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I'm not sure I get the motivation for a Euclidean function having to map to $\mathbb{Z}^{\ge 0}$. E.g. it would seem that $\mathbb{R}^{\ge 0}$ would be a natural choice for a ring of "polynomials" permitting positive real exponents (e.g. $x^{2.5}+x^{0.5}+1$) -- and you can do long division on them too.

Is there something I'm missing?

Abhimanyu Pallavi Sudhir
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    The codomain can be any well-ordered set (so that Euclidean descent terminates). See the paper of Agargun and Fletcher cited here on relationships between variations on the definition of Euclidean rings. – Bill Dubuque May 17 '20 at 17:12
  • And thus by the well-ordering theorem, we can take $\mathbb{R}$ to be the codomain ;) – Fibonacci Cube K May 18 '20 at 09:34
  • Hello, do you know where can one find the paper of Agargun and Fletcher now? The original page seems to be missing… – Censi LI Feb 07 '23 at 08:46
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    @CensiLI the answer linked currently links to https://web.archive.org/web/20200605172616/http://journals.tubitak.gov.tr:80/math/issues/mat-95-19-3/pp-291-299.pdf , which seems to work. – Zachary Barbanell Apr 24 '25 at 07:20

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