I'm trying to disprove that $\forall f: N\rightarrow R^+,\forall g: N\rightarrow R^+, f \in \Omega(g) \iff \lfloor f\rfloor \in \Omega(\lfloor g\rfloor).$
However I need some hints.
I'm trying to disprove that $\forall f: N\rightarrow R^+,\forall g: N\rightarrow R^+, f \in \Omega(g) \iff \lfloor f\rfloor \in \Omega(\lfloor g\rfloor).$
However I need some hints.
I disproved it by a counterexample (two functions) to disprove the implication in the forward direction. If $f(n) = 0.25$ and $g(n) = 1$, then $f \geq 0.25 g(n)$ for all $n$, which shows $f \in \Omega(g)$. However, $\lfloor f(n) \rfloor = 0$, and $\lfloor g(n) \rfloor = 1$, thus $\lfloor f(n) \rfloor$ is not in $\Omega(\lfloor g(n) \rfloor)$.