Floor Defines Equivalence

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 $n \in \Z$:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle x\)	\(\in\)	\(\displaystyle \left[\!\left[{n}\right]\!\right]_{\mathcal R}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \left \lfloor {x}\right \rfloor\)	\(=\)	\(\displaystyle \left \lfloor {n}\right \rfloor = n\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle \exists t \in \left[{0 .. 1}\right): x\)	\(=\)	\(\displaystyle n + t\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \implies\)	\(\displaystyle \)	\(\displaystyle x\)	\(\in\)	\(\displaystyle \left[{n .. n+1}\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Sources

• J.A. Green: Sets and Groups (1965)... (previous)... (next): Exercise $2.2, \ 2.3$
• A.N. Kolmogorov and S.V. Fomin‎: Introductory Real Analysis (1968): $\S 1.4$: Example $3$
• Thomas A. Whitelaw: An Introduction to Abstract Algebra (1978)... (previous)... (next): Exercise $3.1$