Equivalence Classes of Dipper Relation

Theorem
Let $m \in \N$ be a natural number.

Let $n \in \N_{>0}$ be a non-zero natural number.

Let $\RR_{m, n}$ be the dipper relation on $\N$:


 * $\forall x, y \in \N: x \mathrel {\RR_{m, n} } y \iff \begin {cases} x = y \\ m \le x < y \text { and } n \divides \paren {y - x} \\ m \le y < x \text { and } n \divides \paren {x - y} \end {cases}$

The equivalence classes of $\RR_{m, n}$ are:
 * $\eqclass x {\RR_{m, n} } = \begin {cases} \set x & : x < m \\ \set {y \in \N: \paren {m \le y} \land \paren {\exists k \in \Z: x = y + k n} } & : x \ge m \end {cases}$

Thus the classes are:

and it is seen that there are exactly $m + n$ such equivalence classes.

Proof
From Dipper Relation is Equivalence Relation we have that $\RR_{m, n}$ is indeed an equivalence relation.

Let $x < m$.

Then by definition:
 * $y \in \eqclass x {\RR_{m, n} } \iff x = y$

and so:
 * $\eqclass x {\RR_{m, n} } = \set x$

Let $m \le x$.

Let $x \mathrel {\RR_{m, n} } y$.

First we note that $m \le y$.

Let $x = y$.

Then $y \in \eqclass x {\RR_{m, n} }$ immediately, and:
 * $x = y + 0 \times n$

That is:
 * $\exists k \in \Z: x = y + k n$

where in this instance $k = 0$.

Let $x > y$.

We have that:

Let $x < y$.

We have that:

Hence when $m \le x$, we have that:
 * $\eqclass x {\RR_{m, n} } = \set {y \in \N: \paren {m \le y} \land \paren {\exists k \in \Z: x = y + k n} }$