Proof of $n - \lceil \dfrac{n}{2} \rceil = \lfloor \dfrac{n}{2} \rfloor$ (Latex test - MathJax)
The key here is to divide the cases, it is not really difficult at all To proof : $n - \lceil \dfrac{n}{2} \rceil = \lfloor \dfrac{n}{2} \rfloor$ Case 1. $ n = 2k + 1$ $$n - \lceil \dfrac{n}{2} \rceil = 2k + 1 - ( k+1) = \lfloor k + \dfrac{1}{2} \rfloor = k$$ Both the LHS and RHS are $k$ therefore, the equality stands true. Case 2. $ n = 2k$ $$n - \lceil \dfrac{n}{2} \rceil = 2k - (k) = \lfloor \dfrac{2k}{2} \rfloor = k $$ Both the LHS and RHS are $k$ therefore, the equality stands true. Thus, we have shown $n - \lceil \dfrac{n}{2} \rceil = \lfloor \dfrac{n}{2} \rfloor$