# 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.

$\Box$

### Symmetry

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

Thus the floor function is symmetric.

$\Box$

### 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 from Real Number is Floor plus Difference‎:

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

Thus:

$x = n + t_x, z = n + t_z$

and:

$\floor x = \floor z$

Thus the floor function is transitive.

$\Box$

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

$\Box$

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

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

 $\displaystyle x$ $\in$ $\displaystyle \eqclass n {\mathcal R}$ $\displaystyle \leadsto \ \$ $\displaystyle \floor x$ $=$ $\displaystyle \floor n$ $\displaystyle$ $=$ $\displaystyle n$ $\displaystyle \leadsto \ \$ $\displaystyle \exists t \in \hointr 0 1: x$ $=$ $\displaystyle n + t$ $\displaystyle \leadsto \ \$ $\displaystyle x$ $\in$ $\displaystyle \hointr n {n + 1}$

$\blacksquare$