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?
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41$\begingroup$ You are losing the existence of an order compatible with the field. $\endgroup$– SerserCommented Apr 8, 2018 at 20:35
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17$\begingroup$ We loose the total ordering relation. $\endgroup$– userCommented Apr 8, 2018 at 20:35
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2$\begingroup$ @fleablood Wait, we are allowed to interchange powers with negative bases, e.g. ((-2)^3)^2=((-2)^2)^3 $\endgroup$– Юрій ЯрошCommented Apr 8, 2018 at 20:45
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13$\begingroup$ @Angew - You only lose the physicists and engineers once you get to quaternions or octonions, generally. $\endgroup$– Obie 2.0Commented Apr 9, 2018 at 8:32
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5$\begingroup$ Related : math.stackexchange.com/questions/240959/… $\endgroup$– Arnaud D.Commented Apr 9, 2018 at 11:38
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7 Answers
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The most important ones as I see it:
- Naturals to integers: lose well-orderedness, gain "abelian group" (and, indeed, "ring").
- Integers to rationals: lose discreteness, gain "field".
- Rationals to reals: lose countability, gain "Cauchy-complete".
- Reals to complexes: lose a compatible total order, gain the Fundamental Theorem of Algebra.
answered Apr 8, 2018 at 20:36
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1$\begingroup$ @Somos You mean a nontrivial automorphism? I don't really have enough experience to know how interesting that is :) $\endgroup$ Commented Apr 8, 2018 at 21:32
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10$\begingroup$ @NoName not quite, well-ordering means that every nonempty subset has a smallest element. For example, the set $\{0\}\cup (1,2)$ has a smallest element, but is not well-ordered. $\endgroup$– ajdCommented Apr 9, 2018 at 5:10
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7$\begingroup$ Nice and concise answer - while you're at it you could provide a "sneak preview" into Quaternions (Minkowski-ish but no longer Abelian) and Octonions (not even associative, but I never learned about an advantage...) ;) $\endgroup$ Commented Apr 9, 2018 at 7:16
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8$\begingroup$ When 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 $\Bbb S$. $\endgroup$– mbomb007Commented Apr 9, 2018 at 19:36
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3$\begingroup$ Weird. I always thought you gain Dedekind-completeness when moving from the rational numbers to the real numbers. $\endgroup$– Asaf Karagila ♦Commented Apr 10, 2018 at 8:01
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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 multi-valued rather than single-valued. However, this is paid back immediately by the nice differentiability properties of complex differentiable functions.
answered Apr 8, 2018 at 20:35
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4$\begingroup$ IMO “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. $\endgroup$ Commented Apr 9, 2018 at 14:13
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1$\begingroup$ On 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 ). $\endgroup$– asmaierCommented Apr 10, 2018 at 6:54
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$\begingroup$ @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? $\endgroup$– nigel222Commented Apr 10, 2018 at 12:32
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$\begingroup$ I can also ask the question, for which numbers $i,j,k$ the following equation is true: $a^2+b^2+c^2+d^2 = (a+ib+jc+kd)(a-ib-jc-kd)$ . 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. $\endgroup$– asmaierCommented Apr 10, 2018 at 20:32
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1$\begingroup$ @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. $\endgroup$– DorisCommented Apr 11, 2018 at 7:00
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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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3$\begingroup$ More 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). $\endgroup$ Commented Apr 8, 2018 at 20:43
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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.
answered Apr 8, 2018 at 20:37
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3$\begingroup$ And 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 [en.wikipedia.org/wiki/Euclidean_domain] we're mapping into the naturals in order to use their properties. $\endgroup$ Commented Apr 8, 2018 at 20:44
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5$\begingroup$ Another property of the integers which is surprisingly useful is that if $a \ne b$ then $|a-b| \ge 1$. $\endgroup$ Commented Apr 9, 2018 at 5:05
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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 self-conjugate. 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 Cayley-Dickson construction can be thought of as a dimension-doubling 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 1-4 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 self-conjugate 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 polynomial-like 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.
