I was showing someone an example of a proof where contrapositive is the way to go. Clearly, I cooked up the wrong implication. In any event, now I want to complete the proof regardless. And, do so by showing the contrapositive even though I know invoking a contradiction is easier in this case. I want to show, let $a \neq 0 \neq b$, $$ a \not\in \mathbb{Q} \wedge b \in \mathbb{Q} \Rightarrow ab \not\in \mathbb{Q} $$
Then the contrapositive will be, $$ ab \in \mathbb{Q} \Rightarrow a \in \mathbb{Q} \vee b \not\in \mathbb{Q} $$ We start by assuming $ab \in \mathbb{Q}$ then by definition of rationals, $$ ab = \frac{p}{q}, \quad \gcd(p,q)=1, \quad (p,q) \in \mathbb{Z}^2 $$ I am bit confused from here, I tried, $$ a = \frac{p}{bq}, \quad b = \frac{p}{aq} $$ But how does this help me show $a \in \mathbb{Q} \vee b \not\in \mathbb{Q}$?
You assume $ab\in\Bbb Q$ and $b\ne0,$ and want to prove $a \in \mathbb{Q} \vee b \not\in \mathbb{Q},$ i.e. $$b\in \mathbb{Q}\implies a\in \mathbb{Q}.$$ Simply assume moreover that $b\in \mathbb{Q}$ and write $a=\frac{ab}b.$