I'm trying to work through this proof below for homework,
For $k ∈ \Bbb Z$ and $n ∈ \Bbb N$, let $[k]_n$ denote the equivalence class of $k\: (\!\!\bmod n)$. For $a, b ∈ \Bbb N$, prove that $\;a\mid b \implies [k]_b ⊆ [k]_a$.
I believe that since $a$ divides $b$, then there is a relation, $a \sim b$. From here I'm pretty sure you can say that the equivalence class of $a$ is equal to equivalence class of $b$ but I'm having trouble applying modular arithmetic and proving subsets. Any advice?
Note $a \mid b$ means there's an integer $m$ such that
$$b = ma \tag{1}\label{eq1A}$$
Next, for any integer $k$, you have $x \in [k]_{b}$ means for some integer $j$ that
$$x \equiv k \pmod b \implies x - k = jb = j(ma) \implies x \equiv k \pmod a \tag{2}\label{eq2A}$$
This means you also have $x \in [k]_{a}$. As such, this shows $[k]_b \subseteq [k]_a$.