# How to justify $f(n) = O(g(n))$ [duplicate]

The following question is in my homework:

Is the statement $$f(n) = O(g(n))$$ true, when $$f(n) = n/2 + 4$$ and $$g(n) = \sqrt{n} + 2\log_2 n + 3$$?

I understand how $$f(n)$$ is the upper bound of $$g(n)$$. However, I am unsure how to prove it mathematically.

## marked as duplicate by David Richerby, Evil, Discrete lizard♦Sep 12 at 14:45

f(n)=$$\theta(n)$$ while g(n)=O(n^(1/2)) so the statment is not true.
• The answer has been revised, so @YuvalFilmus's comment no longer applies. However, the answer doesn't really prove anything, because it makes assertions at the same level of conceptual difficulty as the thing that's supposed to be proven. It's like proving that $\pi>2$ by saying "Well, $\pi>3$ and $3>2$, QED." Anyone who's prepared to just accept that $\pi>3$ should have just accepted $\pi>2$ in the first place. – David Richerby Nov 16 at 0:24