Let $r \in \mathbb{N}$. Observe the equivalence relation $\sim \subseteq \mathbb{Z} \times \mathbb{Z}$ defined as $x \sim y :\Leftrightarrow (\exists k \in \mathbb{Z})(y=x+kr)$. Show that an operation $'+'$ on $\mathbb{Z}/\sim$ given by $[x]_\sim + [y]_\sim := [x+y]_\sim$ is well-defined.
My attempt at a proof:
Let $x\sim x', y\sim y'$, then we know that $(\exists k_1 \in \mathbb{Z})(x=x'+k_1r)$ and $(\exists k_2 \in \mathbb{Z})(y=y'+k_2r)$. Adding $x$ and $y$, we see that $x+y=x'+y'+(k_1+k_2)r$. Because $\mathbb{Z}$ is closed under addition, $k_1+k_2 \in \mathbb{Z}$, and eventually $x+y\sim x'+y'$.
Unfortunately, I'm not even sure whether I'm on the right track here. As far as I can see, I've shown that adding two elements of (different) equivalence classes yields an element, whose equivalence class is in $\mathbb{Z}/\sim$. Is that sufficient to show that $'+'$ is well-defined?