes, so I'm just trying to make sure I understand the definition.
We write $x \sim y$, when $x-y \in \mathbb{Q}$.
$2\pi -\pi = \pi \notin \mathbb{Q} \implies 2\pi \nsim \pi$
Seems simple enough. However, it doesn't sound right. This is, in my mind, the same thing as saying: every irrational number belongs to a different equivalence class than all other irrational numbers, besides rational translations of itself. That just seems, I don't know, too trivial of a property. I would expect, for example, something like $2\sqrt{5} \sim \sqrt 5$.
Yes, you are right - for this equivalence relationship $2\pi$ and $\pi$ are not equivalent to each other.
There is no one definitive equivalence relationship on any set; we can (and do) study multiple different equivalence relationships depending on the problem we're trying to solve. The particular equiv. rel. you've given here does strike me as on the trivial side, and you don't need to understand anything deeper about it. Although (SPOILER!) this equiv. rel. is used to prove the existence of unmeasurable sets, but you'll only see this if you go on to study measure theory and Lebesgue integration.
BTW, if you wanted $2\sqrt{5} \sim_2 \sqrt{5} $ you could define $$x \sim_2 y \iff x/y \in \mathbb{Q}$$ (for $x,y \in \mathbb{R} \setminus \{0\}$), which would be a second, different equivalence relationship. Of course now $1+\pi$ and $\pi$ are no longer equivalent.