Given $\mathbb Z^-=\{x\in \mathbb Z:x<0\}$ and $T = \mathbb Z^-\times \mathbb N$, let the binary relation $\odot$ be defined as follows:
$$\begin{aligned} (a,b) \odot (c,d) \Longleftrightarrow a \leq c \land b \mid d \end{aligned}$$
Check if the operation $\odot$ sets order, in particular total order and also find minimum, maximum, maximal and minimal elements in $(T, \odot)$ if they exists.
In order for $\odot$ to be a partial order, its reflexivity, antisymmetry and transitivity has to be proved.
Reflexivity
It can be easily shown that $\forall (a,b) \in T$
$$\begin{aligned}(a,b) \odot (a,b) \Longleftrightarrow a \leq a \land b \mid b\end{aligned}$$
so this relation is reflexive.
Antisymmetry
In order for this property to be true the following condition must be valid $\forall (a,b),(c,d) \in T$:
$$\begin{aligned} (a,b) \odot (c,d) \land (c,d) \odot (a,b) \Rightarrow (a,b) = (c,d) \end{aligned}$$
The antisymmetry doesn't apply to this relation because while the following is always true:
$$\begin{aligned} a \leq c \land c \leq a \Rightarrow a = c \end{aligned}$$
we cannot state the same for the following:
$$\begin{aligned} b \mid d \land d \mid b \Rightarrow b = d\end{aligned}$$
because as $b \neq 0 \land d = 0$ (or similarly $b = 0 \land d \neq 0)$ then
$$\begin{aligned}\exists \alpha \in \mathbb Z : \alpha b = d \end{aligned}$$
$$\begin{aligned}\exists \beta \in \mathbb Z : \beta d = b \end{aligned}$$
whereas $\alpha = 0$ but $\nexists \beta \in \mathbb Z : \beta 0 = d$. Hence this relation isn't antisymmetric.
Transitivity
We need to prove
$$\begin{aligned} (a,b) \odot (c,d) \land (c,d) \odot (e,f) \Rightarrow (a,b) \odot (e,f) \end{aligned}$$
it's valid the following:
$$\begin{aligned} a \leq c \land c \leq e \Rightarrow a \leq e \end{aligned}$$
the following is valid as well
$$\begin{aligned} a \mid c \land c \mid f \Rightarrow a \mid f \end{aligned}$$
as $\exists x \in \mathbb Z : ax = c$ and $\exists y \in \mathbb Z : yc = f$
it's safe to say $yxa=f$.
Conclusion: this relation isn't partial order, so in particular isn't a total order.
Is everything ok with my exercise or did I do anything wrong? When it comes to look for maximum, minimum and minimal or maximal elements I feel lost. So I think there's no maximum or minimum because both $\mathbb Z^-$ and $\mathbb N$ are inifinite. Apparently I don't have a clue on how to spot minimal or maximal elements for $T$. Will you please help me out with that?
Best Answer
Actually, it is a partial order. Given natural numbers $a,b$, we can conclude from $a\vert b\wedge b\vert a$ that $a=b.$ Your counterexample doesn't work, because, while everything divides $0$, $0$ only divides itself.
A maximum element would be some $(a,b)\in T$ such that for all $(c,d)\in T$, $c\leq a\wedge d\vert b$. This would require that $a$ be the greatest element of $\Bbb Z^-$ and that $b$ be a natural number that all natural numbers divide. Can you think of what that would be? Note that if there is a maximum element, then it is the only maximal element.
A minimum element would be some $(a,b)\in T$ such that for all $(c,d)\in T$, $c\geq a\wedge b\vert d$. Then $a$ would have to be the least element of $\Bbb Z^-$, which is not possible.
Given any $(a,b)$, we have that $(a-1,b)\odot(a,b)$, so there is always a different element preceding it. Thus, there are no minimal elements, either.
To determine whether or not $\odot$ totally orders $T$, we must determine whether for distinct $(a,b),(c,d)\in T$ we need have $(a,b)\odot(c,d)$ or $(c,d)\odot(a,b)$. It shouldn't be difficult to see that this need not hold (I leave it to you to find an example), so that $\odot$ does not totally order $T$, after all.
Edit: Well done in showing that it isn't a total order, and in finding the maximum element. It is in fact maximum (not just maximal) since for each $(c,d)\in T$ we have $c\leq-1$ and $d\vert 0$. Partial orders may or may not have a maximum or a minimum element, and may or may not have maximal or minimal elements. For an example of a partial order with a maximum element, take any set $A$, any $b\notin A$ and define a partial order $\precsim$ on $A\cup\{b\}$ by $c\precsim c$ for all $c$, and $a\precsim b$ for all $a\in A$. Then $b$ is the maximum element, and each $a\in A$ is incomparable to the others, so minimal, but not minimum if $A$ has multiple elements.
To elaborate on the antisymmetry of your example, let's suppose that $m,n\in\Bbb N$ such that $m\vert n\wedge n\vert m$. Hence, $$m=nv\wedge n=mw$$ for some $v,w\in\Bbb N$. If $m=0$, then so is $n$. If $m\neq 0$, then since $m=nv=mvw$, then $vw=1$, which is only possible for $v=w=1$ since $v,w\in\Bbb N$. Thus, again, $m=n$. Does that clear things up?