# Determine whether the relations are symmetric, antisymmetric, or reflexive.

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

Determine whether they are symmetric, antisymmetric or reflexive.

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

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

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

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

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

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

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,

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?

• Hello. Can you walk us through why you think $R_1$ is symmetric? – Mankind May 10 '15 at 13:43
• Yes of course. Because, $(2,3) \in R$ and $(3,2) \in R$. Isn't so? – Man_Of_Wisdom May 10 '15 at 13:45
• @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. – MPW May 10 '15 at 13:48
• I think, my understanding of relations is flawed. I must review it! – Man_Of_Wisdom May 10 '15 at 13:51
• $(2, 3) \in R_1$ 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. – Arthur May 10 '15 at 13:53

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.

• 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? – Man_Of_Wisdom May 10 '15 at 13:55
• @Man_Of_Wisdom Yes exactly. – mrp May 10 '15 at 13:57
• Then be ready for +15 reputation. :) – Man_Of_Wisdom May 10 '15 at 13:58
• I guess you mean $R_5$ rather than $R_6$ in your last sentence – MPW May 10 '15 at 14:23

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: $(1,4)\in R_6\wedge (4,1)\notin R_6$

• Helpful in context of antisymmetry! – Man_Of_Wisdom May 10 '15 at 14:41

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 $R_6$.

• Can you explain it more lucidly please? – Man_Of_Wisdom May 10 '15 at 13:48
• 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. – MPW May 10 '15 at 13:53