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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

Caleb Stanford
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  • Your proof seems alright to me – Stefan4024 Sep 05 '13 at 02:05
  • I think your fourth paragraph ("By axiom...") is confusing people. I think you mean to assume throughout that $m$ and $n$ are both positive. Otherwise some of the things that you say (like if $m$ negative and $n$ positive, $(-m)n < 0$) are false. Or, perhaps you mean in all your statements to just have $mn > 0$ or $mn < 0$, rather than all the negative signs you include. Currently what you've written is false. – Caleb Stanford Sep 05 '13 at 02:23
  • Is my answer acceptable? If not, let me know what it is you are yet to understand. – Michael Albanese Sep 27 '13 at 19:03
  • Hi Michael, sorry about not getting back to you sooner. Well, I got my assignment back and I was able to prove that if m>0 and n <0 then mn<0 but I was not able to prove that if mn<0 then either M<0 and n>0 or m>0 and n<0, so do you mind help with that? –  Sep 29 '13 at 18:02

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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 $-1$.

Michael Albanese
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  • So, you saying that I should rewrite my last sentence like this, "Hence, if m is negative and n is positive or m is positive and n is negative, then mn<0". That make sense. I did not realize I was implying a double negative. Is this complete then? To me it seems like I am still missing something, but I am not sure what. –  Sep 05 '13 at 02:34
  • Well, if you use the word 'hence', the statement which follows it should be implied by the statement previous to it, but that isn't the case. You've shown $mn > 0$ when $m$ and $n$ have the same sign. Now $m$ and $n$ have different signs, but $-m$ and $n$ have the same sign so $(-m)n > 0$. There are many ways of writing this but the basic idea is to take the case where $m$ and $n$ have different signs and somehow modify it so that you are in a situation with two numbers of the same sign, that way you can apply your earlier results. – Michael Albanese Sep 05 '13 at 03:02