TITLE: Is any relation which contains only one ordered pair transitive?
QUESTION [14 upvotes]: 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 $P$?

REPLY [4 votes]: $R$ is transitive in an empty sense, because in a transitive relation, we want the following $ \forall x,y,z\in A, (x,y)\in R \wedge (y,z)\in R \Rightarrow (x,z)\in R $, but this doesn't guarantee the existence of $3$ pairs in $R$. For simplicity, let's write $ \alpha =[(x,y)\in R \wedge (y,z)\in R],  \beta =[(x,z)\in R] $. So what we want is that $ \alpha \Rightarrow \beta$, but if $ \alpha $ has a truth value of $0$, then according to the truth table of implication, then $\beta$ will always have a truth value of $1$, which means that a relation $R$ is transitive if $ \alpha$ implies $ \beta $ in $R$.
For example in this case, only $(1,2)$ is in $R$ (meaning $ \alpha $ has a truth value of $0$, because we cannot find $2$ distinct pairs in $R$), and therefore, $ \beta $ has a truth value of $1$, therefore, we found that $ \alpha \Rightarrow \beta$, and therefore, $R$ is transitive.