Find some function $f$ such that $ f(2,3) < f(3,2) < f(2,4) < f(4,2) < f(3,4) < f(4,3) < ... < f(13,14) < f(14,13) $ For context, what this question is essentially asking is if there is some simple function by which the following properties emerge:
$$ 1. - \forall a,b\in\mathbb{Z},\; (a > b) \iff (f(a,b) > f(b,a)) $$
$$ 2. - \forall a,b,c\in\mathbb{Z},\; (a > c) \iff (f(a,b) > f(c,b)) $$
$$ 3. - \forall a,b,c\in\mathbb{Z},\; (b > c) \iff (f(a,b) > f(a,c)) $$
$$ 4. - \forall a,b,c,d\in\mathbb{Z},\; (\max(a,b) > \max(c,d)) \implies (f(a,b) > f(c,d)) $$
If this is impossible or too difficult, then I would still be perfectly happy with a simple function (preferably a linear function with integer coefficients) by which the following arises:
$$ f(2,3) < f(3,2) < f(2,4) < f(4,2) < f(3,4) < f(4,3) < ... < f(13,14) < f(14,13) $$
Edit: It doesn't have to be a linear or even an elementary function, although it would be nice.
 A: Yes we can find an explicit function via a weird methodological construction as follows:
Set $r(s) = \frac{e^s}{1+e^s}$ so that $r$ maps $\mathbb{R}$ bijectively with the unit open interval $(0,1)$.
$\textbf{Stage 0:}$ The function $f_{0}(x,y) = x-y$ satisfies condition $1$.
$\textbf{Stage 1:} $ The function $f_{1}(x,y) = x+y+r(f_{0}(x,y))$ satisfies conditions $1,2,3$
$\textbf{Stage 2:} $ The function $f_{2}(x,y) = \max(x,y) + r(f_{1}(x,y))$ satisfies all conditions.
It seems $f_{2}$ does the job.
$\textbf{Edit:}$ It is asked below whether there is a function $f : \mathbb{R}^2 \rightarrow \mathbb{R}$ which satisfies the four conditions listed above for all of $\mathbb{R}^2$ (not just $\mathbb{Z}$). The answer is no. There is no such function which satisfies both property $4$ and $3$ or both property $4$ and $2$ (both situations are symmetric).
$\textbf{Proof of claim:}$
By symmetry, it suffices to prove that there is no such function that satisfies both property $4$ and $3$.
Suppose there is such an $f$ which satisfies conditions $4$ and $3$, set $g(x) := f(x,x)$. Note that $g$ is strictly increasing due to the fourth condition.
$\textbf{Lemma:}$ $$ \forall a \in \mathbb{R} \lim_{x \rightarrow a^{-}}g(x) < g(a) $$
where $\lim_{x \rightarrow a^{-}}$ is the left-hand limit (limit from below). Note that the limit exists as $g$ is strictly increasing.
$\textbf{Proof of Lemma:} $ Suppose for some $a \in \mathbb{R}$ we have $$\lim_{x \rightarrow a^{-}} g(x) \geq g(a)$$
by the strictly increasing property of $g$ we must have
$$\lim_{x \rightarrow a^{-}} g(x) = g(a).$$
Now for all $y \in \mathbb{R}$ and all $\epsilon > 0$ we must have
$$f(a,y) \geq f(a-\epsilon, a-\epsilon) = g(a-\epsilon)$$
by the fourth condition. Thus $$f(a,y) \geq \lim_{x \rightarrow a^{-}} g(x) = g(a) = f(a,a) $$
which is impossible by the third condition for $y < a$. This proves the lemma.
Using this lemma for all $a \in \mathbb{R}$ we can choose $q_{a} \in \mathbb{Q}$ so that $q_{a}$ belongs to the open interval $I_{a} := (\lim_{x \rightarrow a^{-}} g(x), g(a))$. Note that for $a \neq b$ $q_{a} \neq q_{b}$ as $I_{a} \cap I_{b} = \emptyset$. Thus there are uncountably many
rationals which is impossible, this is a contradiction.
