Consider the expression


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

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

$\frac{2n-1}{2}=n-\frac{1}{2}$, so the floor is $n-1$.

  • $\begingroup$ ...of course assuming that $\;n\;$ ranges over the integers, which seems implicit in the OP's choice of that variable name. $\endgroup$ – Marnix Klooster Mar 10 '16 at 5:05

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.