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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?

Red Banana
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Zigu
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  • I think you may want to add a condition like up to isomorphisms or something like are there fields ${F_r}{r\in\mathbb{R}}$ such that $r_1<r_2$ implies $F{r_1}$ is a proper subfield of $F_{r_2}$ – Amr Jun 28 '13 at 23:57
  • Hm. I guess you are right. That is a implicit assumption I was making. Are there countably distinct numerical types. Need to change the title too. – Zigu Jun 28 '13 at 23:58
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    Also "mathematical number types" is very vague – Amr Jun 28 '13 at 23:59
  • Is there a better term for it? I came up with the question on a fly and just typed it out... – Zigu Jun 28 '13 at 23:59
  • If I understand you well, I think you want to ask the second question in my first comment (However I am not sure if I understand you as your terminology was vague) . – Amr Jun 29 '13 at 00:04
  • @Zigu I think you may need to clarify the question a bit. If you are looking for fields, the answers given are appropriate, but you list quaternions and octonions and those a not fields, there are non-commutative see http://en.wikipedia.org/wiki/Quaternion. Also, there are number systems such as http://en.wikipedia.org/wiki/Surreal_number that have infinite numbers as well. – Ryan Sullivant Jun 29 '13 at 07:36

2 Answers2

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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.

Community
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Zev Chonoles
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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.

Erik Vesterlund
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