# What does this floor expression evaluate to?

Consider the expression

$$\left\lfloor\frac{2n-1}{2}\right\rfloor\;:$$

does it make sense that this floor function will evaluate to $n$? Or should it be $n-1$?

• Are we saying anything about n? Is it an integer, real number, etc.? – The Great Duck Mar 10 '16 at 5:19

$\frac{2n-1}{2}=n-\frac{1}{2}$, so the floor is $n-1$.
• ...of course assuming that $\;n\;$ ranges over the integers, which seems implicit in the OP's choice of that variable name. – Marnix Klooster Mar 10 '16 at 5:05