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)`

, from memory.^{2}-> a^{2}+ 2abi - b^{2}But I’m wondering if there are

othermathematical 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.

**Answer**

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)=x2+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 C! One example is the *Quaternions*. Where the complex numbers can be visualized as a plane (i.e. a+bi↔(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 i2=j2=k2=−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.

**Attribution***Source : Link , Question Author : Community , Answer Author : Simon Rose*