1
$\begingroup$

This exercise is given in my textbook and I am trying to solve it.

Determine whether they are symmetric, antisymmetric or reflexive.

$R_1=\{(2,2), (2,3), (2,4), (3,2), (3,3), (3,4)\}$

$R_2=\{(1,1), (1,2), (2,1), (2,2), (3,3), (4,4)\}$

$R_3=\{(2,4), (4,2)\}$

$R_4=\{(1,2), (2,3), (3,4)\}$

$R_5=\{(1,1), (2,2), (3,3), (4,4)\}$

$R_6=\{(1,3), (1,4), (2,3), (2,4), (3,1), (3,4)\}$

My answers:

1- $R_1$ is symmetric.

2-$R_2$ is reflexive, symmetric.

3-$R_3$ is symmetric.

4-$R_4$ is antisymmetric.

5-$R_5$ is reflexive, antisymmetric.

6-$R_6$ is symmetric,

Book's answers:

1-None of the these properties.

2-$R_2$ is reflexive and symmetric.

3-$R_3$ is symmetric.

4-$R_4$ is antisymmetric.

5-$R_5$ is reflexive, symmetric and antisymmetric.

6-None of these properties.

You can see that some of my answers don't match the answers given in book. Is that probably a misprint or I am wrong somewhere?

$\endgroup$
  • $\begingroup$ Hello. Can you walk us through why you think $R_1$ is symmetric? $\endgroup$ – Mankind May 10 '15 at 13:43
  • $\begingroup$ Yes of course. Because, $(2,3) \in R$ and $(3,2) \in R$. Isn't so? $\endgroup$ – Man_Of_Wisdom May 10 '15 at 13:45
  • 2
    $\begingroup$ @Man_Of_Wisdom: But this sort of thing has to be true for every member of the relation, not just one particular member. You have to check them all. $\endgroup$ – MPW May 10 '15 at 13:48
  • 1
    $\begingroup$ I think, my understanding of relations is flawed. I must review it! $\endgroup$ – Man_Of_Wisdom May 10 '15 at 13:51
  • 1
    $\begingroup$ $(2, 3) \in R_1$ and $(3, 2) \in R_1$, which speaks in favour of symmetry, but is not enough to prove it: you have $(2, 4) \in R_1$ and $(4, 2) \notin R_1$. Therefore you have a "witness" against symmery, and therefore $R_1$ is not symmetric. $\endgroup$ – Arthur May 10 '15 at 13:53
0
$\begingroup$

For a relation $R$ to be symmetric, we have to have for all elements in $R$ that if $(x,y) \in R$, then also $(y,x) \in R$. You have found some elements in $R_1$ such that both $(x,y) \in R_1$ and $(y,x) \in R_1$, but for example $(2,4) \in R_1$ but $(4,2) \notin R_1$, hence it it not symmetric because it doesn't satisfy the criterion for every element.

There is also an element in $R_6$ that makes it non-symmetric, can you find it?

As for $R_5$, since every element is of the form $(x,x)$, it also (obviously) holds that $(x,x) \in R_6$, so it is symmetric.

$\endgroup$
  • $\begingroup$ Yes, I can find it now. it is not symmetric because, In particular, $(1,4) \in R$ but $(4,1) \notin R$. Am I right? $\endgroup$ – Man_Of_Wisdom May 10 '15 at 13:55
  • 1
    $\begingroup$ @Man_Of_Wisdom Yes exactly. $\endgroup$ – mrp May 10 '15 at 13:57
  • $\begingroup$ Then be ready for +15 reputation. :) $\endgroup$ – Man_Of_Wisdom May 10 '15 at 13:58
  • 1
    $\begingroup$ I guess you mean $R_5$ rather than $R_6$ in your last sentence $\endgroup$ – MPW May 10 '15 at 14:23
1
$\begingroup$

I preassume that you are dealing with relations on set $\{1,2,3,4\}$. That is not mentioned in your question but is essential information. If it lacks then it can e.g. not be checked wether the relations are reflexive.

1) $R_1$ is not symmetric: $(2,4)\in R_1\wedge (4,2)\notin R_1$

5) $R_5$ is (also) antisymmetric. What makes you think it is not? Can you find a pair $(a,b)$ with $(a,b)\in R_5\wedge (b,a)\in R_5\wedge a\neq b$? If not then it is antisymmetric.

6) $R_6$ is not symmetric: $(1,4)\in R_6\wedge (4,1)\notin R_6$

$\endgroup$
  • $\begingroup$ Helpful in context of antisymmetry! $\endgroup$ – Man_Of_Wisdom May 10 '15 at 14:41
0
$\begingroup$

The book is right. Consider $(2,4)$; what would have to be true if the relation $R_1$ were symmetric? Likewise with $(1,4)$ for $R_6$.

$\endgroup$
  • $\begingroup$ Can you explain it more lucidly please? $\endgroup$ – Man_Of_Wisdom May 10 '15 at 13:48
  • $\begingroup$ Symmetry means something very specific. It means "If $(a,b)$ is in $R$, then $(b,a)$ is also in $R$". So you must check that if you reverse each pair, the resulting pairs are all still in $R$. For example, you know $(2,3)$ is in $R$. Is $(3,2)$ in $R$? Yes, so keep testing. If you exhaust them all with no problem, it is symmetric. But if you find one that fails, it isn't symmetric. You should try the ones that I suggested. $\endgroup$ – MPW May 10 '15 at 13:53

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.