Inference from Inequalities

Theorem

If $ab>0$, then either

1. $a>0$ and $b>0$, or
2. $a<0$ and $b<0$.

Corollary

If $ab<0$, then either

1. $a<0$ and $b>0$, or
2. $a>0$ and $b<0$.

Proof

Suppose the contrary is true that $a>0$.

Choose an arbitrary $\epsilon >0$, for example $\epsilon =\frac{1}{2}a$. Then we have $a>\epsilon >0$.

Becuase the hypothesis $\epsilon >a>0$ (see the initial premise) leads the contradictory conclusion that $a>\epsilon$, we can conclude that the hypothesis is incorrect and we conclude that $a=0$.