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I am looking for functions $f(x,y)$, real arguments, continuous, with the following properties:

  1. $f(m,n) = r$, where $r$ is integer $> 0$ if and only if $m,n$ are integers $> 0$.

  2. $f(m,n) \le f(r,s)$ if and only if $m n \le r s$, with $m,n,r,s$ integers $> 0$.

I would like to know if functions with properties 1 alone, 2 alone, or both together exist, and if they have been studied and given a specific name (for further investigation). In particular, I would be curious to know if relatively simple closed-form functions satisfying both conditions exist, or can exist at all.

If useful to allow existence, the further constraints $m < n, r < s$ can be used.

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    $\begingroup$ Nine versions in one hour does not inspire confidence. $\endgroup$ Commented May 18, 2023 at 12:38

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Such functions as you require do not exist. Your requirements impose that $f(1,1) < f(1,2) < f(2,2)$ (for example, $f(1,1) \leq f(1,2)$ by (2), but $\neg (f(1,2) \leq f(1,1))$ also by (2), so we have $f(1,1) < f(1,2)$). Now consider the continuous function $x \mapsto f(x,x)$ on $[1,2]$: by the intermediate value theorem, there exists $x$ with $1<x<2$ such that $f(x,x) = f(1,2)$. Since $f(1,2)$ is an integer but $f(x,x)$ is not by your requirement (1), this is a contradiction.

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  • $\begingroup$ I am trying to understand. Sorry, I think a got a bit lost in your proof. if you use f(x,x) where the arguments are equal (like moving on a diagonal), how can we apply it to the values [1,2] which are distinct? $\endgroup$ Commented May 18, 2023 at 11:48
  • $\begingroup$ Note that $[1,2]$ is a compact interval, not a point in $\mathbb{R}^2$ $\endgroup$ Commented May 18, 2023 at 12:15
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    $\begingroup$ I see that we are now on version 8 of this question. It took me two minutes to answer the original question, essentially by the same method as @Gro-Tsen. It is trivial to find continuous functions satisfying just 1 or just 2. I vote to close. $\endgroup$ Commented May 18, 2023 at 12:30
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    $\begingroup$ @Jada Yes, $f(x,x)$ is the diagonal value, and $f(1,2)$ is an off-diagonal value. My argument is that because $f(1,1) < f(1,2) < f(2,2)$, by the intermediate value theorem, there must be a diagonal value $f(x,x)$ with $1<x<2$ which is equal to the off-diagonal $f(1,2)$, which is impossible because one is supposed to be an integer and the other isn't, so your two conditions cannot hold simultaneously. $\endgroup$ Commented May 18, 2023 at 13:06
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    $\begingroup$ @Jada: The continuous function $f\colon\mathbb{R}^2\to\mathbb{R}$ which takes a point $P \in \mathbb{R}^2$ to $1 + \min(\frac{1}{2}, d(P,\mathbb{N}_{>0}^2))$ where $d(P,\mathbb{N}_{>0}^2)$ denotes the distance from $P$ to the closest point of the form $(m,n)$ with $m,n>0$ integers, takes the value $1$ exactly on those points, and is otherwise strictly between $1$ and $2$, so it satisfies your condition (1). $\endgroup$ Commented May 18, 2023 at 13:38

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