Prove if $a\ne0$ is rational and $b$ is irrational, then $ab$ is irrational

Question

Prove if $a\ne0$ is rational and $b$ is irrational, then $ab$ is irrational.

any advice on how to prove this? I tried to prove the contrapositive and rearranged $ab = (n/m)(n/m)$ to get $n^2 = m^2(ab)$ but im stuck in a dead end and don't know what to do from here

Answer 1

Try a contradiction proof. Suppose $ab$ is rational, then dividing by the rational $a\ne0$ we get $b$ rational, a contradiction. Thus $ab$ is irrational.

Answer 2

If $ab = \frac{m}{n}$ with $a = \frac{c}{d}$ then $b = \frac{dm}{cn}$ i.e. $b$ is rational a contradiction.

Answer 3

Suppose for the sake of contradiction that $a$ is rational and $ab$ is irrational and $b$ is not irrational. Then we have $a$ and $b$ rational, and $ab$ irrational. Since $a$ and $b$ are rational, we know there are integers $c,d, e, f$ for which $a =\dfrac cd$ and $b =\dfrac ef$. Then $ab =\dfrac {ce}{df}$ , and since both $ce$ and $df$ are integers, it follows that $ab$ is rational.

But this is a contradiction because we started out with $ab$ irrational.

Answer 4

Proof (by Contradiction)

I am not sure if we need to justify why we can divide by a in line 5.