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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,

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

Thanks.

  • 0
    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).2011-11-04
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    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.2011-11-04

3 Answers 3

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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?2011-11-04
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    @Akito, no, you're confusing that with symmetry.2011-11-04
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    @HenningMakholm: So what to do with $(1,2)$? It does not satisfy the condition..2011-11-04
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    @HenningMakholm: I got the point that you made but there is no pair for $(1,2)$2011-11-04
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    @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.2011-11-04
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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