Floor defines Equivalence Relation

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

Let $\left \lfloor {x}\right \rfloor$ be defined as the floor function of $x$.

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

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

Proof
Checking in turn each of the critera for equivalence:

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

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

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 $n \in \Z$: