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Intriguing.
An element of $\mathbb{R}^\odot$ is an equivalence class of pairs of real numbers, where $(a,b)$ and $(c,d)$ are equivalent iff $a d = b c$.
Doesn’t that make $(0,0)$ equivalent to everything?
Re. #2:
Doesn’t that make $(0,0)$ equivalent to everything?
Apparently, it does:
Although $0/0$ doesn’t simplify, it’s an absorbing element for addition: $0 /0 + x = 0 /0$.
We also write $\infty$ for $1 : 0$ and $\bot$ for $0 : 0$; these are the only elements of $\mathbb{R}^\odot$ that don't come from $\mathbb{R}$.
Re #2, #3: I looked in the supplied link, and apparently the actual equivalence relation is not what is written, but rather (see page 4)
(x,y)~(x’,y’) if and only if there are s,s’∈S such that (sx,sy)=(s’x’,s’y’),
where S in this case is the set of nonzero real numbers.
So 0/0=a/b if and only if a=b=0.
See here: https://web.archive.org/web/20200521074722if_/https://www2.math.su.se/reports/2001/11/2001-11.pdf
So this looks more like the projective line with an added 0/0.
Curiously, it does form a variety of algebras.
Yes, that's a mistake to just say $a d = b c$. The more general situation is as Dmitri quoted, where $S$ can be any submonoid of the original commutative rig, but I didn't get that far last night. I wanted to give the motivating example (the one that gives the concept its name), and the equivalence relation simplifies a lot if you start with a field and let $S$ be the monoid of nonzero elements, but it doesn't simplify quite as much as I wrote!
Incidentally, I really want to get to the topology of these, which is not in the reference by Carlström (and possibly not anywhere). Wheels are typically not Hausdorff, and in fact the specialization preorder on a wheel can be defined from the algebraic structure alone (I think). In $\mathbb{R}^\odot$, for example, $0/0$ is the only closed point and is a specialization of every other point.
Another useful interpretation to keep in mind is the following.
Recall that for a ring $R$, one can interpret a fraction $x/y$ as being a partially defined rational function on $Spec(R)$, defined on the complement of the closed set $V(y)$
Similarly for the wheel of fractions on $R$, one can interpret $x/y$ as a partially defined projective number, defined on the complement of $V(x, y)$.
There’s some neat sense in which wheel arithmetic subsumes some amount of bookkeeping regarding to where things are well-defined.
Yeah, there's a thing from projective geometry where we allow a point at infinity; this lets us divide nonzero numbers by zero. But there's also a thing from domain theory where we give every type an extra element to catch errors when things are undefined, and this lets us divide zero by zero too. A wheel has both of these.
The motivation for wheels looks the same as that for meadows. It would be interesting to see how these are related.
How do I interpret “$(a,b)=(c,d)$ iff $a d = b c$, and additionally $(a,b)\neq (0,0)$ iff $(c,d) \neq (0,0)$” as a definition of an equivalence relation?
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