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Subadditivity and Natural Logs

  1. Sep 13, 2009 #1
    1. The problem statement, all variables and given/known data
    Show that the natural logarithm is not subadditive.
    You could use ln(1/2+1/3)[tex]\leq[/tex]ln(1/2)+ln(1/3), but mathematicians view all such numerical evidence as an invalid proof.

    2. Relevant equations
    ln(a+b)[tex]\leq[/tex]ln(a)+ln(b)


    3. The attempt at a solution
    ln(1/2+1/2)[tex]\leq[/tex]ln(1/2)+ln(1/2)

    Well, my real question is what does the "mathematicians view all such numerical evidence as an invalid proof" mean? I am pretty sure this needs to be a proof by counter-example, which involves numerical evidence. What am I missing?
     
    Last edited: Sep 13, 2009
  2. jcsd
  3. Sep 13, 2009 #2
    Hint:

    [tex]ln(a)+ln(b) \geq ln(a+b)[/tex]

    [tex]ln(ab) \geq ln(a+b)[/tex]

    a>0, b>0
     
  4. Sep 13, 2009 #3
    So, I am guessing that hint entails:

    ab[tex]\geq[/tex]a+b

    _______
    I have two other ideas, but I have tried every combination of them to get them both to be true..... Nothing I try seems make it work.

    I am thinking that the ln(ab) and ln(a+b) have to not be irrational for this to be a valid proof? Is my logic correct?

    Also, I think a and b have to be some combination of 1 and ex....

    Am I getting warm at all?
     
  5. Sep 16, 2009 #4
    Well, I got frustrated and decided that a qualitative answer about the properties of a natural log is as good as a quantitative one, so:

    My proof:

    Show by counter example:

    a = 1
    b = x, 0<x<1

    We know lnx is negative and ln1 = 0 by the properties of a logarithm.

    Thus,

    ln(a+b) has to be positive, because 1 + x, 0<x<1 is greater than one.

    and

    ln(a) + ln(b) has to be negative because ln(1) = ln(a) = 0 and ln(b) = ln(x) 0<x<1 = (negative number)

    Furthermore,

    ln(a+b)>ln(a)+ln(b)

    Therefore, the natural log is not subadditive.

    Works in my mind.
     
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