Floor defines Equivalence Relation

Theorem
Let $$\mathcal{R}$$ be the relation defined on $$\mathbb{R}$$ such that $$\forall x, y, \in \mathbb{R}: \left({x, y}\right) \in \mathcal{R} \iff \left \lfloor {x}\right \rfloor = \left \lfloor {y}\right \rfloor$$.

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

Proof
Checking in turn each of the critera for equivalence:

Reflexive
$$\forall x \in \mathbb{R}: \left \lfloor {x}\right \rfloor = \left \lfloor {x}\right \rfloor$$.

Symmetric
$$\forall x, y \in \mathbb{R}: \left \lfloor {x}\right \rfloor = \left \lfloor {y}\right \rfloor \Longrightarrow \left \lfloor {y}\right \rfloor = \left \lfloor {x}\right \rfloor$$.

Transitive
Let $$\left \lfloor {x}\right \rfloor = \left \lfloor {y}\right \rfloor, \left \lfloor {y}\right \rfloor = \left \lfloor {z}\right \rfloor$$.

Let $$n = \left \lfloor {x}\right \rfloor = \left \lfloor {y}\right \rfloor = \left \lfloor {z}\right \rfloor$$, 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 Floor plus Difference‎.

Thus $$x = n + t_x, z = n + t_z$$ and $$\left \lfloor {x}\right \rfloor = \left \lfloor {z}\right \rfloor$$.


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