## Prove taht MN < 0

Sequences, counting (including probability), logic and truth tables, algorithms, number theory, set theory, etc.

### Prove taht MN < 0

Hi,

I need to prove that mn < 0 if and only if m > 0 and n < 0 or m < 0 and n > 0.

I know that I can say that if m > 0 and n > 0 then mn > 0 or if m < 0 and n < 0 then mn > 0, therefore if m < 0 and n > 0 then mn < 0 or if m > 0 and n < 0 then mn < 0. I can reference an axiom and a corollary from my text book to show this but I am not sure if I should do anything else, or if there is a better way to complete this exercise.

Any help or insight is appreciated.

Tony
tonyc1970

Posts: 15
Joined: Fri Aug 02, 2013 4:58 pm

### Re: Prove taht MN < 0

tonyc1970 wrote:I need to prove that mn < 0 if and only if m > 0 and n < 0 or m < 0 and n > 0.

You need then to prove two things:

i) If m < 0 and n > 0 or, in the alternative, if m > 0 and n < 0, then mn < 0.
ii) If mn < 0, then m < 0 and n > 0 or else m > 0 and n < 0.

tonyc1970 wrote:I know that I can say that if m > 0 and n > 0 then mn > 0 or if m < 0 and n < 0 then mn > 0...

Agreed. (I am assuming that you have valid logical backing for this statement.)

tonyc1970 wrote:...therefore if m < 0 and n > 0 then mn < 0 or if m > 0 and n < 0 then mn < 0.

I do not see how this follows. You need to show (i), which may be stated as "If P (or P'), then Q". You say that you can show "If not-P, then not-Q". But are these logically equivalent?

Also, what progress have you made on proving (ii)? Thank you.
nona.m.nona

Posts: 241
Joined: Sun Dec 14, 2008 11:07 pm

### Re: Prove taht MN < 0

Hi,

Here is what I came up with.

By axiom O.1 (p.14 of the study guide), if m and n a positive, then mn > 0. By corollary 1.14 (p. 15 of the study guide), if m and n are negative, then (-m)(-n) > 0. Therefore, if both m and n are either positive or negative, it follows that if m is negative and n is positive or m is positive and n is negative, then (-m)(n) < 0 is equal to [-(mn) < 0] and (m)(-n) < 0 is also equal to [-(mn) < 0]. Hence, mn < 0, if and only if m > 0 and n < 0 or m < 0 or n > 0.

Thanks,

Tony
tonyc1970

Posts: 15
Joined: Fri Aug 02, 2013 4:58 pm

### Re: Prove taht MN < 0

tonyc1970 wrote:...if both m and n are either positive or negative, it follows that if m is negative and n is positive....

The above seems to say "If they have the same sign, then it follows that they might have opposite signs". Your meaning is unclear here. Perhaps it might be helpful to use shorter statements to clarify your steps and reasoning.
nona.m.nona

Posts: 241
Joined: Sun Dec 14, 2008 11:07 pm

### Re: Prove taht MN < 0

Hi,

What I am trying to show is that if mn > 0 and (-m)(-n) > 0 then it must be the case that (-m)(n) < 0 and (m)(-n) < 0.

My study guide would break this down even more by taking (-m)(n) < 0 and stating that this is equal to -(m)(n) < 0. This basically says make the product of m * n and make it negative.

I hope this helps.

tonyc1970

Posts: 15
Joined: Fri Aug 02, 2013 4:58 pm

### Re: Prove taht MN < 0

The 1st reply split this into the 2 parts. Which part are you trying to do now?
buddy

Posts: 86
Joined: Sun Feb 22, 2009 10:05 pm