# Can we extend the real numbers by using hexagonal coordinates on a plane?

What if we used three axes on a plane like in the picture below? Then we could define any point on a plane using three numbers:

$$P=(a,b,c)$$

Three numbers seems excessive, however we do not need to use negative numbers that way.

$$a,b,c \geq 0$$

Of course, such coordinates in general will not be unique, since:

$$0=(0,0,0)=(1,1,1)=(2,2,2)=...$$

We can define a canonical form for each number this way:

$$m=\min(a,b,c)$$

$$P=(a,b,c)-(m,m,m)$$

This is not much different from using '$-$' sign in the usual complex plane.

The important point is that negative real numbers have the canonical form:

$$(-a,0,0)=(0,a,a)$$

On the left picture you can see how the numbers with integer coordinates form hexagonal cells. However, if we can use the easily scalable cells on the right side to define fractional coordinates. The arithmetics is simple:

$$(a_1,a_2,a_3) \pm (b_1,b_2,b_3)=(a_1\pm b_1,a_2 \pm b_2,a_3 \pm b_3)$$

We can always get rid of any negatives by transforming the result to canonical form.

The multiplication law can only be defined as follows:

$$(a_1,a_2,a_3) \cdot (b_1,b_2,b_3)=(a_1 b_1 + a_2 b_3 + a_3 b_2, a_3 b_3+a_1 b_2+a_2 b_1,a_2 b_2 + a_1 b_3+a_3 b_1)$$

If we let $a$ be the real part of the number, then this rule agrees well with the rules for real numbers. It's also commutative, associative and transitive.

The conjugate of any number is defined:

$$P^*=(a,b,c)^*=(a,c,b)$$

Then we can see that multiplication by its conjugate gives the square of the norm:

$$||P||^2=(a,b,c) \cdot (a,c,b)=(a^2 + b^2+c^2, bc+ac+ab,bc + ab+ac)=$$

$$=(a^2 + b^2+c^2-bc-ac-ab,0,0)=a^2 + b^2+c^2-bc-ac-ab$$

The inverse of the number is introduced this way:

$$\frac{1}{P}=\frac{1}{(a,b,c)}=\frac{(a,c,b)}{a^2 + b^2+c^2-bc-ac-ab}$$

Finally let's introduce the units:

$$(a,b,c) = (a,0,0)+(0,b,0)+(0,0,c)=a+b \mathbb{j}+c \mathbb{k}$$

Using the multiplication law above we can show that:

$$\mathbb{j}^2=\mathbb{k}$$

$$\mathbb{k}^2=\mathbb{j}$$

$$\mathbb{j k}=\mathbb{k j}=1$$

These are just cube roots of unity:

$$\mathbb{j}^3=1$$

$$\mathbb{k}^3=1$$

The units are not linearly independent!

$$1+\mathbb{j}+\mathbb{k}=(1,1,1)=0$$

The imaginary unit will be:

$$\mathbb{i}=\frac{1}{\sqrt{3}}(\mathbb{j}-\mathbb{k})=\frac{1}{\sqrt{3}}(0,1,-1)=\frac{1}{\sqrt{3}}(1,2,0)$$

So this number field works as well as the complex number field,doesn't it? Why is it never introduced in textbooks? Isn't it useful, especially with its relation to the hexagonal coordinate system?

• You might be onto something! – ploosu2 Feb 13 '16 at 23:53
• You might find the Eisenstein integers interesting. It is a similar system, but with the coordinates integer. – Marc Paul Feb 13 '16 at 23:59
• There is no "nice" algebraic structure between the complex numbers and the quaternions, which is what your "three axis" discussion seems to seek out. – hardmath Feb 13 '16 at 23:59
• @hardmath, no, that's not it, the number field I described is isomorphic to complex number field – Yuriy S Feb 14 '16 at 0:06

## 1 Answer

It depends on what context you want to think of these as.

If you want to look at $\mathbb{Z}[\zeta_3]$, this is a structure that is studied. It's studied own it's own, and is known as the Eisenstein Integers, and as an example of $\mathbb{Z}[\zeta_p]$ which has a lot of interesting properties as a collection of rings. These are called Cyclotomic Fields.

If you want to look at it as $\mathbb{R}[\zeta_3]$, then it is the case that it is the same field as $\mathbb{R}[\zeta_4]=\mathbb{C}$. The expression as $\mathbb{C}$ is usually considered better because $\mathbb{R}[i]$ has an orthogonal basis. In circumstances where a hexagonal grid is relevant, it is used.

• "In circumstances where a hexagonal grid is relevant, it is used." Do you happen to know any examples? – Marc Paul Feb 14 '16 at 0:06
• Examples of when a hexagonal grid is relevant, or examples where this expression is used because a hexagonal grid is relevant? It gets used in chemistry to describe certain crystals. – Stella Biderman Feb 14 '16 at 0:08