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For example, $$R = \{ (1,1),(1,2),(2,1),(2,2) \} \quad\text{for}\quad A = \{1,2,3\}.$$

This relation is symmetric and transitive.

I understand that the relation is symmetric, but my brain does not have a clear concept how this is transitive. First, this is symmetric because there is $(1,2) \to (2,1)$.

However for transitive, there is $(1,1)$ and $(1,2)$ but there is no "another" $(1,2)$ in the relation technically or does that $(1,2)$ imply the same thing? So for example, $(1,1)\land (1,2)\to (1,2)$? Is this why it's transitive?

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3 Answers

up vote 6 down vote accepted

There is no need to have multiple copies of the ordered pair to satisfy transitivity (indeed, there shouldn't be, since a relation is a set).

Transitivity requires that if $(a,b)$ and $(b,c)$ are present in the relation, then so is $(a,c)$. The fact that $a = b$ in your particular example doesn't change that. You simply notice that $(1,1)$ is present and $(1,2)$ is present, so transitivity demands that $(1,2)$ be present. You've already noted its presence in the relation, so there's nothing to check.

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Ok thank you for clarifying that. –  Aaron Nov 12 '12 at 22:01
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In mathematics we have three types of relation those are:

  • Reflexive

This is the relation type by which an elements of a set are mapped to itself, that is $aEa$ then $aRa$.

  • Symmetric

This is the relation whereby $(a,b)Ea$ that is $aRb$ implies $bRa$.

  • Transitive

This is the relation whereby $aRb$ and $bRc$ which implies $aRc$.

In addition when we talk about EQUIVALENT RELATION is when the both above are equal.

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In your relation, R, $1\to2$ and $2 \to 1$. That is, $1 \sim 2 \sim 1$. Note $(1, 1)\in R$. Alternatively, $2 \to 1 \to 2$; that is, $2 \sim 1 \sim 2.$ Note $(2, 2) \in R$.

There are no other relations to worry about, since, having established the relation is reflexive, we have $(1, 1)$, from which it is evident that $1\sim 1 \sim 1$ and for $(2,2)$ it is evident that $2 \sim 2\sim 2$.

The relation $R$ is therefore transitive.

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