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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.

  1. $xR_ky \Longleftrightarrow k\:|\:(x+y)$
  2. $xS_ky \Longleftrightarrow k\:|\:(x-y)$
  3. $xT_ky \Longleftrightarrow x - y = k$

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.

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  • $\begingroup$ Your argument for "symmetric" starts out well but then goes too far. Yes, commutativity shows that $xR_ky\iff yR_kx$ but that certainly doesn't show that $k$ always divides $x+y$. $\endgroup$ – lulu Dec 15 '18 at 21:04
  • $\begingroup$ For transitive: note that, for all $k$, $1R_k(k-1) $ and $(k-1)R_k1$ but $1R_k1\iff k\in \{1,2\}$. So unless $k$ is $1$ or $2$ transitivity fails. Can you check $k=1,2$? $\endgroup$ – lulu Dec 15 '18 at 21:09
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    $\begingroup$ Final hint: your argument for $(2)$ is much too hasty. If $k$ divides $x-y$ then it also divides $y-x$. $\endgroup$ – lulu Dec 15 '18 at 21:11
  • $\begingroup$ I'm sorry for deleting my comment, but I accidentaly sent it before it was ready. Both $x$ and $y$ are $\mathbb{N}$, thus their sum is $\mathbb{N}$. And if $k = 1$, it divides any natural number. $\endgroup$ – whiskeyo Dec 15 '18 at 21:11
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    $\begingroup$ Exactly correct, Well, I'd phrase it as "the equivalence classes are the Even numbers and the Odd numbers". $\endgroup$ – lulu Dec 16 '18 at 0:06
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Hint.

$(a)R_k$ is transitive for $k=1$, so we only need to check for $k=2$. If $x+y$ is even, either both $x,y$ are even or they are both odd. If $y$ is even, so is $z$ because $y+z$ is even. Similarly, if $y$ is odd, so is $z$. This means $x,z$ are either both even or both odd, or that $x\ R_2\ z$.

$(b)$ The equivalence of $S_k$ does not necessitate $x-y=y-x$. It only necessitates if $k|\ x-y$, then $k|\ y-x$, which is always true. In fact, $S_k$ is an equivalence relation $\forall k\in\Bbb N$.

$(c)$ Try $x-y=y-x=k=0$.

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  • $\begingroup$ I proved that $R_k$ from example (a) is equivalence relation for $k\in \{1,2\}$ and in (b) that it is true $\forall k \in \mathbb{N}$. When it comes to relation $xT_ky \Leftrightarrow x-y=k$ I know that it is reflexive for $k=0$ and symmetric for $k=0$, but I am not sure how to determine is transitive. According to earlier conclusion (that $k=0$), I think that it is enough to check if it works for $T_0$, which means $(x-y=0 \wedge y-z=0) \Rightarrow x-z=0$, the only case when it works is when $x=y=z$ - does it say that $T_0$ is not equivalence relation (because it works only in one case)? $\endgroup$ – whiskeyo Dec 16 '18 at 18:36
  • $\begingroup$ $T_0$ is just the equality relation. It is transitive. By definition, if $T_0$ is transitive, then $x\ T_0\ y, y\ T_0\ z\implies x\ T_0\ z$. Now, $x\ T_0\ y\implies x=y; y\ T_0\ z\implies y=z\ \therefore x=z$ or $x\ T_0\ z$. $\endgroup$ – Shubham Johri Dec 17 '18 at 6:38
  • $\begingroup$ The fact that it only works in one case means the equivalence classes of $T_0$ are singleton. They are $\{1\},\{2\},\{3\}...$ $\endgroup$ – Shubham Johri Dec 17 '18 at 6:42

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