# Find classes of relation equivalence

$$S=\{3n+1:n∈N\} = \{1,4,7,10,...\}$$ and relation is defined as:

$$(x,y) ∈ ρ \text{ def }⇔ 4|(x + 3y)$$

I need to prove that relation is relation of equivalence (that means that it is reflexive, symmetric and transitive.) I know how to do that, and once I prove that it is relation of equivalence, I need to find the equivalence classes.

My question is: For example: class of $$1$$ is defined as $$1=\{x∈S : x\text{ is related to }1\} = \{x∈S: 4|(x + 3*1)\} = \{3n+1:n∈N\text{ and }4|(3n+1)+3\text{ or just }4|(3n+3)\}$$ ???

Well, let's say that the equivalence class of $$1$$ is $$\bar 1=\{x\in S \colon (x,1)\in \rho\}=$$$$=\{x\in S\colon 4|x+3\cdot 1\}=\{x\in \mathbb N_0\colon x=3n+1, 4|(3n+1)+3\}.$$
So $$\bar 1$$ is the set of natural numbers such that $$x=3n+1$$, $$(n\in \mathbb N_0)$$ and $$4|3n+4$$.
Since this is the same that $$4|3n$$, this implies $$4|n$$, so $$n=4k$$ with $$k\in \mathbb N_0$$.
So the elements of $$\bar 1$$ are those of the form $$3n+1=12k+1$$ with $$k\in \mathbb N_0$$, that is, $$\bar 1=\{1,13,25,37,\ldots\}$$ (note that, in fact, $$\bar 1\subset S$$) and you can figure out what (and how many) other equivalence classes there are.
• No, because $\{0,4,8,16,\ldots\}\not\subset S$. – Alejandro Nasif Salum Nov 28 '18 at 18:43
• You can do the same math for $4$ instead of $1$, and you would get$$\bar 4=\{4,16,28,40,\ldots\}.$$ Also $$\bar7=\{7,19,31,43,\ldots\}$$ and $$\bar{10}=\{10,22,34,46,\ldots\},$$ and those are the only four equivalence classes in $S$. – Alejandro Nasif Salum Nov 29 '18 at 21:17