Verification of simple inequality proof

  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?

    Thanks,
    -GeoMike-
     
    Last edited: Jul 31, 2006
  2. jcsd
  3. 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. 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. Office_Shredder

    Office_Shredder 4,500
    Staff Emeritus
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    Gold Member

    Proof by reverse is great, but you have to show that the steps can logically be reversed at the end
     
  6. HallsofIvy

    HallsofIvy 40,214
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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. Thank you for the replies! :smile:

    -GeoMike-
     
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