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Real number is often used to represent a point in a 1-dimensional number line. Real numbers are written as $a$ where $a \in \mathbb{R}$.

Complex number is often used to represent a point in a 2-dimensional plane. Complex numbers are written as $a + bi$ where $a,b \in \mathbb{R}$ and $i^2 = -1$.

What about higher dimensional space?

We can extend the concept of complex number, and write a point in a 3-dimensional space as $a + bi + cj$ where $a,b,c \in \mathbb{R}$, but how do we define i and j?

Certainly, this also applies to 4-dimensional space. We can write a point in 4D as $a + bi + cj + dk$, how do we define $i,j,k$?

How about even higher n-dimensional space? Is there a system to define i,j,k,... that will extend to arbitrary dimensions? Is there a name for it?

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Very related: Is there a third dimension of numbers? – Rahul Sep 12 '11 at 21:25
This is a very important question! You should read the wikipedia page on quaternions, which give the 4D solution. From there you will find links to discussion of other dimensions. The answers are intricate and rather surprising! – Oliver Sep 12 '11 at 21:42
What kind of structure do you want on these numbers? – Qiaochu Yuan Sep 12 '11 at 22:09

It depends upon what structure you want. In $\mathbb{R}^n$ we do not have the operation of multiplying two elements, as we do on $\mathbb{C}$. If you want that, you probably want a normed division algebra, which are available in four and eight dimensions, but no more.

If you just want a vector space they are basis vectors but you can't multiply two elements.

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For four dimensions there are the quaternions, which saw widespread use to represent 3D vectors for several decades in the 1800s until people became comfortable the idea of adding things (vectors) that cannot be multiplied ... – Henning Makholm Sep 12 '11 at 21:22

Just for kicks, here's a direct argument that you can't extend $\mathbb{C}$ to something three dimensional where you can still multiply.

Start with your reals, $i$, and $j$ as before. The question is, what should $ij$ be?

Since it will be something in your three dimensional thing, you can express it as $ij = a + bi + cj$ for some real numbers $a$, $b$, and $c$. Multiplying everything on the left by $i$ gives $$-j = ai - b + cij $$ and re-expanding $ij$ then gives \begin{align*} -j &= ai -b +c(a+bi + cj) \\ &=ai-b+ac+cbi+c^2j\\ &= (ac-b) + (a+bc)i +c^2 j\end{align*} and now equation coefficients gives $ -1 = c^2$, contradicting the fact that $c$ was real.

Thus, if you extend to a three dimensional setting, you are not allowed to multiply $i$ and $j$. To me, this makes the quaternions (a 4 dimensional setting mentioned in the other answers) much more surprising!

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You should totally look about Quaternions,Octonions and all other sets defined as the same way as them. Quaternions are used to represent numbers in a 4 dimensional space whereas Octonions are used in 8 dimensional space. You could also look to my question and it's answer about numbers defined on a 4 dimensional space.

EDIT : you can also add the Sedenions to the list, numbers on a 16 dimensional space. These kind of numbers, for $n$ dimensions can be written as follows : $$\alpha_1 +\alpha_2e_1+\alpha_3e_2+...+\alpha_{n}e_{n-1}$$ where $\alpha_n\in\Bbb R$ and $e_n^2=-1$ and for all number a and b, $a≠b, e_a≠e_b$. All the $e_n$s are said to be the elementary units of the set. So you can represent these numbers in a $n$ dimensional space using the the coefficients $\alpha_n$. For complex numbers, $n=2$ and $e_1=i$. For Quaternions, $n=4$ and $e_1=i, e_2=j, e_3=k$.

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Over 20 years ago I found the way to generalize the complex numbers to higher dimensions.

Basically it is very simple: on \mathbb{R}^3 you define an imaginary unit j via stating j^3 = -1.

Your complex numbers now look like X = x + yj + zj^2.

During the last 1.5 years I have picked up studying this stuff again, some of the results can be found at a page on my website:

Compared to the complex plane the 3D complex numbers form a far more complicated structure; so not only university freshmen can learn from it, the same goes for retired math professors...

A nice starter exercise would be the next:

Use the above definition of 3D complex numbers (with j^3 = -1) and proof that the equation X^2 = -1 has no solution inside \mathbb{R}^3

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If we take complex number in 3d plan than we found 2 condition, let we take 3 axis x y and z:

  1. x is real and yand z are imaginary, that is: $x+i(y+z)$

  2. x and z is real but y is imaginary, that is: $x+z+iy$

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This is a mapping from $\mathbb{R}^3\mapsto\mathbb{C}$, but it does not define any structure on $\mathbb{R}^3$. – robjohn Nov 2 '12 at 13:17

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