# In a Venn diagram, where are other number sets located?

I remember of this image I've learned at school:

I've heard about other number (which I'm not really sure if they belong to a new set) such as quaternions, p-adic numbers. Then I got three questions:

• Are these numbers on a new set?
• If yes, where are these sets located in the Venn diagram?
• Is there a master Venn diagram where I can visualize all sets known until today?

Note: I wasn't sure on how to tag it.

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You wrote: I wasn't sure on how to tag it. I've added the number-systems tag, which seems reasonable to me. –  Martin Sleziak Oct 18 '12 at 6:50
@MartinSleziak Yep. Thank you. –  Voyska Oct 18 '12 at 7:13
The diagram suggests that there are other real numbers besides rational and irrational :) Anyway, while $\mathbb{Q}_p$'s would intersect your diagram (and each other) only in $\mathbb{Q}$ (and $\mathbb{Z}_p$ in rationals without $p$ in denominator), it is important to consider all (non-canonical!) field embeddings (you can embed $\mathbb{Q}_p$ to $\mathbb{C}$ if you wish). You should also consider algebraic numbers (in $\mathbb{C}$ AND in $\mathbb{Q}_p$'s and their extensions). –  user8268 Oct 18 '12 at 7:39
It also suggests that "whole number" means something definite, which is distinct from both the naturals and the integers, when in fact "whole number" is a horrible phrase that means either the naturals (with or without $0$) or the integers. –  Chris Eagle Oct 18 '12 at 7:44
@ChrisEagle: That's the term in Hebrew, "whole numbers". –  Asaf Karagila Oct 18 '12 at 7:45

This Venn diagram is quite misleading actually.

For example, the irrationals and the rationals are disjoint and their union is the entire real numbers. The diagram makes it plausible that there are real numbers which are neither rational nor irrational. One could also talk about algebraic numbers, which is a subfield of $\mathbb C$, which meets the irrationals as well.

As for other number systems, let us overview a couple of the common ones:

1. Ordinals, extend the natural numbers but they completely avoid $\mathbb{Z,Q,R,C}$ otherwise.
2. $p$-adic numbers extend the rationals, in some sense we can think of them as subset of the complex numbers, but that is a deep understanding in field theory. Even if we let them be on their own accord, there are some irrational numbers (real numbers) which have $p$-adic representation, but that depends on your $p$.
3. You can extend the complex numbers to the Quaternions (and you can even extend those a little bit).
4. You could talk about hyperreal numbers, but that construction does not have a canonical model, so one cannot really point out where it "sits" because it has many faces and forms.
5. And ultimately, there are the surreal numbers. Those numbers extend the ordinals, but they also include $\mathbb R$.

Now, note that this diagram is not very... formal. It is clear it did not appear in any respectable mathematical journal. It is a reasonable diagram for high-school students, who learned about rationals and irrationals, and complex numbers.

I would never burden [generic] high-school kids with talks about those number systems above.

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Extend the quaternions a little bit? You mean octonions? –  Voyska Oct 18 '12 at 18:41
@Gustavo: Yes, and Sedenions. Both, however, are non-commutative and non-associative. So everything behaves much worse than expected to in "normal number systems". –  Asaf Karagila Oct 18 '12 at 18:59

Strictly speaking $\mathbb{R}$ is not a subset of $\mathbb{C}$, rather it is isomorphic to a subfield of $\mathbb{C}$. Same for $\mathbb{Q}\subseteq\mathbb{R}$. Now $\mathbb{Z}$ is also isomorphic to a subring of $\mathbb{Q}$, not a proper subset. Whenever you have two algebraic structures $A$ and $B$ with respect to same binary operations, it may be possible to 'identify' $A$ with some subset of of $B$, that means to show an isomorphism between $A$ and a subset of $B$ with respect to the defined opereations; it this case you can write $A\subseteq B$, in some loose sense.

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I appreciate this answer, but it is needlessly pedantic for most purposes. –  Austin Mohr Oct 18 '12 at 8:06
If you really want to, those are actual subsets and not just embeddings. –  Asaf Karagila Oct 18 '12 at 8:13
I would be interested to see your proof that $\mathbb{R} \not\subset \mathbb{C}$. I suspect it might use definitions that are not universally agreed upon. (Of course, so would a proof that $\mathbb{R} \subset \mathbb{C}$.) –  Trevor Wilson Oct 18 '12 at 16:57
@Trevor: Many people consider $\mathbb C$ to be defined as a quotient of $\mathbb R[x]$, so its elements are equivalent classes. We can easily identify those corresponding to real numbers, but it's not the same. Similarly for $\mathbb Q$ in $\mathbb R$, and similarly for $\mathbb Z$ and so on. However from a set theoretical point of view, you can always just fix $\mathbb C$ and declare that the other objects are those subsets, which is what I meant in my comment. Either way, I agree with Austin that this answer is not useful here. –  Asaf Karagila Oct 18 '12 at 19:02
@AsafKaragila Good point. I was thinking that we could define the others as subsets of $\mathbb{C}$, but forgetting that in defining $\mathbb{C}$ we probably defined $\mathbb{R}$ along the way in a manner that is incompatible with this new definition. –  Trevor Wilson Oct 18 '12 at 19:06