Direct proof of a contraposition If the product of two integers is not divisible by some integer $n$, then neither integer is divisible by $n$.
$Q1$. State the contrapositive statement:
$A1$. If either $a$ or $b$ is divisible by $n$ then $ab$ is divisible by $n$.
$Q2$. Do a direct proof of the contrapositive statement.
$A2$. Isn't the answer to $Q1$ also the proof? How else do you word the proof for division?
 A: Answering a question like this comes down to remembering the definition of what you're trying to prove, and then showing that the definition holds.
In this case, we'd like to show that "If either a or b is divisible by n then ab is divisible by n".  The definition of divisibility is as follows:
We say that a number $m$ is divisible by $n$ if there exists an integer $q$ such that $m = qn$
Now, suppose that either $a$ or $b$ is divisible by $n$.
If $a$ is divisible by $n$, then $a=qn$ for some integer $q$, which means that 
$$
ab=(qn)b=(qb)n
$$
which means that $ab$ is divisible by $n$. 
Similarly, if $b$ is divisible by $n$, then $b=qn$ for some integer $q$, which means that
$$
ab = a(qn)=(qa)n
$$
which means that $ab$ is divisible by $n$.  Thus, our proof is complete.

In practice, the part of the proof starting with "Similarly..." would be abbreviated by saying something like the following:

If $b$ is divisible by $n$, then $ab=ba$, and by the previous argument, $ba$ is divisible by $n$.  Thus, $ab$ is divisible by $n$.

OR

If $b$ is divisible by $n$, then we can simply switch the labels $a$ and $b$ to find that, by the above argument, $ab$ is divisible by $n$

OR

If $b$ is divisible by $n$, then by symmetry, we conclude $ab$ is divisible by $n$.

This is a nice way to avoid having to go through the same sequence of steps multiple times.
A: Your contrapositive is pretty much correct (just be careful not to use an exclusive-or; we technically want to use an inclusive-or). I would word it as:

If $a,b,n \in \Bbb Z$ such that $a$ is divisible by $n$ or $b$ is divisible by $n$, then $ab$ is also divisible by $n$.

For a proof, note that $ab$ is divisible by $a$ and $ab$ is divisible by $b$. Hence, regardless of whether $a$ is divisible by $n$ or $b$ is divisible by $n$, it follows by the transitivity of the divisibility relation that $ab$ is also divisible by $n$, as desired.
A: Just use the definition: if $a$ is divisible by $n$, then there exists $c$ such that $a=nc$. Can you use this to finish the proof?
A: $$a = nc \longrightarrow ab=ncb.$$
