# Floor or Ceiling when getting results for N after doing Big O and Big Theta?

I'm working on my data structures homework and proving Big O and Big Theta for a couple questions. After choosing my values for C and getting to the end I'm left with the fraction 1/300 for $N_0$ for both Big O and Big Theta. I understood from lecture that we actually take the ceiling of the fraction for Big O, making it 1. My question is, do I also take the ceiling of Big Theta, or would it be the floor? Since Big Theta is a lower-bound.

The reason we take floor or ceiling at all in this context is that the definition asks for an integer $N_0$. Now the definitions of big $O$ and big $\Theta$ are both of the form
There exists a real constant $C > 0$ and an integer constant $N_0$ such that for all $n \geq N_0$, the following holds: ...
Let us denote what should hold by $P(n)$. If you have shown that $P(n)$ holds for all $n \geq N_0$, then it follows that it holds for all $n \geq \lceil N_0 \rceil$ as well, since if $n \geq \lceil N_0 \rceil$ then also $n \geq N_0$. It doesn't necessarily follow that $P(n)$ holds for all $n \geq \lfloor N_0 \rfloor$, since if $N_0$ is not an integer, $\lfloor N_0 \rfloor \not \geq N_0$.