In my previous post I discussed the trirational numbers, which are representations of complex numbers that can also represent ratios of three numbers. Now let's look at the generalization that represents ratios of \(n\) numbers: the \(n\)-rational numbers, or more generally the multirational numbers.
\[r\unicode{x25B6}d_1\unicode{x25B6}d_2\unicode{x25B6}\dots\unicode{x25B6}d_{n-1}\]
Here, \(r\in\Bbb{R}\) is called the numerator while \(d_1,\dots,d_{n-1}\in\Bbb{R}_{>0}\) are the denominators of the \(n\)-rational number.
We can say that a trirational number \(a\unicode{x25B6}b\unicode{x25B6}c\) is a \(3\)-rational number, while a fraction \(\frac{a}{b}\) is a \(2\)-rational number that can be written as \(a\unicode{x25B6}b\) (if \(b\) is positive) or \((\operatorname{sgn}(b)*a)\unicode{x25B6}{|b|}\).
Ratios and the scaling rule
When you scale each component of a ratio by the same amount, the ratio stays the same. The same applies to multirationals because of this identity:
\[x\unicode{x25B6}x\unicode{x25B6}\dots\unicode{x25B6}x=1\]
This property allows fractions and trirational numbers to act as homogeneous forms for real numbers and complex numbers, respectively.
We'll go back to this concept of homogeneous forms near the end of this post, but first let's look at \(n\)-rational operations without referencing the ambient algebra.
Multiplication
As with fractions and trirationals, multiplication is component-wise for any two \(n\)-rational numbers:
\[\left(\begin{matrix}r \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_1 \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_2 \\[-5pt]
\unicode{x25BC} \\[-5pt]
\vdots \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_{n-1} \\[-5pt]
\end{matrix}\right)*
\left(\begin{matrix}r' \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_1' \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_2' \\[-5pt]
\unicode{x25BC} \\[-5pt]
\vdots \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_{n-1}' \\[-5pt]
\end{matrix}\right)=
\begin{matrix}r*r' \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_1*d_1' \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_2*d_2' \\[-5pt]
\unicode{x25BC} \\[-5pt]
\vdots \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_{n-1}*d_{n-1}' \\[-5pt]
\end{matrix}\]
Multiplicative inverse
There are two ways to write the 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:
\[\left(\begin{matrix}
r \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_1 \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_2 \\[-5pt]
\unicode{x25BC} \\[-5pt]
\vdots \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_{n-1} \\[-5pt]
\end{matrix}\right)^{-1}\quad=\quad
\begin{matrix}
{\operatorname{sgn}(r)}\left(\prod_{a=1}^{n-1}d_{a}\right) \\
\unicode{x25BC} \\[-5pt]
{|r|}\frac{\prod_{a=1}^{n-1}d_{a}}{d_1} \\
\unicode{x25BC} \\[-5pt]
{|r|}\frac{\prod_{a=1}^{n-1}d_{a}}{d_{2}} \\
\unicode{x25BC} \\[-5pt]
\vdots \\[-5pt]
\unicode{x25BC} \\[-5pt]
{|r|}\frac{\prod_{a=1}^{n-1}d_{a}}{d_{n-1}} \\
\end{matrix}\]
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 to ensure that the resulting denominators are positive.
Scalar Multiplication
There are two significant multiplication rules when it comes to multirationals and any single real number \(x\). The first one is this:
\[x*
\left(\begin{matrix}r \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_1 \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_2 \\[-5pt]
\unicode{x25BC} \\[-5pt]
\vdots \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_{n-1} \\[-5pt]
\end{matrix}\right)\quad=\quad\left(\begin{matrix}
x \\[-5pt]
\unicode{x25BC} \\[-5pt]
1 \\[-5pt]
\unicode{x25BC} \\[-5pt]
1 \\[-5pt]
\unicode{x25BC} \\[-5pt]
\vdots \\[-5pt]
\unicode{x25BC} \\[-5pt]
1 \\[-5pt]
\end{matrix}\right)*
\left(\begin{matrix}r \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_1 \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_2 \\[-5pt]
\unicode{x25BC} \\[-5pt]
\vdots \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_{n-1} \\[-5pt]
\end{matrix}\right)\quad=\quad
\begin{matrix}x*r \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_1 \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_2 \\[-5pt]
\unicode{x25BC} \\[-5pt]
\vdots \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_{n-1} \\[-5pt]
\end{matrix}\]
The second one is the same scaling rule we already know:
\[\left(\begin{matrix}
x \\[-5pt]
\unicode{x25BC} \\[-5pt]
x \\[-5pt]
\unicode{x25BC} \\[-5pt]
x \\[-5pt]
\unicode{x25BC} \\[-5pt]
\vdots \\[-5pt]
\unicode{x25BC} \\[-5pt]
x \\[-5pt]
\end{matrix}\right)*
\left(\begin{matrix}r \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_1 \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_2 \\[-5pt]
\unicode{x25BC} \\[-5pt]
\vdots \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_{n-1} \\[-5pt]
\end{matrix}\right)\quad=\quad
\begin{matrix}r \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_1 \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_2 \\[-5pt]
\unicode{x25BC} \\[-5pt]
\vdots \\[-5pt]
\unicode{x25BC} \\[-5pt]
d_{n-1} \\[-5pt]
\end{matrix}\]
Addition
General addition
Special cases
When only their numerators differ:
When they can be simplified to fractions:
\begin{align}(r_1\unicode{x25B6}d_1\unicode{x25B6}d_1\unicode{x25B6}\dots\unicode{x25B6}d_1)+(r_2\unicode{x25B6}d_2\unicode{x25B6}d_2\unicode{x25B6}\dots\unicode{x25B6}d_2)&=\frac{r_1}{d_1}+\frac{r_2}{d_2}=\frac{r_1d_2+r_2d_1}{d_1d_2} \\[10pt]
&=(r_1d_2+r_2d_1)\unicode{x25B6}d_1d_2\unicode{x25B6}d_1d_2\unicode{x25B6}\dots\unicode{x25B6}d_1d_2
\end{align}
Additive inverse
Multirationals w/ different sizes
We already know that \(a\unicode{x25B6}b\unicode{x25B6}b=a/b=a\unicode{x25B6}b\), so for example,
\begin{align}(x\unicode{x25B6}y\unicode{x25B6}z)*(a\unicode{x25B6}b)&=(x\unicode{x25B6}y\unicode{x25B6}z)*(a\unicode{x25B6}b\unicode{x25B6}b) \\[5pt]
&=xa\unicode{x25B6}yb\unicode{x25B6}zb
\end{align}
In fact, fractions can always be represented as an \(n\)-rational number for any \(n\): If the fraction's denominator is positive then the numerator stays the same and the \((n-1)\) denominators are all just the same original denominator. So for example, \(5/9\) as a \(7\)-rational is \(5\unicode{x25B6}9\unicode{x25B6}9\unicode{x25B6}9\unicode{x25B6}9\unicode{x25B6}9\unicode{x25B6}9\). If the fraction's denominator is negative, then we can just move the negative sign to the numerator and use the same process.
As for the general case of converting an arbitrary \(a\)-rational to a \(b\)-rational number, this is a lot trickier, and will involve finding an algebra where both \(a\)-rational and \(b\)-rational numbers exist. Which leads us to the next section.
Homogeneous forms of unital algebras
As fractions can be seen as homogeneous coordinates of the real number line and trirationals as those of the complex plane, it is natural to wonder which algebras other multirational numbers are a homogeneous form of. For example, which algebra contains an element that can be represented by the \(4\)-rational homogeneous form \(2\unicode{x25B6}3\unicode{x25B6}2\unicode{x25B6}3\)? Let's discuss this concept in more detail.
Roots of Unity
\(n\)-rational numbers can be evaluated by using the following exponential product:
\[r\unicode{x25B6}d_1\unicode{x25B6}d_2\unicode{x25B6}\dots\unicode{x25B6}d_{n-1}=r^1*d_1^s*d_2^{s^2}*\dots*d_{n-1}^{s^{n-1}}\]
where \({\sum_{a=0}^{n-1}s^a}=0\) so that \(x\unicode{x25B6}x\unicode{x25B6}\dots\unicode{x25B6}x=1\) (the scaling rule for multirationals).
It might be tempting to assume that \(\{1,s,s^2,\dots,s^{n-1}\}\) are the \(n\)th roots of unity on the complex plane. While this assumption is correct for fractions and trirationals, it does not work for \(n\)-rationals with \(n>3\). For example, for \(n=4\), we can easily see that \(2*3^i*2^{-1}*3^{-i}=1=1*1^i*1^{-1}*1^{-i}\neq{2\unicode{x25B6}3\unicode{x25B6}2\unicode{x25B6}3}\), because the ratio \((1:1:1:1)\) is not the same as the ratio \((2:3:2:3)\).
The problem is that these roots of unity in the complex plane contain elements that are negatives or conjugates of each other. This is fine for fractions and trirationals:
\[a/a=a^1*a^{-1}=a^{1-1}=a^{0}=1=(1:1)\sim(a:a)\]
\begin{align}
a\unicode{x25B6}b\unicode{x25B6}b&=a^1*b^{\frac{-1+i\sqrt{3}}{2}}*b^{\frac{-1-i\sqrt{3}}{2}} \\
&=a^1*b^{\frac{-1+i\sqrt{3}-1-i\sqrt{3}}{2}}=a^1*b^{-\frac{2}{2}} \\
&=a^1*b^{-1}=a/b\sim(a:b)\sim(a:b:b)
\end{align}
But for \(n>3\), if two of the exponents are negatives of each other, something like this might happen:
\begin{align}a^1*\dots*b^{s^k}*\dots*b^{-s^k}*\dots*d^{s^{n-1}}&=a^{1}*\dots*b^{s^k-s^k}*\dots*d^{s^{n-1}} \\
&=a^{1}*\dots*b^{0}*\dots*d^{s^{n-1}}=a^{1}*\dots*1*\dots*d^{s^{n-1}} \\
&=a^{1}*\dots*1^{s^k}*\dots*1^{-s^k}*\dots*d^{s^{n-1}}\sim(a:\dots:1:\dots:1:\dots:d) \\
&\not\sim(a:\dots:b:\dots:b:\dots:d)
\end{align}
\begin{align}
a^1*\dots*b^{s^j}*\dots*b^{\overline{s^j}}*\dots*d^{s^{n-1}}&=a^{1}*\dots*b^{s^j+\overline{s^j}}*\dots*d^{s^{n-1}}=a^{1}*\dots*b^{2*\operatorname {Re}(s^j)}*\dots*d^{s^{n-1}} \\
&=(a*b^{2*\operatorname{Re}(s^j)})^{1}*\dots*1^{s^j}*\dots*1^{\overline{s^j}}*\dots*d^{s^{n-1}}\sim((a*b^{2*\operatorname{Re}(s^j)}):\dots:1:\dots:1:\dots:d) \\
&\not\sim(a:\dots:b:\dots:b:\dots:d)
\end{align}
\(\boldsymbol{\Bbb{X}_{n}}\) Algebras
Let us denote the aforementioned family of algebras as \(\Bbb{X}_n\), where \(n\) is the number of dimensions in the algebra. As previously implied, \(\Bbb{X}_1=\Bbb{R}\) and \(\Bbb{X}_2=\Bbb{C}\).
Elements in any such \(\Bbb{X}_{n}\) can be written in Cartesian form using an orthonormal basis with \(n\) elements, so multirationals in \(\Bbb{X}_{n}\) can be converted to a Cartesian form with \(n\) terms. Being able to convert multirationals to a different form is essential in studying less trivial aspects like addition.
In the next part of this series, we will try to look for and describe this \(\Bbb{X}_n\) family of unital algebras.
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