$A= \Bbb R \\ R=\{(x,y)\in\Bbb R^2|x-y\in \Bbb Z\}$

Determine if the relation is (a)reflexive, (b)symmetric, (c)transitive, (d)anti-reflexive, (e)anti-symmetric, (f)asymmetric, (g)equivalence relation.

This is what I did but I'm not sure:

It is reflexive: $\forall x:(x,x)\in R:x-x=0\in\Bbb Z$

It is symmetric: $\forall x,y\in\Bbb R:xRy\in\Bbb Z\Rightarrow yRx\in\Bbb Z$

It is transitive: $\forall a,b,c\in \Bbb R:(aRb \ and \ bRc)\in \Bbb Z \Rightarrow aRc $

It isn't anti-reflexive: $(1,1)\in \Bbb R$

It is anti-symmetric: $aRa\in \Bbb Z \Rightarrow a=a$

It isn't asymmetric because it symmetric.

There is an equivalence relation.

Is it correct ? Thanks.

  • 1
    $\begingroup$ Yes, it is correct but once you've proved it is ref., sym. and trans. you do not need to prove it isn't anti-whatever. $\endgroup$ – DonAntonio Dec 14 '13 at 11:17
  • 3
    $\begingroup$ What you have for anti-symmetry is not correct. $3-5\in \mathbb{Z}$ and $5-3\in \mathbb{Z}$, but $5\neq 3$. $\endgroup$ – Malice Vidrine Dec 14 '13 at 11:23
  • $\begingroup$ @MaliceVidrine are you sure that it's for all x ? NVM, it is for all. They didn't say that in class, I had to check in wiki. $\endgroup$ – GinKin Dec 14 '13 at 11:25
  • 1
    $\begingroup$ @GinKin, I think Malice meant that in order to prove antisymmetry you must prove that $\;aRb\;\wedge\;bRa\implies a=r\;$ , which of course isn't fulfilled in this case. What you wrote is wrong. $\endgroup$ – DonAntonio Dec 14 '13 at 11:31
  • $\begingroup$ $xRy\in\mathbb Z$ is nonsensical. $\endgroup$ – Carsten S Dec 14 '13 at 12:10

Most of your answers are correct, but the justifications given are a little confusing. In general, you should offer a genuine proof. For example:

It is reflexive.

Proof. Let $x \in \mathbb{R}$ be fixed but arbitrary. Then $x-x=0$. Thus $x-x \in \mathbb{Z}.$ So $xRx.$

Anyway, your answers for "reflexive", "symmetric" and "transitive" are correct.

The claim that $R$ is anti-symmetric is incorrect. Observe that $0R1$ and $1R0$, but it does not follow that $0=1$.

Also, if a relation on a non-empty domain is reflexive, then its not anti-reflexive (exercise!). So that answer is also correct. Along a similar vein, the only relation that is both symmetric and asymmetric is the always-false relation. But since $0R0$, the given relation $R$ is not always false. So it cannot be asymmetric. Therefore, that answer is also correct.

Edit. By the way, defining $R$ via set-builder notation is imo confusing. I would suggest defining $R$ as the unique subset of $\mathbb{R}^2$ such that:

$$\forall x,y \in \mathbb{R} : xRy \;\leftrightarrow\;x-y \in \mathbb{Z}.$$

From the above form, it is obvious that any time $xRy$ is written down, we may deduce $x-y \in \mathbb{Z}$, and any time $x-y \in \mathbb{Z}$ is written down, we may deduce $xRy$.

  • $\begingroup$ Yes I do get confused by the set notation quite often. Can I always replace the | with "iff" like you did ? $\endgroup$ – GinKin Dec 14 '13 at 12:45
  • 1
    $\begingroup$ @GinKin, yeah. The definition of $\{x \in X \mid P(x)\}$ is that its the unique set $A$ such that for all $x \in X$ we have $x \in A \leftrightarrow P(x)$. $\endgroup$ – goblin Dec 14 '13 at 13:41

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