Spotting maximum, minimum, maximal and minimal elements in a poset Let $(\mathbb N^* \times \mathbb N^*, \varphi)$ be a poset defined as follows:
$$\begin{aligned} (a,b)\varphi(c,d)\Leftrightarrow ab<cd \text{ or } (a,b) = (c,d)\end{aligned}$$
Check if $\varphi$ is a total order and determine maximum, minimum, maximal and minimal elements.
It's easy to prove that $\varphi$ is not total because as we consider $(a,b),(c,d) \in \mathbb N^* \times \mathbb N^* : a = d \text{ and } b = c$ then
$$\begin{aligned} ab \nless cd \text{, } cd \nless ab \text{ and } (a,b) \neq (c,d)\end{aligned}$$
What I am having a very hard time with is spotting maximum, minimum, maximal and minimal elements. In my opinion the poset doesn't have any maximum or maximal elements as $\mathbb N^* \times \mathbb N^*$ is infinite, but it does have a minimal element which is $(1,1)$. Given that $\nexists (\varepsilon, \varepsilon') : 1 = \varepsilon' \text{ and } 1 = \varepsilon$ and $(1,1) \neq (\varepsilon, \varepsilon')$ then $(1,1)$ is comparable to all elements in $\mathbb N^* \times \mathbb N^*$, does this make it also the minimum? 

I think your reasoning is sound enough.
If there is a maximal element, $(c,d)$ say, then  $(a,b) \varphi (c,d)$ for all other elements $(a,b)$ in $\mathbb{N}^* \times \mathbb{N}^*$ which implies that:
\begin{equation}
ab < cd \text{ or } (a,b)=(c,d)
\end{equation}
But then $(c,d) \varphi (c+1,d)$ since $cd < (c+1)d$ contradicting the maximality of $(c,d)$.
Finally, $(1,1)$ is the minimal element as $1 \cdot 1=1$ is the smallest a product of two elements in $\mathbb{N}^*$ can be. That is, any other element $(e,f)$ has $e>1$ or $f>1$ In the first case, $(1,1) \varphi (e,f)$ since $1\cdot 1 < e \leq ef$. The second case is similar.
Also I like your profile: "Just a guy who hopes to live through his algebra exam." +1 for having realistic goals, haha.
