Is any relation which contains only one ordered pair transitive?

I need clarification.

Let $A=\{1,2,3\}$ be a set and $R=\{(1,2)\}$ be a relation on $A$.

Is it a Transitive relation? I am confused because some text books say $R$ is transitive if it contains only one ordered pair.

I am not able to explain why $R$ can said to be transitive in the above case.

A relation is said to be transitive if $(a,b) \in R$ and $(b,c) \in R$ then $(a,c) \in R $.

If P then Q.

$P: (a,b) \in R$ and $(b,c) \in R$ and $Q:(a,c) \in R$

But here only one condition of $P$ is satisfied. According to some source, if second condition i.e, $(b,c) \in R$ does not exist, $R$ is said to be transitive. Can we say $R$ is transitive? Or do we need both conditions of $P$?

The transitive condition is true vacuously. That's like saying "All the women in the car are on fire" is true, when a man is in the car alone.

Yes, it is a transitive relation, vacuously so. That is, there are no counter examples in the relation that violate transitivity.

Transitivity requires that

If $(P)$: $(i)$ $(a, b) \in R\;$ AND $(ii)$ $(b, c) \in R$, (conditions)

THEN $(Q)$: it must follow that $(a, c) \in R$ (consequent)

Since $(P)$, the conditions (i) and (ii), will never both be realized/satisfies since the only element in $R$ is $(1, 2)$, we have that the implication $(P) \implies (Q)$ is vacuously true.

NOTE: We can equivalently define transitivity as a property that HOLDS UNLESS there exists a case (counterexample) for which both conditions in $(P)$ are met, but the consequent $(Q)$ is false (does not hold.)