Hint with floors and ceiling proof Prove that $$\forall n \in \mathbb{Z},\left \lceil \frac{n-1}{2} \right\rceil = \left\lfloor \frac{n}{2} \right\rfloor$$
I decided to approach it by extending floor and ceiling definition and got
$$ \frac{n-1}{2} \leq \left\lceil \frac{n-1}{2} \right\rceil = \left\lfloor \frac{n}{2} \right\rfloor \leq \frac{n}{2} $$
And now I am stuck with $$\frac{n}{2} = \frac{n-1}{2} $$
which only made it worse.
Any advice or hint on how I can approach this proof differently?

Let $n = 2m + 1$ for some $m \in \mathbb{Z}$. In this case, $\displaystyle \left\lceil \frac{n-1}{2} \right\rceil = m$. Similarly, $\displaystyle \left\lfloor \frac{n}{2} \right\rfloor = \left\lfloor m + \frac{1}{2} \right\rfloor = m$. Consequently, $\displaystyle \left\lceil \frac{n-1}{2} \right\rceil = \left\lfloor \frac{n}{2} \right\rfloor$.
We can similarly consider the case $n = 2m$. For this case, $\displaystyle \left\lfloor \frac{n}{2} \right\rfloor = m$ and $\displaystyle \left\lceil \frac{n-1}{2} \right\rceil = \left\lceil m - \frac{1}{2} \right\rceil = m$. Once again, we can conclude $\displaystyle \left\lceil \frac{n-1}{2} \right\rceil = \left\lfloor \frac{n}{2} \right\rfloor$, giving us the required result.
