Floor defines Equivalence Relation

Theorem
Let $x \in \R$ be a real number.

Let $\floor x$ denote the floor function of $x$.

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

Then $\mathcal R$ is an equivalence, and $\forall n \in \Z$, the $\mathcal R$-class of $n$ is the half-open interval $\hointr n {n + 1}$.

Proof
Checking in turn each of the critera for equivalence:

Reflexivity

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

Thus the floor function is reflexive.

Symmetry

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

Thus the floor function is symmetric.

Transitivity
Let $\floor x = \floor y$ and $\floor y = \floor z$.

Let $n = \floor x = \floor y = \floor z$, 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 Floor plus Difference‎.

Thus $x = n + t_x, z = n + t_z$ and $\floor x = \floor z$.

Thus the floor function is transitive.

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

Now we show that the $\mathcal R$-class of $n$ is the interval $\hointr n {n + 1}$.

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