Is this relation transitive, reflexive, symmetric?

Question

I am having a hard time identifying transitive relations. I think I understand those that are symmetric, but do correct me if I'm wrong.

For a set $S = \{0,1,2,3,4\}$ and a relation $Z = \{(0,2),(2,2),(2,3),(3,4)\}$ I have found:

I think it is not reflexive because there is no loop including 1.

I am struggling with this one, but I think it is transitive.

I believe this is not symmetric as there is no $(2,0), (3,2)$ or $(4,3)$.

Any help is much appreciated, I don't seem to be able to get my head around what I feel like are likely to be really simple concepts.

Answer 1

It is not transitive because it has (2,3) and (3,4) but doesn't have (2,4).

Correct, it's not symmetric for the reason you noted.

Also correct, it's not reflexive. To be reflexive all (a,a) must be part of the relation (for any a). That doesn't hold here apparently.

Answer 2

It is not reflexive because $(0,0)\notin Z$; Is is not symmetric because of the reasons you stated. For instance, $(0,2)\in Z$ but $(2,0)\notin Z$. It is not transitive; for example, $(0,2)\in Z$ and $(2,3)\in Z$ but $(0,3)\notin Z$

Answer 3

This set is not transitive because $(2,3),(3,4) \in Z $ but $ (2,4)\notin Z$. The reason you provided for not-symmetric is right.

Answer 4

You are correct about symmetry and reflexivity.

For transitivity, we need that every time there are two pairs of the form $(a,b)$ and $(b,c)$ in the relation, it must also be the case that $(a,c)$ is in the relation. The relation you have given is not transitive, can you find two pairs $(a,b)$ and $(b,c)$ in your relation such that $(a,c)$ is not in the relation?

Since you seem to think of reflexivity in terms of "loops", perhaps the following way of thinking about transitivity will be helpful: If we consider a relation $R$ as a directed graph, where there is an arrow from $a$ to $b$ if $(a,b) \in R$, then

• reflexivity means that there is a loop on every node,
• symmetry means that every time there is an arrow from a node to another node, then there is also an arrow going the other way, and
• transitivity means that every time there is an arrow from $a$ to $b$ and also from $b$ to $c$, then there must also be an arrow going directly from $a$ to $c$.