Answer
See the solution.
Work Step by Step
This statement is true, and we will prove it by contradiction.
First, note that the statement is a conditional sentence, and we can rewrite it as "(If $a$ is a nonzero rational number $\land$ $b$ is an irrational number) $\rightarrow$ ($ab$ is irrational)." Then to prove it by contradiction, we will assume this statement is false, and show that this assumption leads to a contradiction.
Now, for this conditional sentence to be false, the antecedent must be true and the consequent must be false. So these are the assumptions/suppositions we will make to begin our proof, and show that they lead to a contradiction.
$Proof.$
Assume the statement is false. That is, assume $a$ is a nonzero rational number, $b$ is an irrational number, and $ab$ is rational. Then since $a$ is a nonzero rational number, $a=\frac{m}{n}$, where $m$ and $n$ are both nonzero integers. (Note, that by the definition of 'rational number', $m$ and $n$ are integers with $n$ nonzero, and because $a$ is nonzero, $m$ must be nonzero too.) Also, since $ab$ is rational, $ab=\frac{p}{q}$, where $p$ and $q$ are integers and $q\neq0$. Since $a\neq 0$, we may divide both sides of $ab=\frac{p}{q}$ by $a$. This gives $b=\frac{p}{qa}=\frac{p}{q\frac{m}{n}}=\frac{pn}{qm}$, where $pn$ and $qm$ are integers and $qm\neq 0$. (Note $qm\neq 0$ since $q$ and $m$ are both nonzero.) But this means $b$ is rational, which contradicts $b$ being irrational. Thus, the statement must be true.$_\Box$
