I read a proof about ceiling function equivalence relation as linked here. The theorem to prove is:
Let $\mathcal R$ be the relation defined on $\mathbb{R}$ such that: $\forall x, y \in \mathbb{R}: \left({x, y}\right) \in \mathcal R \iff \left \lceil {x}\right \rceil = \left \lceil {y}\right \rceil$
where $\left \lceil {x}\right \rceil$ is the ceiling of $x$.
Then $\mathcal R$ is an equivalence.
Proving reflexitivity and symmetry is trivial, but I have questions about proving transitivity.
According to definition, we should just proof that if $\left \lceil {x}\right \rceil = \left \lceil {y}\right \rceil$ and $\left \lceil {y}\right \rceil = \left \lceil {z}\right \rceil$, then $\left \lceil {x}\right \rceil = \left \lceil {z}\right \rceil$. In the proof, the author took extra steps to prove this. But isn't $\left \lceil {x}\right \rceil = \left \lceil {y}\right \rceil = \left \lceil {z}\right \rceil$ trivial? Are there any reason why the author went those extra steps?
Yes, it is trivial. Actually, it is so trivial that it has nothing to do with the ceiling function. If $f\colon X\longrightarrow Y$ is any function and if you define a binary relation $\sim$ on $X$ by $x\sim y$ if and only if $f(x)=f(y)$, then $\sim$ is an equivalence relation.