1
$\begingroup$

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

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.

$\endgroup$
  • 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
3
$\begingroup$

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 $xRy$.

$\endgroup$
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.