Why $R = \{(1,2),(1,3)\}$ is transitive relation?

Question

As I understand the definition of Transitive Relation, $R$ should have $(2,3)$ to make it a transitive relation. I do not understand why this case is still a transitive relation without $(2,3)$. Thanks in advance.

Answer 1

Transitive means $R(a, b) $ and $R(b, c) $ implies $R(a, c) $. In your example, since $2 \ne1$, the conditions $R(a, b) $ and $R(b, c) $ never happen, so it is transitive.

Answer 2

The definition of transitive is that if you have $xRy$ and $yRz$ you have $xRz$. In this case if we want to apply the axiom we have to choose $x=1$ and $y=1$ as all the pairs of our relation have $1$ as the first element. However we do not have $1R1$, so there is no way to make the antecedent true. That makes the relation transitive because the implication never fails.

If we added $(2,3)$ to the relation it would still be transitive. We could now satisfy the antecedent by choosing $x=1,y=2,z=3$ and the consequent would be true because we have $1R3$

Answer 3

Notice that $a R b$ and $b R c$ is impossible.

[If $a R b$ then $a=1$ and $b\ne 1$. But if $b\ne 1$ then we can not have $b R c$.]

If $a R b$ and $b R c$ is impossible than $a R b$ and $b R c\implies anything$ is vacuously true.

Another way to look at it is: For a function to NOT be transitive you must have an example where $a Rb$ and $b Rc$ but $a\not R c$. In this relation you can't find any such counter example because there is no case where $a Rb$ and $b Rc$.

Addendum:

As I understand the definition of Transitive Relation, R should have (2,3) to make it a transitive relation.

Not quite.

Transitivity says: If you have the first two parts of a jigsaw chain, you must have the third last part. If you have $(a,b)$ and $(b,c)$ you must have $(a,c)$.

It does NOT say if you hae the two ends of the jigsaw chain, that you must have the midddle. If you have $(a,b)$ and $(a,c)$ there is no reason you should assume you must also have the middle piece $(b,c)$.

If Adam passes the bucket Bob, and Bob passes it to Carol, then the bucket goes from Adam to Carol.

But if Adam somethimes passes the bucket to Bob, and Adam sometimes passes the bucket to Carol, there is no reason to assume Bob has to pass the bucket to Carol. Bob can just sit on the bucket for all we know.

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$A$ and $B\implies C$. Means if you have $A$ and $B$ then you must also have $C$. It doesn't mean if you have $A$ and $C$ then you need $B$. If $(1,3)$ is the first term, and $(2,3)$ is the middle term and $(1,3)$ is the last term, the transitivity says: If you have the FIRST and SECOND you must have the THIRD. It does NOT say, if you have the FIRST and THIRD you must have the SECOND.