Transitive Relations: Special case Suppose I have a set $A = \{1,2,3,4\}$ and I define a relation $R$ on $A$ as $R = \{(1,2),(3,4)\}$. Is this relation transitive or not?
My book says a relation is transitive if $(x,y) \in R$ and $(y,z) \in R$ implies $(x,z) \in R$ for all $x,y,z \in R$. I cannot conclude what that means for the example above.
So is a relation called transitive or not when there are no ordered pairs like $(x,y),(y,z)$ that belong to $R$?
 A: Sure, the binary relation $R = \{(1,2), (3,4)\}$ on the set $A = \{1,2,3,4\}$ is transitive.
Indeed, the transitive property is vacuously true for the relation $R$: since there are no $x,y,z$ in $A$ such that $(x,y) \in R$ and $(y,z) \in R$, there is nothing to check.
A: Yes, your relation is indeed transitive.  The nonexistence of any triples $x,y,z$ such that $(x,y),(y,z)$ are both in your relation does not matter.  Being transitive merely says that if such a triple existed then you must also have $(x,z)$ in the relation.
Reworded, a relation is transitive iff for any "directed path" (if they even exist in the first place) you can take from one element to another you can also get there by using a single step instead.  Since all of your directed paths are of length $1$ anyways, there is nothing more to check.
A: Or one can determine the truth value of the proposition : $\forall x,y,z \in X : ( x \rm{R} y \land y\rm{R} z  )$ $\implies$ $x\rm{R}z $. 
Let P= $( x \rm{R} y \land y\rm{R} z  )$ ,    Q= $x\rm{R}z$ 
Suppose we have $ x\rm{R}y$, so its value is true, the truth value of $ y\rm{R}z$ is always false. So the truth value of $P$ is false, based on the truth value table of $\land$. 
On the other hand, the truth value of $Q$ is false, so the truth value of $P\implies Q$ will be true, based on the truth value of $ \implies$.It means the definition holds, so the transitive property of the relation is satisfied.
