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Homework Help: Verification of simple inequality proof

  1. Jul 31, 2006 #1
    The question given is:
    If a < b, prove that a < (a+b)/2 < b

    The book had a different proof than the one I came up with. I understand the book's proof, I just want to know if my proof is also ok.

    I did the following:

    a < (a+b)/2 < b
    2a < a+b < 2b
    a < b < 2b-a
    a-b < 0 < b-a

    Since it was given that a < b, a-b must be less than 0, and b-a must be greater than zero, so the inequality a < (a+b)/2 < b is true if a < b.

    Is this ok?

    Last edited: Jul 31, 2006
  2. jcsd
  3. Jul 31, 2006 #2
    It looks a bit like you did the proof in reverse. It seems that you started with a<(a+b)/2<b and then got to a<b rather than starting with a<b and proving a<(a+b)/2<b like the problem seems to want.
  4. Jul 31, 2006 #3
    really you should show the direction of the implications you're using, so what you mean is actually

    [tex]a<(a+b)/2<b \Longleftarrow 2a<(a+b)<2b \Longleftarrow a<b<2b-a \Longleftarrow a-b<0<b-a \Longleftarrow a<b,[/tex]

    and not the other way. It doesn't matter much here since all the inequalities there are equivalent (so the implications work both ways, ie. in fact [itex]a<b \Longleftrightarrow a<(a+b)/2<b[/itex]).
    Last edited by a moderator: Aug 1, 2006
  5. Jul 31, 2006 #4


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    Proof by reverse is great, but you have to show that the steps can logically be reversed at the end
  6. Aug 1, 2006 #5


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    That's sometimes called "synthetic proof". It's often used to prove trig identitities. Start with what you want to prove and work back to an obviously true statement. It's valid as long as every stepe is reversible,.
  7. Aug 1, 2006 #6
    Thank you for the replies! :smile:

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