Trirational numbers in the complex plane (and generalizations)

Question

(Note: I've posted my own answer, slightly redefining trirationals to be composed of reals instead of integers and addressing the problems pointed out here. Please take note of this while reading my original question.)

Definition: By trirational number, I mean a number that can represent a ratio of three integers (e.g. $2:3:5$) in the same way rational numbers represent a ratio of two integers. I will express trirational numbers in this form: $a\unicode{x25B6}b\unicode{x25B6}c$.

Let $t$ be a trirational number that represents the triple integer ratio $a:b:c$. We'll define a trirational number to be

$$t=a\unicode{x25B6}b\unicode{x25B6}c\overset{\text{def}}{=}a*b^{\omega}*c^{\omega^2}$$

where $\omega$ and $\omega^2$ are the two primitive third roots of unity. This is analogous to how a rational number representing the ratio $a:b$ can be expressed as $a*b^{-1}$ where $-1$ is of course the primitive second root of unity. Since $\omega$ and $\omega^2$ have non-zero imaginary parts, $t$ is not confined to the real number line.

In order for $t$ to be a proper representation of $a:b:c$, it must represent the fact that $a:b:c=xa:xb:xc$ for any integer $x$. This is satisfied given that $1+\omega+\omega^2=0$, and therefore

$$(xa)\unicode{x25B6}(xb)\unicode{x25B6}(xc)=(xa)*(xb)^{\omega}*(xc)^{\omega^2}=x^{1+\omega+\omega^2}t=t$$

The ternary operation used to generate trirational numbers is analogous to division in certain ways:

• $x\unicode{x25B6}x\unicode{x25B6}x=1$
• $x\unicode{x25B6}1\unicode{x25B6}1=x$
• $0\unicode{x25B6}x\unicode{x25B6}y=0$
• $x\unicode{x25B6}0\unicode{x25B6}y$ and $x\unicode{x25B6}y\unicode{x25B6}0$ are both undefined.
• Just as $(x/y)^{-1}=y/x$, we have $(x\unicode{x25B6}y\unicode{x25B6}z)^{\omega}=z\unicode{x25B6}x\unicode{x25B6}y$, $\\(x\unicode{x25B6}y\unicode{x25B6}z)^{\omega^2}=y\unicode{x25B6}z\unicode{x25B6}x$, and $\\(x\unicode{x25B6}y\unicode{x25B6}z)^{-1}=~(x\unicode{x25B6}y\unicode{x25B6}z)^{\omega+\omega^2}=zy\unicode{x25B6}xz\unicode{x25B6}yx$

Other properties:

• $x\unicode{x25B6}y\unicode{x25B6}y=x/y$, meaning all rational numbers are trirational numbers. Take note that both $x\unicode{x25B6}y\unicode{x25B6}1\neq x/y$ and $x\unicode{x25B6}1\unicode{x25B6}y\neq x/y$ unless $y=1$.
• $y\unicode{x25B6}x\unicode{x25B6}y=(x/y)^{\omega}$, meaning two rational numbers $\frac{a}{c}$ and $\frac{b}{c}$ can be combined into a trirational number $a\unicode{x25B6}b\unicode{x25B6}c$ via $\frac{a}{c}*(\frac{b}{c})^{\omega}=(a\unicode{x25B6}c\unicode{x25B6}c)*(c\unicode{x25B6}b\unicode{x25B6}c)=a\unicode{x25B6}b\unicode{x25B6}c$
• $y\unicode{x25B6}y\unicode{x25B6}x=(x/y)^{\omega^2}$, meaning two rational numbers $\frac{a}{b}$ and $\frac{c}{b}$ can be combined into a trirational number $a\unicode{x25B6}b\unicode{x25B6}c$ via $\frac{a}{b}*(\frac{c}{b})^{\omega^2}=(a\unicode{x25B6}b\unicode{x25B6}b)*(b\unicode{x25B6}b\unicode{x25B6}c)=a\unicode{x25B6}b\unicode{x25B6}c$

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Problems with generalization:

At first it seemed likely to me that a generalized $n$-rational number representing a ratio of $n$ integers $a_1:a_2:a_3:...:a_n$ would be of the form $$a_1*a_2^{\omega_1}*a_3^{\omega_2}*...*a_n^{\omega_{n-1}}$$

where ${\omega_1}$ up to ${\omega_{n-1}}$, along with $1$, are the $n$th roots of unity. But problems arise even for $n=4$.

The fourth roots of unity in the complex plane are $1$, $i$, $-1$, and $-i$, which give us this candidate form for 4-rational numbers:

$$a*b^i*c^{-1}*d^{-i}=\frac{a}{c}*(\frac{b}{d})^i$$

This might seem promising because $1+i-1-i=0$, but it still fails as a representation of ratios of four integers. For example, $2:5:4:10$ in the above form would simplify to $\frac{1}{2}*(\frac{1}{2})^i$ even though $1:1:2:2$ is not an equivalent ratio. Even worse, something like $2:7:2:7$ in this form ends up as $1$!

It looks like we'll need to look in higher dimensions for a solution.

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Updates:

As per @Stinking Bishop's advice I'll just quickly define trirational multiplication:

$$(a_1\unicode{x25B6}b_1\unicode{x25B6}c_1)*(a_2\unicode{x25B6}b_2\unicode{x25B6}c_2)\overset{\text{def}}{=}a_1a_2\unicode{x25B6}b_1b_2\unicode{x25B6}c_1c_2$$

Unfortunately, I still don't have a general definition of trirational addition. I actually can't find anything online about adding ratios of three numbers. I even tried looking into projective spaces as @Qiaochu Yuan's suggested, but it seems addition is also a problem there.

So in most cases, I need to convert the trirational addends to complex numbers before I could add them. These are the only (trivial) exceptions:

$$(a_1\unicode{x25B6}b\unicode{x25B6}c)+(a_2\unicode{x25B6}b\unicode{x25B6}c)\overset{\text{def}}{=}(a_1+a_2)\unicode{x25B6}b\unicode{x25B6}c$$ $$(a_1\unicode{x25B6}b_1\unicode{x25B6}b_1)+(a_2\unicode{x25B6}b_2\unicode{x25B6}b_2)\overset{\text{def}}{=}(a_1b_2+a_2b_1)\unicode{x25B6}b_1b_2\unicode{x25B6}b_1b_2$$

Aside from those, I usually don't even know how to convert the complex sum in the general case back to trirational form!

As for generalizing trirational numbers (i.e. $n$-rational numbers), I now see that it's impossible to do for $n>3$ in the complex plane using roots of unity due to simplification problems:

For any $n$th root of unity, if $n$ is even then you get $n$th root pairs that are negatives of each other, so terms could cancel out: $a^{\omega_0}*a^{-\omega_0}=1$. This is fine for $n=2$ (the rational numbers), but for $n>2$ it could break the number's representation of ratios of $n$ numbers, as we saw above with the fourth roots of unity.

If $n$ is odd then you get $n$th root pairs that are conjugates of each other so they devolve into real numbers: $a^{\omega_1}*a^{\overline {\omega_1}}=a^{2*\operatorname {Re} (\omega_1)}$. This is okay for $n=3$ (the trirationals) but for $n>3$ it could once again break the representation. A representation of say a ratio of five numbers like $a:b:c:c:b$ that evaluates to a real number is not a very good representation, and is not what a 5-rational number should be.

Later I will try to find a solution for 4-rational and 5-rational numbers outside of the complex plane.

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Questions:

• While division can be interpreted as a scaling down of a number (when the divisor is a scalar $d>1$) or a rotation in the opposite direction (when the divisor has a non-zero imaginary part), is there also a geometric interpretation that could help visualize the ternary $\unicode{x25B6}\unicode{x25B6}$ operation?

• Aside from the trivial case of rational numbers (where $a/b=a\unicode{x25B6}b\unicode{x25B6}b$), are there trirational numbers that, when you're given their polar or rectangular form, you can derive or even estimate the trirational form $a\unicode{x25B6}b\unicode{x25B6}c$ via a simple algorithm?

• Does anyone have other ideas for trirational number addition? Perhaps there's another clue I'm missing? In real life we can combine two collections of the same three kinds of things in different ratios, so my intuition tells me a general trirational addition formula should exist.

Answer 1

One formal complaint that you may need to resolve first. What if $b$ or $c$ is negative? Say, what is $(-1)^\omega$?

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Now, to my main point:

I would pay less attention to what "trirational numbers" are, and would pay more attention to how they behave.

For example, what is $(a\unicode{x25B6}b\unicode{x25B6}c)+(a'\unicode{x25B6}b'\unicode{x25B6}c')$ (addition)? What is $(a\unicode{x25B6}b\unicode{x25B6}c)\times (a'\unicode{x25B6}b'\unicode{x25B6}c')$ (multiplication)?

Note that you need to define those operations so that the "scaling" does not affect the result. I believe you probably want:

$$(a\unicode{x25B6}b\unicode{x25B6}c)\times (a'\unicode{x25B6}b'\unicode{x25B6}c')\overset{\text{def}}{=}(aa')\unicode{x25B6}(bb')\unicode{x25B6}(cc')$$

which is ok, because when you "scale" the factors, the product will be scaled and so represents the same "trinomial number". Can you do the same for addition?

This mimics the construction of rational numbers. We define that $(a/b)\times(a'/b')\overset{\text{def}}{=}(aa')/(bb')$ but also $(a/b)+(a'/b')\overset{\text{def}}{=}(ab'+a'b)/(bb')$, and both of those operations play well with scaling, and with each other!

Once you have defined your operations, investigate the properties of your construction. Do you get a structure of a ring? Is this ring a field?

You may find out that the answers to the above questions is "yes", and that, in addition, you can see this as a subfield of $\mathbb C$. It may even happen that the map $a\unicode{x25B6}b\unicode{x25B6}c\to ab^\omega c^{\omega^2}$ is an injective ring or field homomorphism. Or you may reach a completely different/opposite conclusion.

The bottom line is that, whether the mapping to complex numbers helps or not, the algebraic construction is viable without any reference to complex numbers, and the monomorphism into $\mathbb C$ may well be purely incidental. But, you need to define addition and multiplication and make sure that they "play well" with scaling. (And ideally with each other, so that you get a nice algebraic structure with good properties.) I think you have done that for multiplication but not (yet) for addition.

So for "quadrinomial numbers", start with defining the operations on the quartuplets of integers, the way you want them, but paying attention that "scaling" does not affect the results. That will produce an algebraic structure that you may well call "quadrinomial numbers". Then try to establish a link with $\mathbb C$ (if such a link exists. It may not.)

Answer 2

If we redefine $a\unicode{x25B6}b\unicode{x25B6}c$ such that $a\in\Bbb{R}$ and $b,c\in\Bbb{R}_{>0}$ instead of being integers, we get the following nifty conversions between the complex polar form and the trirational form:

$$re^{i\theta}=re^{\frac{\theta}{\sqrt{3}}}\unicode{x25B6}e^{\frac{2\theta}{\sqrt{3}}}\unicode{x25B6}1=re^{\frac{-\theta}{\sqrt{3}}}\unicode{x25B6}1\unicode{x25B6}e^{\frac{-2\theta}{\sqrt{3}}}=r\unicode{x25B6}e^{\frac{\theta}{\sqrt{3}}}\unicode{x25B6}e^{\frac{-\theta}{\sqrt{3}}}$$

$$a\unicode{x25B6}b\unicode{x25B6}c=\frac{a}{\sqrt{bc}}*e^{i\operatorname{ln}\left(\frac{b}{c}\right)\frac{\sqrt{3}}{2}}$$

Through this (and Euler's formula) we can convert trirationals to complex rectangular forms, add them, then convert them back to trirational form. A general addition formula for the trirational form can then be derived from this, though it's quite complicated so I will leave it as an exercise to the reader.

Since the denominators $b$ and $c$ are always positive, the additive inverse is $(-a)\unicode{x25B6}b\unicode{x25B6}c$ while the multiplicative inverses are ${\operatorname{sgn}(a)}bc\unicode{x25B6}{|a|}c\unicode{x25B6}{|a|}b$ as well as $a^{-1}\unicode{x25B6}b^{-1}\unicode{x25B6}c^{-1}$ for any non-zero trirational number.

I think it's clear that this definition of trirational numbers is a field isomorphic to $\Bbb{C}$.

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As for generalizations, an $n$-rational number $r\unicode{x25B6}d_{1}\unicode{x25B6}d_{2}\unicode{x25B6}\dots\unicode{x25B6}d_{n-1}$, where the numerator $r\in\Bbb{R}$ and the denominators $d_{1},\dots,d_{n-1}\in\Bbb{R}_{>0}$ is equal to the following:

$$r*d_{1}^{\omega_{n}}*d_{2}^{\omega_{n}^{2}}*\dots*d_{n-1}^{\omega_{n}^{n-1}}$$

where $\{1,\omega_{n},\dots,\omega_{n}^{n-1}\}$ is a cyclic group with affinely independent elements from an algebra with $n-1$ dimensions and where

$$\sum_{x=0}^{n-1}\omega_{n}^{x}=0$$

The affine independence of this cyclic group means that for $n>2$, none of the elements in the group would be negatives of each other. This solves the problem we saw with roots of unity in the complex plane for any even $n>3$, where opposite elements caused havoc in the ratio representation.

The convex hull of this cyclic group is a regular $(n-1)$-simplex centered at the origin. I wrote about algebras that contain such cyclic groups in this question, which I eventually answered myself, detailing things like the orthonormal basis and matrix representations of numbers in such algebras. This is vital when studying $n$-rational numbers, especially their addition and subtraction.

Regarding division, there are two kinds of multiplicative inverse for a non-zero $n$-rational number. Here's one:

$$(r\unicode{x25B6}d_1\unicode{x25B6}d_2\unicode{x25B6}\dots\unicode{x25B6}d_{n-1})^{-1}=r^{-1}\unicode{x25B6}d_1^{-1}\unicode{x25B6}d_2^{-1}\unicode{x25B6}\dots\unicode{x25B6}d_{n-1}^{-1}$$

i.e. we simply take each term's multiplicative inverse. Or we could do it like this:

$$(r\unicode{x25B6}d_1\unicode{x25B6}d_2\unicode{x25B6}\dots\unicode{x25B6}d_{n-1})^{-1}={\operatorname{sgn}(r)}\left(\prod_{a=1}^{n-1}d_{a}\right)\unicode{x25B6}{|r|}\frac{\prod_{a=1}^{n-1}d_{a}}{d_1}\unicode{x25B6}{|r|}\frac{\prod_{a=1}^{n-1}d_{a}}{d_{2}}\unicode{x25B6}\dots\unicode{x25B6}{|r|}\frac{\prod_{a=1}^{n-1}d_{a}}{d_{n-1}}$$

i.e., each term is replaced by the product of all the other terms, except the numerator retains its original sign, and it is the absolute value of $r$ that's used in the denominators. This ensures that the resulting denominators are positive and that every non-zero $n$-rational number has a multiplicative inverse.

Multiplication is component-wise, so multiplying an $n$-rational number to its first inverse will immediately turn all terms to $1$, while multiplying it to its second inverse will render all terms equal to each other and make the whole thing reducible to $1$.