Prove that $mn < 0$ if and only if exactly one of $m,n$ is positive I need to prove that $mn <  0$ if and only if $m > 0$ and $n < 0$ or $m < 0$ and $n > 0$.
So I need to prove two cases:
1. If $m < 0$ and $n > 0$ or, in the alternative, if $m > 0$ and $n < 0$, then $mn < 0$.
2. If $mn < 0$, then $m < 0$ and $n > 0$ or else $m > 0$ and $n < 0$.
So far I have, 
By axiom O.1 (from the study guide), if $m$ and $n$ are positive , then $mn > 0$, and by corollary 1.14 (study guide), if $m$ and $n$ are negative, then $(-m)(-n) > 0$. Hence, if $m$ is negative and $n$ is positive or $m$ is positive and $n$ is negative, then $(-m)n < 0$ and $m(-n) < 0$.
I am not sure if this I am on the right track or not, but at this point I am just completely stumped. Any help is welcome.
Thanks,
Tony

It looks OK up until the last sentence. You say that if $m$ is negative and $n$ is positive or $m$ is positive and $n$ is negative $(-m)n < 0$ and $m(-n) < 0$. This is false; these quantities are positive, not negative. I think I know what you meant, but you need to rewrite this part.
Note, if $m$ is negative, don't write $-m$ to refer to that same number. That's like saying $-1$ is negative and then using $-(-1) = 1$ in place of $-1$.
