Floor defines Equivalence Relation

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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 $n \in \Z$:

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

$\blacksquare$


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