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Motivation

NaN as a value exists in IEEE floating-point numbers. Since every operation involving NaN has NaN as the outcome, IEEE floating-point numbers are not fields. I want to define a new algebraic structure so NaN could be encapsulated.

Definitions

A field $(F,+,\times)$ is NaNable if there exists a binary operation $\sim$ such that $(F \cup \{\text{NaN}\}, \sim, +)$ is a field, where $\text{NaN}$ is the identity element of $\sim$.

$(F \cup \{\text{NaN}\}, \sim, +,\times)$ is a NaNification of $(F,+,\times)$, and is a NaN-field.

Examples

Examples of NaNable finite fields include $\mathbb{Z}_2$, $\mathbb{Z}_3$, $\mathbb{Z}_7$, and $\mathbb{Z}_{31}$. The operation tables of NaNification of $\mathbb{Z}_3$ is:

$$ \begin{array}{c|cccc} \sim & \text{NaN} & 0 & 1 & 2 \\ \hline \text{NaN} & \text{NaN} & 0 & 1 & 2 \\ 0 & 0 & \text{NaN} & 2 & 1 \\ 1 & 1 & 2 & \text{NaN} & 0 \\ 2 & 2 & 1 & 0 & \text{NaN} \\ \end{array} $$

$$ \begin{array}{c|cccc} + & \text{NaN} & 0 & 1 & 2 \\ \hline \text{NaN} & \text{NaN} & \text{NaN} & \text{NaN} & \text{NaN} \\ 0 & \text{NaN} & 0 & 1 & 2 \\ 1 & \text{NaN} & 1 & 2 & 0 \\ 2 & \text{NaN} & 2 & 0 & 1 \\ \end{array} $$ $$ \begin{array}{c|ccc} \times & 0 & 1 & 2 \\ \hline 0 & 0 & 0 & 0 \\ 1 & 0 & 1 & 2 \\ 2 & 0 & 2 & 1 \\ \end{array} $$

Note that multiplication involving $\text{NaN}$ is not defined yet.

Questions

  1. Is there a notable example of NaNable infinite field?

  2. Is there a dedicated name for $\sim$? Not as a symbol, but as an operation?

  3. It turns out that if every multiplication involving $\text{NaN}$ is defined as $\text{NaN}$, distributivity is satisfied in the NaN-field. Does this mean I should define it accordingly?

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Dannyu NDos
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    Any operation involving NaN should (morally) return NaN, so I would say yes to (3). Is there an intuitive interpretation of what $\sim$ should mean? – Jordan Barrett May 28 '20 at 05:10
  • @JordanMitchellBarret NaN is some kind of singularity, and $\sim$ just is to compromise the singularity. – Dannyu NDos May 28 '20 at 05:22
  • Can you not just define the operations the same way for any finite field (and then presumably for $\mathbb{Z}$ too)? – Jordan Barrett May 28 '20 at 05:27
  • @JordanMitchellBarrett I don't think so. – Dannyu NDos May 28 '20 at 05:33
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    The multiplicative group of a field has the property that, for every positive integer $n$, the equation $x^n=1$ has at most $n$ solutions. It follows that a NaNable field, the equation $nx=0$ has at most $n$ solutions. This implies a NaNable field of positive characteristic is a prime field. So if $(F\cup{NaN},\sim,+,\times)$ is an infinite NaN field, then $(F,+,\times)$ is characteristic $0$, and it follows that $(F\cup{NaN},\sim,+)$ has no non-trivial roots of unity, hence has characteristic $2$. – Julian Rosen May 28 '20 at 22:30
  • @JulianRosen I follow you, but I think $F = \mathbb{Z}_2$ is the only exception on the "characteristic 2" part, where the characteristic is actually 3. – Dannyu NDos May 28 '20 at 23:10
  • This is a very interesting idea, but what is the motivation for this $~$ thing, intuitively? Just to make it a field? – Mike Battaglia Feb 17 '24 at 01:35

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