Problem:
Show that the relation $x R y$ iff $x \leq y$ is a poset over the set of integers $\mathbb{Z}$
My work:
I know that to show the relation is a poset or a post order, I have to show the relation is reflexive, transitive, and anti-symmetric.
For reflexive, suppose $a$ is an integer in $\mathbb{Z}$, then $a$ is by definition $\leq a$. Therefore $(a,a)$ is an ordered pair in the relation or $a R a$.
For anti-symmetric, suppose $a$ and $b$ are some integers in $\mathbb{Z}$. If $a R b$, $a \leq b$. The only way $b R a$ or $b \leq a$ is if $b = a$, in that case $b \leq a$. If $b>a$, then $a$ cannot be $\geq b$. We proved that if $a R b$ and
$b R a$, then $a =b$.
I had trouble with transitive. I knew this was transitive right away because lets say that
$1 \leq 2$, $2 \leq 3$, then you can definitely conclude that $1 \leq 3$. How would you mathematically explain this? If $a \leq b$, $b \leq c$, then $a \leq c$. Is there some mathematical way to say $a \leq c$? I would just say by intuition.
That's all you need to say, really.
For any integer triplet $(a, b, c)$, if $a\leq b$ and $b\leq c$, then $a\leq c$. Thus $(\Bbb Z, \leq)$ is transitive.
$$\forall (a,b,c)\in\Bbb Z^3\;\Big((a\leq b)\wedge (b\leq c) \;\to\; (a\leq c)\Big)$$
If you like, you can demonstrate this by considering the four cases: