Ceiling defines Equivalence Relation

Theorem
Let $\RR$ be the relation defined on $\R$ such that:
 * $\forall x, y, \in \R: \tuple {x, y} \in \RR \iff \ceiling x = \ceiling y$

where $\ceiling x$ is the ceiling of $x$.

Then $\RR$ is an equivalence, and $\forall n \in \Z$, the $\RR$-class of $n$ is the half-open interval $\hointl {n - 1} n$.

Proof
Checking in turn each of the criteria for equivalence:

Reflexivity

 * $\forall x \in \R: \ceiling x = \ceiling x$

Thus the ceiling function is reflexive.

Symmetry

 * $\forall x, y \in \R: \ceiling x = \ceiling y \implies \ceiling y = \ceiling x$

Thus the ceiling function is symmetric.

Transitivity
Let:
 * $\ceiling x = \ceiling y$
 * $\ceiling y = \ceiling z$

Let:
 * $n = \ceiling x = \ceiling y = \ceiling z$

which follows from transitivity of $=$.

Thus:
 * $x = n - t_x, y = n - t_y, z = n - t_z: t_x, t_y, t_z \in \hointr 0 1$

from Real Number is Ceiling minus Difference‎.

So:
 * $x = n - t_x$

and:
 * $z = n - t_z$

and so:
 * $\ceiling x = \ceiling z$

Thus the ceiling function is transitive.

Thus we have shown that $\RR$ is an equivalence.

Now we show that the $\RR$-class of $n$ is the interval $\hointl {n - 1} n$.

Defining $\RR$ as above, with $n \in \Z$:

Also see

 * Floor defines Equivalence Relation