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Inequality proof

  1. Aug 29, 2013 #1
    1. The problem statement, all variables and given/known data
    Prove

    If ## 0 \leq a < b ## and ## 0 \leq c < d ## then ## ac < bd ##

    3. The attempt at a solution

    not sure how to even start on this,

    was thinking if a = 0 or c = 0, then ac = 0, but bd > 0 (which is given) so bd > ac
    however this seems like I'm cheating because they give you bd > ac, so I don't think it's valid
     
  2. jcsd
  3. Aug 29, 2013 #2

    verty

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    They don't give you ac < bd, that is what you must show.

    What you have done so far is correct but you still have to show it is true when 0 < a, 0 < c.

    Hint: you were probably given the rule that, for 0 < c, a < b → ac < bc. Use it.
     
  4. Aug 29, 2013 #3
    Yes we were given that, but I couldn't proceed how to use it. I used ac < bd to get to where I am right now.
     
  5. Aug 29, 2013 #4

    tiny-tim

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    no, you can't use ac < bd !!

    use verty's :smile: hint

    (second hint: if you could prove ac < bc, what would you still need to get ac < bd ? :wink:)
     
  6. Aug 31, 2013 #5
    I'm sorry, I'm trying but I just don't see how to proceed.

    For the proof of c>0 and a<b prove ac<bc it was very simple, all I had to use was (b-a).c > 0, but when there are two inequality signs I don't know how to deal with it. Is multiplying by b or d to each of the inequalities allowed? For instance, if 0<=a<b and 0 <=c<d is it true that 0<=bc < bd?
     
  7. Aug 31, 2013 #6

    tiny-tim

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    hi phospho! :smile:
    that's right! … if you can prove ac < bc and bc < bd,

    then obviously you have ac < bc < bd, ie ac < bd

    sooo … now prove ac < bc and bc < bd (separately) :wink:
     
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