# Q: Is there a number set that is “above” complex numbers?

Physicist: Yes, but they don’t fix problems the way the complex numbers do.

The nice thing about real numbers (which includes basically every number you might think of: 0, 1, π, -5/2, …) is that no matter how you add, subtract, multiply, or divide (other than 0) them together, you always get another real number.  This property is called being “closed”.  A mathematician would say “the real numbers are closed under addition, because any real numbers added together always give you another real number”.

Closed-ness is comforting to have, because it means that when you’re doing basic math, no matter how you jump you’ll always have somewhere to land.  Mathematically speaking.

However!  When you’re doing square roots the real numbers are not closed.  When you ask “$\sqrt{x}=?$” what you mean is “$x=?^2$“.  For example, to find $\sqrt{4}=?$, you just answer the question, $?^2 = 4$, and find that the answers are ? = 2 or -2.  But if you try the same thing with $\sqrt{-4}$ you’ll be trying to answer the question $?^2 = -4$, which doesn’t have any answers (try it).  To “solve” this problem Euler decided to make up a new “number” called “$i$“, with the property that $i^2=-1$, and complex numbers were born.

Every complex number can be written “A+Bi”, where A and B are regular numbers. Notice that when B=0 you’ve got a regular, real number, so the complex numbers include the real numbers.

So here’s where the question comes into play.  $i$ may patch the problem with $\sqrt{-1}$, but does it just give rise to a new problem when you try to figure out what $\sqrt{i}$ is?  Turns out: no!

$\sqrt{i} = \pm(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i)$

You can check this by squaring it:

$\begin{array}{ll}\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i\right)^2\\\smallskip=\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i\right)\left(\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}i\right)\\\smallskip=\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}i^2+\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}i+\frac{1}{\sqrt{2}}\frac{1}{\sqrt{2}}i\\\smallskip=\frac{1}{2}+\frac{1}{2}i^2+\frac{1}{2}i+\frac{1}{2}i\\\smallskip=\frac{1}{2}-\frac{1}{2}+\frac{1}{2}i+\frac{1}{2}i \\\smallskip=i\end{array}$

Weirdly enough, there is absolutely no combination of roots/exponentiations or multiplications/divisions or additions/subtractions that can break out of complex numbers.  Where the closed-ness of real numbers fail, complex numbers hold strong.  This is one of the important aspects of the “fundamental theorem of algebra“.  You can tell mathematicians think it’s important, they don’t call just anything the “fundamental theorem of whatever”.

Finally, here’s the answer, there are a lot of (infinite) number-systems bigger than the complex numbers that contain the complex numbers in the same way that complex numbers contain the real numbers.  However, they’re not “needed”.

The smallest number system that’s bigger than the complex numbers is the “quaternions”.  The real numbers can be built from “1” and then seeing what you can get from any combination of adds, multiplies, etc. (and then filling in the gaps).  Complex numbers can be built the same way, starting with “1” and “i”.  Quaternions are built out of 1, i, j, and k.  i, j, and k all do basically the same thing that i does in complex numbers; $i^2=j^2=k^2=-1$.  In addition, ij=k, jk=i, ki=j, and if you flip the order you flip the sign, so ji=-k.

Quaternions don’t “patch holes” that the complex numbers have, but they do help with some very complicated problems that other number-systems can’t handle easily.  To head off the next obvious question; yes, there are even larger number-systems, like the “octonians“, and inventing ever higher systems is easy enough.

The graph is a picture from here.

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### 10 Responses to Q: Is there a number set that is “above” complex numbers?

1. Alyssa says:

“To “solve” this problem Euclid decided to make up a new “number” called “i“, with the property that i^2 = -1, and complex numbers were born.”
Shouldn’t that be “Euler”? Or is my math history hilariously wrong?

2. The Physicist says:

It’s hilariously right!
That was a massive brain-fart on my part.

3. One particularly widespread usage of Quaternions is for handling 3D rotations in computer graphics. Using individual rotational matrices for each rotational axis can result in gimbal lock, cutting out one rotational axis. Quaternions, when used properly, can represent all possible 3 dimensional rotations in the form “rotate x degrees around an arbitrary axis N”, thus eliminating the possibility of gimbal lock (since the rotation is represented as a single transformation) and usually being more computationally efficient.

There is another plane of numbers that are orthogonal to the complex numbers (or at least form their own distinct number plane from complex numbers) known as dual numbers, which have found use (or perhaps abuse) in rendering fractals in graphics (by utilizing their automatic differentiating properties), and have their own equivalent quaternions. Wikipedia claims they are used in physics, so perhaps you are already familiar with them, but regardless, I thought I would mention a few examples of real-world problems that can be elegantly solved with number systems that would otherwise be of questionable practical use.

4. Complex numbers weren’t originally needed to solve quadratic equations, but higher order ones.

If a polynomial has no real roots, then it was interpreted that it didn’t have any roots (they had no need to fabricate a number field just to force solutions). However, a notion of complex numbers was needed to solve real roots of polynomials of order >2. That is, it was used as an artifact of accounting in the same way that negative numbers are used. For example, the equation y^3=p*y+q (a “depressed cubic”) was known to have a simple formula for calculating roots:

y=((q/2+sqrt((q/2)^2-(p/3)^2)))^(1/3)+((q/2-sqrt((q/2)^2-(p/3)^2)))^(1/3)

(Found by del Ferro and Tartaglia). Consider the depressed cubic y^3=15y+4. It is easy to check that this cubic has a real solution at y=4. However, when plugging in p=15 and q=4 in the displayed formula, you get an expression involving complex numbers. Of course, the imaginary part vanishes after the arithmetic is done so that one obtains the real solution y=4.

As far as how quaternions aren’t “needed,” that is a highly subjective (if not plain wrong) assessment. Maxwell used quaternions to develop classical electromagnetism as we know it. And in many ways the quaternion formulation may be superior than its vector counterpart.

5. Will says:

Heh, I remember trying to get my head around Quarternions during my Graphics Design course.

I eventually concluded it was more trouble than it was worth and decided to leave them alone in their black box to do my rotations 😛

6. sha says:

hi
thanks
hossein from iran

7. Mike says:

FYI, according to a Princeton University blog, Euler popularized complex numbers, but before him Descartes first used the term “imaginary” in 1637, and before that Bombelli defined them in 1572, and before that Cardano invented them in around 1545 (and long before that Heron of Alexandria first observed them in the first century AD).

8. Zacharie says:

Ok but I do believe thats not quite correct, for example, 2^(sqrt(2)) = ? Now from what I know this number cannot be written in the form a + bi, because it is non-algebraic. But I’m probably wrong since this isn’t my field of maths, so yeah.

9. SuperSupermario24 says:

@Zacharie:
a and b don’t have to be algebraic, you can easily have e + πi if you wanted