Ceiling defines Equivalence Relation

Theorem
Let $$\mathcal{R}$$ be the relation defined on $$\R$$ such that:
 * $$\forall x, y, \in \R: \left({x, y}\right) \in \mathcal{R} \iff \left \lceil {x}\right \rceil = \left \lceil {y}\right \rceil$$

Then $$\mathcal{R}$$ is an equivalence, and $$\forall n \in \Z$$, the $\mathcal{R}$-class of $$n$$ is the interval $$\left({n-1 \, . \, . \, n}\right]$$.

Proof
Checking in turn each of the critera for equivalence:

Reflexive

 * $$\forall x \in \R: \left \lceil {x}\right \rceil = \left \lceil {x}\right \rceil$$.

Symmetric

 * $$\forall x, y \in \R: \left \lceil {x}\right \rceil = \left \lceil {y}\right \rceil \Longrightarrow \left \lceil {y}\right \rceil = \left \lceil {x}\right \rceil$$.

Transitive
Let $$\left \lceil {x}\right \rceil = \left \lceil {y}\right \rceil, \left \lceil {y}\right \rceil = \left \lceil {z}\right \rceil$$.

Let $$n = \left \lceil {x}\right \rceil = \left \lceil {y}\right \rceil = \left \lceil {z}\right \rceil$$, which follows from transitivity of $$=$$.

Thus $$x = n - t_x, y = n - t_y, z = n - t_z: t_x, t_y, t_z \in \left[{0 \,. \, . \, 1}\right)$$ from Real Number is Ceiling minus Difference‎.

Thus $$x = n - t_x, z = n - t_z$$ and $$\left \lceil {x}\right \rceil = \left \lceil {z}\right \rceil$$.

Thus we have shown that $$\mathcal{R}$$ is an equivalence.


 * Now we show that the $\mathcal{R}$-class of $$n$$ is the interval $$\left({n-1 \, . \, . \, n}\right]$$.

Defining $$\mathcal{R}$$ as above, with $$n \in \Z$$:

$$ $$ $$ $$