Guidelines for finding maximum, minimum, maximal and minimal elements of a poset What I am struggling the most these days is determining maximum, minimum, maximal and minimal elements of a poset. I realize I'm often misled by the definition of total order given by the well known $\leq$. What I am looking for is a general rule that will help in spotting the above mentioned elements, but let me give you an example of what I mean.
Let $(a,b) \in \mathbb Z \times \mathbb Z$ and $d(a,b) \in \mathbb N$ the only non-negative GCD of the pair. Let also $(\mathbb Z \times \mathbb Z, \pi)$ be defined as follows:
$$\begin{aligned} (a,b)\pi (c,t) \Leftrightarrow (a,b) = (c,t) \text{ or } d(a,b) < d(c,t)\end{aligned}$$
it's easy enough showing $(\mathbb Z \times \mathbb Z, \pi)$ is not a total order: let's take $(a,b) \neq (c,t) \in \mathbb Z \times \mathbb Z : d(a,b) = d(c,t)$ then $(a,b) \not{\pi} (c,t)$ and $(c,t) \not{\pi} (a,b)$, so these two elements are not comparable.
Now the question: what are the minimal and maximal elements? Is there a minimum or a maximum? 
I have found that $d(a,b) = 0 \Leftrightarrow (a,b) = (0,0)$.
In order for $d(a,b)$ to be a minimal element does it have to be the GCD for the least number of $(a,b)$ elements or just the least value? As $(0,0)$ is the only element with a null GCD does it automatically make $(0,0)$ a minimum?
Similarly when we come to the maximal element/maximum are we looking for the greatest GCD value or the greatest number of elements with the same GCD?
As $\mathbb N$ is infinite I'd say there aren't maximal elements and in particular there's no maximum. 
Will you please shed light on my confusion? 
 A: The ordering is based directly on the GCD, so it’s the actual GCD that matters, not the number of pairs with a given GCD.


*

*For any $\langle a,b\rangle\in\Bbb Z\times\Bbb Z$, $\langle 0,0\rangle\pi\langle a,b\rangle$: either $\langle a,b\rangle=\langle 0,0\rangle$, or $\gcd(a,b)>0=\gcd(0,0)$. This shows that $\langle 0,0\rangle$ is the minimum element of the poset (and hence if of course also minimal).

*Let $\langle a,b\rangle\in\Bbb Z\times\Bbb Z$. If $n=|a|+|b|+1$, then $\gcd(a,b)<n=\gcd(n,n)$, so  $\langle a,b\rangle\pi\langle n,n\rangle$. Clearly $\langle n,n\rangle\ne\langle a,b\rangle$, so $\langle n,n\rangle$ is strictly larger than $\langle a,b\rangle$ with respect to $\pi$. This shows that no element of $\Bbb Z\times\Bbb Z$ is maximal, which of course implies that the poset also has no maximum element. Note that it isn’t enough to know that $\Bbb N$ is infinite: to show that the poset has no maximal elements, you must show that there are pairs $\langle a,b\rangle$ with arbitrarily large GCDs. 
