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 $P$?
 A: $R$ 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. 
A: Yes, the relation will be transitive, but for vacuous reasons. There are no pairs $(a,b)$ and $(b,c)$ in $R$, so the statement
$$\forall (a,b),(c,d)\in R (b=c \Rightarrow (a,d)\in R)$$
does in fact hold.
A: 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.
A: 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.)
