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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?

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  • $\begingroup$ Hello. Can you walk us through why you think $R_1$ is symmetric? $\endgroup$
    – Mankind
    May 10, 2015 at 13:43
  • $\begingroup$ Yes of course. Because, $(2,3) \in R$ and $(3,2) \in R$. Isn't so? $\endgroup$ May 10, 2015 at 13:45
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    $\begingroup$ @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, 2015 at 13:48
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    $\begingroup$ I think, my understanding of relations is flawed. I must review it! $\endgroup$ May 10, 2015 at 13:51
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    $\begingroup$ $(2, 3) \in R_1$ 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, 2015 at 13:53

3 Answers 3

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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.

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  • $\begingroup$ 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? $\endgroup$ May 10, 2015 at 13:55
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    $\begingroup$ @Man_Of_Wisdom Yes exactly. $\endgroup$
    – mrp
    May 10, 2015 at 13:57
  • $\begingroup$ Then be ready for +15 reputation. :) $\endgroup$ May 10, 2015 at 13:58
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    $\begingroup$ I guess you mean $R_5$ rather than $R_6$ in your last sentence $\endgroup$
    – MPW
    May 10, 2015 at 14:23
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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: $(1,4)\in R_6\wedge (4,1)\notin R_6$

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  • $\begingroup$ Helpful in context of antisymmetry! $\endgroup$ May 10, 2015 at 14:41
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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 $R_6$.

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  • $\begingroup$ Can you explain it more lucidly please? $\endgroup$ May 10, 2015 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, 2015 at 13:53

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