# Rational numbers with odd denominator

Notation. By $\mathbb{Z}_{(2)}$, I mean the localization of $\mathbb{Z}$ at the prime ideal $(2).$ So basically, this is obtained by adjoining a multiplicative inverse for every positive prime number distinct from $2$. Equivalently, we can think of $\mathbb{Z}_{(2)}$ as the set of rational numbers with odd denominator. So: $\mathbb{Z}_{(2)} \subseteq \mathbb{Q}.$ This is a local ring, of course, and judging by how often my commutative algebra lecturer used the phrase "local ring" during those lectures I mostly didn't understand, that's probably important.

Background. Although localization is usually thought of as an "advanced" topic, despite this, I recently recently that $\mathbb{Z}_{(2)}$ actually arises in high school math. In particular, given $x \in \mathbb{R}_{\neq 0}$ and $q \in \mathbb{Z}_{(2)}$, we can make good sense of the expression $x^q.$ Such expressions make sense, in particular even if $x$ is negative. Of course, we can (and often do) replace $q \in \mathbb{Z}_{(2)}$ with $q \in \mathbb{Q}$ or even $q \in \mathbb{R}$, but notice this forces us to assume that $x$ is positive to compensate. There's also a semiring $\mathbb{N}_{(2)}$ obtained by localizing the natural numbers at the prime ideal $(2)$. More concretely: $\mathbb{N}_{(2)} = \{q \in \mathbb{Z}_{(2)} : q \geq 0\}.$

This gives us three exponentiation functions, all denoted $x,q \mapsto x^q$.

\begin{align*} \mathbb{R} \times \mathbb{N}_{(2)} &\rightarrow \mathbb{R} \\ \mathbb{R}_{\neq 0} \times \mathbb{Z}_{(2)} &\rightarrow \mathbb{R}_{\neq 0} \\ \mathbb{R}_{> 0} \times \mathbb{R}_{\color{white}{(2)}} &\rightarrow \mathbb{R}_{> 0} \end{align*}

Question. I'd like to learn a bit more about the number systems $\mathbb{Z}_{(2)}$ and $\mathbb{N}_{(2)}$ from the point of view of these exponentiation functions. For instance, does the localness of $\mathbb{Z}_{(2)}$ tell us anything important about the function $\mathbb{R}_{\neq 0} \times \mathbb{Z}_{(2)} \rightarrow \mathbb{R}_{\neq 0}$ denoted $x,q \mapsto x^q$? If so, I'd like to know about this. Also, I'm interested in "elementary" things that I can teach to high school students about $\mathbb{Z}_{(2)}$ and $\mathbb{N}_{(2)},$ that do not require a lot of abstract mathematics.

Links or references preferred, but direct explanations and further thoughts are also welcome.

• I don't think there's anything specific about local rings that make your exponentiation well-defined. I think it's more about the number $2$ which happens to be both prime and equal to the degree of the algebraic closure of $\mathbb{R}$ over $\mathbb{R}$. For example if you replace $\mathbb{Z}_{(2)}$ with $\mathbb{Z}_{(3)}$ the maps you have don't work anymore – Jay Sep 8 '16 at 0:49
• @Jay, can you elaborate on the algebraic closure thing? – goblin Sep 8 '16 at 1:48
• So you can define $r^{1/n}$ for $r \in \mathbb{R}$ if $x^n - r$ has a (canonical) root in $\mathbb{R}$. This doesn't happen iff $n$ is even in which case there are either two roots or none depending on if $r > 0$ or $r < 0$ (excluding $r = 0$). This is because $\mathbb{R}$ is not algebraically closed but once you adjoin $\sqrt{-1}$ it is – Jay Sep 8 '16 at 2:48
• @Jay, still not following. My thinking is that we can define $r^{1/n}$ without any special care or thought iff $x^n-r$ has a unique root. The moment you go to $\mathbb{C}$, suddenly there's always $n$ roots, and this ceases to work. In some sense, the problems get worse when we go to $\mathbb{C}$, not better. (I'm sure that with enough knowledge of complex analysis, these "problems" actually end up being blessings in disguise, though. Complex analysis is just magical like that.) – goblin Sep 8 '16 at 13:20
• I'm not saying things become better in $\mathbb{C}$. I'm saying that the number $2$ is special in your setting because it's equal to $[\overline{\mathbb{R}} : \mathbb{R}]$. The only $n$ for which you $\textit{can't}$ define $r^{1/n}$ is $n$ even i.e. you $\textit{can}$ define $r^q$ for $q \in \mathbb{Z}_{(2)}$. Also if $k$ is any field such that $[\overline{k} : k] < \infty$ then $[\overline{k} : k] = 2$ (I think this is called the Artin-Schreier theorem) so it's a special thing about $2$ – Jay Sep 8 '16 at 20:28