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I've been teaching my 10yo son some (for me, anyway) pretty advanced mathematics recently and he stumped me with a question. The background is this.

In the domain of natural numbers, addition and multiplication always generate natural numbers, staying in the same domain.

However subtraction of a large number from a smaller one needs to "escape" into the domain of integers, and division may result in escape to the real domain (like 3 / 5 -> 0.6).

It was a simple step from there to taking the square root of a negative number, hence requiring the escape into the complex domain, such as 4+7i.

He quite easily picked up that each of these domains was a superset of another, natural -> integer -> real -> complex.

However, he then asked if an operation on a complex number would require yet another escape, a question I had to investigate. Now, it turns out that the square root of a complex number is simply another complex number along the lines of mathematical distribution: (a+bi)2 -> a2 + 2abi - b2, from memory.

But I'm wondering if there are other mathematical operations performed on complex numbers (or any of its subset domains) that can't be represented within the complex domain.

Apologies if I've used the wrong terms, it's been about 30 years since I did University level math.

Brian Minton
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    Except for dividing a number by $0$ and raising $0$ to the power of $0$, any operation on any complex number yields a complex number. – barak manos Jan 07 '15 at 08:01
  • However, there exist the domains of superreal, hyperreal and surreal numbers, and I would think (but I can't guarantee for superreals and surreals) they can be extended to contain $\mathbb {C} $. – Vincenzo Oliva Jan 07 '15 at 08:08
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    There are also the Quarternions ($z=a+ib+jc+kd\equiv(a,b,c,d)$), Octonions, and so on, each of which is a generalisation of the complex numbers. Each generalisation, however, loses some nice property of the "previous" set of numbers, e.g. commutatvity, and so on. I don't recall the exact details. – gone Jan 07 '15 at 08:10
  • Yes, there do exist domains larger than the complex numbers. For example the [Quaternions](http://en.wikipedia.org/wiki/Quaternion). They are a superset of the complex numbers just in the same way as complex numbers are a proper superset of the reals. Of course there are operations within the Quaternions that would take you out of the complex numbers but in my opinion these operations don't come as naturally as the sqare root of negative numbers. – Chris Jan 07 '15 at 08:13
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    pbs is talking about the [Cayley-Dickson construction](http://en.wikipedia.org/wiki/Cayley%E2%80%93Dickson_construction): this is (one of) the construction(s) that builds the complex numbers from the real numbers. It can be repeated to build the quaternions (from the complex numbers), and then the octonions (from the quaternions). And as pbs said, after each step, one loses an algebraic property. – Jesse Madnick Jan 07 '15 at 08:14
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    Congratulations for making your tutoring engaging enough to provoke your son's question. – guest Jan 07 '15 at 08:25
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    I was going to criticize for using the unclear "larger than," but then I saw that not only did the body specify what you meant, in terms of proper supersets, but it is overall a very well-formed and -presented question. Also, if your 10-year-old is already learning about complex numbers (and asking this sort of insightful question), then he might soon be ready for a discussion about what is meant by one set being "larger" than another. For example, $\Bbb{N}\subset\Bbb{Z}\subset\Bbb{Q}$, but they are all the same size (cardinality). This counter-intuitive fact is a great point of inquiry. – KSmarts Jan 07 '15 at 18:42
  • Since no one has mentioned it yet, I'll also link to Riemann surfaces, which expand the complex plane in ways so that you can, for example, take a sensible natural logarithm. http://en.wikipedia.org/wiki/Riemann_surface – MartianInvader Jan 07 '15 at 23:54
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    There is a little bit of imposture. From integers to rational numbers, from rational to algebraic numbers and from real to complex numbers — it is so to say "algebraic way to enlarge a domain". But there is a step from algebraic numbers (or from rational numbers) to real numbers which is not, it is more like "topological way to enlarge". – sas Jan 08 '15 at 02:15
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    The answer to this question—that for any algebraic operation, the solution is still a complex number—is known as the Fundamental Theorem of Algebra. It was Gauss's favorite thing. – Nick Matteo Jan 08 '15 at 15:59
  • @KSmarts, I modified the title, hope that's better. –  Jan 09 '15 at 02:06
  • @barakmanos *Any* operation? How do you know that `flambozzle(1 + 3i)` is a complex number? (where `flambozzle` is defined as `flambozzle(x) = x + jx^2`, so it's actually not a complex number) – user253751 Jan 09 '15 at 06:41
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    One minor point: division does not escape from the integers to the real numbers, but only from the integers to the rationals. The real numbers are larger than the rationals. – David Conrad Jan 09 '15 at 20:57
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    And for more precision, don't forget the algebraic numbers. natural → integer (subtraction) → rational (division) → algebraic (roots)... Beyond that it gets tangled; you can have algebraic complex numbers, whose real & imaginary components still don't reach the transcendentals. – Foo Bar Jan 10 '15 at 23:19
  • I'm surprised nobody has talked about projective numbers (and projective or inversive geometry), or the extended real numbers. (I would have, but I haven't found the time yet) –  Jan 11 '15 at 17:55
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    I'd like to thank everyone for educating me and mine. My son _sort_ of gets the quaternions and octonions (though he originally suspected they were eight-sided onions) but, of course, his next question was "where are the hexadecions?". I think that's probably enough for him to be getting on with for now, school's not even teaching him yet that square roots have _two_ solutions and I don't want him beaten up by his peers :-) Cheers, all. –  Jan 18 '15 at 11:49
  • Try and solve the equation $xy-yx=1$ in the complex numbers... – JP McCarthy Jun 08 '15 at 14:43
  • @barak manos this is simply wrong. First, $0^0=1$, second, $\log 0$ is not a complex number. – Anixx Feb 13 '21 at 16:53

12 Answers12

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First of all, kudos for introducing these ideas to your 10 year-old! That's an excellent way to get them interested in mathematics at an early age.

As to your question: The short answer is no. Any algebraic operation that you can do of the sort you're describing will yield a complex number. This is due to the fact that they are algebraically closed. What this means is the following.

One way that you can show that you can find a larger domain than the real numbers is by looking at the polynomial $$ f(x) = x^2 + 1 $$ You can easily see that there are no solutions to the equation $f(x) = 0$ in the real numbers (just as the equation $x + 1 = 0$ has no solutions in the positive integers). So we must escape to a larger domain, the complex numbers, in order to find solutions to this equation.

A domain (to use your term) being algebraically closed means that every polynomial with coefficients in that domain has solutions in that domain. The complex numbers are algebraically closed, so no matter what polynomial-type expression that you write down, it will have as solution a complex number.

Now, this isn't to say that there aren't larger domains than $\mathbb{C}$! One example is the Quaternions. Where the complex numbers can be visualized as a plane (i.e. $a + bi \leftrightarrow (a, b)$), the quaternions can be visualized as a four-dimensional space. These are given by things that look like $$ a + bi + cj + dk $$ where $i, j, k$ all satisfy $i^2 = j^2 = k^2 = -1$, and moreover $ij = -ji = k$. The interesting fact about the quaternions is that they are non-commutative. That is, the order in which we multiply matters!

There are also Octonions, which are even weirder, and are an 8-dimensional analogue.

Anyhow, the answer is in the end that it sort of depends. In most senses, the complex numbers are as far as you can go in a relatively natural way. But we can still look at bigger domains if we want, but we have to find other ways to build them.

Simon Rose
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    I was rather hoping the answer was a simple "no" :-) Maybe I'll leave it a while before I introduce him to this stuff, not least because it's going to take _me_ several years to digest it :-) –  Jan 07 '15 at 08:25
  • Fair enough! As a simple answer, "no" is perfectly fine and correct, from the context of your question. – Simon Rose Jan 07 '15 at 08:26
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    You might want to mention also sedenions, and hypercomplex numbers, namely numbers of the form $a+b\varepsilon + (c+\varepsilon) i $, where $ a, b, c $ are real and $\varepsilon $ is an infinitesimal. – Vincenzo Oliva Jan 07 '15 at 08:29
  • There's also the question of how you cut the complex plane (https://en.wikipedia.org/wiki/Branch_point) to turn an expression with multiple solutions into a function with a single value. This is in a loose sense an "escape", and does build surfaces with more structure than just a boring old complex plane. – Steve Jessop Jan 07 '15 at 10:55
  • ... but I wouldn't recommend just throwing that in there off-hand when teaching: it gets quite hard quite quickly and it's very easy to make confusing mistakes. That might be my own bias speaking though, since it's something I was exposed to but never pursued. – Steve Jessop Jan 07 '15 at 11:04
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    Solutions of a polynomial?.. Shouldn't one use "root" or "zero" term, not "solution"? A polynomial isn't an equation or a problem to have a solution. – Ruslan Jan 07 '15 at 14:02
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    Dividing by 0 allows you to escape the complex domain, does it not? That is, it has no solution in the complex domain, but by extending to the Riemann sphere we find a solution. I mean, we lose a bunch of nice properties, but it seems to fit in the spirit of the question. – Aaron Dufour Jan 07 '15 at 14:48
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    Root may have been a better choice, fair, but I was trying as much as possible to shy away from technical language. As for considering the Riemann Sphere... I'd say that seems a little bit different than what is suggested, which is algebraic operations. And saying that $1/0 = \infty$ is not something that I'd like to toss around willy-nilly to people teaching 10-year olds, since this is more likely to be confusing than not. – Simon Rose Jan 07 '15 at 15:00
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    See also the [Fundamental Theorem of Algebra](https://en.wikipedia.org/wiki/Fundamental_theorem_of_algebra) – imallett Jan 10 '15 at 06:08
  • Maybe "solution of a polynomial equation" is better? – Paŭlo Ebermann Oct 07 '17 at 18:20
  • Bicomplex numbers (tessenaries) are commutative, so why did you chose quaternions that are non-commutative? – Anixx Feb 13 '21 at 16:55
  • @Anixx - the questioner did not define "domain" in a particular way, so of course it is up to me to interpret it however I would like. I assumed that part of what makes these extensions still useful is the ability to perform division, which you cannot universally do in the bicomplex numbers. – Simon Rose Feb 15 '21 at 08:17
  • Are complex numbers closed under super-roots and hyper-roots and so on operations? Because that's how we came out of real and landed on complex! – Sourav Kannantha B Jan 21 '22 at 12:49
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There are reasons to escape even from the complex number system, but perhaps not as nice and fruitful reasons as in the development from $\mathbb N$ to $\mathbb C$.

I recon any equation or system of equations in $\mathbb C$ without solutions, will have some solution for some extension of $\mathbb C$.

$z=z+1$ will have a solution in $\mathbb C^*=\mathbb C \cup \{\infty\}$, with some new arithmetic laws.

While

$ \left\{ \begin{array}{l} x^2=-1 \\ y^2=-1 \\ z^2=-1 \\ xy=z \end{array} \right. $

will have solutions in the Quaternions.

But both these cases of extensions demands significant changes of the rules. In the first case, e.g. the cancelling law has to be modified and in the second case commutativity is lost.


I'll guess almost no system of relations of the type where all parenthesis are stated, which has no solutions in $\mathbb C$ (that is, leads to contradiction in $\mathbb C$), can generate an "escape" in an associative algebraic extension of $\mathbb C$.

More to read on Wikipedia: Hypercomplex numbers

Lehs
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    Actually, if you define the inverse of zero, that's one route of escape. Then if you think of complex geometry as a 2D plane embedded in higher space, there are rotation operations that get you out. Geometric Algebra is an n-dimensional generalization from real through beyond octonions; and it's actually conceptually simpler in some ways. Conformal Geometric Algebra deals with what the inverse of zero actually means. http://slehar.wordpress.com/2014/07/24/geometric-algebra-conformal-geometry/ – Rob Jan 09 '15 at 00:04
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    I'd actually say you should consider the quaternion case a bit more seriously. If you think about it, each time you "escape" to a new number system, they are more generalized and you lose specifics. From natural numbers to integers you lose the notion of a smallest or "first" number. Integers to rational: you no longer have a finite "number of numbers" between any two points. From rational to real you lose the notion that all variables terminate. Real to complex, you lose the notion of greater than. Quaternion: commutativity is lost. Octonion: Associativity is lost; sedenions:zero divisors – user507974 Jan 11 '15 at 07:49
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    @user507974: So at some point do you lose everything? – user541686 Jan 11 '15 at 10:33
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    If my (very rudimentary) understanding of sedenions is right, it's already really hard to do math with it since multiplying two nonzero terms together can give you zero results. Flipping that logic around, not being able to say that x or y are zero when their product is zero is pretty debilitating for solving anything that isn't basic addition. There isn't much I can think of left for you to do at point. – user507974 Jan 11 '15 at 10:39
  • Bicomplex numbers (tessenaries) are commutative, so why did you chose quaternions that are non-commutative? – Anixx Feb 13 '21 at 16:56
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I can think of hyperreal numbers and surreal numbers.

The idea is to add infinitely large and infinitely small numbers to the reals. If you have seen the debate whether $0.999...$ and $1$ are the same, you can understand it is a natural concept to introduce. :-)

Hyperreal numbers were used in the elaboration of calculus but were abandoned in favor of the concept of limits. Surreal numbers are a "larger" set that has applications in game theory. A superset of complex numbers would be the surcomplex numbers.

Florian F
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There are lots of "bigger" domains. I want to point something out here:

Once you get to the level of the complex numbers, it becomes unclear what a number is. You can't put the complex numbers into any order that preserves the complex algebraic properties. And you can't really use them to count things either. In a funny way, the complex numbers are representations of rotations and scalings. That is definitely a funny kind of number.

So, with that in mind, you might want to start talking about linear algebra. Vectors and matrices are generalized numbers that represent linear transformations, like rotations, scaling, symmetric "flipping".

nomen
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    Well, I gather you could sort them using real as the primary key and imaginary as the secondary. Or you could sort them based on distance from the origin if you map them to a 2d graph (but then `7+9i` is "equal" to `9+7i` which is clearly ridiculous now that I think about it). But I see your point. I _do_ wonder how many people thought that way when negative numbers were introduced :-) –  Jan 08 '15 at 02:26
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    2D Graph example that is even more ridiculous: -1 is "equal" to 1, i, and -i – k_g Jan 09 '15 at 01:39
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Good question!

I like many of the answers so far. I'm a little surprised at one idea that's missing. The extensions you and your son are exploring don't really get to the real numbers.

Starting from the natural numbers, subtraction tells you to invent (discover?) the negative numbers. Division gets you to the rational numbers. If you want to find square roots (or, more generally, solve polynomial equations whose coefficients are the numbers you already have) you get to the algebraic numbers, some of which are complex. But none of those extensions take you to the transcendental numbers, like $\pi$. Those come about when you try to make sense of the claim that every point on a geometric line should have a numerical coordinate (once you've chosen points for 0 and 1).

Ethan Bolker
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The short answer is this:

  • Yes, the complex numbers are closed for polynomials using the algebraic operations of addition, multiplication, subtraction, and division.
  • Yes, there are larger domains than the complex numbers, such as the quaternions and octonians.

I think I have to share this video, which wonderfully describes the relationship between the real numbers, the complex numbers, the quaternions, and the octonians.

https://www.youtube.com/watch?v=uw6bpPldp2A

TL;DR:

  • The naturals, integers, and reals are not closed under normal algebra.
  • The complex numbers are, and are the simplest closed domain to include the natural numbers.
  • The quaternions are a closed superset of the complex numbers, but are not commutative.
  • The octonians are a closed superset of the quaternions, but are not associative.
Kendall Frey
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This is how I feel the number system developed not sure whether it is completely true

When we had Natural Numbers we wanted to solve Linear Equations but unfortunately we needed to solve equations of the form $ax+b=0$. But unfortunately not all equations have solutions. So they expanded the number system slowly from Natural Numbers to Integers and then to Rational Numbers.

Then something interesting happened, the extension of Rationals to Reals is I believe motivated more due to completion of space rather than algebraic closure. Then we expanded it complex numbers so that we can deal with $\sqrt {-1}$. Here I find it surprising that adjoining a single root of polynomial $x^2+1=0$ we are able to solve all the polynomials(this is a consequence of Fundamental Theorem of Algebra). So if you take any polynomial with coefficients in $\mathbb{C}$ it will have all its roots in $\mathbb{C}$ . So if operations you mean addition ,subtraction, multiplication, taking $n^{th}$ root or taking limits it will still be Complex

happymath
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    The extension of the rationals to the reals was motivated by geometry, although I suppose you could twist this into being about completion. – Kevin Arlin Jan 07 '15 at 08:55
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    Getting beyond the rationals in the first place is because of equations like $x^2 = 2$, which we think should have a solution because geometry tells us the solution is the length of a side of a particular triangle, but has no solution in the rationals any more than $2x = 1$ has an integer solution. But going all the way to the reals from the rationals is more than just adding solutions to algebraic equations, it is indeed about completeness. – Steve Jessop Jan 07 '15 at 11:00
  • @SteveJessop when I meant that it is motivated by completion I thought so because there are lot more numbers than just algebraic numbers though $x^2=2$ told us there are more but we added lot more than the algebraic closure suggested – happymath Jan 08 '15 at 05:07
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The Quaternions, Octonions etc mentioned above are not generally interesting for their arithmetic properties, but are very significant for other reasons, because they are at the root of a number of exceptional structures in mathematics, such as Lie Algebras with exceptional symmetry.

[I haven't repeated any notes on their construction]

The one proviso on this is that the quaternion product $$(a+bi+cj+dk)(w+xi+yj+zk)=$$$$(aw-bx-cy-dz)+(ax+bw+cz-dy)i+(ay-bz+cw+dx)j+(az+by-cx+dw)k$$ gives us, on taking norms, the identity $$(a^a+b^2+c^2+d^2)(w^2+x^2+y^2+z^2)=$$$$(aw-bx-cy-dz)^2+(ax+bw+cz-dy)^2+(ay-bz+cw+dx)^2+(az+by-cx+dw)^2$$

which can be used in the proof that every positive integer is a sum of four square numbers.

I wouldn't recommend the linked notes by John Baez to a 10-year-old, but the introduction gives an indication of the wider mathematical interest in this question.

Mark Bennet
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When my kids were around this age and we were talking algebra and numbers, they found infinities (and all the ways you can get to them from the reals and integers) fascinating. Any 10 y/o can identify with the idea of 'repeating forever' ;-) And it allows you to explore the ways that 'infinity' is, and is not, a 'number'. The paradoxes of countable infinities (e.g. that the even numbers have the same cardinality as the whole numbers) were a source of fascination for me as a child.

Quaternions & similar would have been meaningless to my kids, as they did not have familiarity with the underlying concepts and uses - nothing to ground them to.

The schools were not very friendly to math-forward kids. They'd come home and tell me how boring 'math' was - which turned out to mean, arithmetic practice. I told them "mathematics is an infinite palace. Arithmetic is just the parking lot out front."

Spike0xff
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There are the dual numbers, numbers of the form $a+b\epsilon$ (that's the Greek letter epsilon), where $a$ and $b$ are real (or complex) numbers.

The defining property of these is that $\epsilon\neq0$, but $\epsilon^2=0$. Granted, these are a lot less useful than complex numbers (especially since there's no such thing as $\frac1\epsilon$), which is why they are less well known.

By the way: Even though you can't always divide by $\epsilon$, you can cancel it out of an equation:$$a\epsilon=b\epsilon\iff a=b\quad(a,b\in\mathbb R)$$

By the way, here's a slightly interesting thing: If $p(x)$ is a polynomial (and $x$ is real), than $p(x+\epsilon)=p(x)+p'(x)\epsilon$, where $p'(x)$ is another polynomial that depends on $p$. (Explicitly: If $p(x)=a_nx^n+a_{n-1}x^{n-1}+\dotsb+a_1x+a_0$, then $p'(x)=na_nx^{n-1}+(n-1)a_{n-1}x^{n-2}+\dotsb+a_1$. If you know calculus, you'll recognize this as the derivative of $p$.)
You can use this to prove some interesting properties of the $'$ thing:

$(p\pm q)'=p'\pm q'$

$(pq)'=pq'+p'q$

Using the related fact that $p(x+a\epsilon)=p(x)+p'(x)a\epsilon$ (prove this!), you can prove that:

$(p(q(x))'=q'(x)p'(q(x))$

Akiva Weinberger
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I'm wondering if there are other mathematical operations performed on complex numbers (or any of its subset domains) that can't be represented within the complex domain.

Maybe it's a bit of a stretch, but I am surprised this operation, which opens up a whole new world, has not been mentioned yet:

$$\int (\cdot) dx$$

Let me expand a bit. I will more or less copy-paste the same idea three times to highlight the similarity in ideas.

  • When you play with $\mathbb N_0$, you notice that a) you can add and multiply, b) there is a mildly interesting "successor" operator $S: \mathbb N_0 \rightarrow \mathbb N_0$ with $S(0)=1, S(3)=4$, $S(17) =18$ etc. You notice that almost all numbers have predecessors (are in the image of $S$), but $0$ is not. So you just $$\text{invent a new thing, called } -1,$$ such that $S(-1)=0$. It feels a bit like cheating. You wonder if this $-1$ also needs a predecessor. However, since also you want to compatibly extend your rules of addition and multiplication, you'll notice that you'll need $-2 := 2(-1), -3:= 3(-1)$ etc. anyway, and it turns out that if you define that $-2$ as predecessor of $-1$, $-3$ as predecessor of $-2$ etc., indeed, suddenly you have constructed a new structure which is a) still closed under addition and multiplication, and b) now your operator $S$ is surjective. You have just invented $\mathbb Z$.

  • When you play with $\mathbb Z$, you notice that a) you can add, subtract, and multiply, and b) for each $k \in \mathbb Z$, there is a mildly interesting "scaling" operator $M_k: \mathbb Z \rightarrow \mathbb Z$ which is linear and satisfies $M_k(1)=k$. You notice that for most $k$, some numbers are in the image of $M_k$, but many, including $1$, are not. So you just $$\text{invent a new thing, called } 1/k,$$ such that $M_k(1/k)=1$. It feels a bit like cheating. You wonder whether $1/k$ also needs a new preimage under $M_k$. However, since also you want to compatibly extend your rules of addition, subtraction, and multiplication, you'll notice that you'll need $(1/k)^2$, $(1/k)^3$ etc. anyway, which give more preimages under $M_k$. It turns out you have constructed a new structure which is a) still closed under addition, subtraction, and multiplication, and b) now your operator $M_k$ is surjective. You have just invented the localisation $\mathbb Z_{(k)} = \mathbb Z [1/k]$. If you do this for all $M_k$ (all prime $k$ suffice), you have invented $\mathbb Q$.

  • When you play with $\mathbb R$, you notice that a) you can add, subtract, and multiply, and b) there is a mildly interesting "square" operator $\square: \mathbb R \rightarrow \mathbb R$ with $\square(-2.5)= 6.25, \square(\pi)=\pi^2$, $\square(17) =289$ etc. You notice that many numbers have square roots (are in the image of $\square$), but many, including $-1$, are not. So you just $$\text{invent a new thing, called } i,$$ such that $\square(i)=-1$. It feels a bit like cheating. You wonder whether $i$ also needs a square root. However, since also you want to compatibly extend your rules of addition, subtraction, and multiplication, you'll notice that you'll need $3+4i$, $-\pi-7.43i$, $\frac12\sqrt2(1+i)$ etc. anyway, and it turns out that indeed, suddenly you have constructed a new structure which is a) still closed under addition, subtraction, and multiplication, and b) now your operator $\square$ is surjective. You have just invented $\mathbb C$.

  • When you play with $\mathbb C$, you notice that a) you can add, subtract, and multiply, and b) there is a "differentiation" operator $D: \mathbb C \rightarrow \mathbb C$ which is linear and satisfies $D(ab) = aD(b)+bD(a)$. You notice that to be honest, this $D$ is rather dull as it just sends every number to $0$, i.e. any nonzero number does not have a preimage under $D$. So you just $$\text{invent a new thing, called } x,$$ such that $D(x)=1$. It feels a bit like cheating. You wonder whether $x$ also needs a new preimage under $D$. However, since also you want to compatibly extend your rules of addition, subtraction, and multiplication, you'll notice that you'll need $x^2$, $x^3$ etc. anyway, and you find that $D(\frac12 x^2)=x, D(\frac13 x^3)=x^2$ etc. It turns out you have constructed a new structure which is a) still closed under addition, subtraction, and multiplication, and b) now your operator $D$ is surjective. You have just invented the polynomial ring $\mathbb C[x]$. And if you also want to extend division, you will invent the field of rational functions $\mathbb C(x)$.

So I think this is at least an interesting additional perspective: that in each case we interpret the idea of the extension as making a certain operator surjective by "inventing" preimages for it. Certainly the operator $D$ feels a bit forced, as it is only obvious in hindsight that what we want to find preimages for is not just "the map that sends everything to $0$" but "the map which happens to send everything so far to $0$ because it satisfies the product rule of derivations." To further highlight the analogy, I should have e.g. defined the operator $S$ as "a map that satisfies $S(a)S(b) = S(ab)+a+b$ (which just happens to necessarily be the succesor function $S(n)=n+1$ on $\mathbb Z$).

The other thing that I glossed over here is the technical details of what it means to "compatibly" extend in each case.

As a last interesting remark, the construction of the polynomial ring as stated works over any characteristic $0$ field in place of $\mathbb C$; but if you do that over e.g. $\mathbb Z$, you'll find that you have to construct more than just the polynomials (because e.g. $x$ still does not have a preimage under $D$ in $\mathbb Z[x]$), and this time you more or less invent a divided power structure.

Torsten Schoeneberg
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Not exactly a superset, but you may want to look at homogenous coordinates (often used in projective geometry). Help you deal with what would otherwise be infinite numbers, and, in particular, points (and other geometric constructs) at infinity.

Think of a point infinitely far away, at the "edge" of the R2 plane. Say it's in the direction of the vector (3,2). You might want to identify it by the coordinates (3 infinities, 2 infinities) - sorry, I don't know how to type the lying figure 8. Now that might be intuitively nice, but it's arithmetically useless, as 3 infinities = 2 infinities = infinity. In homogeneous coords, your point becomes {3:2:0}, and you can do all sorts of maths with it.

Please send me your email address; I have some graphic materials you might find useful.

Xirdal
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    Learn how to paste the graphics here, if you are stuck find other examples of answers with graphics (and click on "edit" to see how they embed the graphics), of course don't actually edit anything – Squirtle Jan 13 '15 at 22:40