Check for which $k$ given relations on set $\mathbb{N}$ are reflexive, symmetric or transitive. For these relations, that are equivalence relations, describe their equivalence classes.
- $xR_ky \Longleftrightarrow k\:|\:(x+y)$
- $xS_ky \Longleftrightarrow k\:|\:(x-y)$
- $xT_ky \Longleftrightarrow x - y = k$
For the first example $xR_ky \Longleftrightarrow k\:|\:(x+y)$, I tried to do it like this:
- check if the relation $R_k$ is reflexive, so $xR_kx \Longleftrightarrow k\:|\:(x+x)\equiv k\:|\:2x$ - from that we get that $2x$ is always divisible if $k=1$ or $k=2$
- check if the relation $R_k$ is symmetric, so $\Big[xR_ky \Longleftrightarrow k\:|\:(x+y)\Big] \Longrightarrow \Big[yR_kx \Longleftrightarrow k\:|\:(y+x)\Big]$, which is true, because addition is commutative. From that we can conclude, that $\exists n\in\mathbb{N} : x+y=k\cdot n$.
- check if the relation $R_k$ is transitive, so $(xR_ky \wedge yR_kz) \Rightarrow xR_kz$.
Thus $\exists n\in\mathbb{N} : x+y=k\cdot n$ and $\exists m\in\mathbb{N} : y+z=k\cdot m$.
And this is the first moment that I got stuck and I am not sure what to do next. Though I suppose that it is going to be an equivalence relation, but what would its equivalence classes look like?
When it comes to examples (2) and (3) I can easily say that they are not equivalence relations, as in both cases $x-y \neq y-x$, which tells us that the relation is not symmetric.