$$A = \{1, 2, 3, 4\}$$
Is the relation x > y on $A$ transitive?
If yes please explain why it is so.
There will be the pair (2,1) as 2 > 1. However y being 1 will have no other pair (y,z) as 1 is the smallest number in the set $A$. So the condition
for the relation to be transitive:
$\forall x,y,z \in A : x R y \land y R z \implies x R z$.
will return false as there will be no $y R z$ pair to satisfy the whole condition.
Is my thinking right?
On
Its transitive. The relation $>$ is defined on the natural numbers as $x>y$ if there a natural number $a>0$ such that $x=y+a$.
Thus if $x>y$ and $y>z$, then there are $a,b>0$ such that $x=y+a$ and $y=z+b$. Therefore, $x=z+(a+b)$ with $a+b>0$. Hence, $x>z$.
As another question of the OP (see comments) if a relation is transitive, its inverse relation is also transitive. Inverse of $>$ is $<$ and inverse of $\geq$ is $\leq$.
The relation $\geq$ is also transitive. Its definition: $x\geq y$ iff there is a natural number $a\geq 0$ such that $x=a+y$.
No, you have universal quantification. If you have pair $(x, y)$ in relation, you don't necessary need to have a pair $(y, z)$ also in relation - you just need if there is such pair to also have $(x, z)$ in relation.
The transitivity condition requires you to have pair $(x, z)$ only if there already are pairs $(x, y)$ and $(y, z)$. It doesn't require anything about $(x, y)$ and $(y, z)$.
And this relation is transitive. You can see it by either writing all $6$ of its elements directly, or by noting that it's just restriction of usual transitive order on integer numbers to set $A$.