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Math Help - Prove ab > 0 (basic proof)

  1. #1
    Member mybrohshi5's Avatar
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    Prove ab > 0 (basic proof)

    Suppose a,b are elements of the Real numbers and a*b > 0

    Then either a>0 and b>0 OR a<0 and b<0


    I figured the easiest way to do this proof is using trichotomy, but it seems like there should be a way to compress what I have into fewer steps (but maybe not).

    So this is what I have:

    assume b>0

    if a = 0 then 0*b > 0 is false since 0 = 0

    if a < 0 then a*b > 0 is false since (ab < 0) the inequality would flip when multiplying by a value less than 0

    if a > 0 then a*b > 0 is true

    assume b < 0

    if a = 0 then 0*b > 0 is false since 0=0

    if a < 0 then a*b > 0 is true

    if a > 0 then a*b > 0 is false since (ab < 0) the inequality would flip when multiplying by a value less than 0

    Thus we see when b > 0 and a > 0 then ab >0, and when b < 0 and a < 0 then ab > 0

    Any criticism on my proof would be greatly appreciated

    Thanks!
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  2. #2
    MHF Contributor

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    Re: Prove ab > 0 (basic proof)

    You left out the case that b= 0.
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