$0\neq a\in \mathbb Q, b\in \mathbb R \setminus\mathbb Q \text{ (b is irrational)}$ Prove that $\frac a b$ is irrational.
From defintion $a=\frac m n$ such that $m,n\in \mathbb Z, n\neq 0$.
Take the contrapositive: suppose $\frac m {nb}\in \mathbb Q$ prove $\frac m n\notin \mathbb Q$.
Immediate contradiction from defining $m,n\in \mathbb Z, n\neq 0$. Thus $\frac m {nb}$ is irrational.
Well I'm not sure I'm using the contrapositive right and I tried to combine contrapositive with proof by contradiction but I have a feeling I'm wrong...
Answer
You are definitely mixing up contrapositive and contradiction. It is quite easy to do as a lot of proofs that can be done with one can be done with the other. Many students make this mistake early on in their careers. What you should say is that "Suppose $\dfrac{m}{nb}\in\Bbb Q$, then $b\in \Bbb Q$." The reason is that you want to prove "if $b\in \Bbb R\setminus \Bbb Q$, then $\dfrac{m}{nb}\in\Bbb R\setminus \Bbb Q$." The contrapositive of this statement is the statement I gave above because we want to negate each portion (and reverse the direction). The negation of $\dfrac{m}{nb}\in \Bbb R\setminus \Bbb Q$ is $\dfrac{m}{nb}\in \Bbb Q$, and likewise for the other. Philosophically, your argument is okay but you need to fix it up a little so that you're not mixing up contrapositive and contradiction.
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