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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?

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

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

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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.

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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.

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