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I am having trouble finding if the following relation is transitive or intransitive. I would be very thankful if someone could help me out by explaining transitivity and its rules with regard to this example,

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


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Beware terminology: You mean whether the relation you show is transitive, not whether the set is. (There is something called "transitive" for sets in general, but that is something almost entirely different from what you're speaking about here). – Henning Makholm Nov 4 '11 at 15:07
Is your problem that you don't know what "transitive" means, or is it that you have a definition, but don't understand how to apply it? If the latter, then please quote the definition and share some thoughts about how it might apply to your example. – Henning Makholm Nov 4 '11 at 15:10
up vote 3 down vote accepted

A relation on a set is transitive if, when we have (a,b) and (b,c), we have also (a,c).

So here, we just check some cases. 4 is on on its own - so we don't really care about 4.

Let's look at 1 and 3. We have (1,3) and (3,1) - do we have (1,1)? Yes, we do. Similarly, we have (3,1) and (1,3), and we also have (3,3). These are all the relations on 3.

But then we consider 1 and 2...

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$(1,2)$ is there but $(2,1)$ is not there, so its intransitive, right? – Fahad Uddin Nov 4 '11 at 15:30
@Akito, no, you're confusing that with symmetry. – Henning Makholm Nov 4 '11 at 15:35
@HenningMakholm: So what to do with $(1,2)$? It does not satisfy the condition.. – Fahad Uddin Nov 4 '11 at 16:37
@HenningMakholm: I got the point that you made but there is no pair for $(1,2)$ – Fahad Uddin Nov 4 '11 at 16:47
@Akito, a single pair cannot possibly fail to satisfy the condition. The condition is "when we have (a,b) and (b,c) ...", so the only way the condition can fail to hold is if we can find two pairs such that one is (a,b) and the other is (b,c) yet there is no (a,c) in the relation. – Henning Makholm Nov 4 '11 at 17:02

Transitivity says that if you have $aRb$ and $bRc$ you also have $aRc$. When your relation is a given set of pairs, you have to look through and see if it is true. If $a=2$, there is no $b$ such that $aRb$, so we satisfy the requirement. If $a=4$, we must have $b=4$ and then $c=4$ and we do get $aRc$. When $a$ is $1$ or $3$, you have more possibilities for $b$ and have to follow each path to see where it leads.

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here 3R1 and 1R2 but it is false that 3R2 so it is not transitive

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