$S=\{3n+1:n∈N\} = \{1,4,7,10,...\}$ 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 $$1=\{x∈S : x\text{ is related to }1\} = \{x∈S: 4|(x + 3*1)\} = \{3n+1:n∈N\text{ and }4|(3n+1)+3\text{ or just }4|(3n+3)\}$$ ???
Please help, thank you!
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.