1. Sep 13, 2009

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)$$\leq$$ln(1/2)+ln(1/3), but mathematicians view all such numerical evidence as an invalid proof.

2. Relevant equations
ln(a+b)$$\leq$$ln(a)+ln(b)

3. The attempt at a solution
ln(1/2+1/2)$$\leq$$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. Sep 13, 2009

### njama

Hint:

$$ln(a)+ln(b) \geq ln(a+b)$$

$$ln(ab) \geq ln(a+b)$$

a>0, b>0

3. Sep 13, 2009

So, I am guessing that hint entails:

ab$$\geq$$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?

4. Sep 16, 2009

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.