Let's Invent the Triplex Numbers

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Recommends Anixx · 2022-02 · Mathematics Stack Exchange · on “Non-Existence of Three-Dimensional Numbers”

There can be 3-dimensional commutative hypercomplex number systems. I will give three examples. 3D numbers. Example 1 Take $\mathbb{R}^3$ with Hadamard product. In other words, triplets of numbers with element-wise multiplication. Now assign $(1,1,1)=1,(-1,1,1)=j, (1,1,-1)=k$ . A number would be written in the form $a+bj+ck$ . Algebraically it will be a commutative ring with zero divisors (hence, not a field, but that's OK). For instance $(j-1)(k-1)=0$ . Here is a Mathematica code to experiment with: Unprotect[Power]; Power[0, 0] = 1; Protect[Power]; $Pre = (# /. {j -> {-1, 1, 1}, k -> {1, 1, -1}}) /. {x_, y_, z_} -> x/2 + z/2 + (j (y - x))/2 + (k (y - z))/2 &; Using this code one can see th

Cites J.G. · 2017-12 · Mathematics Stack Exchange · on “How many "super imaginary" numbers are there?”

Your $h$ -based number system is called split-complex numbers, and what you called $h$ is in my experience usually called $j$ (although, as @user14972 notes, $h$ is sometimes used). A related system introduces an $\epsilon$ satisfying $\epsilon^2=0$ , and this gives dual numbers. Linear transformations guarantee these two systems and complex numbers are the only ways to extend $\mathbb{R}$ to a $2$ -dimensional commutative associative number system satisfying certain properties. However: The Cayley-Dickson construction allows you to go from real numbers to complex numbers and thereafter double the dimension as often as you like by adding new square roots of $-1$ , taking you to quaternions,

Cites Mark S. · 2022-04 · Mathematics Stack Exchange · on “Unconventional hypercomplex numbers”

@SouravKannanthaB I think "an example for an algebra whose base is not defined in the powers of 2" would be the "triplex numbers" that Angel [LINK] mentioned in their answer [LINK] and that I mentioned as an example of a polynomial quotient ring, as defined by "$i^3=1$", and which are featured in this YouTube video [this video] that Angel referred to.

Cites Anixx · 2022-05 · Mathematics Stack Exchange · on “Is there a third dimension of numbers?”

There can be 3-dimensional commutative hypercomplex number systems. I will give t̶h̶r̶e̶e̶ four examples. 3D numbers. Example 1 Take $\mathbb{R}^3$ with Hadamard product. In other words, triplets of numbers with element-wise multiplication. Now assign $(1,1,1)=1,(-1,1,1)=j, (1,1,-1)=k$ . A number would be written in the form $a+bj+ck$ . Algebraically it will be a commutative ring with zero divisors (hence, not a field, but that's OK). For instance $(j-1)(k-1)=0$ . Here is a Mathematica code to experiment with: Unprotect[Power]; Power[0, 0] = 1; Protect[Power]; $Pre = (# /. {j -> {-1, 1, 1}, k -> {1, 1, -1}}) /. {x_, y_, z_} -> x/2 + z/2 + (j (y - x))/2 + (k (y - z))/2 &; Using this code one

Cites Anixx · 2022-05 · Mathematics Stack Exchange · on “From algebraic point of view, what are the similarities and differences between these two hypercomplex systems?”

I would like to know how compare from the algebraic point of view these two 3-dimensional hypercomplex number systems. 3-dimensional split-complex numbers Take $\mathbb{R}^3$ with Hadamard product. In other words, triplets of numbers with element-wise multiplication. Now assign $(1,1,1)=1,(-1,1,1)=j, (1,1,-1)=k$ . A number would be written in the form $a+bj+ck$ . Algebraically it will be a commutative ring with zero divisors (hence, not a field, but that's OK). For instance $(j-1)(k-1)=0$ . Here is a Mathematica code to experiment with: Unprotect[Power]; Power[0, 0] = 1; Protect[Power]; $Pre = (# /. {j -> {-1, 1, 1}, k -> {1, 1, -1}}) /. {x_, y_, z_} -> x/2 + z/2 + (j (y - x))/2 + (k (y - z)

Cites Anixx · 2022-05 · Mathematics Stack Exchange · on “Name for the reals augmented with an $x$ such that $x^2 = x$”

Such objects are called idempotents. Split-complex numbers have two idempotents, $1/2+j/2$ and $1/2-j/2$ . Triplex numbers [this video] also have idempotents, for instance, $\frac{1+i+j}2$ and $\frac{2-i-j}2$ (totally 8). Also, consider the one-point or two-point compactifications of real numbers (the later is called "extended real line", $\overline{\mathbb{R}}$ ). In both, there is one idempotent, $\infty$ , because $\infty^2=\infty$ .

Cites Anixx · 2022-05 · Mathematics Stack Exchange · on “Is this 3-dimensional split-complex numbers system viable? What are the properties?”

I wonder whether you have the same point of view about the second system described here: [LINK] and in this video: [this video] (triplex numbers)

Cites Francis Ocoma · 2022-09 · Mathematics Stack Exchange · on “Highly polytopic algebras”

( Update: I ended up posting two answers for this, each addressing different things. I feel the second one [LINK] is more deserving to be accepted if nobody posts a better answer.) Aside from the complex numbers, the triplex numbers [this video] , and their isomorphisms, I'd like to know which other unital algebras with dimensions greater than one follow these rules: The algebra must contain non-trivial finite cyclic groups under multiplication aside from $\{-1,1\}$ . This means the algebra has more than two "roots of unity". Given any such cyclic group, the elements of the group must form some or all the vertices of a single regular convex polytope instance in the underlying vector space of the

Cites Anixx · 2023-07 · Mathematics Stack Exchange · on “Is there any where I can find info about Triplex Numbers?”

Triplex numbers (as described in this video [this video] to which OP refers) have the following properties. Here, $1=\left( \begin{array}{ccc} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \\ \end{array} \right)$ $j=\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \\ \end{array} \right)$ $k=\left( \begin{array}{ccc} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{array} \right)$ Triplex numbers are isomorphic to $\mathbb{R}\times\mathbb{C}$ via isomorphism: $$\left( \begin{array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ 1 & 0 & 0 \\ \end{array} \right) \to (1, e^{2 i \pi/3}), \left( \begin{array}{ccc} 0 & 0 & 1 \\ 1 & 0 & 0 \\ 0 & 1 & 0 \\ \end{array} \right) \to (1, e^{-2 i \pi/3});$$ $$(x,a+bi)\to\frac{1}