Why is $\{(1,2),(3,4)\}$ a transitive relation?

Question

I am having a hard time understanding why $\{(1,2),(3,4)\}$ is a transitive relation to $S = \{1,2,3,4\}$.

My understanding is for this to be transitive, $1$ would have to relate to $4$ which is not in the relation. If $x = 1$, $y = 2$ and $y = 3$ and $z = 4$ then x!Rz. I know I'm missing something in my understanding of transitive relationships.

Answer 1

A relation $R$ on a set $S$ is said to be transitive iff it fulfills the following:

For any $x, y, z$ in $S$, if $xRy$ and $yRz$, then $xRz$.

Now, in your case we cannot find $x, y, z$ such that $xRy$ and $yRz$. So what does the above definition even mean in this case?

This is a phenomenon (or perhaps a convention) in mathematics and logic called vacuous truth. If you have an "if-then" statement, and whatever follows the "if" is false, then the statement itself is considered true, no matter what follows after "then".

It might be a bit difficult to wrap your head around at first. I offer two explanations / interpretations, and hope that at least one of them can help you.

1. An "if-then" statement is a promise. What I mean by that is that to prove an "if-then" statement true, then you have to guarantee that every time someone comes up to you with a case demonstrating that whatever follows "if" is true, you have to demonstrate that whatever follows the "then" is true.

   Take, for instance, "If it's raining outside, then the ground is wet." Someone points out and tells you that it's raining, you point to the ground and say it's wet. Promise kept. If you can show that you can always keep that promise, then the "if-then" statement is considered true.

   In this specific case, no one can ever come forward with a case of "$xRy$ and $yRz$", so you will be able to keep your promise. Therefore $R$ is transitive.

2. Look at the contrapositive. The contrapositive of any statement is equivalent to the statement in itself. The contrapositive (or perhaps a contrapositive) of the definition of transitivity is

   For any $x, y, z \in S$, if $x\not R z$, then either $x \not R y$ or $y \not R z$ (where the slash means "not", so $x \not R y$ is equivalent to $(x, y) \notin R$).

   This is a bit of work to check, but hopefully the logic itself is a bit more intuitive.

Answer 2

Transitive relations are vacuously true. Thus if there is nothing to prove that a relationship is not transitive then it is inherently transitive. In the relation you mention here there is no $xRy$ and $yRz$ in the relation that makes the statement $xRy\land yRz\Rightarrow xRz$ false. Thus the relation is transitive.