Equivalence classes modulo 7 are pairwise disjoint

Question

Where do I got from here? I really really have no idea.

Answer 1

If $y \in A_n$, then $y \equiv n \pmod{7}$, so there exists $k \in \mathbb{Z}$ such that $y - n = 7k$. If $y \in A_m$, then $y \equiv m \pmod{7}$, so there exists $l \in \mathbb{Z}$ such that $y - m = 7l$. Hence,

\begin{align*} (y - n) - (y - m) & = 7k - 7l\\ m - n & = 7(k - l)\\ \end{align*}

Since the integers are closed under subtraction $k - l \in \mathbb{Z}$. Therefore, $m \equiv n \pmod{7} \Rightarrow A_n = A_m$.