How to determine the floor of $\frac {x} {2}$

Question

I am trying to determine if the bound is sharp for the theorem that states the connectedness of a graph is at most the floor of $2m/n$ where $m$ is the size of the graph $G$ and $n$ is the order. I am trying to do this for the graph of the complete bipartite where it is $K_{\frac{n}{2},\frac{n}{2}}$. I get up the the point of simplifying $\kappa (G)\leq\frac{2m}{n}$ but I am not sure how to find the floor of $\frac {n}{2}$ (which is up to where I simplify it) since this is using variables. Any suggestions? ( I also am not sure if I am using the correct $m$ as I have said this to be $\frac{n^2}{4}$)

Answer 1

In your example, $\frac{n}{2}$ must be an integer, because otherwise $K_{\frac{n}{2}, \frac{n}{2}}$ does not make sense. This means we do not need to round down to the nearest integer when we take the floor function (because $\frac{n}{2}$ is already an integer), and $\lfloor \frac{n}{2} \rfloor = \frac{n}{2}$.