Why does $\frac{a}{b}<0$ imply $ab<0$?

Question

I'm not sure if this was asked before, but my question is: why does $\frac{a}{b}<0$ imply $ab<0$? How do you prove it both intuitively and rigorously(using math)? I think I understand it intuitively: it's becuase for $\frac{a}{b}$ to be negative, exactly one of $a$ or $b$ has to be negative. For $ab$ to be negative, exactly one of $a$ or $b$ has to be negative. That means that these two imply each other. But how would I prove this rigorously? If I multiply both sides of $\frac{a}{b}<0$ by $b$, first of all, I don't know whether $b$ is positive or negative so I don't know which way the inequality sign is facing, and second, even if we did know that it flipped or didn't flip, we would only get $a<0$ or if the sign didn't flip $a>0$. Do I split it into cases then(case 1: $b<0$ and case 2: $b>0$)? It seems like it would work but there might be a more slicker way of proving it?

Answer 1

Multiply both sides by $b^2$ which is always positive and hence doesn't flip the inequality sign.

Answer 2

Of course, assuming $b \neq 0$ :

$$\frac{ab}{a/b} = b^2$$

is positive, so the numerator and denominator have to have the same sign : ergo if one is negative, so is the other. (Which proves a stronger statement to the one you propose).

---

This approach neither breaks $b$ into cases, nor starts by supposing $\frac ab$ negative and proceeding.

Answer 3

If you know one, you can obtain the other by multiplying/dividing by $b^2$, which is possible since that will always be a positive quantity. Hence, it suffices to show that any one of these statements is true.

Answer 4

For real numbers $a,b$: $$\frac{a}{b} <0 \implies \frac{ab}{b^2} <0 \implies ab <0,$$ as $b^2$ is positive definite which can be transferred to RHS without changing the sign of inequality.