I need to prove this via either direct proof, or contrapositive.
Unsure of the best way to approach this.
if $a \equiv b\mod n$ and $c \equiv d\mod n$, then $ac \equiv bd\mod n$
So far I have:
Suppose $a \equiv b\mod n$ and $c \equiv d\mod n$, then it follows that $n|(a-b)$ and $n|(c-d)$ but I am unsure on where to go from here.
Since $n \mid (a - b)$ and $n \mid (c - d)$, we know that there exist $x,y \in \mathbb Z$ such that: \begin{align*} nx &= a - b \\ ny &= c - d \end{align*} But then observe that: \begin{align*} ac - bd &= ac - bc + bc - bd \\ &= c(a - b) + b(c - d) \\ &= c(nx) + b(ny) \\ &= n(\underbrace{cx + by}_{\in~\mathbb Z}) \end{align*} Thus, we conclude that $n \mid (ac - bd)$ so that $ac \equiv bd \pmod n$, as desired. $~~\blacksquare$
Hint. Another way to rewrite the statement $$a\equiv b\pmod n$$ is $$a=b+nt\quad\hbox{for some integer $t$}.$$ See if that helps.
