Are mathematical relations intrinsically transitive? Here's the question: 
Let there be a set A = {1,2,3}.
Let relation R in set A be defined as R = {(1,2),(3,3)}
My textbook says that the relation is neither reflexive nor symmetric but transitive.
I was not quite sure of this so I rechecked the definition of a transitive relation.
My maths textbook defines the transitive property as follows:-

A relation R in a set A is called transitive if $(a_1,a_2),(a_2,a_3) \in R$ implies
  that $(a_1,a_3) \in R$, for all $a_1,a_2,a_3 \in A$

Now the relation R in the question contains (1,2) but does not contain the $(a_2,a_3)$ pair. 
Is R still transitive because relations are supposed to be intrinsically transitive and the lack of the $(a_2,a_3)$ pair remove the need for the relation R to contain the $(a_1,a_3)$ pair? 
Can someone explain the reasoning behind this.
P.S. I have checked about 10 questions with the title "Is this relation transitive" to make sure that this is not a repeat question. I apologize if I have written a duplicate question.

If there weren't any two pairs $(a_1,a_2)$ and $(a_2,a_3)$ both belonging to $R$, then the implication
$$(a_1,a_2)\in R \wedge (a_2,a_3) \in R \quad \implies \quad (a_1,a_3)\in R$$
would be true because of the antecedent being always false.
Note however that this is not the case, since you have one case to analize:
$$(3,3) \in R \wedge (3,3) \in R$$
(here $a_1=a_2=a_3=3$); if $(a_1,a_3)=(3,3)$ would happen not to belong to $R$ that would be enough to say that $R$ is not transitive.
However, it is 'also' true that $(3,3)\in R$, so having analyzed every posible case which could disprove the implication, $R$ is transitive.
