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

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?

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

• oh ok thanks i will accept this answer when the time limit is up. Jul 7 '20 at 4:00
• I can't believe a simple answer like this slipped my mind. Jul 7 '20 at 4:01
• Happens to the best of us :) Jul 7 '20 at 4:04

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.

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.

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.