Transitive Relations

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.

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.

OK see here, we call a relation R transitive if $(a,b)\in R$ and $(b,c)\in R$ then $(a,c)\in R$. That is if a related to $b$, we will search whether $b$ transit any other element $c$ then only we are going to check whether $a$ related to $c$ if $b$ won't transit we by default consider it as transitive. Now in your example $1$ related to $1$ only and $1$ transit no other element. So there will be no transition from $1$ to any other element except $1$. Therefore your given relation is transitive. @ Uddhav