I'm a little confused on what to answer to this: Determine whether the function floor(x) is big omega of x.
The above function holds for integers but not for real numbers. According to the definition of big omega x could be either real or integer so I'm unsure what is the right answer although I've seen that the solution is that it does hold according to some online solutions. Could somebody explain the correct answer for this? Could I say on my answer that it holds for integers and not for reals? I'm taking a basic course on discrete math so I'm relatively new to this concept, thanks for your help.
Best Answer
You can take $ \text{floor}(x) >\frac{x}{2} = \Omega(x)$
EDIT: OK, so you want to show that some function $f(x)$ is bounded from the below by $cg(x)$, i.e. the function multiplied by some positive constant. in this specific case you have $f(x) = \text{floor} \ x$ and $g(x) = x$. The job is to show that for $\exists \ c \ \text{s.t.} \forall \ x >a>0$ the inequality $f(x) > c x = \Omega(x)$ holds. One such constant (although by far not the only one) is $\frac{1}{2}$, because clearly $\forall \ x > 1 \ \text{floor} \ x > \frac{x}{2}$.