Is the set of all distinct mathematical number types countable? I was reading this article where the author explains that there are numbers outside the complex set and that you can arbitrarily generate new types using the same method as he described to generate the quaternions.
My question is: Is the set of all mathematical number types countable?
IE (1,2,3,...) -> (0, integers, rationals, reals, complex, quaternions, ...)
Using the way he generates to count, you might be able to set up an injection with elements used to generate the set (1 then 1, i then 1, i, j, k, and so on) and there are finite number of operations used to generate the other numbers (or at least in his article he mentioned a finite set of operations.).
Is this right or is there a quick counter example?
 A: Consider $\mathbb{Q}$, the field of rational numbers. For any set $S$, we can form the field $\mathbb{Q}(S)$, the field of rational functions in the indeterminates $S$. This has transcendence degree $|
S|$ over $\mathbb{Q}$. If two such extensions $\mathbb{Q}(S)$ and $\mathbb{Q}(T)$ are isomorphic, then they must have the same transcendence degree, so that $|S|=|T|$. Thus, we can conclude that for any possible set cardinality $\alpha$, there is an extension of $\mathbb{Q}$ of transcendence degree $\alpha$. But the collection of all possible set cardinalities is so large, it is not even a set itself, but rather a proper class (see here on Wikipedia). Thus, there are certainly uncountably many non-isomorphic fields containing $\mathbb{Q}$, but there are more than that; in fact, there is no "number" of isomorphism classes, there are simply too many.
Thus, we have shown that the collection of isomorphism classes of characteristic zero fields is a proper class, and hence as well, the collection of isomorphism classes of all fields is a proper class.
A: The set of transcendental numbers, call it $T$, is uncountable. Thus if $u$ is transcendental, the set $\{\mathbb{Q}(u)\}_{u \in T}$ of fields in uncountable. Of course, there could be some relation between transcendentals (I don't know) that would lead to this cardinality $|T|$ being countable.
edit: ok now the topic changed entirely, nevermind.
