# Transitive Relations

For example, R = { (1,1),(1,2),(2,1),(2,2) } for 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) => (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)^(1,2)=>(1,2) ?? is this why it's transitive?

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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

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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Remember that transitivity means that if a is related to b, and b is related to c, then a is related to c.

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