# Prove/disprove these two claims

1. Mar 20, 2010

### Melodia

1. The problem statement, all variables and given/known data
I'm having trouble understanding this pair of claims:

2. Relevant equations

3. The attempt at a solution

So g is some function whose result is greater or equal to 0... but I don't really get what the rest of the claim means.
Could someone please explain them to me?

2. Mar 20, 2010

### sutupidmath

$$\mbox{ If } f(n)= O(g(n)), \mbox{ then it means that the function f(n) is of order at most g(n)}.$$

$$\mbox{ That is, by def., f(n) is big oh of g(n) if ther exists some positive constant C such that}$$

$$|f(n)|\leq C|g(n)|$$

Considering this, then what the problem is asking is that if f+g is of order at most h, then each of f and g must be of order at most h.

I haven't worked any of these problems out myself, but a proof by contradiction looks promising.

3. Mar 21, 2010

### Melodia

Ok thanks for clearing it up.
So basically it's saying:
If f(n) + g(n) $$\leq$$ c[h(n)]
Then f(n) $$\leq$$ c[h(n)] and g(n) $$\leq$$ c[h(n)]
The tricky part is that f(n) could be negative right? Since it says "$$\forall$$ f". Then f(n) could be larger than (f + g)(n)? Or does "$$\forall$$ f, g $$\in$$ F" mean "for all f and g that are in F"?

Last edited: Mar 21, 2010
4. Mar 21, 2010

### sutupidmath

Look at the definition of f:N-->R+. Yes ti means for all f, g that are in F.

5. Mar 21, 2010

### Melodia

Oh ok, then the answer would be obvious right? I mean if both f and g has to be positive, then either f or g has to be smaller than f + g; then they have to be in O(h) as well.