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Please see the below equation.

As the title suggests I have been trying to work out if the following Binary Relation is Transitive.

I have been able to conclude that the relation is Anti-symmetric and Reflexive due to having loops at the end of each vertex and not having a Symmetric nature.

Let $X = \{1, 2, 3, 4\}$

Consider binary relation $R = \{(1,1),(2,1),(2, 2),(3, 1),(3, 3),(4, 1), (4, 2), (4, 4)\}$

Graham Kemp
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Kenny Boy
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3 Answers3

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Remember the definition: $R$ is transitive iff for any $a, b, c$, if $(a, b)$ and $(b, c)$ are in $R$, then so is $(a, c)$. If $R$ is not transitive, you should be able to find a counterexample.

Of course, this makes showing that $R$ is transitive a bit messy: you have to convince yourself that no counterexample exists. For general $R$ this can be quite complicated, but for this $R$ it's not too bad. Make a list of all the triples $a, b, c$ such that $(a, b)$ and $(b, c)$ are each in $R$; now check each of these.

Noah Schweber
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    Plus, the reflexive pairs $(a,a)$ can be skipped in the transitive check, which reduces the remaining combinations even further. – dxiv Oct 14 '16 at 00:32
  • Yes, I can now see it is, thank you all for your explanations and I can see that this is a partial order. – Kenny Boy Oct 14 '16 at 00:36
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The relationship is transitive. With a case this small it is easily enough to trace through all the examples and check that $ (a,b) $ and $ (b,c) $ implies $(a,c)$.

Take each pair of pairs if they match up check for the completion of it.

Q the Platypus
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The key to this is to look for a counter example: a bridge without end in the relation $$\text{If }\exists a\,\exists b\,\exists c\,[~(a,b){\in}R \wedge (b,c){\in}R \wedge (a,c){\notin}R~]\text{ then the relation is not transitive}$$

Of course this means you have to check exhaustively to ensure you have not missed any.

I can't find any.

Graham Kemp
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