Given $c \in \mathbb{R}$ and $m,n \in \mathbb{N}$, is it possible to combine the inequations
- $n < m$
- $\frac{8}{c} < m$
into one inequation?
I argued that due $n < m$ and $\frac{8}{c} < m$, there has to be one $n$ between $m$ and $\frac{8}{c}$, being $n = \left\lceil \frac{8}{c} \right\rceil$, so that $m > n > \frac{8}{c}$, but I'm not sure if that is correct.
Context
$\forall c \in R^{+} \exists n \in \mathbb{N}: \forall m \in \mathbb{N} \land m > n: |f(m)|< c*|g(m)) \implies f \in \mathcal{o}(g)$
So, this is about the landau-notation of $f(x)$ raises slower than $g(x)$, the task is to show that a specific case ($f(x) = 8x$ and $g(x) = x^2$) complies to this definition.
Note: I already solved it until I encounter both inequations together and want to show that there is an $n$ for all cases.