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Before I can get to grips with something rigorously I need to understand it in an intuitive way. I was trying to get my head around the supernatural numbers. Looking at the definition it looks like they could extend the conventional numbers, by placing them in union with something analogous to the residues of the p-adics. Is this a fair description?

Martin Sleziak
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Robert Frost
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  • Are supernatural numbers anything more that elements of ${\mathbb N}^{\mathbb N}$ for which you look at the componentwise mulitplication? – Nikolaj-K Jun 28 '18 at 20:31

2 Answers2

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The set of supernatural numbers is an extension to the set of natural numbers:

  • The factorization of a natural number must contain a finite number of primes
  • The factorization of a supernatural number may contain an infinite number of primes
barak manos
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To better comprehend supernatural (or Steinitz) numbers it is illuminating to examine the various contexts in which they arise naturally. For example, see the equivalent conditions $(c)-(i)$ listed below (excerpted from here).

Note $ $ Said descriptions are independent of the first three introductory paragraphs on locally simple algebras, so you can skip them if such algebras are not of interest (start reading at the paragraph "One can observe that ...")

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Bill Dubuque
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  • Wow that is some pretty heavy reading! – Robert Frost Aug 15 '16 at 15:06
  • @RobertFrost I suggest you start first with relation to subgoups of the additive group or rationals (d), which is perhaps the simplest. – Bill Dubuque Aug 15 '16 at 15:11
  • Ok I will read. In the p-adics, $p^1$ is represented as a 1 carried up from the 1st digit. $p^2$ is a $1$ carried up from the 2nd digit and so on. If there's some arbitrarily large number $a$ such that $n\leq a\forall n\in\mathbb{N}$ and we say $0=\overline9+1$ then zero in the p-adics is equivalent to $p^a$. Zero in the p-adics therefore would seem to be equal to a 1 carried up from the $(a-1)^{th}$ digit to the $a^{th}$. The p-adics are therefore equivalent modulo $p^{a-1}$... – Robert Frost Aug 15 '16 at 15:37
  • Based on my (incorrect?) interpretation of the supernaturals, these extend the naturals by exactly these numbers $p^x:x\geq a$, which are the modulus under which the p-adics are equivalent. I think that's my question better formulated. is that a correct intuitive description of them? – Robert Frost Aug 15 '16 at 15:37
  • @RobertFrost In what context did you encounter supernaturals? – Bill Dubuque Aug 15 '16 at 15:59
  • I just stumbled upon the fact that they extend the notion of compound numbers except with possibly infinite powers of each prime, and I'm interested in p-adic arithmetic so I was hoping for a quick, intuitive understanding of how they might be related and useful. My intuition says $2^\infty$ is supernatural and all $2$-adics are equivalent $\pm 2^{\infty}$. – Robert Frost Aug 15 '16 at 16:17
  • @RobertFrost Could you please give some links to the expositions that you seek to better understand. – Bill Dubuque Aug 15 '16 at 16:37
  • It's my own exposition. My logic is this. When we move, in the p-adics, from $\overline9$ to $0$, by the addition of $1$, we might imagine we have carried the $(p-1)*p^{\infty}$ to $p^{\infty+1}$, so the $p$-adics are equivalent $p^{\infty}$ where $p^{\infty}$ is a supernatural number. And that is why $p$-adics are so powerful. – Robert Frost Aug 22 '16 at 13:20
  • I am just deleting. Reposting is too hard because the copy/paste loses MathJax formatting. – Oscar Lanzi Apr 24 '24 at 13:52