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?

I'd try splitting it into $n$ even and $n$ odd.
If $n$ is even then $n = 2k$ with $k$ an integer, and
$$\lceil\frac{n-1}{2}\rceil = \lceil\frac{2k-1}{2}\rceil = \lceil k - (1/2)\rceil = k.$$
You asked for just a hint so does this help?
