As I know when you move to "bigger" number systems (such as from complex to quaternions) you lose some properties (e.g. moving from complex to quaternions requires loss of commutativity), but does it hold when you move for example from naturals to integers or from reals to complex and what properties do you lose?

40You are losing the existence of an order compatible with the field. – Serser Apr 08 '18 at 20:35

17We loose the [total ordering relation](https://math.stackexchange.com/questions/487997/totalorderingoncomplexnumbers). – user Apr 08 '18 at 20:35

You also lose $a^{bc} = (a^b)^c = (a^c)^b$ but you lost tat with negative bases. However as we no longer have order, we can't specify a condition that $a$ be real. We lose that for $a \in \mathbb R; a > 1$ that $\log_a x$ is a single value function. – fleablood Apr 08 '18 at 20:40

2@fleablood Wait, we are allowed to interchange powers with negative bases, e.g. ((2)^3)^2=((2)^2)^3 – Юрій Ярош Apr 08 '18 at 20:45

13@Angew  You only lose the physicists and engineers once you get to quaternions or octonions, generally. – Obie 2.0 Apr 09 '18 at 08:32

5Related : https://math.stackexchange.com/questions/240959/whatdowelosepassingfromtherealstothecomplexnumbers – Arnaud D. Apr 09 '18 at 11:38

Simplicity?????? – Sentinel Apr 09 '18 at 15:02

[Related](https://math.stackexchange.com/q/86434/206300), possibly a duplicate. – mbomb007 Apr 09 '18 at 19:30

[Also related](https://english.stackexchange.com/q/234607/125966) – mbomb007 Apr 09 '18 at 19:38

@ЮрійЯрош No, we are not allowed to interchange powers with negative bases even in just the reals: $({(1)}^2)^½=1$ but $({(1)}^½)^2$ is undefined. – hvd Apr 10 '18 at 05:29

@hvd Yeah, you're right rational powers are undefined for negative bases. So we actually lost interchanging of powers with negative bases moving from integers to rationals. But with integer powers and negative bases we can interchange powers. – Юрій Ярош Apr 10 '18 at 12:01

Possibly related: https://math.stackexchange.com/questions/641809/whatspecificalgebraicpropertiesarebrokenateachcayleydicksonstagebeyon and https://math.stackexchange.com/questions/114864/whyarepropertieslostinthecayleydicksonconstruction – pregunton Apr 10 '18 at 12:23

@ЮрійЯрош Actually, I think interchanging of powers was lost even earlier, when moving from nonnegative integers to all integers: consider $(0^{1})^0$ vs $(0^0)^{1}$. But yeah, there are still plenty of situations where powers can be interchanged. – hvd Apr 10 '18 at 12:38

@hvd Yeah, you're right. And if you consider 0 as natural number, then interchanging of powers doesn't even appear. – Юрій Ярош Apr 10 '18 at 12:40
7 Answers
The most important ones as I see it:
 Naturals to integers: lose wellorderedness, gain "abelian group" (and, indeed, "ring").
 Integers to rationals: lose discreteness, gain "field".
 Rationals to reals: lose countability, gain "Cauchycomplete".
 Reals to complexes: lose a compatible total order, gain the Fundamental Theorem of Algebra.
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1@Somos You mean a nontrivial automorphism? I don't really have enough experience to know how interesting that is :) – Patrick Stevens Apr 08 '18 at 21:32

Yes, well, conjugation carries over to quaternions and octonions and is closely associated with the norm. – Somos Apr 09 '18 at 00:48

1+1, but "wellordering" means there's a smallest element, right? I had to think that out for a bit. – No Name Apr 09 '18 at 04:59

9@NoName not quite, wellordering means that every nonempty subset has a smallest element. For example, the set $\{0\}\cup (1,2)$ has a smallest element, but is not wellordered. – ajd Apr 09 '18 at 05:10


6Nice and concise answer  while you're at it you could provide a "sneak preview" into Quaternions (Minkowskiish but no longer Abelian) and Octonions (not even associative, but I never learned about an advantage...) ;) – Tobias Kienzler Apr 09 '18 at 07:16

@Somos You are referring to multiplying by the [complex conjugate](https://xkcd.com/849/), right? ;) – chtz Apr 09 '18 at 16:16

@chtz Yes, multiplying a number by its conjugate gives the norm. That is, the sum of squares of the component coefficients. Of course, for real numbers, the conjugate is just the number itself. – Somos Apr 09 '18 at 18:59

1@TobiasKienzler Agreed. For completeness, I'd like to see what is lost from Complexes to Quaternions to Octonions. If you read the [Wikipedia page for Octonions](https://en.wikipedia.org/wiki/Octonion), there may be some advantages/uses described. They "satisfy a weaker form of associativity, namely they are alternative... Additionally, octonions have applications in fields such as string theory, special relativity, and quantum logic." – mbomb007 Apr 09 '18 at 19:26

7When moving from Complex $\Bbb C$ to Octonions $\Bbb O$, the numbers gain the property of being "onions". This is lost upon moving to the Sedenions $\Bbb S$. – mbomb007 Apr 09 '18 at 19:36


1Strictly speaking, from rationals to reals you *regain* Cauchy completeness. The integers are Cauchycomplete, for the reason that any integer Cauchy sequence is eventually constant. – celtschk Apr 10 '18 at 06:53

1Weird. I always thought you gain Dedekindcompleteness when moving from the rational numbers to the real numbers. – Asaf Karagila Apr 10 '18 at 08:01


1@mbomb007 At least they remain "ions", though no sign of charge is mentioned... – Tobias Kienzler Apr 10 '18 at 13:22

@ypercubeᵀᴹ They are the "complex numbers". They aren't referred to as "complexes" AFAIK, unlike the integers, rationals, reals, quaternions, etc. – mbomb007 Apr 10 '18 at 14:01

@mbomb007 the answer has (bullet 4): *"Reals to **complexes**: lose a ..."* I wondered because I've seen such use. – ypercubeᵀᴹ Apr 10 '18 at 14:29

@ypercubeᵀᴹ That's one person. If you search for it, you will find few relevant results because "complexes" is a word for many things. It's not used colloquially. – mbomb007 Apr 10 '18 at 14:37
The most important property you loose when moving from real to complex numbers is definitely the notion of an order, i.e. $\mathbb{R}$ is an ordered field whereas $\mathbb{C}$ is not. This follows from the following proposition (Abstract Algebra by P.A. Grillet):
A field $F$ can be ordered if and only if $1$ is not a sum of squares of elements of $F$.
Moreover, we have to be careful in defining certain functions due to the fact that now even standard functions turn out to be multivalued rather than singlevalued. However, this is paid back immediately by the nice differentiability properties of complex differentiable functions.
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4IMO “the nice differentiability properties” are a huge _downside_ of complex numbers. Differentiability becomes such a strong property that it's almost impossible to define a differentiable function, short of writing down a single closed expression for it. – leftaroundabout Apr 09 '18 at 14:13

1On a side note: Should an entity which cannot be ordered still be called number? I sometimes feel a better name would be Duonions ( in analogy to Quaternions ). – asmaier Apr 10 '18 at 06:54

@asmaier but complex numbers arise as solutions to problems involving only real numbers, such as roots of polynomials. If the solutions of $x^2=1$ are numbers, how can the solutions of $x^2=1$ not be? – nigel222 Apr 10 '18 at 12:32

I can also ask the question, for which numbers $i,j,k$ the following equation is true: $a^2+b^2+c^2+d^2 = (a+ib+jc+kd)(aibjckd)$ . Answer: The "numbers" must obey the following relations: $ijk=1$ and $ik=k$, $ji=k$ , etc. But we don't call $i,j,k$ numbers (these symbols don't even commute), we call them quaternions. So it is very well possible to write down equations using numbers, which cannot be solved by ordinary numbers. – asmaier Apr 10 '18 at 20:32

@asmaier The name *duonions* is actually quite funny (makes me cry). But arguing what is and what is not a number is, at least I think, a topic for another post and there are a few post on SE covering this issue. However, I am with you, and calling the reals "the numbers" and nothing else. – TheGeekGreek Apr 10 '18 at 21:22

@leftaroundabout That may be true. Nonetheless, I think that it is slightly more of a gain than of a loss. However, you comment made me think about real differentiable functions which are not standard ones like polynomials, rational or trigonometric/exponential functions and their compositions. So you claimed that it is almost impossible to define differentiable functions, how would they look in the real case? Because I think the ones I mentioned are the most used ones and for many things sufficient. – TheGeekGreek Apr 10 '18 at 21:28


1@asmaier There isn't a definition of "number" as far as I know, so you can call quaternions "quaternionic numbers" if you like, in analogy with "complex numbers". An important distinction though is that $\mathbb{C}/\mathbb{Q}$ is a field extension, whereas the quaternions are only a $\mathbb{Q}$algebra. So if you think of "numbers" as a field, then that includes complexes and excludes quaternions. – Doris Apr 11 '18 at 07:00
Losing order is the most important but we also lose that if $b > 1$ then $b^z$ is injective so $\ln z$ (or $\log_b z$) is no longer a function but an equivalence class.
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3More generally, Riemann surfaces become a thing. I'd argue that this is something gained, rather than lost (though of course if you project onto the plane then you lose information again). – Patrick Stevens Apr 08 '18 at 20:43
Comparing naturals to integers, there is a smallest natural (0 or 1) which often makes solving problems easier.
Comparing reals to complex, you can always compare reals but there is no complete ordering of the complex numbers.
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3And oftentimes in proofs involving the complex numbers people will take the modulus of their numbers in order to use the order structure on the reals to prove what they want to, and similarly during proofs concerning integers, they'll take absolute values in order to use the fact that every set of naturals has a least element. Actually, this reasoning goes even farther because in any Euclidean Domain [https://en.wikipedia.org/wiki/Euclidean_domain] we're mapping into the naturals in order to use their properties. – Jordan Hardy Apr 08 '18 at 20:44

4Another property of the integers which is surprisingly useful is that if $a \ne b$ then $ab \ge 1$. – marty cohen Apr 09 '18 at 05:05
The biggest thing you lose when you move from the reals to the complex numbers is the ordering. You can, of course, find some ordering on the complex numbers, but the ordering will have nothing to do with the algebraic structure.
On the real numbers, if $a < b$, and $c$ is positive, then $ac < bc$. And for any numbers at all we have $a < b$ if and only if $a + c < b + c$. You can't build an ordering on the complex numbers that obeys the same properties (You can prove this. The easiest way is to assume i is positive, and find a contradiction, then assume i is negative and find a contradiction).
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Many people have said the crucial thing is order. I'd defend another fact we lose: that a number is selfconjugate. It's not obvious this matters, since we don't bother defining a "conjugate" of a real. But there's a reason I bring it up. The CayleyDickson construction can be thought of as a dimensiondoubling operation on algebras with involution, where if we start with $\mathbb{R}$ we must equip to it the identity map as our "conjugation". Suppose we double from $A$ to $A^\prime$; then there's a nice equivalence between certain properties of $A$ and others of $A^\prime$. As noted in propositions 14 here:
 The involution on $A^\prime$ is distinct from the identity;
 Iff this is true also of $A$, $A^\prime$ doesn't commute;
 Iff that is true also of $A$, $A^\prime$ doesn't associate;
 Iff that is true also of $A$, $A^\prime$ doesn't alternate.
In other words, the fact that $\mathbb{C}$ isn't selfconjugate explains all the famous later losses of nice properties, such as $\mathbb{H}$ not commuting.
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$\mathbb C$ is more rigid. Holomorphic functions are analytics. This means complex manifolds are polynomiallike and more akin to algebraic varieties than to real manifolds. By contrast $\mathbb R$ manifolds can be glued together using functions like $\exp(1/x^2)$ that can $C^\infty$smoothly transition from one function to another.
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