# Describing Distinct Equivalence Classes of a Relation

Suppose that $R$ is a relation on the set of complex numbers $\mathbb{C}$. The relation $R$ is defined as follows:

For any two complex numbers $w,z \in \mathbb{C}$, $$w R z \Leftrightarrow \operatorname{Re}(w)-\operatorname{Re}(z)=k, \quad \operatorname{Im}(w)-\operatorname{Im}(z) = l \sqrt{2}$$

for some integers $k,l \in \mathbb{Z}$.

Question: Describe the distinct equivalence classes of $R$.

How can I visualise the equivalence classes graphically?

• Try re-writing as follows: \begin{align*} w R z & \iff \operatorname{Re}(w) - \operatorname{Re}(z) \in \mathbb{Z}, \operatorname{Im}(w) - \operatorname{Im}(z) \in \sqrt{2} \mathbb{Z} \end{align*} Both of these should bear resemblance to an equivalence relation familiar from topology and/or abstract algebra. – AJY Nov 28 '15 at 16:57
• Better expression: $wRZ \iff w - z \in \mathbb{Z} + \sqrt{2} i \mathbb{Z}$ – AJY Nov 28 '15 at 17:03
• @AJY Requesting you to consolidate and put it as a nice answer. There are many from the community who could benefit, and this question does not remain 'Unanswered' officially – Shailesh Nov 28 '15 at 17:10
• ... or a + bi R c + di <=> a + bi = (a + k) + (b + root(2)l)i – fleablood Nov 28 '15 at 17:10
• I'm not sure what "visualize graphically" means exactly, but one can consider the distinct representative numbers as all the numbers w where $0 \le Re(w) < 1$ and $0 \le Im(w) < \sqrt(2)$ and any other $a + bi R (a - \lfloor a \rfloor) + \sqrt 2 [\frac {b}{\sqrt 2} - \lfloor \frac {b}{\sqrt 2} \rfloor ]i$. But I'm not sure that is what the OP was asking. – fleablood Nov 28 '15 at 17:19

So we know that $$w R z \iff \operatorname{Re}(w)-\operatorname{Re}(z)=k, \quad \operatorname{Im}(w)-\operatorname{Im}(z) = l \sqrt{2}$$
for some integers $k,l \in \mathbb{Z}$. This can be re-written as $$w R z \iff \operatorname{Re}(w)-\operatorname{Re}(z) \in \mathbb{Z}, \quad \operatorname{Im}(w)-\operatorname{Im}(z) \in \sqrt{2} \mathbb{Z}$$
Visually, this means that in the same way $\mathbb{Z}^{2}$ makes a "grid" (more specifically a lattice, as you don't have the sides of the rectangle, only the vertices) of $\mathbb{R}^{2}$, each of these equivalence classes makes a grid of the complex plane, where each equivalence class is a lattice that makes rectangles which are of length $1$ along the real axis, and length $\sqrt{2}$ along the imaginary axis. In other words, you can make the equivalence class $[z]$ by starting at $z$, gridding the plane with $1 \times \sqrt{2}$ rectangles (such that $z$ is a vertex of one of these rectangles), and then keeping only the vertices of the rectangles.
EDIT: In case you're unfamiliar, when given sets $A, B$, we define $A + B : = \{ a + b : a \in A, b \in B \}$. Similarly, we define $A + b : = \{ a + b : a \in A \}$.