Why the relation is called TRANSITIVE where the required information is missing Consider a relation $R$ = {$(a, b), (c, d)$}.
This relation is transitive.
Since transitive pair $(a, b)$ and $(b, c)$ is absent, We cannot prove that this relation is TRANSITIVE or NOT TRANSITIVE.
What we say is that it is transitive "by vacuity“ or "vacuously true".
My questions are:-


*

*Why did we define vacuous statements as true rather than false?

*Cannot we say that there are not enough facts available to determine the transitivity of the relation? Is it necessary that a relation must be called either TRANSITIVE or NOT TRANSITIVE?


Thanks.

Edit(added 11:27 PM, 23 March 20): I know that the relation in my question is transitive and I also know how. That is not my question. My question is, here we have an if-then statement, $P⇒Q$, (the condition of transitivity), now when the transitive pair (a, b) and (b, c) is absent, P becomes false, then it implies every Q statement. It means that this implies (i) relation is transitive, as well as (ii) relation is not transitive. Then why do we define this type of relation  as transitive (vacuously true) rather than not transitive (vacuously false)?

Edit: I am surprised, why this question is marked as duplicate. I saw the mentioned question and observed that they are different. The mentioned question is interested in whether the relation is transitive or not, whereas in my question I have admitted from the beginning that the relation is transitive and questioned the logic behind it. Do you people think they are same? 
 A: A relation R on a set S is transitive IF whenever there are pairs $(x, y), (y, z)\in R$, THEN $(x, z)\in R$.  It doesn't disqualify a relation from being transitive in cases where there ARE no pairs for which $(x, y), (y, z) \in R$.
Put differently, relation $R$ on a set $S$ is transitive, unless there are elements such that $(x, y) \in S$ AND $(y, z)\in S$ but $(x, z) \notin S$.
In your example, $(a, b) \in S$, but $(b, c) \notin X$, and $(c, d) \in S$, but $(d, a) \notin S$, and $(a, c)\notin S,$ even though $(c, d)\in S$.  So  it is transitive trivially, because there are no pairs $(x, y)\in S$ AND $(y, z)\in S$, where $(x, z) \notin S$. 
A: The answer to your question has roots in mathematical logic. The statement of  the form you are referring here is a conditional statement where you must have a certain hypothesis to check for the conclusion. Mathematically, it is $P\implies Q$ form  where $P$ is antecedent (hypothesis) and $Q$ is consequent (conclusion). The truth value  of $P\implies Q$ is $T$ (true) in the following cases:
$(1)$ when both $P$ and $Q$ are true
$(2)$when $P$ is false (irrespective of the truth value of $Q$)
Case (2) says that whenever the hypothesis is false (doesn't hold),the conditional $P\implies Q$ always holds.
A: By definition a relation $R$ is not transitive if we can find pairs $(x,y),(y,z)\in R$ such that $(x,z)\notin R$.
Looking at the relation in your question we observe that such pairs cannot be found.
Conclusion: the relation is not not transitive, or equivalently is transitive.

Edit (concerning your second question):
In logic where "excluded middle" is absent a statement is true or false. Applying that to the statement "$R$ is transitive" we find that this is a true statement ($R$ is transitive) or a false statement ($R$ is not transitive).
A: *

*If a person is not in a situation falling under the conditions of a rule, vacuously, this person is OK with this rule, and what this person does is vacuously right ( as to this rule). 


Suppose I define the property " safe driver" as 

" ($x$ is a safedriver) iff  ( if $x$ is driving a car at time $t$  , then $x$
  is not drunk at time $t$)".

Is my friend Peter, who has no car licence and has never driven a car, a " safe driver"? 
Yes he is, vacuously. 
The reason is that the condition of the rule " $x$ is driving a car at time $t$ " is always false when applied to Peter. So Peter can never be in a situation where (1) the condition of the rule holds and (2) Peter does not obey what the rule commands in that situation. 


*

*In the same way, the condition of the rule defining a " transitive relation " is false for the relation you are considering. The condition is : having ( at least) two ordered pairs such that the second element of one pair is also the first element of the other. 

*So, this relation cannot possibly violate the rule, and, therefore, is " transitive" vacuously. 
