# Is there a “homomorphism/isomorphism” (conceptually) between $\mathbb{C - R}$ and $\mathbb{R - Q}$

The Background: I was thinking about ways to conceptualize the way rational and irrational numbers interact within the real line while helping somebody in an elementary real analysis class. It came to my mind that one way of thinking about this was to represent an irrational number x as:

$$r \in \mathbb{Q}$$ $$i \in \mathbb{R} - \mathbb{Q}$$ $$r + i = x$$

As a notational simplification from this point let $\mathbb{R} - \mathbb{Q} := \mathbb{I}$ and also: $\mathbb{C} - \mathbb{R} = \hat{\mathbb{C}}$ I then noticed that there were some similarities (and differences between this set under real-number addition and multiplication and C when the real line was deleted) mainly:

$$\forall a,b \in \mathbb{I}$$ $$\forall z,w \in \hat{\mathbb{C}}$$

$$a + b = r_a + r_b +i_a + i_b = r_c + i_c = c \in \mathbb{R}$$ similar to: $$z + w = \mathcal{Re}(z) +\mathcal{Re}(w) + i(\mathcal{Im}(z) +\mathcal{Im}(w)) = u \in \mathbb{C}$$ and $$ab = (r_a + i_a)(r_b + i_b) = r_ar_b + r_ai_b+ i_ar_b + i_ai_b = r_c + i_c \in \mathbb{R}$$ similar to: $$zw = \mathcal{Re}(z)\mathcal{Re}(w) - \mathcal{Im}(z)\mathcal{Im}(w) + i(\mathcal{Re}(z)\mathcal{Im}(w) + \mathcal{Re}(w)\mathcal{Im}(z)) = u \in {\mathbb{C}}$$

The primary difference I noticed between this and complex number arithmetic was that the value $i_ai_b$ could be rational or irrational, and the sign change.

It also seemed to me that there would be a way to compensate for these differences in some homomorphism/isomorphism between $\hat{\mathbb{C}}$ and $\mathbb{I}$ or alternatively maybe just one quadrant of $\hat{\mathbb{C}}$

I am aware that multiplication and addition lack closure (that is the product/sum of two rationals can be irrational, and the product/sum of two complex numbers can be real), and that there is also no additive or multiplicative identity present in either set (which is why I think such a homomorphism/isomorphism may exist in the first place). I figure if such an homomorphism/isomorphism does exist then it will make explaning the lack of these properties in $\mathbb{I}$ easier when talking to student with some experience with complex algebra, but little analysis background. (as well as provide an interesting little tidbit to toss around)

The Question: Has anyone seen a construction that of a homomorphism/isomorphism between $\mathbb{C - R}$ and $\mathbb{R - Q}$, or can anybody point me in the right direction for developing one? Alternatively is there a counter-argument that disproves this possibility that I'm missing somewhere?

I'm also looking for more of a conceptual notion of homomorphism/isomorphism than a particular one (since these sets don't have nice algebraic properties, or the identity elements that make such a construction a homomorphism in traditional definitions).

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As you point out, $\mathbb{C} \setminus \mathbb{R}$ and $\mathbb{R} \setminus \mathbb{Q}$ do not have algebraic structure, so the only kind of homomorphism I can think of would be topological, i.e., a continuous function. Every continuous function $\mathbb{C} \setminus \mathbb{R} \to \mathbb{R} \setminus \mathbb{Q}$ has only one or two points in its range, so this doesn't seem very interesting. There are lots of continuous functions in the other direction, but that doesn't seem very interesting either. I don't think "all non-fields have structural similarities" is a very good heuristic. –  Trevor Wilson Sep 11 '12 at 1:59
They are isomorphic as sets, and not much else. –  Alex Becker Sep 11 '12 at 2:04
@TrevorWilson, What about with a different topology on them? For example a Zariski topology. –  Keivan Sep 11 '12 at 3:02
@Keivan I did not think about that. Does something interesting happen? –  Trevor Wilson Sep 11 '12 at 3:12
One other criticism: splitting a complex number z into Re(z) and Im(z) is natural and unique, but splitting a real number x into r_x and i_x (which I think is your notation for "rational part" and "irrational part") is neither natural nor unique. (For example, what's to stop $\sqrt{2}$ having "rational part" 50 and "irrational part" $\sqrt{2}-50$?) –  Billy Sep 11 '12 at 6:19

I think the similarities you point to are far too superficial for any "homomorphism" between the two in any usual sense of the word (surely not as rings, groups, monoids, etc.) However, this does fit into a more general formulation of a pattern. Namely, for many submonoids $S\subset M$,

$$\begin{array}{c|c|c}\bullet & S & M\setminus S \\ \hline S & S & M\setminus S \\ \hline M\setminus S & M\setminus S & \rm n/a\end{array}\tag{\star}$$

is a graded partial multiplication table (I am making that phrase up) which encodes where a product of two elements each of $S$ or $M\setminus S$ independently will end up; $\rm n/a$ stands for not applicable because in general a product of two elements from $M\setminus S$ may be in $S$ or $M\setminus S$.

(This does not apply to every type of monoid, but it does to most "naturally-appearing" ones; I am having a hard time pinning down precise conditions for $(\star)$ to be valid.)

Furthermore, every element of $M$ may be written as a word consisting of elements alternately from $S$ and $M\setminus S$ (not in general uniquely). If $M$ is commutative in particular then we may go further and say every element of $M$ may be written as $a+b$ with $a\in S$ and $b\in M\setminus S$ (again, not in general uniquely).

As a very particular corollary the same table $(\star)$ applies to both the addition and multiplication operations in a commutative ring, and in particular in fields, and in particular for $\Bbb R\subset\Bbb C,~\Bbb Q\subset\Bbb R$.

Upon further reflection, I think there may be a way to phrase this as you want: abstractly $M$ can be "quotiented" by $S$ to yield the partial algebra $(\{S,M\setminus S\},\bullet)$ (where $(M\setminus S)\bullet(M\setminus S)$ is undefined). Furthermore, we can quotient in both addition and multiplication operations on a commutative ring structure, creating another, more complicated partial algebra. We can say that the partial algebras $(\{\Bbb R,\Bbb C-\Bbb R\},+,\times)$ and $(\{\Bbb Q,\Bbb R\setminus\Bbb Q\},+,\times)$ are isomorphic!

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You cannot expect anything in the way of a topological homeomorphism: $\mathbb{R-Q}$ is totally disconnected, while$\mathbb{C-R}$ has two connected components.

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Just to add another point of view, I would not expect any kind of significant morphism, basically because the two constructions leading from $\mathbb Q$ to $\mathbb R$ and from $\mathbb R$ to $\mathbb C$ are radically different.

The first is, in my opinion, the most complex one to understand and to come up with, you need to introduce Cauchy sequences, set up an equivalence relation and so on. The second construction is just a degree two field extension, or more generally an algebraic closure.

Trying to esemplify things one little bit more, one could think going from $\mathbb Q$ to $\mathbb R$ as "let's fill in the gaps between rational numbers so we get a nice connected line" and going from $\mathbb R$ to $\mathbb C$ as "I want every polynomial to have a root". As you can see, these requests are deeply different in nature, so again one can't expect any meaningful morphism.

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