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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.

Bill Dubuque
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Arjun Vyavaharkar
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    I don't think your header is good. The abstract question looks interesting. The specific one is a triviality, since you provide no restrictions on $f$. But, really, the abstract question looks intriguing. What is the source? – lulu Feb 19 '22 at 01:43
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    I actually came up with it because of a programming project. Specifically, I wanted to rank each and every single full house hand in poker from the best full house to the worst (hence why it starts at 2 and ends at 14). – Arjun Vyavaharkar Feb 19 '22 at 04:18
  • I also included the last, more specific example in the header because I unfortunately cannot include all 4 properties in the title :P – Arjun Vyavaharkar Feb 19 '22 at 04:20

1 Answers1

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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.

acreativename
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