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Simple question on inequality

  1. Jan 8, 2016 #1
    Let a>0. It is also true that a+b>0. Can we prove that b>0 always?
    My attempt
    b>-a.... but 0>-a.

    therefore min (b,0)>-a
    case 1: b <0.
    If, b <-a, then a <-b
    so a+b <b-b
    so, a+b <0.
    Contradiction
    Hence b>0.
    Is my proof right?
     
  2. jcsd
  3. Jan 8, 2016 #2

    Samy_A

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    No.

    a=20, b=-5
    a>0
    a+b=15>0

    But b<0.

    So where is the mistake in your proof?
     
  4. Jan 8, 2016 #3

    Svein

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    No.
     
  5. Jan 8, 2016 #4
    Yeah....I get it. b>-a...and b. an be between 0 and -a
     
  6. Jan 8, 2016 #5
    Yup, I did not consider between 0 and -a
     
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