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Proving Inequalities

  1. Sep 29, 2009 #1
    Ok... So today, someone asked me a simple question: Why do two negatives become a positive number when multiplied together? This is intuitively basic, but not as easy to prove (unless there's some simple proof that I didn't think of). This was the basic proof that I came up with:

    Let a, b, c > 0 and a, b, c [tex]\in[/tex] R.

    a > 0

    a/(-b) < 0/(-b)

    a/(-b) < 0

    a/(-b)(-c) > 0/(-b)(-c)

    a/(-b)(-c) > 0

    Since a/(-b)(-c) is greater than 0, and a is positive, then (-b)(-c) must also be positive. Therefore, multiplying by two negatives will produce a positive.

    I know my proof isn't very good, but I am not a mathematician. I thought that I came up with a semi-decent way to show that multiplying two negatives together will produce a positive number. Now, the next question that person asked me was: Well then why does the inequality flip over when you divide by a negative? Again, it's intuitively obvious but I do not know how to prove it. Does anyone know how I could prove that or a better way to prove what I just attempted to prove?

    P.S. I meant to put this in general math because it would be more appropriate there, but I accidently posted it here and cannot delete it. So maybe a moderator can move it...
    Last edited: Sep 29, 2009
  2. jcsd
  3. Sep 30, 2009 #2


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    Is this a/((-b)(-c)) or (a/(-b))(-c)?

    Of course, you use "if c< 0 and a> b then ac< bc" extensively. That doesn't seem to me to be more obvious than "negative times negative is positive" itself. That would be proved by "If c< 0 then (-c)> 0. If also a> b then a(-c)> b(-c) or -(ab)> -(ac). Now add ab and ac to both sides: ab+ ac- (ab)= ac> ab+ ac- (ac)= ab." Immediately from that follows "negative times negative equals positive": if a< 0 and b< 0, then ab> 0b= 0.
  4. Sep 30, 2009 #3
    Doesn't matter... It works out the same...

    Thanks for the advice... I'll keep that in mind when trying to expand my current proof...
  5. Oct 1, 2009 #4
    Why did you divide?

    Couldn't you get the same thing with

    -a < 0
    (-a)(-b) > 0(-b) = 0

  6. Oct 1, 2009 #5
    I divided because it doesn't really make a difference to the proof and it's easier on me. I wanted to show that a/((-b)(-c)) > 0. Since the numerator, by definition, is greater than 0, then the denominator must be greater than 0. I thought it would be an easy way to look at it since you're looking at a, and (-b)(-c) separately. I wanted to organize it so that you could look to the top of the fraction for one and the bottom for the other...
  7. Oct 2, 2009 #6
    24/((-8)(-3)) < (24/(-8))(-3) so I don't understand why you say it doesn't matter what order you preform the operations.
  8. Oct 2, 2009 #7
    Ah, okay ... I get why you did it, but I still prefer my way ;), hehe. I guess it just depends how you want to see it ... my way shows that the product of 2 negatives is greater than 0.

    As for ramsey2879, I think Whatever123 meant that it doesn't matter in the sense that they are both greater than zero ... but if you consider Whatever123's response to my question, then really, it should be a / ((-b)(-c)), so that the denominator is a product of two negatives :)
  9. Oct 2, 2009 #8
    I'm saying that it doesn't matter in my proof because I do not care what the actual value is; I just care about if it's greater or less than 0. I'm just worried about the signs, not the value. So it does work out the same for my proof.
  10. Oct 2, 2009 #9
    Yes. That was an original typo mistake that I simply missed. I updated it when I responded to you.
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