Combining two inequations

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.

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Yes, $m>\max(n,8/c)$. – AD. Nov 23 '10 at 21:50
$m>n>\frac{8}{c}$ cannot hold because of two cases, those being $\frac{8}{c}=n$ and $n<\frac{8}{c}$. – Ralth Nov 23 '10 at 21:58
I think I see your misunderstanding. You know that $m>\frac{8}{c}$ implies what you want, but you need want a natural number. Take the next largest one, $n = \lceil\frac{8}{c} \rceil$ as you said before. Remember, for a given $c$, you only have to find a single $n$ that will work and any will do. – Ralth Nov 23 '10 at 22:50
Thanks Ralth, that helped to see my mistake. – Femaref Nov 24 '10 at 13:12

Nope. What if $m=2$, $c=7$?

There's really no way of finding out the what the relationship b/w $n$ and $8/c$ is without additional information.

Also, the task of finding an $n$ that satisfies certain properties is very different from the task of determining whether $n$ satisfies certain properties. (It's the difference between "there exists" and "for all").

If I understand you correctly, what you actually need is to find an $n$ such that $n<m \Rightarrow \frac8c<m$. Then $n = \lfloor \frac8c \rfloor$ should satisfy that.
So, you need to show that for every $c>0 \in \mathbb{R}$ there exists $n \in \mathbb{R}^+$ such that $|f(x)|<c|g(x)|$ for all $x>n$. Given a $c>0$, let $n=\frac{8}{c}$.
Then... $x>n \implies |f(x)|<c|g(x)|$