I came across the following exercise:

"Let $p$ be a prime number, and $d: \Bbb Z\times\Bbb Z\to[0,+\infty)$ be a function defined by $d_p(x,y)=p^{-\max\{m\in\Bbb N\,:\,p^m\,|\,x-y\}}$. Prove that $d_p$ is a metric on $\Bbb Z$."

The pipe symbol meaning "divides".

The paper also has a solution, but my question is about something else: Won't the set $\{m\in\Bbb N\,:\,p^m\,|\,x-y\}$ potentially be empty, e.g. for $p=3$, $x=2$, $y=4$? Is there some definition of maximum by which the empty set has one?

(Note that I'm assuming $0\notin\Bbb N$, because another exercise in the same paper defines a set $\{{1\over n}\,:\,n\in \Bbb N \}$.)

  • 2
    $\begingroup$ $0$ should be included in the natural numbers. I normally assume this as convention, but some others like to start from one. In this case, I think that zero is included in the natural numbers, so in the case you are describing the answer would be zero. $\endgroup$ Feb 22, 2017 at 10:43
  • $\begingroup$ @астон вілла олоф мэллбэрг: that was what I thought as well, but elsewhere in the paper it does seem like ℕ is not meant to include zero, as far as I can tell. Isn't 0∉ℕ the more common convention? $\endgroup$
    – Simplex
    Feb 22, 2017 at 10:47
  • 1
    $\begingroup$ Yes, $0 \notin N$ is the more common convention (over here,I believe they call the set containing zero the "whole numbers"). But surely, $0$ must be included, for if it is not, then your point is valid. $\endgroup$ Feb 22, 2017 at 10:49
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    $\begingroup$ Then there is an inconsistency in the paper, isn't it? Certainly you must go back and review the paper. You are absolutely right, both these definitions cannot co-exist in this paper. $\endgroup$ Feb 22, 2017 at 10:57
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    $\begingroup$ I'd say the author(s) got a little sloppy and didn't realize they were using inconsistent notation. They surely meant to allow $m=0$ in the definition of $d_p(x,y)$. $\endgroup$ Feb 22, 2017 at 11:53


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