two-dimensional number system without negative numbers

Question

Is there any existing literature for the number system that looks like this?

Like the complex number system, this system exists on a plane. But instead of $i$ and $-1$, it has two numbers-- called $q$ and $d$ here-- that form two new families of numbers moving away from the origin at 120° and 240°, respectively, relative to the ray representing the non-negative real numbers. Three rays (the non-negative real numbers, the $q$ numbers, and the $d$ numbers) divide the plane equally into three regions (instead of four regions as in the regular complex numbers).

Instead of subtraction, this number system has two new operations: One to move 120° up by $x$ units (by adding $xq$) and the other to move 120° down by $x$ units (by adding $xd$). We can still technically do subtraction, though: $x-y = x+yq+yd$

The division operation $x/y = x*y^{-1}$ is also equal to $x*y^q*y^d$.

So we have $x + xq + xd = 0$ and $x * x^q * x^d = 1$.

Multiplication works as follows: \begin{array}{r|rrr} & 1 & q & d & \\ \hline 1 & 1 & q & d & \\ q & q & d & 1 & \\ d & d & 1 & q & \\ \end{array}

Of course, q and d exist in the complex number system as well:

$q=-0.5+i\frac{\sqrt{3}}{2}$

$d=-0.5-i\frac{\sqrt{3}}{2}$

Edit: Thanks to Will's comment, I learned that the number I called $q$ here is actually the Eisenstein unit $\omega$, while $d$ is $\omega^2$. Cool!

Edit: As Berci noted in the comments, if $a,b,c>0$, then $a + bq + cd$ can be rewritten as $a' + b'q + c'd$ where at least one of $a'$, $b'$, and $c'$ is $0$. This normalized form can be obtained by subtracting the minumum of $\{a, b, c\}$ from $a$, $b$, and $c$. Let's call this normalized form the canonical form.

If we restrict ourselves to canonical forms, I noticed that there's an analogue of the superellipse concept in this number system:

Any $x + yq + zd$ that is in canonical form is on a superellipse analogue if $x^n + y^n + z^n = 1$. (Let's not deal with semi-diameters for now.)

For $n=1$, the superellipse analogue is a triangle instead of a diamond.

For $n=2$, the superellipse analogue looks more like a Reuleaux triangle than a circle.

--

I'm not a mathematician, so that's all I could come up with given my very limited knowledge, and I would really like to know more. I am particularly interested in knowing how this can be generalized in higher dimensions. Notice that for any real number $x$, the points $x$, $xq$, and $xd$ form an equilateral triangle on the plane. I reckon you could generalize the system for any $n$-dimensional space by constructing a regular $n$-simplex centered at the origin and constructing $n+1$ rays emanating from the origin and passing through all vertices, but I don't know how multiplication would look like there aside from the vague idea that you could translate it from an appropriate Cayley–Dickson algebra.

Answer 1

Recall that the first octant in 3D is the set of points $(x,y,z)$ such that $x\geq 0$, $y\geq 0$, and $z\geq 0$. This intersects the XY, XZ, and YZ planes in quadrants. In particular:

• for XY: the set of points such that $x\geq 0$, $y\geq 0$, and $z=0$.
• for XZ: the set of points such that $x\geq 0$, $y=0$, and $z\geq 0$.
• for YZ: the set of points such that $x=0$, $y\geq0$, and $z\geq 0$.

What seems to be going on is that you're taking these three quadrants and projecting them onto the plane described by the equation $x+y+z=0$. Every point in this plane comes from the projection of exactly one point in the union of the three quadrants. The quadrants share one point in common (the origin), and every pair of quadrants shares a ray.

The projection formula onto this plane is given by the function $$ \pi(x,y,z) = (\tfrac{2}{3}x-\tfrac{1}{3}y-\tfrac{1}{3}z,-\tfrac{1}{3}x+\tfrac{2}{3}y-\tfrac{1}{3}z,-\tfrac{1}{3}x-\tfrac{1}{3}y+\tfrac{2}{3}z)$$ All this is doing is subtracting $\tfrac{1}{3}(x+y+z)$ from each coordinate to make the sum of the resulting coordinates equal $0$. The formula for each quadrant reduces to \begin{align*} \pi(x,y,0) &= (\tfrac{2}{3}x-\tfrac{1}{3}y,-\tfrac{1}{3}x+\tfrac{2}{3}y,-\tfrac{1}{3}x-\tfrac{1}{3}y) \\ \pi(x,0,z) &= (\tfrac{2}{3}x-\tfrac{1}{3}z,-\tfrac{1}{3}x-\tfrac{1}{3}z,-\tfrac{1}{3}x+\tfrac{2}{3}z) \\ \pi(0,y,z) &= (-\tfrac{1}{3}y-\tfrac{1}{3}z,\tfrac{2}{3}y-\tfrac{1}{3}z,-\tfrac{1}{3}y+\tfrac{2}{3}z) \end{align*} And for each ray: \begin{align*} \pi(x,0,0) &= (\tfrac{2}{3}x,-\tfrac{1}{3}x,-\tfrac{1}{3}x) \\ \pi(0,y,0) &= (-\tfrac{1}{3}y,\tfrac{2}{3}y,-\tfrac{1}{3}y) \\ \pi(0,0,z) &= (-\tfrac{1}{3}z,-\tfrac{1}{3}z,\tfrac{2}{3}z) \\ \end{align*} And of course the origin: \begin{align*} \pi(0,0,0) &= (0,0,0) \end{align*} Given a point on the $x+y+z=0$ plane, can we see which point it's the projection of? The only points that project to $(x,y,z)$ are points of the form $(x+c,y+c,z+c)$. All we need to do is let $c$ be the least number such that either $x+c=0$, $y+c=0$, or $z+c=0$. From this, we get your idea:

A way to parameterize a 2-dimensional plane is the set of points $(x,y,z)$ with $x\geq 0$, $y\geq 0$, and $z\geq 0$ such that either $x=0$, $y=0$, or $z=0$. We can realize this plane as the plane with the equation $x+y+z=0$ by taking $(x,y,z)$ and subtracting the average $\tfrac{1}{3}(x+y+z)$ from each coordinate.

Notice that there is a 3-fold rotational symmetry from rotating 3D space 120 degrees about the line $x=y=z$. This corresponds to the 120 degree rotation you noticed in your system of coordinates.

This suggests a generalization to higher dimensions. For example, to work with 3D, you can think of it as the set of points in 4D satisfying $x+y+z+w=0$. Then, with your system we can instead think of the points $(x,y,z,w)$ such that $x,y,z,w\geq 0$ and at least one of these numbers is $0$. To project onto the $x+y+z+w=0$ plane, we just subtract $\tfrac{1}{4}(x+y+z+w)$ from each coordinate. This time there is tetrahedral symmetry. Rather than three axes, you'll have four axes pointing in the directions of the vertices of a regular tetrahedron.

Even more generally, what we're doing is taking a convex polyhedron (the octant in 3D for example) and a hyperplane (a space of one less dimension, the plane $x+y+z=0$ for example) with the property that the projection of the boundary of the polyhedron onto the hyperplane is a bijection. Each face of the polyhedron gives a different region of the hyperplane this way, and you can use the plane to give the faces of the polyhedron a system of coordinates.