Anisotropic Four-Dimensional Spaces of Real Numbers

Abstract

This article constructs all the anisotropic spaces of four-dimensional numbers in which the commutative and associative operations of addition and multiplication are defined. In this case, so-called “zero divisors” appear in these spaces. The structures of zero divisors in each space are described and their properties are investigated. It is shown that there are two types of zero divisors and they form a two-dimensional subspace of the four-dimensional space. A space of 4 × 4 matrices is constructed that is isomorphic to the space of four-dimensional numbers. The concept of the spectrum of a four-dimensional number is introduced and a bijective mapping between four-dimensional numbers and their spectra is constructed. Thanks to this, methods for solving linear and quadratic equations in four-dimensional spaces are developed. It is proven that a quadratic equation in a four-dimensional space generally has four roots. The concept of the spectral norm is introduced in the space of four-dimensional numbers and the equivalence of the spectral norm to the Euclidean norm is proved.

1. Introduction

Attempts at the generalization of the concept of number were made in mathematics repeatedly. The very first, which turned out to be successful, is the concept of a complex number, which can be represented by an ordered pair of real numbers, the set of such numbers forming a field, like a set of real numbers. Complex numbers have proven to be extremely useful in both theoretical and applied applications: many problems that seemed inaccessible have been given a transparent treatment from the point of view of complex analysis and have been solved in general terms. An example is the fundamental theorem of algebra on the existence of at least one root of a non-constant polynomial with complex coefficients in the field of complex numbers. Complex analysis plays an important role in plane problems of mathematics and mechanics.

The next generalization of the concept of number was the hypercomplex four-dimensional numbers proposed by William Hamilton, quaternions [1], which found important applications in physics [2]. However, it should be noted that quaternions, due to the lack of commutativity of multiplication, do not form a field, and for this reason their application is limited. Due to the non-commutativity of multiplication, it was not possible to construct a full-fledged four-dimensional mathematical analysis that would generalize the one-dimensional and two-dimensional analogues.

One can also note the Cayley–Dickson construction (also known as the Cayley–Dickson process or procedure) [3,4]. In this operation, an algebra over a field (or ring) is constructed in an iterative manner, the dimension of which doubles at each iteration. Using this construction, one can sequentially obtain the following extensions from real numbers: complex numbers, quaternions, octonions, and so on. They are called Cayley–Dickson algebras. These composition algebras are useful in theoretical and mathematical physics.

Each time the Cayley–Dickson construction is applied, some symmetry of the real field disappears: first the order, then successively such properties of multiplication as commutativity, associativity, and alternativity. At the same time, up to the octonions, the property of no zero divisors is preserved.

The well-known Frobenius theorem [5] states that it is impossible to further extend the concept of a number without sacrificing some arithmetic property. At the beginning of the 21st century, the Kazakh mathematician M. M. Abenov developed a four-dimensional mathematics [6], different from quaternions, in which the multiplication of numbers is associative and commutative, but one has to deal with zero divisors. Abenov managed to construct a coherent theory, which he called four-dimensional mathematics, and demonstrated some of its applications to solving problems in hydrodynamics. Later, together with M. B. Gabbasov, he obtained other four-dimensional spaces of numbers. In his work [7], Gabbasov and his co-authors obtained analytical solutions to the Cauchy problem for a mathematical model of filtration theory in a three-dimensional non-stationary case. The present article serves the purpose of generalizing this theory to the so-called anisotropic case, when different dimensions have different scales, while the commutativity and associativity of multiplication are preserved. It is important to note that the proposed vector spaces of four-dimensional numbers are a natural generalization of the spaces of one-dimensional and two-dimensional (complex) numbers.

2. Various Anisotropic Spaces of Four-Dimensional Numbers

Let us consider space of four-dimensional numbers $x=\left(x_1,x_2,x_3,x_4\right)$, where $x_i\in R, i=1,2,3,4$.

Two numbers $x=\left(x_1,x_2,x_3,x_4\right)$ and $y=\left(y_1,y_2,y_3,y_4\right)$ are considered as equal if $x_i=y_i, i=1,2,3,4$.

Let us enter addition and subtraction operations as coordinate-wise addition and subtraction which are associative and commutative.

Let us enter multiplication operation so that it is associative and commutative. Let us give four real numbers $\alpha ,\beta ,\gamma ,\delta$ such that $\alpha ·\beta ·\gamma ·\delta \ne 0$. We will call the anisotropic product of numbers $x=\left(x_1,x_2,x_3,x_4\right)$ and $y=\left(y_1,y_2,y_3,y_4\right)$ a number $z=x·y=\left(z_1,z_2,z_3,z_4\right)$, where

$$\begin{array}{c}\begin{array}{c}{z}_1=\alpha x_1{y}_1+{\displaystyle \frac{\gamma \delta }{\alpha }}{x}_2{y}_2+{\displaystyle \frac{\beta \delta }{\alpha }}{x}_3{y}_3+{\displaystyle \frac{\beta \gamma }{\alpha }}{x}_4{y}_4\\ z_2=\alpha x_2{y}_1+\alpha x_1{y}_2+\beta x_4{y}_3+\beta x_3{y}_4\end{array}\\ \begin{array}{c}{z}_3=\alpha x_3{y}_1+\gamma x_4{y}_2+\alpha x_1{y}_3+\gamma x_2{y}_4\\ z_4=\alpha x_4{y}_1+\delta x_3{y}_2+\delta x_2{y}_3+\alpha x_1{y}_4\end{array}\end{array}$$

(1)

Further, instead of the anisotropic product, we will often use just “product” or $\left(\alpha ,\beta ,\gamma ,\delta \right)$-product.

Theorem 1.

The entered operation of multiplication meets the following conditions:

(1)

$x·y=y·x$ (commutativity of multiplication);

(2)

$\left(x·y\right)·z=x·\left(y·z\right)$(associativity of multiplication);

(3)

$\left(x+y\right)·z=x·z+y·z$ (associativity of multiplication concerning addition),

for any $x,y,z\in R^4$.

Proof. 

This is carried out by direct check. □

We will call $\left(x_1,0,0,0\right)$ the real number.

It follows from ratios (1) that at multiplication of real number a by four-dimensional number, there is the coordinate-wise multiplication on $\alpha ·a$, that is, $\left(a,0,0,0\right)·\left(x_1,x_2,x_3,x_4\right)=\left(\alpha a{x}_1,\alpha a{x}_2,\alpha a{x}_3,\alpha a{x}_4\right)$.

The following four numbers are called basic numbers: $J_1=\left({\displaystyle \frac{1}{\alpha }},0,0,0\right)$, $J_2=\left(0,{\displaystyle \frac{1}{\alpha }},0,0\right)$, $J_3=\left(0,0,{\displaystyle \frac{1}{\alpha }},0\right)$, $J_4=\left(0,0,0,{\displaystyle \frac{1}{\alpha }}\right)$.

Let us construct the multiplication table of basic numbers (

Table 1. Products of basic numbers for general case.

	$J_1$	$J_2$	$J_3$	$J_4$
$J_1$	$J_1$	$J_2$	$J_3$	$J_4$
$J_2$	$J_2$	${\displaystyle \frac{\gamma \delta }{{\alpha }^3}}{J}_1$	${\displaystyle \frac{\delta }{{\alpha }^2}}{J}_4$	${\displaystyle \frac{\gamma }{{\alpha }^2}}{J}_3$
$J_3$	$J_3$	${\displaystyle \frac{\delta }{{\alpha }^2}}{J}_4$	${\displaystyle \frac{\beta \delta }{{\alpha }^3}}{J}_1$	${\displaystyle \frac{\beta }{{\alpha }^2}}{J}_2$
$J_4$	$J_4$	${\displaystyle \frac{\gamma }{{\alpha }^2}}{J}_3$	${\displaystyle \frac{\beta }{{\alpha }^2}}{J}_2$	${\displaystyle \frac{\beta \gamma }{{\alpha }^3}}{J}_1$

):

Accordingly, any four-dimensional number $x=\left(x_1,x_2,x_3,x_4\right)$ can be represented as an expantion by basic numbers $x=x_1·J_1+x_2·J_2+x_3·J_3+x_4·J_4$.

The following four numbers are called unit numbers: $I_1=\left(1,0,0,0\right)$, $I_2=\left(0,1,0,0\right)$, $I_3=\left(0,0,1,0\right)$, $I_4=\left(0,0,0,1\right)$. For any number $x=\left(x_1,x_2,x_3,x_4\right)$ we have $x·I_1=\left(\alpha x_1,\alpha x_2,\alpha x_3,\alpha x_4\right)$.

Let $x=\left(x_1,x_2,x_3,x_4\right)$ be a four-dimensional number. Let us consider together with it the following numbers:

$$x^{\left(2\right)}=\left(x_1,x_2,-x_3,-x_4\right)=x_1{J}_1+x_2{J}_2-x_3{J}_3-x_4{J}_4,$$

$$x^{\left(3\right)}=\left(x_1,-x_2,x_3,-x_4\right)=x_1{J}_1-x_2{J}_2+x_3{J}_3-x_4{J}_4,$$

$$x^{\left(4\right)}=\left(x_1,-x_2,-x_3,x_4\right)=x_1{J}_1-x_2{J}_2-x_3{J}_3+x_4{J}_4.$$

Let us calculate the product $x·x^{\left(2\right)}·x^{\left(3\right)}·x^{\left(4\right)}$:

$$x·x^{\left(2\right)}=\left(\alpha x_1^2+{\displaystyle \frac{\gamma \delta }{\alpha }}{x}_2^2-{\displaystyle \frac{\beta \delta }{\alpha }}{x}_3^2-{\displaystyle \frac{\beta \gamma }{\alpha }}{x}_4^2, 2\alpha x_1{x}_2-2\beta x_3{x}_4, 0, 0\right),$$

$$\begin{array}{rl}\hfill x·x^{(2)}·x^{(3)}=(& {\alpha }^2{x}_1^3-\gamma \delta x_1{x}_2^2-\beta \delta x_1{x}_3^2-\beta \gamma x_1{x}_4^2+2\frac{\beta \gamma \delta }{\alpha }{x}_2{x}_3{x}_4,{\alpha }^2{x}_1^2{x}_2-\gamma \delta x_2^3+\beta \delta x_2{x}_3^2+\beta \gamma x_2{x}_4^2\hfill \\ & -2\alpha \beta x_1{x}_3{x}_4,{\alpha }^2{x}_1^2{x}_3+\gamma \delta x_2^2{x}_3-\beta \delta x_3^3+\beta \gamma x_3{x}_4^2-2\alpha \gamma x_1{x}_2{x}_4,2\alpha \delta x_1{x}_2{x}_3-\beta \delta x_3^2{x}_4\hfill \\ & -{\alpha }^2{x}_1^2{x}_4-\gamma \delta x_2^2{x}_4+\beta \gamma x_4^3),\hfill \end{array}$$

$$\begin{array}{cc}\hfill x·x^{\left(2\right)}·x^{\left(3\right)}·x^{\left(4\right)}& \hfill \\ \hfill & =({\alpha }^3{x}_1^4+{\displaystyle \frac{{\gamma }^2{\delta }^2}{\alpha }}{x}_2^4+{\displaystyle \frac{{\beta }^2{\delta }^2}{\alpha }}{x}_3^4+{\displaystyle \frac{{\beta }^2{\gamma }^2}{\alpha }}{x}_4^4-2\alpha \gamma \delta x_1^2{x}_2^2-2\alpha \beta \delta x_1^2{x}_3^2-2\alpha \beta \gamma x_1^2{x}_4^2\hfill \\ \hfill & -2{\displaystyle \frac{\beta \gamma {\delta }^2}{\alpha }}{x}_2^2{x}_3^2-2{\displaystyle \frac{\beta {\gamma }^2\delta }{\alpha }}{x}_2^2{x}_4^2-2{\displaystyle \frac{{\beta }^2\gamma \delta }{\alpha }}{x}_3^2{x}_4^2+8\beta \gamma \delta x_1{x}_2{x}_3{x}_4, 0, 0, 0).\hfill \end{array}$$

Thus, $x·x^{\left(2\right)}·x^{\left(3\right)}·x^{\left(4\right)}$ is the real number.

Definition 1.

The real number

$$\begin{gathered} \left|x\right|=({\alpha }^4{x}_1^4+{\gamma }^2{\delta }^2{x}_2^4+{\beta }^2{\delta }^2{x}_3^4+{\beta }^2{\gamma }^2{x}_4^4-2{\alpha }^2\gamma \delta x_1^2{x}_2^2-2{\alpha }^2\beta \delta x_1^2{x}_3^2-2{\alpha }^2\beta \gamma x_1^2{x}_4^2-2\beta \gamma {\delta }^2{x}_2^2{x}_3^2 \\ -2\beta {\gamma }^2\delta x_2^2{x}_4^2-2{\beta }^2\gamma \delta x_3^2{x}_4^2+8\alpha \beta \gamma \delta x_1{x}_2{x}_3{x}_4{)}^{{\displaystyle \frac{1}{4}}} \end{gathered}$$

(2)

is called the symplectic module of four-dimensional number $x=\left(x_1,x_2,x_3,x_4\right)$.

At the same time, we assume that numbers $\alpha ,\beta ,\gamma ,\delta$ are set so that expression under the radical is non-negative. As we will see below, such cases are possible, that is, there are such values $\alpha ,\beta ,\gamma ,\delta$ that the four-degree form in the right part of (2) is positively defined for any $x_i\in R,i=1,2,3,4$.

Definition 2.

The number $x^{*}=x^{\left(2\right)}·x^{\left(3\right)}·x^{\left(4\right)}$ is called the conjugate number to number  $x$. Then

$$x·x^{*}·I_1={\left|x\right|}^4·J_1.$$

(3)

Direct calculation gives the conjugate number $x^{*}=\left(x_1^{*},x_2^{*},x_3^{*},x_4^{*}\right)$ to four-dimensional number $x=\left(x_1,x_2,x_3,x_4\right)$:

$$x_1^{*}=x_1\left({\alpha }^2{x}_1^2-\gamma \delta x_2^2-\beta \delta x_3^2-\beta \gamma x_4^2\right)+{\displaystyle \frac{2\beta \gamma \delta }{\alpha }}{x}_2{x}_3{x}_4,$$

$$x_2^{*}=x_2\left(-{\alpha }^2{x}_1^2+\gamma \delta x_2^2-\beta \delta x_3^2-\beta \gamma x_4^2\right)+2\alpha \beta x_1{x}_3{x}_4,$$

$$x_3^{*}=x_3\left(-{\alpha }^2{x}_1^2-\gamma \delta x_2^2+\beta \delta x_3^2-\beta \gamma x_4^2\right)+2\alpha \gamma x_1{x}_2{x}_4,$$

$$x_4^{*}=x_4\left(-{\alpha }^2{x}_1^2-\gamma \delta x_2^2-\beta \delta x_3^2+\beta \gamma x_4^2\right)+2\alpha \delta x_1{x}_2{x}_3.$$

The conjugate numbers to basic numbers are as follows:

$$J_1^{*}=J_1, J_2^{*}={\displaystyle \frac{\gamma \delta }{{\alpha }^3}}·J_2, J_3^{*}={\displaystyle \frac{\beta \delta }{{\alpha }^3}}·J_3, J_4^{*}={\displaystyle \frac{\beta \gamma }{{\alpha }^3}}·J_4.$$

The conjugate numbers to unit numbers are as follows:

$$I_1^{*}={\alpha }^2{I}_1, I_2^{*}={\displaystyle \frac{\gamma \delta }{\alpha }}·I_2, I_3^{*}={\displaystyle \frac{\beta \delta }{\alpha }}·I_3, I_4^{*}={\displaystyle \frac{\beta \gamma }{\alpha }}·I_4$$

Let $x=\left(x_1,x_2,x_3,x_4\right)$ be a four-dimensional number with nonzero symplectic module. Then there exists a unique $x^{-1}$, called the inverse to x, so that $x·x^{-1}=x^{-1}·x=J_1$. Multiplying both parts of the last equality by $x^{*}·I_1$, we obtain $x^{-1}·x·x^{*}·I_1=x^{*}·I_1$ or ${\left|x\right|}^4{x}^{-1}=x^{*}·I_1$. Let us multiply both parts of this equality by number $\left({\displaystyle \frac{1}{\alpha ·{\left|x\right|}^4}},0,0,0\right)$, so that, taking into account that $\left({\displaystyle \frac{1}{\alpha ·{\left|x\right|}^4}},0,0,0\right)·{\left|x\right|}^4=\left({\displaystyle \frac{1}{\alpha ·{\left|x\right|}^4}},0,0,0\right)·\left({\left|x\right|}^4,0,0,0\right)=I_1$, we obtain $x^{-1}·I_1=x^{*}·\left({\displaystyle \frac{1}{\alpha ·{\left|x\right|}^4}},0,0,0\right)·I_1$. Reducing both parts of equality by $I_1$, we have

$$x^{-1}={\displaystyle \frac{1}{\alpha {\left|x\right|}^4}}·x^{*}=\left({\displaystyle \frac{x_1^{*}}{{\left|x\right|}^4}},{\displaystyle \frac{x_2^{*}}{{\left|x\right|}^4}},{\displaystyle \frac{x_3^{*}}{{\left|x\right|}^4}},{\displaystyle \frac{x_4^{*}}{{\left|x\right|}^4}}\right)$$

(4)

We then define division operation of four-dimensional numbers as

$${\displaystyle \frac{y}{x}}=y·x^{-1}={\displaystyle \frac{1}{\alpha {\left|x\right|}^4}}{x}^{*}y \mathrm{if} \left|x\right|\ne 0$$

We investigate values of constants $\alpha ,\beta ,\gamma ,\delta$ for which the symplectic module (2) is non-negative for any $x\in R^4$. For this purpose, we consider various cases of signs of these constants.

• Case $\beta >0$, $\gamma >0$, $\delta >0$.

This case coincides with the general case considered above. In this case, the symplectic module of four-dimensional number (2) can be represented in the following form:

$$\begin{array}{cc}\hfill {\left|x\right|}^4={\alpha }^4{x}_1^4& +{\gamma }^2{\delta }^2{x}_2^4+{\beta }^2{\delta }^2{x}_3^4+{\beta }^2{\gamma }^2{x}_4^4-2{\alpha }^2\gamma \delta x_1^2{x}_2^2-2{\alpha }^2\beta \delta x_1^2{x}_3^2-2{\alpha }^2\beta \gamma x_1^2{x}_4^2-2\beta \gamma {\delta }^2{x}_2^2{x}_3^2\hfill \\ \hfill & -2\beta {\gamma }^2\delta x_2^2{x}_4^2-2{\beta }^2\gamma \delta x_3^2{x}_4^2+8\alpha \beta \gamma \delta x_1{x}_2{x}_3{x}_4\hfill \\ \hfill & =\left[{\left(\alpha x_1-\sqrt{\beta \delta }{x}_3\right)}^2-{\left(\sqrt{\gamma \delta }{x}_2-\sqrt{\beta \gamma }{x}_4\right)}^2\right]\left[{\left(\alpha x_1+\sqrt{\beta \delta }{x}_3\right)}^2-{\left(\sqrt{\gamma \delta }{x}_2+\sqrt{\beta \gamma }{x}_4\right)}^2\right],\hfill \end{array}$$

or in the following forms:

$${\left|x\right|}^4=\left[{\left(\alpha x_1-\sqrt{\gamma \delta }{x}_2\right)}^2-{\left(\sqrt{\beta \delta }{x}_3-\sqrt{\beta \gamma }{x}_4\right)}^2\right]\left[{\left(\alpha x_1+\sqrt{\gamma \delta }{x}_2\right)}^2-{\left(\sqrt{\beta \delta }{x}_3+\sqrt{\beta \gamma }{x}_4\right)}^2\right],$$

and

$${\left|x\right|}^4=\left[{\left(\alpha x_1-\sqrt{\beta \gamma }{x}_4\right)}^2-{\left(\sqrt{\gamma \delta }{x}_2-\sqrt{\beta \delta }{x}_3\right)}^2\right]\left[{\left(\alpha x_1+\sqrt{\beta \gamma }{x}_4\right)}^2-{\left(\sqrt{\gamma \delta }{x}_2+\sqrt{\beta \delta }{x}_3\right)}^2\right].$$

As can be seen from these formulas, the symplectic module of number in this case is not positively defined, so that is not the module. If the module of number is not well-defined, then further constructions do not make sense.

2.

Case $\beta <0$, $\gamma >0$, $\delta >0$.

In this case, we will replace in definition of multiplication of numbers (1) $\beta$ with $-\beta$ and we will consider $\beta >0$ subsequently. Next, multiplication of two four-dimensional numbers is defined by the following formulas:

$$\begin{array}{c}\begin{array}{c}{z}_1=\alpha x_1{y}_1+{\displaystyle \frac{\gamma \delta }{\alpha }}{x}_2{y}_2-{\displaystyle \frac{\beta \delta }{\alpha }}{x}_3{y}_3-{\displaystyle \frac{\beta \gamma }{\alpha }}{x}_4{y}_4\\ z_2=\alpha x_2{y}_1+\alpha x_1{y}_2-\beta x_4{y}_3-\beta x_3{y}_4\end{array}\\ \begin{array}{c}{z}_3=\alpha x_3{y}_1+\gamma x_4{y}_2+\alpha x_1{y}_3+\gamma x_2{y}_4\\ z_4=\alpha x_4{y}_1+\delta x_3{y}_2+\delta x_2{y}_3+\alpha x_1{y}_4\end{array}\end{array}$$

(5)

Next, multiplication table of basic numbers takes the following form:

In this case, the module of four-dimensional number (2) can be transformed as follows:

$$\begin{gathered} {\left|x\right|}^4={\alpha }^4{x}_1^4+{\gamma }^2{\delta }^2{x}_2^4+{\beta }^2{\delta }^2{x}_3^4+{\beta }^2{\gamma }^2{x}_4^4-2{\alpha }^2\gamma \delta x_1^2{x}_2^2+2{\alpha }^2\beta \delta x_1^2{x}_3^2+2{\alpha }^2\beta \gamma x_1^2{x}_4^2+2\beta \gamma {\delta }^2{x}_2^2{x}_3^2+ \\ 2\beta {\gamma }^2\delta x_2^2{x}_4^2-2{\beta }^2\gamma \delta x_3^2{x}_4^2-8\alpha \beta \gamma \delta x_1{x}_2{x}_3{x}_4=\left[{\left(\alpha x_1-\sqrt{\gamma \delta }{x}_2\right)}^2+{\left(\sqrt{\beta \delta }{x}_3-\sqrt{\beta \gamma }{x}_4\right)}^2\right]· \\ \left[{\left(\alpha x_1+\sqrt{\gamma \delta }{x}_2\right)}^2+{\left(\sqrt{\beta \delta }{x}_3+\sqrt{\beta \gamma }{x}_4\right)}^2\right]. \end{gathered}$$

(6)

As shown in

Table 2. Products of basic numbers in Case 2.

	$J_1$	$J_2$	$J_3$	$J_4$
$J_1$	$J_1$	$J_2$	$J_3$	$J_4$
$J_2$	$J_2$	${\displaystyle \frac{\gamma \delta }{{\alpha }^3}}{J}_1$	${\displaystyle \frac{\delta }{{\alpha }^2}}{J}_4$	${\displaystyle \frac{\gamma }{{\alpha }^2}}{J}_3$
$J_3$	$J_3$	${\displaystyle \frac{\delta }{{\alpha }^2}}{J}_4$	$-{\displaystyle \frac{\beta \delta }{{\alpha }^3}}{J}_1$	$-{\displaystyle \frac{\beta }{{\alpha }^2}}{J}_2$
$J_4$	$J_4$	${\displaystyle \frac{\gamma }{{\alpha }^2}}{J}_3$	$-{\displaystyle \frac{\beta }{{\alpha }^2}}{J}_2$	$-{\displaystyle \frac{\beta \gamma }{{\alpha }^3}}{J}_1$

, the basic numbers $J_1$ and $J_2$ are real, and basic numbers $J_3$ and $J_4$ are imaginary. At the same time, the module of number $\left(x_1,0,x_3,0\right)$ is defined by the formula $\left|x\right|=\sqrt{{\alpha }^2{x}_1^2+\beta \delta x_3^2}$, that is, the number $\left(x_1,0,x_3,0\right)$ can be accepted as the imaginary.

The conjugate number $x^{*}=\left(x_1^{*},x_2^{*},x_3^{*},x_4^{*}\right)$ to the number $x=\left(x_1,x_2,x_3,x_4\right)$ is defined by the following formula:

$$x_1^{*}=x_1\left({\alpha }^2{x}_1^2-\gamma \delta x_2^2+\beta \delta x_3^2+\beta \gamma x_4^2\right)-{\displaystyle \frac{2\beta \gamma \delta }{\alpha }}{x}_2{x}_3{x}_4,$$

$$x_2^{*}=x_2\left(-{\alpha }^2{x}_1^2+\gamma \delta x_2^2+\beta \delta x_3^2+\beta \gamma x_4^2\right)-2\alpha \beta x_1{x}_3{x}_4,$$

$$x_3^{*}=x_3\left(-{\alpha }^2{x}_1^2-\gamma \delta x_2^2-\beta \delta x_3^2+\beta \gamma x_4^2\right)+2\alpha \gamma x_1{x}_2{x}_4,$$

$$x_4^{*}=x_4\left(-{\alpha }^2{x}_1^2-\gamma \delta x_2^2+\beta \delta x_3^2-\beta \gamma x_4^2\right)+2\alpha \delta x_1{x}_2{x}_3.$$

The conjugate numbers to basic numbers have the following form:

$$J_1^{*}=J_1, J_2^{*}={\displaystyle \frac{\gamma \delta }{{\alpha }^3}}·J_2, J_3^{*}=-{\displaystyle \frac{\beta \delta }{{\alpha }^3}}·J_3, J_4^{*}=-{\displaystyle \frac{\beta \gamma }{{\alpha }^3}}·J_4.$$

The conjugate numbers to unit numbers have the following form:

$$I_1^{*}={\alpha }^2{I}_1, I_2^{*}={\displaystyle \frac{\gamma \delta }{\alpha }}·I_2, I_3^{*}=-{\displaystyle \frac{\beta \delta }{\alpha }}·I_3, I_4^{*}=-{\displaystyle \frac{\beta \gamma }{\alpha }}·I_4.$$

The conjugate number to the imaginary number $\left(x_1,0,x_3,0\right)$ has the form ${\displaystyle \frac{{\left|x\right|}^2}{\alpha }}\left(x_1,0,-x_3,0\right)$, which easily follows from the last formulas.

Let us denote the received space of four-dimensional numbers by $M_2\left(\alpha ,\beta ,\gamma ,\delta \right)$, where the index 2 stands for the number of the considered case, and we will call it anisotropic space of four-dimensional numbers.

Let us associate with each four-dimensional number $x=\left(x_1,x_2,x_3,x_4\right)$ some matrix $F\left(x\right)$ of the following form:

$$F\left(x\right)=\left(\begin{array}{cc}\begin{array}{cc}\alpha x_1& {\displaystyle \frac{\gamma \delta }{\alpha }}{x}_2\\ \alpha x_2& \alpha x_1\end{array}& \begin{array}{cc}-{\displaystyle \frac{\beta \delta }{\alpha }}{x}_3& -{\displaystyle \frac{\beta \gamma }{\alpha }}{x}_4\\ -\beta x_4& -\beta x_3\end{array}\\ \begin{array}{cc}\alpha x_3& \gamma x_4\\ \alpha x_4& \delta x_3\end{array}& \begin{array}{cc}\alpha x_1& \gamma x_2\\ \delta x_2& \alpha x_1\end{array}\end{array}\right)$$

(7)

The mapping $F:x\to F\left(x\right)$ is one-to-one and surjection. Indeed, for two different numbers, $x$ and $y$, there correspond different matrices, and for any matrix of the specified form, it is possible to find the corresponding four-dimensional number.

Next, multiplication of two four-dimensional numbers $x=\left(x_1,x_2,x_3,x_4\right)$ and $y=\left(y_1,y_2,y_3,y_4\right)$ in the space $M_2\left(\alpha ,\beta ,\gamma ,\delta \right)$ can be represented in the form $x·y=F\left(x\right)·y$, where the multiplication sign in the left part is understood in the sense of (5), and multiplication sign in the right part is treated as multiplication of the matrix by the vector.

Thus, we define alternative definition of multiplication of four-dimensional numbers by the matrix (7).

The inverse number to four-dimensional number $x=\left(x_1,x_2,x_3,x_4\right)$ in the sense of multiplication is defined by Formula (4), if $\left|x\right|\ne 0$. Next, we define division operation of four-dimensional numbers as $\frac{y}{x}=y\cdot x^{-1}=\frac{1}{\alpha |x{|}^4}{x}^{\ast }y$, if $\left|x\right|\ne 0$.

We will call space $M_2\left(1,1,1,1\right)$ isotropic space $M_2$. In isotropic space, the multiplication of two numbers is defined by equalities

$$\begin{array}{c}\begin{array}{c}{z}_1=x_1{y}_1+x_2{y}_2-x_3{y}_3-x_4{y}_4\\ z_2=x_2{y}_1+x_1{y}_2-x_4{y}_3-x_3{y}_4\end{array}\\ \begin{array}{c}{z}_3=x_3{y}_1+x_4{y}_2+x_1{y}_3+x_2{y}_4\\ z_4=x_4{y}_1+x_3{y}_2+x_2{y}_3+\alpha x_1{y}_4\end{array}\end{array},$$

(8)

and the matrix $F\left(x\right)$ has the form

$$F\left(x\right)=\left(\begin{array}{cc}\begin{array}{cc}{x}_1& x_2\\ x_2& x_1\end{array}& \begin{array}{cc}-x_3& -x_4\\ -x_4& -x_3\end{array}\\ \begin{array}{cc}{x}_3& x_4\\ x_4& x_3\end{array}& \begin{array}{cc}{x}_1& x_2\\ x_2& x_1\end{array}\end{array}\right)$$

(9)

The symplectic module of four-dimensional number in isotropic space $M_2$ has the form

$$\left|x\right|=\sqrt[4]{\left[{\left(x_1-x_2\right)}^2+{\left(x_3-x_4\right)}^2\right]·\left[{\left(x_1+x_2\right)}^2+{\left(x_3+x_4\right)}^2\right]}.$$

(10)

3.

Case $\beta >0$, $\gamma <0$, $\delta >0$.

In this case, we will replace in definition of multiplication of numbers (1) $\gamma$ with $-\gamma$ and we will consider $\gamma >0$ subsequently. Next, multiplication of two four-dimensional numbers is defined as follows:

$$\begin{array}{c}\begin{array}{c}{z}_1=\alpha x_1{y}_1-{\displaystyle \frac{\gamma \delta }{\alpha }}{x}_2{y}_2+{\displaystyle \frac{\beta \delta }{\alpha }}{x}_3{y}_3-{\displaystyle \frac{\beta \gamma }{\alpha }}{x}_4{y}_4\\ z_2=\alpha x_2{y}_1+\alpha x_1{y}_2+\beta x_4{y}_3+\beta x_3{y}_4\end{array}\\ \begin{array}{c}{z}_3=\alpha x_3{y}_1-\gamma x_4{y}_2+\alpha x_1{y}_3-\gamma x_2{y}_4\\ z_4=\alpha x_4{y}_1+\delta x_3{y}_2+\delta x_2{y}_3+\alpha x_1{y}_4\end{array}\end{array}.$$

(11)

The multiplication table of basic numbers is given in the following table.

In this case, the module of four-dimensional number (2) can be transformed as follows:

$$\begin{gathered} {\left|x\right|}^4={\alpha }^4{x}_1^4+{\gamma }^2{\delta }^2{x}_2^4+{\beta }^2{\delta }^2{x}_3^4+{\beta }^2{\gamma }^2{x}_4^4+2{\alpha }^2\gamma \delta x_1^2{x}_2^2-2{\alpha }^2\beta \delta x_1^2{x}_3^2+2{\alpha }^2\beta \gamma x_1^2{x}_4^2+2\beta \gamma {\delta }^2{x}_2^2{x}_3^2- \\ 2\beta {\gamma }^2\delta x_2^2{x}_4^2+2{\beta }^2\gamma \delta x_3^2{x}_4^2-8\alpha \beta \gamma \delta x_1{x}_2{x}_3{x}_4=\left[{\left(\alpha x_1-\sqrt{\beta \delta }{x}_3\right)}^2+{\left(\sqrt{\gamma \delta }{x}_2-\sqrt{\beta \gamma }{x}_4\right)}^2\right]· \\ \left[{\left(\alpha x_1+\sqrt{\beta \delta }{x}_3\right)}^2+{\left(\sqrt{\gamma \delta }{x}_2+\sqrt{\beta \gamma }{x}_4\right)}^2\right]. \end{gathered}$$

(12)

As shown in

Table 3. Products of basic numbers in Case 3.

	$J_1$	$J_2$	$J_3$	$J_4$
$J_1$	$J_1$		$J_3$	$J_4$
$J_2$	$J_2$	$-{\displaystyle \frac{\gamma \delta }{{\alpha }^3}}{J}_1$	${\displaystyle \frac{\delta }{{\alpha }^2}}{J}_4$	$-{\displaystyle \frac{\gamma }{{\alpha }^2}}{J}_3$
$J_3$	$J_3$	${\displaystyle \frac{\delta }{{\alpha }^2}}{J}_4$	${\displaystyle \frac{\beta \delta }{{\alpha }^3}}{J}_1$	${\displaystyle \frac{\beta }{{\alpha }^2}}{J}_2$
$J_4$	$J_4$	$-{\displaystyle \frac{\gamma }{{\alpha }^2}}{J}_3$	${\displaystyle \frac{\beta }{{\alpha }^2}}{J}_2$	$-{\displaystyle \frac{\beta \gamma }{{\alpha }^3}}{J}_1$

, the basic numbers $J_1$ and $J_3$ are real, and basic numbers $J_2$ and $J_4$ are imaginary. Therefore, in this case, as the imaginary, we will accept number $\left(x_1,x_2,0,0\right)$. The module of the imaginary is defined by the formula $\left|x\right|=\sqrt{{\alpha }^2{x}_1^2+\gamma \delta x_2^2}$.

The conjugate number $x^{*}=\left(x_1^{*},x_2^{*},x_3^{*},x_4^{*}\right)$ to number $x=\left(x_1,x_2,x_3,x_4\right)$ is defined by the following formula:

$$x_1^{*}=x_1\left({\alpha }^2{x}_1^2+\gamma \delta x_2^2-\beta \delta x_3^2+\beta \gamma x_4^2\right)-{\displaystyle \frac{2\beta \gamma \delta }{\alpha }}{x}_2{x}_3{x}_4,$$

$$x_2^{*}=x_2\left(-{\alpha }^2{x}_1^2-\gamma \delta x_2^2-\beta \delta x_3^2+\beta \gamma x_4^2\right)+2\alpha \beta x_1{x}_3{x}_4,$$

$$x_3^{*}=x_3\left(-{\alpha }^2{x}_1^2+\gamma \delta x_2^2+\beta \delta x_3^2+\beta \gamma x_4^2\right)-2\alpha \gamma x_1{x}_2{x}_4,$$

$$x_4^{*}=x_4\left(-{\alpha }^2{x}_1^2+\gamma \delta x_2^2-\beta \delta x_3^2-\beta \gamma x_4^2\right)+2\alpha \delta x_1{x}_2{x}_3.$$

The conjugate numbers to basic numbers have the following form:

$$J_1^{*}=J_1, J_2^{*}=-{\displaystyle \frac{\gamma \delta }{{\alpha }^3}}·J_2, J_3^{*}={\displaystyle \frac{\beta \delta }{{\alpha }^3}}·J_3, J_4^{*}=-{\displaystyle \frac{\beta \gamma }{{\alpha }^3}}·J_4.$$

The conjugate numbers to unit numbers have the following form:

$$I_1^{*}={\alpha }^2{I}_1, I_2^{*}=-{\displaystyle \frac{\gamma \delta }{\alpha }}·I_2, I_3^{*}={\displaystyle \frac{\beta \delta }{\alpha }}·I_3, I_4^{*}=-{\displaystyle \frac{\beta \gamma }{\alpha }}·I_4.$$

The conjugate number to the imaginary number $x=\left(x_1,x_2,0,0\right)$ has the form ${\displaystyle \frac{{\left|x\right|}^2}{\alpha }}\left(x_1,-x_2,0,0\right)$.

Let us denote the received anisotropic space of four-dimensional numbers, similarly to the previous case, by $M_3\left(\alpha ,\beta ,\gamma ,\delta \right)$.

The matrix F(x) from (7) in this case looks as follows:

$$\left(x\right)=\left(\begin{array}{cc}\begin{array}{cc}\alpha x_1& -{\displaystyle \frac{\gamma \delta }{\alpha }}{x}_2\\ \alpha x_2& \alpha x_1\end{array}& \begin{array}{cc}{\displaystyle \frac{\beta \delta }{\alpha }}{x}_3& -{\displaystyle \frac{\beta \gamma }{\alpha }}{x}_4\\ \beta x_4& \beta x_3\end{array}\\ \begin{array}{cc}\alpha x_3& -\gamma x_4\\ \alpha x_4& \delta x_3\end{array}& \begin{array}{cc}\alpha x_1& -\gamma x_2\\ \delta x_2& \alpha x_1\end{array}\end{array}\right)$$

(13)

The mapping $F:x\to F\left(x\right)$ is one-to-one and onto, and multiplication of two four-dimensional numbers $x=\left(x_1,x_2,x_3,x_4\right)$ and $y=\left(y_1,y_2,y_3,y_4\right)$ in space $M_3\left(\alpha ,\beta ,\gamma ,\delta \right)$ can be represented in the form $x·y=F\left(x\right)·y$. Division operation is defined similarly to Case 2.

We will call space $M_3\left(1,1,1,1\right)$ isotropic space $M_3$. In isotropic space, the multiplication of two numbers is defined by the following equalities:

$$\begin{array}{c}\begin{array}{c}{z}_1=x_1{y}_1-x_2{y}_2+x_3{y}_3-x_4{y}_4\\ z_2=x_2{y}_1+x_1{y}_2+x_4{y}_3+x_3{y}_4\end{array}\\ \begin{array}{c}{z}_3=x_3{y}_1-x_4{y}_2+x_1{y}_3-x_2{y}_4\\ z_4=x_4{y}_1+x_3{y}_2+x_2{y}_3+\alpha x_1{y}_4\end{array}\end{array}$$

(14)

and the matrix $F\left(x\right)$ has the following form:

$$F\left(x\right)=\left(\begin{array}{cc}\begin{array}{cc}{x}_1& -x_2\\ x_2& x_1\end{array}& \begin{array}{cc}{x}_3& -x_4\\ x_4& x_3\end{array}\\ \begin{array}{cc}{x}_3& -x_4\\ x_4& x_3\end{array}& \begin{array}{cc}{x}_1& -x_2\\ x_2& x_1\end{array}\end{array}\right)$$

(15)

The symplectic module of four-dimensional number in isotropic space $M_3$ has the following form:

$$\left|x\right|=\sqrt[4]{\left[{\left(x_1-x_3\right)}^2+{\left(x_2-x_4\right)}^2\right]·\left[{\left(x_1+x_3\right)}^2+{\left(x_2+x_4\right)}^2\right]}$$

(16)

It is isotropic space $M_3$ that is considered in [6].

4.

Case $\beta >0$, $\gamma >0$, $\delta <0$.

In this case, we will replace in definition of multiplication of numbers (1) $\delta$ with $-\delta$ and we will consider $\delta >0$ subsequently. Next, multiplication of two four-dimensional numbers is defined by the following formulas:

$$\begin{array}{c}\begin{array}{c}{z}_1=\alpha x_1{y}_1-{\displaystyle \frac{\gamma \delta }{\alpha }}{x}_2{y}_2-{\displaystyle \frac{\beta \delta }{\alpha }}{x}_3{y}_3+{\displaystyle \frac{\beta \gamma }{\alpha }}{x}_4{y}_4\\ z_2=\alpha x_2{y}_1+\alpha x_1{y}_2+\beta x_4{y}_3+\beta x_3{y}_4\end{array}\\ \begin{array}{c}{z}_3=\alpha x_3{y}_1+\gamma x_4{y}_2+\alpha x_1{y}_3+\gamma x_2{y}_4\\ z_4=\alpha x_4{y}_1-\delta x_3{y}_2-\delta x_2{y}_3+\alpha x_1{y}_4\end{array}\end{array}$$

(17)

The multiplication table of basic numbers takes the form given in the next table.

In this case, the module of four-dimensional number (2) can be transformed as follows:

$$\begin{gathered} {\left|x\right|}^4={\alpha }^4{x}_1^4+{\gamma }^2{\delta }^2{x}_2^4+{\beta }^2{\delta }^2{x}_3^4+{\beta }^2{\gamma }^2{x}_4^4+2{\alpha }^2\gamma \delta x_1^2{x}_2^2+2{\alpha }^2\beta \delta x_1^2{x}_3^2-2{\alpha }^2\beta \gamma x_1^2{x}_4^2-2\beta \gamma {\delta }^2{x}_2^2{x}_3^2+ \\ 2\beta {\gamma }^2\delta x_2^2{x}_4^2+2{\beta }^2\gamma \delta x_3^2{x}_4^2-8\alpha \beta \gamma \delta x_1{x}_2{x}_3{x}_4=\left[{\left(\alpha x_1-\sqrt{\beta \gamma }{x}_4\right)}^2+{\left(\sqrt{\gamma \delta }{x}_2-\sqrt{\beta \delta }{x}_3\right)}^2\right]· \\ \left[{\left(\alpha x_1+\sqrt{\beta \gamma }{x}_4\right)}^2+{\left(\sqrt{\gamma \delta }{x}_2+\sqrt{\beta \delta }{x}_3\right)}^2\right]. \end{gathered}$$

(18)

As can be seen from

Table 4. Products of basic numbers in Case 4.

	$J_1$	$J_2$	$J_3$	$J_4$
$J_1$	$J_1$	$J_2$	$J_3$	$J_4$
$J_2$	$J_2$	$-{\displaystyle \frac{\gamma \delta }{{\alpha }^2}}{J}_1$	$-{\displaystyle \frac{\delta }{\alpha }}{J}_4$	${\displaystyle \frac{\gamma }{\alpha }}{J}_3$
$J_3$	$J_3$	$-{\displaystyle \frac{\delta }{\alpha }}{J}_4$	$-{\displaystyle \frac{\beta \delta }{{\alpha }^2}}{J}_1$	${\displaystyle \frac{\beta }{\alpha }}{J}_2$
$J_4$	$J_4$	${\displaystyle \frac{\gamma }{\alpha }}{J}_3$	${\displaystyle \frac{\beta }{\alpha }}{J}_2$	${\displaystyle \frac{\beta \gamma }{{\alpha }^2}}{J}_1$

, the basic numbers $J_1$ and $J_4$ are real, and basic numbers $J_2$ and $J_3$ are imaginary. In this case, as the imaginary number, we will accept number $\left(x_1,x_2,0,0\right)$. The module of the imaginary is defined by the formula $\left|x\right|=\sqrt{{\alpha }^2{x}_1^2+\gamma \delta x_2^2}$.

The conjugate number $x^{*}=\left(x_1^{*},x_2^{*},x_3^{*},x_4^{*}\right)$ to number $x=\left(x_1,x_2,x_3,x_4\right)$ is defined by the following formula:

$$x_1^{*}=x_1\left({\alpha }^2{x}_1^2+\gamma \delta x_2^2+\beta \delta x_3^2-\beta \gamma x_4^2\right)-{\displaystyle \frac{2\beta \gamma \delta }{\alpha }}{x}_2{x}_3{x}_4,$$

$$x_2^{*}=x_2\left(-{\alpha }^2{x}_1^2-\gamma \delta x_2^2+\beta \delta x_3^2-\beta \gamma x_4^2\right)+2\alpha \beta x_1{x}_3{x}_4,$$

$$x_3^{*}=x_3\left(-{\alpha }^2{x}_1^2+\gamma \delta x_2^2-\beta \delta x_3^2-\beta \gamma x_4^2\right)+2\alpha \gamma x_1{x}_2{x}_4,$$

$$x_4^{*}=x_4\left(-{\alpha }^2{x}_1^2+\gamma \delta x_2^2+\beta \delta x_3^2+\beta \gamma x_4^2\right)-2\alpha \delta x_1{x}_2{x}_3.$$

Respectively, the conjugate numbers to basic numbers have the following form:

$$J_1^{*}=J_1, J_2^{*}=-{\displaystyle \frac{\gamma \delta }{{\alpha }^3}}·J_2, J_3^{*}=-{\displaystyle \frac{\beta \delta }{{\alpha }^3}}·J_3, J_4^{*}={\displaystyle \frac{\beta \gamma }{{\alpha }^3}}·J_4.$$

The conjugate numbers to unit numbers have the following form:

$$I_1^{*}={\alpha }^2{I}_1, I_2^{*}=-{\displaystyle \frac{\gamma \delta }{\alpha }}·I_2, I_3^{*}=-{\displaystyle \frac{\beta \delta }{\alpha }}·I_3, I_4^{*}={\displaystyle \frac{\beta \gamma }{\alpha }}·I_4.$$

The conjugate number to the imaginary number $x=\left(x_1,x_2,0,0\right)$ has the form ${\displaystyle \frac{{\left|x\right|}^2}{\alpha }}\left(x_1,-x_2,0,0\right)$.

Let us denote the obtained anisotropic space of four-dimensional numbers by $M_4\left(\alpha ,\beta ,\gamma ,\delta \right)$.

The matrix $F\left(x\right)$ from (7) in this case looks as follows:

$$F\left(x\right)=\left(\begin{array}{cc}\begin{array}{cc}\alpha x_1& -{\displaystyle \frac{\gamma \delta }{\alpha }}{x}_2\\ \alpha x_2& \alpha x_1\end{array}& \begin{array}{cc}-{\displaystyle \frac{\beta \delta }{\alpha }}{x}_3& {\displaystyle \frac{\beta \gamma }{\alpha }}{x}_4\\ \beta x_4& \beta x_3\end{array}\\ \begin{array}{cc}\alpha x_3& \gamma x_4\\ \alpha x_4& -\delta x_3\end{array}& \begin{array}{cc}\alpha x_1& \gamma x_2\\ -\delta x_2& \alpha x_1\end{array}\end{array}\right)$$

(19)

It is also easily proven that the mapping $F:x\to F\left(x\right)$ is one-to-one and onto. Alternative definitions of multiplication and division operations are defined, as well as those in the previous cases.

We will call space $M_4$(1,1,1,1) isotropic space $M_4$. In isotropic space, the multiplication of two numbers is defined by the following equalities:

$$\begin{array}{c}\begin{array}{c}{z}_1=x_1{y}_1-x_2{y}_2-x_3{y}_3+x_4{y}_4\\ z_2=x_2{y}_1+x_1{y}_2+x_4{y}_3+x_3{y}_4\end{array}\\ \begin{array}{c}{z}_3=x_3{y}_1+x_4{y}_2+x_1{y}_3+x_2{y}_4\\ z_4=x_4{y}_1-x_3{y}_2-x_2{y}_3+\alpha x_1{y}_4\end{array}\end{array},$$

(20)

and the matrix $F\left(x\right)$ has the following form:

$$F\left(x\right)=\left(\begin{array}{cc}\begin{array}{cc}{x}_1& -x_2\\ x_2& x_1\end{array}& \begin{array}{cc}-x_3& x_4\\ x_4& x_3\end{array}\\ \begin{array}{cc}{x}_3& x_4\\ x_4& -x_3\end{array}& \begin{array}{cc}{x}_1& x_2\\ -x_2& x_1\end{array}\end{array}\right)$$

(21)

The symplectic module of four-dimensional number in isotropic space $M_4$ has the following form:

$$\left|x\right|=\sqrt[4]{\left[{\left(x_1-x_4\right)}^2+{\left(x_2-x_3\right)}^2\right]·\left[{\left(x_1+x_4\right)}^2+{\left(x_2+x_3\right)}^2\right]}$$

(22)

5.

Case $\beta <0$, $\gamma <0$, $\delta >0$.

In this case, multiplication of two four-dimensional numbers is defined by the following formulas:

$$\begin{array}{c}\begin{array}{c}{z}_1=\alpha x_1{y}_1-{\displaystyle \frac{\gamma \delta }{\alpha }}{x}_2{y}_2-{\displaystyle \frac{\beta \delta }{\alpha }}{x}_3{y}_3+{\displaystyle \frac{\beta \gamma }{\alpha }}{x}_4{y}_4\\ z_2=\alpha x_2{y}_1+\alpha x_1{y}_2-\beta x_4{y}_3-\beta x_3{y}_4\end{array}\\ \begin{array}{c}{z}_3=\alpha x_3{y}_1-\gamma x_4{y}_2+\alpha x_1{y}_3-\gamma x_2{y}_4\\ z_4=\alpha x_4{y}_1+\delta x_3{y}_2+\delta x_2{y}_3+\alpha x_1{y}_4\end{array}\end{array}.$$

(23)

At the same time, further, we consider that $\beta >0$, $\gamma >0$, $\delta >0$.

The multiplication table of basic numbers takes the form given in

Table 5. Products of basic numbers in Case 5.

	$J_1$	$J_2$	$J_3$	$J_4$
$J_1$	$J_1$	$J_2$	$J_3$	$J_4$
$J_2$	$J_2$	$-{\displaystyle \frac{\gamma \delta }{{\alpha }^3}}{J}_1$	${\displaystyle \frac{\delta }{{\alpha }^2}}{J}_4$	$-{\displaystyle \frac{\gamma }{{\alpha }^2}}{J}_3$
$J_3$	$J_3$	${\displaystyle \frac{\delta }{{\alpha }^2}}{J}_4$	$-{\displaystyle \frac{\beta \delta }{{\alpha }^3}}{J}_1$	$-{\displaystyle \frac{\beta }{{\alpha }^2}}{J}_2$
$J_4$	$J_4$	$-{\displaystyle \frac{\gamma }{{\alpha }^2}}{J}_3$	$-{\displaystyle \frac{\beta }{{\alpha }^2}}{J}_2$	${\displaystyle \frac{\beta \gamma }{{\alpha }^3}}{J}_1$

.

In this case, the module of four-dimensional number (2) can be transformed as follows:

$$\begin{array}{cc}\hfill {\left|x\right|}^4={\alpha }^4{x}_1^4+& {\gamma }^2{\delta }^2{x}_2^4+{\beta }^2{\delta }^2{x}_3^4+{\beta }^2{\gamma }^2{x}_4^4+2{\alpha }^2\gamma \delta x_1^2{x}_2^2+2{\alpha }^2\beta \delta x_1^2{x}_3^2-2{\alpha }^2\beta \gamma x_1^2{x}_4^2-2\beta \gamma {\delta }^2{x}_2^2{x}_3^2\hfill \\ \hfill & +2\beta {\gamma }^2\delta x_2^2{x}_4^2+2{\beta }^2\gamma \delta x_3^2{x}_4^2+8\alpha \beta \gamma \delta x_1{x}_2{x}_3{x}_4\hfill \\ \hfill & =\left[{\left(\alpha x_1-\sqrt{\beta \gamma }{x}_4\right)}^2+{\left(\sqrt{\gamma \delta }{x}_2+\sqrt{\beta \delta }{x}_3\right)}^2\right]·\left[{\left(\alpha x_1+\sqrt{\beta \gamma }{x}_4\right)}^2+{\left(\sqrt{\gamma \delta }{x}_2-\sqrt{\beta \delta }{x}_3\right)}^2\right]\hfill \end{array}$$

(24)

In this case, the module of the imaginary number is defined by the formula $\left|x\right|=\sqrt{{\alpha }^2{x}_1^2+\gamma \delta x_2^2}$ because as imaginary numbers we take $\left(x_1,x_2,0,0\right)$, as well as in Case 4.

Notice that if modules of four-dimensional numbers in Cases 2, 3 and 4 are expressed similarly, then the number module in this fifth case significantly differs from the previous cases.

The conjugate number $x^{*}=\left(x_1^{*},x_2^{*},x_3^{*},x_4^{*}\right)$ to number $x=\left(x_1,x_2,x_3,x_4\right)$ is defined by the following formula:

$$x_1^{*}=x_1\left({\alpha }^2{x}_1^2+\gamma \delta x_2^2+\beta \delta x_3^2-\beta \gamma x_4^2\right)+{\displaystyle \frac{2\beta \gamma \delta }{\alpha }}{x}_2{x}_3{x}_4,$$

$$x_2^{*}=x_2\left(-{\alpha }^2{x}_1^2-\gamma \delta x_2^2+\beta \delta x_3^2-\beta \gamma x_4^2\right)-2\alpha \beta x_1{x}_3{x}_4,$$

$$x_3^{*}=x_3\left(-{\alpha }^2{x}_1^2+\gamma \delta x_2^2-\beta \delta x_3^2-\beta \gamma x_4^2\right)-2\alpha \gamma x_1{x}_2{x}_4,$$

$$x_4^{*}=x_4\left(-{\alpha }^2{x}_1^2+\gamma \delta x_2^2+\beta \delta x_3^2+\beta \gamma x_4^2\right)+2\alpha \delta x_1{x}_2{x}_3.$$

The conjugate numbers to basic numbers have the following form:

$$J_1^{*}=J_1, J_2^{*}=-{\displaystyle \frac{\gamma \delta }{{\alpha }^3}}·J_2, J_3^{*}=-{\displaystyle \frac{\beta \delta }{{\alpha }^3}}·J_3, J_4^{*}={\displaystyle \frac{\beta \gamma }{{\alpha }^3}}·J_4.$$

The conjugate numbers to unit numbers have the following form:

$$I_1^{*}={\alpha }^2{I}_1, I_2^{*}=-{\displaystyle \frac{\gamma \delta }{\alpha }}·I_2, I_3^{*}=-{\displaystyle \frac{\beta \delta }{\alpha }}·I_3, I_4^{*}={\displaystyle \frac{\beta \gamma }{\alpha }}·I_4.$$

The conjugate number to the imaginary number $x=\left(x_1,x_2,0,0\right)$ has the form ${\displaystyle \frac{{\left|x\right|}^2}{\alpha }}\left(x_1,-x_2,0,0\right)$.

Let us denote the obtained space by $M_5\left(\alpha ,\beta ,\gamma ,\delta \right)$.

The matrix $F\left(x\right)$ from (7) in this case looks as follows:

$$F\left(x\right)=\left(\begin{array}{cc}\begin{array}{cc}\alpha x_1& -{\displaystyle \frac{\gamma \delta }{\alpha }}{x}_2\\ \alpha x_2& \alpha x_1\end{array}& \begin{array}{cc}-{\displaystyle \frac{\beta \delta }{\alpha }}{x}_3& {\displaystyle \frac{\beta \gamma }{\alpha }}{x}_4\\ -\beta x_4& -\beta x_3\end{array}\\ \begin{array}{cc}\alpha x_3& -\gamma x_4\\ \alpha x_4& \delta x_3\end{array}& \begin{array}{cc}\alpha x_1& -\gamma x_2\\ \delta x_2& \alpha x_1\end{array}\end{array}\right)$$

(25)

It is also easily proven that mapping $F:x\to F\left(x\right)$ is one-to-one and onto. Alternative definitions of multiplication and division operations are defined as in the previous cases.

We will call space $M_5\left(1,1,1,1\right)$ isotropic space $M_5$. In isotropic space, the multiplication of two numbers is defined by the following equalities:

$$\begin{array}{c}\begin{array}{c}{z}_1=x_1{y}_1-x_2{y}_2-x_3{y}_3+x_4{y}_4\\ z_2=x_2{y}_1+x_1{y}_2-x_4{y}_3-x_3{y}_4\end{array}\\ \begin{array}{c}{z}_3=x_3{y}_1-x_4{y}_2+x_1{y}_3-x_2{y}_4\\ z_4=x_4{y}_1+x_3{y}_2+x_2{y}_3+\alpha x_1{y}_4\end{array}\end{array},$$

(26)

and the matrix $F\left(x\right)$ has the following form:

$$F\left(x\right)=\left(\begin{array}{cc}\begin{array}{cc}{x}_1& -x_2\\ x_2& x_1\end{array}& \begin{array}{cc}-x_3& x_4\\ -x_4& -x_3\end{array}\\ \begin{array}{cc}{x}_3& -x_4\\ x_4& x_3\end{array}& \begin{array}{cc}{x}_1& -x_2\\ x_2& x_1\end{array}\end{array}\right).$$

(27)

The symplectic module of four-dimensional number in isotropic space $M_5$ has the following form:

$$\left|x\right|=\sqrt[4]{\left[{\left(x_1-x_4\right)}^2+{\left(x_2+x_3\right)}^2\right]·\left[{\left(x_1+x_4\right)}^2+{\left(x_2-x_3\right)}^2\right]}.$$

(28)

In [5,6], properties of isotropic space $M_5$ are investigated.

6.

Case $\beta <0,\gamma >0,\delta <0$.

In this case, multiplication of two four-dimensional numbers is defined by the following formulas:

$$\begin{array}{c}\begin{array}{c}{z}_1=\alpha x_1{y}_1-{\displaystyle \frac{\gamma \delta }{\alpha }}{x}_2{y}_2+{\displaystyle \frac{\beta \delta }{\alpha }}{x}_3{y}_3-{\displaystyle \frac{\beta \gamma }{\alpha }}{x}_4{y}_4\\ z_2=\alpha x_2{y}_1+\alpha x_1{y}_2-\beta x_4{y}_3-\beta x_3{y}_4\end{array}\\ \begin{array}{c}{z}_3=\alpha x_3{y}_1+\gamma x_4{y}_2+\alpha x_1{y}_3+\gamma x_2{y}_4\\ z_4=\alpha x_4{y}_1-\delta x_3{y}_2-\delta x_2{y}_3+\alpha x_1{y}_4\end{array}\end{array}.$$

(29)

Here, we consider that $\beta >0,\gamma >0,\delta >0$.

The multiplication table of basic numbers is provided in the

Table 6. Products of basic numbers in Case 6.

	$J_1$	$J_2$	$J_3$	$J_4$
$J_1$	$J_1$	$J_2$	$J_3$	$J_4$
$J_2$	$J_2$	$-{\displaystyle \frac{\gamma \delta }{{\alpha }^3}}{J}_1$	$-{\displaystyle \frac{\delta }{{\alpha }^2}}{J}_4$	${\displaystyle \frac{\gamma }{{\alpha }^2}}{J}_3$
$J_3$	$J_3$	$-{\displaystyle \frac{\delta }{{\alpha }^2}}{J}_4$	${\displaystyle \frac{\beta \delta }{{\alpha }^3}}{J}_1$	$-{\displaystyle \frac{\beta }{{\alpha }^2}}{J}_2$
$J_4$	$J_4$	${\displaystyle \frac{\gamma }{{\alpha }^2}}{J}_3$	$-{\displaystyle \frac{\beta }{{\alpha }^2}}{J}_2$	$-{\displaystyle \frac{\beta \gamma }{{\alpha }^3}}{J}_1$

.

In this case, the module of four-dimensional number (2) can be transformed as follows:

$$\begin{gathered} {\left|x\right|}^4={\alpha }^4{x}_1^4+{\gamma }^2{\delta }^2{x}_2^4+{\beta }^2{\delta }^2{x}_3^4+{\beta }^2{\gamma }^2{x}_4^4+2{\alpha }^2\gamma \delta x_1^2{x}_2^2-2{\alpha }^2\beta \delta x_1^2{x}_3^2+2{\alpha }^2\beta \gamma x_1^2{x}_4^2+2\beta \gamma {\delta }^2{x}_2^2{x}_3^2- \\ 2\beta {\gamma }^2\delta x_2^2{x}_4^2+2{\beta }^2\gamma \delta x_3^2{x}_4^2+8\alpha \beta \gamma \delta x_1{x}_2{x}_3{x}_4=\left[{\left(\alpha x_1-\sqrt{\beta \delta }{x}_3\right)}^2+{\left(\sqrt{\gamma \delta }{x}_2+\sqrt{\beta \gamma }{x}_4\right)}^2\right]· \\ \left[{\left(\alpha x_1+\sqrt{\beta \delta }{x}_3\right)}^2+{\left(\sqrt{\gamma \delta }{x}_2-\sqrt{\beta \gamma }{x}_4\right)}^2\right]. \end{gathered}$$

(30)

In this case, the module of the imaginary number is defined by the formula $\left|x\right|=\sqrt{{\alpha }^2{x}_1^2+\gamma \delta x_2^2}$.

Definition of the module is somewhat similar to Case 5.

The conjugate number $x^{*}=\left(x_1^{*},x_2^{*},x_3^{*},x_4^{*}\right)$ to number $x=\left(x_1,x_2,x_3,x_4\right)$ is defined by the following formula:

$$x_1^{*}=x_1\left({\alpha }^2{x}_1^2+\gamma \delta x_2^2-\beta \delta x_3^2+\beta \gamma x_4^2\right)+{\displaystyle \frac{2\beta \gamma \delta }{\alpha }}{x}_2{x}_3{x}_4,$$

$$x_2^{*}=x_2\left(-{\alpha }^2{x}_1^2-\gamma \delta x_2^2-\beta \delta x_3^2+\beta \gamma x_4^2\right)-2\alpha \beta x_1{x}_3{x}_4,$$

$$x_3^{*}=x_3\left(-{\alpha }^2{x}_1^2+\gamma \delta x_2^2+\beta \delta x_3^2+\beta \gamma x_4^2\right)+2\alpha \gamma x_1{x}_2{x}_4,$$

$$x_4^{*}=x_4\left(-{\alpha }^2{x}_1^2+\gamma \delta x_2^2-\beta \delta x_3^2-\beta \gamma x_4^2\right)-2\alpha \delta x_1{x}_2{x}_3.$$

The conjugate numbers to basic numbers have the following form:

$$J_1^{*}=J_1, J_2^{*}=-{\displaystyle \frac{\gamma \delta }{{\alpha }^3}}·J_2, J_3^{*}={\displaystyle \frac{\beta \delta }{{\alpha }^3}}·J_3, J_4^{*}=-{\displaystyle \frac{\beta \gamma }{{\alpha }^3}}·J_4.$$

The conjugate numbers to unit numbers have the following form:

$$I_1^{*}={\alpha }^2{I}_1, I_2^{*}=-{\displaystyle \frac{\gamma \delta }{\alpha }}·I_2, I_3^{*}={\displaystyle \frac{\beta \delta }{\alpha }}·I_3, I_4^{*}=-{\displaystyle \frac{\beta \gamma }{\alpha }}·I_4.$$

The conjugate number to the imaginary number $x=\left(x_1,x_2,0,0\right)$ has the form ${\displaystyle \frac{{\left|x\right|}^2}{\alpha }}\left(x_1,-x_2,0,0\right)$.

Let us denote the obtained space by $M_6\left(\alpha ,\beta ,\gamma ,\delta \right)$.

The matrix $F\left(x\right)$ from (7) in this case looks as follows:

$$F\left(x\right)=\left(\begin{array}{cc}\begin{array}{cc}\alpha x_1& -{\displaystyle \frac{\gamma \delta }{\alpha }}{x}_2\\ \alpha x_2& \alpha x_1\end{array}& \begin{array}{cc}{\displaystyle \frac{\beta \delta }{\alpha }}{x}_3& -{\displaystyle \frac{\beta \gamma }{\alpha }}{x}_4\\ -\beta x_4& -\beta x_3\end{array}\\ \begin{array}{cc}\alpha x_3& \gamma x_4\\ \alpha x_4& -\delta x_3\end{array}& \begin{array}{cc}\alpha x_1& \gamma x_2\\ -\delta x_2& \alpha x_1\end{array}\end{array}\right).$$

(31)

Obviously, mapping $F:x\to F\left(x\right)$ is one-to-one and onto, and defines alternative definition of multiplication of four-dimensional numbers in space $M_6\left(\alpha ,\beta ,\gamma ,\delta \right)$.

We will call space $M_6\left(1,1,1,1\right)$ isotropic space $M_6$. In isotropic space, the multiplication of two numbers is defined by the following equalities:

$$\begin{array}{c}\begin{array}{c}{z}_1=x_1{y}_1-x_2{y}_2+x_3{y}_3-x_4{y}_4\\ z_2=x_2{y}_1+x_1{y}_2-x_4{y}_3-x_3{y}_4\end{array}\\ \begin{array}{c}{z}_3=x_3{y}_1+x_4{y}_2+x_1{y}_3+x_2{y}_4\\ z_4=x_4{y}_1-x_3{y}_2-x_2{y}_3+\alpha x_1{y}_4\end{array}\end{array},$$

(32)

and the matrix $F\left(x\right)$ has the following form:

$$F\left(x\right)=\left(\begin{array}{cc}\begin{array}{cc}{x}_1& -x_2\\ x_2& x_1\end{array}& \begin{array}{cc}{x}_3& -x_4\\ -x_4& -x_3\end{array}\\ \begin{array}{cc}{x}_3& x_4\\ x_4& -x_3\end{array}& \begin{array}{cc}{x}_1& x_2\\ -x_2& x_1\end{array}\end{array}\right)$$

(33)

The symplectic module of four-dimensional number in isotropic space $M_6$ has the following form:

$$\left|x\right|=\sqrt[4]{\left[{\left(x_1-x_3\right)}^2+{\left(x_2+x_4\right)}^2\right]·\left[{\left(x_1+x_3\right)}^2+{\left(x_2-x_4\right)}^2\right].}$$

(34)

7.

Case $\beta >0,\gamma <0,\delta <0$.

In this case, multiplication of two four-dimensional numbers is defined by the following formulas:

$$\begin{array}{c}\begin{array}{c}{z}_1=\alpha x_1{y}_1+{\displaystyle \frac{\gamma \delta }{\alpha }}{x}_2{y}_2-{\displaystyle \frac{\beta \delta }{\alpha }}{x}_3{y}_3-{\displaystyle \frac{\beta \gamma }{\alpha }}{x}_4{y}_4\\ z_2=\alpha x_2{y}_1+\alpha x_1{y}_2+\beta x_4{y}_3+\beta x_3{y}_4\end{array}\\ \begin{array}{c}{z}_3=\alpha x_3{y}_1-\gamma x_4{y}_2+\alpha x_1{y}_3-\gamma x_2{y}_4\\ z_4=\alpha x_4{y}_1-\delta x_3{y}_2-\delta x_2{y}_3+\alpha x_1{y}_4\end{array}\end{array}.$$

(35)

Here, we consider that $\beta >0,\gamma >0,\delta >0$.

The multiplication table of basic numbers takes the form given in

Table 7. Products of basic numbers in Case 7.

	$J_1$	$J_2$	$J_3$	$J_4$
$J_1$	$J_1$	$J_2$	$J_3$	$J_4$
$J_2$	$J_2$	${\displaystyle \frac{\gamma \delta }{{\alpha }^3}}{J}_1$	$-{\displaystyle \frac{\delta }{{\alpha }^2}}{J}_4$	$-{\displaystyle \frac{\gamma }{{\alpha }^2}}{J}_3$
$J_3$	$J_3$	$-{\displaystyle \frac{\delta }{{\alpha }^2}}{J}_4$	$-{\displaystyle \frac{\beta \delta }{{\alpha }^3}}{J}_1$	${\displaystyle \frac{\beta }{{\alpha }^2}}{J}_2$
$J_4$	$J_4$	$-{\displaystyle \frac{\gamma }{{\alpha }^2}}{J}_3$	${\displaystyle \frac{\beta }{{\alpha }^2}}{J}_2$	$-{\displaystyle \frac{\beta \gamma }{{\alpha }^3}}{J}_1$

.

In this case, the module of four-dimensional number (2) can be transformed as follows:

$$\begin{gathered} {\left|x\right|}^4={\alpha }^4{x}_1^4+{\gamma }^2{\delta }^2{x}_2^4+{\beta }^2{\delta }^2{x}_3^4+{\beta }^2{\gamma }^2{x}_4^4-2{\alpha }^2\gamma \delta x_1^2{x}_2^2+2{\alpha }^2\beta \delta x_1^2{x}_3^2+2{\alpha }^2\beta \gamma x_1^2{x}_4^2+2\beta \gamma {\delta }^2{x}_2^2{x}_3^2+ \\ 2\beta {\gamma }^2\delta x_2^2{x}_4^2-2{\beta }^2\gamma \delta x_3^2{x}_4^2+8\alpha \beta \gamma \delta x_1{x}_2{x}_3{x}_4=\left[{\left(\alpha x_1-\sqrt{\gamma \delta }{x}_2\right)}^2+{\left(\sqrt{\beta \delta }{x}_3+\sqrt{\beta \gamma }{x}_4\right)}^2\right]· \\ \left[{\left(\alpha x_1+\sqrt{\gamma \delta }{x}_2\right)}^2+{\left(\sqrt{\beta \delta }{x}_3-\sqrt{\beta \gamma }{x}_4\right)}^2\right]. \end{gathered}$$

(36)

Here, the module of the imaginary number $x=\left(x_1,0,x_3,0\right)$ is defined by the formula $\left|x\right|=\sqrt{\left|{\alpha }^2{x}_1^2+\beta \delta x_3^2\right|}$.

The conjugate number $x^{*}=\left(x_1^{*},x_2^{*},x_3^{*},x_4^{*}\right)$ to number $x=\left(x_1,x_2,x_3,x_4\right)$ is defined by the following formula:

$$x_1^{*}=x_1\left({\alpha }^2{x}_1^2-\gamma \delta x_2^2+\beta \delta x_3^2+\beta \gamma x_4^2\right)+{\displaystyle \frac{2\beta \gamma \delta }{\alpha }}{x}_2{x}_3{x}_4,$$

$$x_2^{*}=x_2\left(-{\alpha }^2{x}_1^2+\gamma \delta x_2^2+\beta \delta x_3^2+\beta \gamma x_4^2\right)+2\alpha \beta x_1{x}_3{x}_4,$$

$$x_3^{*}=x_3\left(-{\alpha }^2{x}_1^2-\gamma \delta x_2^2-\beta \delta x_3^2+\beta \gamma x_4^2\right)-2\alpha \gamma x_1{x}_2{x}_4,$$

$$x_4^{*}=x_4\left(-{\alpha }^2{x}_1^2-\gamma \delta x_2^2+\beta \delta x_3^2-\beta \gamma x_4^2\right)-2\alpha \delta x_1{x}_2{x}_3.$$

The conjugate numbers to basic numbers have the following form:

$$J_1^{*}=J_1, J_2^{*}={\displaystyle \frac{\gamma \delta }{{\alpha }^3}}·J_2, J_3^{*}=-{\displaystyle \frac{\beta \delta }{{\alpha }^3}}·J_3, J_4^{*}=-{\displaystyle \frac{\beta \gamma }{{\alpha }^3}}·J_4.$$

The conjugate numbers to unit numbers have the following form:

$$I_1^{*}={\alpha }^2{I}_1, I_2^{*}={\displaystyle \frac{\gamma \delta }{\alpha }}·I_2, I_3^{*}=-{\displaystyle \frac{\beta \delta }{\alpha }}·I_3, I_4^{*}=-{\displaystyle \frac{\beta \gamma }{\alpha }}·I_4.$$

The conjugate number to the imaginary number $\left(x_1,0,x_3,0\right)$ has the form ${\displaystyle \frac{{\left|x\right|}^2}{\alpha }}\left(x_1,0,-x_3,0\right)$, which easily follows from the last formulas.

Let us denote the obtained anisotropic space of four-dimensional numbers by $M_7\left(\alpha ,\beta ,\gamma ,\delta \right)$.

The matrix $F\left(x\right)$ from (7) in this case looks as follows:

$$F\left(x\right)=\left(\begin{array}{cc}\begin{array}{cc}\alpha x_1& {\displaystyle \frac{\gamma \delta }{\alpha }}{x}_2\\ \alpha x_2& \alpha x_1\end{array}& \begin{array}{cc}-{\displaystyle \frac{\beta \delta }{\alpha }}{x}_3& -{\displaystyle \frac{\beta \gamma }{\alpha }}{x}_4\\ \beta x_4& \beta x_3\end{array}\\ \begin{array}{cc}\alpha x_3& -\gamma x_4\\ \alpha x_4& -\delta x_3\end{array}& \begin{array}{cc}\alpha x_1& -\gamma x_2\\ -\delta x_2& \alpha x_1\end{array}\end{array}\right).$$

(37)

We will call space $M_7\left(1,1,1,1\right)$ isotropic space $M_7$. In isotropic space, the multiplication of two numbers is defined by the following equalities:

$$\begin{array}{c}\begin{array}{c}{z}_1=x_1{y}_1+x_2{y}_2-x_3{y}_3-x_4{y}_4\\ z_2=x_2{y}_1+x_1{y}_2+x_4{y}_3+x_3{y}_4\end{array}\\ \begin{array}{c}{z}_3=x_3{y}_1-x_4{y}_2+x_1{y}_3-x_2{y}_4\\ z_4=x_4{y}_1-x_3{y}_2-x_2{y}_3+\alpha x_1{y}_4\end{array}\end{array},$$

(38)

and the matrix $F\left(x\right)$ has the following form:

$$F\left(x\right)=\left(\begin{array}{cc}\begin{array}{cc}{x}_1& -x_2\\ x_2& x_1\end{array}& \begin{array}{cc}{x}_3& -x_4\\ -x_4& -x_3\end{array}\\ \begin{array}{cc}{x}_3& x_4\\ x_4& -x_3\end{array}& \begin{array}{cc}{x}_1& x_2\\ -x_2& x_1\end{array}\end{array}\right).$$

(39)

The symplectic module of four-dimensional number in isotropic space $M_7$ has the following form:

$$\left|x\right|=\sqrt[4]{\left[{\left(x_1-x_3\right)}^2+{\left(x_2+x_4\right)}^2\right]·\left[{\left(x_1+x_3\right)}^2+{\left(x_2-x_4\right)}^2\right]}$$

(40)

8.

Case $\beta <0,\gamma <0,\delta <0$.

In this case, multiplication of two four-dimensional numbers is defined by the following formulas:

$$\begin{array}{c}\begin{array}{c}{z}_1=\alpha x_1{y}_1+{\displaystyle \frac{\gamma \delta }{\alpha }}{x}_2{y}_2+{\displaystyle \frac{\beta \delta }{\alpha }}{x}_3{y}_3+{\displaystyle \frac{\beta \gamma }{\alpha }}{x}_4{y}_4\\ z_2=\alpha x_2{y}_1+\alpha x_1{y}_2-\beta x_4{y}_3-\beta x_3{y}_4\end{array}\\ \begin{array}{c}{z}_3=\alpha x_3{y}_1-\gamma x_4{y}_2+\alpha x_1{y}_3-\gamma x_2{y}_4\\ z_4=\alpha x_4{y}_1-\delta x_3{y}_2-\delta x_2{y}_3+\alpha x_1{y}_4\end{array}\end{array}.$$

Here, we consider that $\beta >0,\gamma >0,\delta >0$.

In this case, the module of four-dimensional number (2) can be transformed as follows:

$$\begin{array}{cc}\hfill {\left|x\right|}^4={\alpha }^4{x}_1^4+& {\gamma }^2{\delta }^2{x}_2^4+{\beta }^2{\delta }^2{x}_3^4+{\beta }^2{\gamma }^2{x}_4^4-2{\alpha }^2\gamma \delta x_1^2{x}_2^2-2{\alpha }^2\beta \delta x_1^2{x}_3^2-2{\alpha }^2\beta \gamma x_1^2{x}_4^2-2\beta \gamma {\delta }^2{x}_2^2{x}_3^2\hfill \\ \hfill & -2\beta {\gamma }^2\delta x_2^2{x}_4^2-2{\beta }^2\gamma \delta x_3^2{x}_4^2-8\alpha \beta \gamma \delta x_1{x}_2{x}_3{x}_4\hfill \\ \hfill & =\left[{\left(\alpha x_1+\sqrt{\beta \delta }{x}_3\right)}^2-{\left(\sqrt{\gamma \delta }{x}_2-\sqrt{\beta \gamma }{x}_4\right)}^2\right]\left[{\left(\alpha x_1-\sqrt{\beta \delta }{x}_3\right)}^2-{\left(\sqrt{\gamma \delta }{x}_2+\sqrt{\beta \gamma }{x}_4\right)}^2\right],\hfill \end{array}$$

or as follows:

$${\left|x\right|}^4=\left[{\left(\alpha x_1+\sqrt{\gamma \delta }{x}_2\right)}^2-{\left(\sqrt{\beta \delta }{x}_3-\sqrt{\beta \gamma }{x}_4\right)}^2\right]\left[{\left(\alpha x_1-\sqrt{\gamma \delta }{x}_2\right)}^2-{\left(\sqrt{\beta \delta }{x}_3+\sqrt{\beta \gamma }{x}_4\right)}^2\right],$$

or as follows:

$${\left|x\right|}^4=\left[{\left(\alpha x_1+\sqrt{\beta \gamma }{x}_4\right)}^2-{\left(\sqrt{\gamma \delta }{x}_2-\sqrt{\beta \delta }{x}_3\right)}^2\right]\left[{\left(\alpha x_1-\sqrt{\beta \gamma }{x}_4\right)}^2-{\left(\sqrt{\gamma \delta }{x}_2+\sqrt{\beta \delta }{x}_3\right)}^2\right]$$

For this case, the symplectic module is not non-negatively defined form.

Thus, we have six varying spaces of four-dimensional numbers in which operation of multiplication and the corresponding modules of numbers are defined by various formulas. Further, we investigate properties of these spaces.

3. Degenerate Numbers in Spaces of Four-Dimensional Numbers

Definition 3.

The four-dimensional number is called nondegenerate if $\left|x\right|>0$, and degenerate if $\left|x\right|=0$.

We investigate solutions of the equation $\left|x\right|=0$ or degenerate numbers. This equation has explicit solutions, thanks to which we can describe the general structure of degenerate numbers in all spaces $M_i\left(\alpha ,\beta ,\gamma ,\delta \right),i=2,3,4,5,6,7$.

As it follows from definition of the module of number (6) in space $M_2\left(\alpha ,\beta ,\gamma ,\delta \right)$, there are two types of degenerate numbers, namely, numbers of the kinds $\left(c_1,{\displaystyle \frac{\alpha }{\sqrt{\gamma \delta }}}{c}_1,c_2,\sqrt{{\displaystyle \frac{\delta }{\gamma }}}{c}_2\right)$ and $\left(c_1,-{\displaystyle \frac{\alpha }{\sqrt{\gamma \delta }}}{c}_1,c_2,-\sqrt{{\displaystyle \frac{\delta }{\gamma }}}{c}_2\right)$ for any real $c_1$ and $c_2$.

We will call numbers of the kind $\left(c_1,{\displaystyle \frac{\alpha }{\sqrt{\gamma \delta }}}{c}_1,c_2,\sqrt{{\displaystyle \frac{\delta }{\gamma }}}{c}_2\right)$ degenerate numbers of the first type, and numbers $\left(c_1,-{\displaystyle \frac{\alpha }{\sqrt{\gamma \delta }}}{c}_1,c_2,\sqrt{-{\displaystyle \frac{\delta }{\gamma }}}{c}_2\right)$ degenerate numbers of the second type.

Obviously, the only degenerate number belonging both to the first and to the second type is the number 0 = (0,0,0,0). Let us denote the set of all degenerate numbers of the first type by $O_I$, and the set of all degenerate numbers of the second type by $O_{II}$.

The module of a number in space $M_3\left(\alpha ,\beta ,\gamma ,\delta \right)$ is defined by Formula (12). Therefore, in this space, there are also two types of degenerate numbers, $\left(c_1,c_2,{\displaystyle \frac{\alpha }{\sqrt{\beta \delta }}}{c}_1,\sqrt{{\displaystyle \frac{\delta }{\beta }}}{c}_2\right)$ and $\left(c_1,c_2,-{\displaystyle \frac{\alpha }{\sqrt{\beta \delta }}}{c}_1,-\sqrt{{\displaystyle \frac{\delta }{\beta }}}{c}_2\right)$, for any real $c_1$ and $c_2,$ which respectively we will call degenerate numbers of the first type and degenerate numbers of the second type, respectively, and we will also denote them by $O_I$ and $O_{II}$.

Number 0 = (0,0,0,0) is the only degenerate number belonging to both types.

In space $M_4\left(\alpha ,\beta ,\gamma ,\delta \right)$, the set of degenerate numbers of the first type $O_I$ consists from the numbers $\left(c_1,c_2,\sqrt{{\displaystyle \frac{\gamma }{\beta }}}{c}_2,{\displaystyle \frac{\alpha }{\sqrt{\beta \gamma }}}{c}_1\right)$, and the set of degenerate numbers of the second type $O_{II}$ consists of the numbers $\left(c_1,c_2,-\sqrt{{\displaystyle \frac{\gamma }{\beta }}}{c}_2,-{\displaystyle \frac{\alpha }{\sqrt{\beta \gamma }}}{c}_1\right)$ for any real $c_1$ and $c_2$ that follows from definition of symplectic module (18).

Similarly, in space $M_5\left(\alpha ,\beta ,\gamma ,\delta \right)$, the set of degenerate numbers of the first type $O_I$ consists of the numbers $\left(c_1,c_2,-\sqrt{{\displaystyle \frac{\gamma }{\beta }}}{c}_2,{\displaystyle \frac{\alpha }{\sqrt{\beta \gamma }}}{c}_1\right)$, and the set of degenerate numbers of the second type $O_{II}$ consists of the numbers $\left(c_1,c_2,\sqrt{{\displaystyle \frac{\gamma }{\beta }}}{c}_2,-{\displaystyle \frac{\alpha }{\sqrt{\beta \gamma }}}{c}_1\right)$ for any real $c_1$ and $c_2$.

In space $M_6\left(\alpha ,\beta ,\gamma ,\delta \right)$, the symplectic module of number is determined by Formula (30). Therefore, the set of degenerate numbers of the first type $O_I$ consists of the numbers $\left(c_1,c_2,{\displaystyle \frac{\alpha }{\sqrt{\beta \gamma }}}{c}_1,-\sqrt{{\displaystyle \frac{\delta }{\beta }}}{c}_2\right)$, and the set of degenerate numbers of the second type $O_{II}$ consists of the numbers $\left(c_1,c_2,-{\displaystyle \frac{\alpha }{\sqrt{\beta \gamma }}}{c}_1,\sqrt{{\displaystyle \frac{\delta }{\beta }}}{c}_2\right)$ for any real $c_1$ and $c_2$.

In space $M_7\left(\alpha ,\beta ,\gamma ,\delta \right)$, we obtain from the equation $\left|x\right|=0$ that there are two types of degenerate numbers, namely, numbers of the kind $\left(c_1,{\displaystyle \frac{\alpha }{\sqrt{\gamma \delta }}}{c}_1,c_2,-\sqrt{{\displaystyle \frac{\delta }{\gamma }}}{c}_2\right)$ and of the kind $\left(c_1,-{\displaystyle \frac{\alpha }{\sqrt{\gamma \delta }}}{c}_1,c_2,\sqrt{{\displaystyle \frac{\delta }{\gamma }}}{c}_2\right)$ for any real $c_1$ and $c_2$, which are degenerate numbers of the first and second types, respectively.

Theorem 2.

Degenerate numbers in spaces $M_i\left(\alpha ,\beta ,\gamma ,\delta \right),i=2,3,4,5,6,7$, have the following properties:

(1)

If  $x,y\in O_I$, then $x+y\in O_I$,$x-y\in O_I$, $x·y\in O_I$.

(2)

If $x,y\in O_{II}$, then $x+y\in O_{II}$, $x-y\in O_{II}$, $x·y\in O_{II}$.

(3)

If $x\in O_I,y\in O_{II}$, then $x·y=0=\left(0,0,0,0\right)$.

(4)

If $x\in O_I$, $y\notin O_I{\displaystyle \cup }{O}_{II}$, then $x·y\in O_I$.

(5)

If  $x\in O_{II}$, $y\notin O_I{\displaystyle \cup }{O}_{II}$, then $x·y\in O_{II}$.

The proof easily follows from definitions of addition and multiplication in the corresponding spaces.

This theorem describes properties of degenerate numbers in spaces $M_i\left(\alpha ,\beta ,\gamma ,\delta \right),i=2,3,4,5,6,7$. In particular, it follows from the first two properties that sets $O_I$ and $O_{II}$ are subspaces of spaces $M_i\left(\alpha ,\beta ,\gamma ,\delta \right),i=2,3,4,5,6,7$.

The third property claims that points of subspaces $O_I$ and $O_{II}$ are zero divisors. Moreover, as will become obvious from further discussions, there are no other zero divisors. Such obvious description of structure of zero divisors makes it possible to control influence of zero divisors on various mathematical structures. For example, to build the full-fledged calculus, the differential and integral calculus and other constructions.

The last approvals of the theorem say that at multiplication of any four-dimensional number by degenerate number, we will always receive degenerate number.

4. Range of Four-Dimensional Numbers

To each four-dimensional number $x\in M_i,i=2,3,\dots ,7$, we compared some matrix of F (x), determined, respectively, by Formulas (7), (13), (19), (25), (31) and (37), by means of which multiplication of two numbers can be reduced to multiplication of this matrix by the vector determined by the second multiplier.

Theorem 3.

The set of matrices F(x) is closed with respect to matrix operations of addition, subtraction and multiplication, and also multiplication of the matrix by the scalar. The inverse matrix to the nonsingular matrix has the same form.

This can be proven by direct check.

Theorem 4.

For each space $M_i\left(\alpha ,\beta ,\gamma ,\delta \right),i=2,3,4,5,6,7$, mapping F: x → F(x) for any four-dimensional numbers $x,y$  has the following properties:

(1)

$F\left(x\pm y\right)=F\left(x\right)\pm F\left(y\right)$;

(2)

$F\left(cx\right)=cF\left(x\right)$ for any $c\in R$;

(3)

$F\left(xy\right)=F\left(x\right)F\left(y\right)$;

(4)

$F\left(x^{-1}\right)=F^{-1}\left(x\right)$;

(5)

$\det \left(F\left(x\right)\right)={\left|x\right|}^4$;

(6)

$\det \left(F\left(x\right)\pm F\left(y\right)\right)={\left|x\pm y\right|}^4$;

(7)

$\det \left(F\left(\alpha x\right)\right)={\left|\alpha x\right|}^4$;

(8)

$\det \left(F\left(x\right)F\left(y\right)\right)={\left|xy\right|}^4$;

(9)

$\det \left(F^{-1}\left(x\right)\right)={\left|x^{-1}\right|}^4$, where x is nondegenerate number.

Proof. 

Let us prove the theorem for space $M_2\left(\alpha ,\beta ,\gamma ,\delta \right)$. For other spaces, the proof is carried out in a similar way. □

Properties (1) and (2) are obvious. Let us prove property (3).

$$F\left(x\right)F\left(y\right)=\left(\begin{array}{cc}\begin{array}{cc}\alpha x_1& {\displaystyle \frac{\gamma \delta }{\alpha }}{x}_2\\ \alpha x_2& \alpha x_1\end{array}& \begin{array}{cc}-{\displaystyle \frac{\beta \delta }{\alpha }}{x}_3& -{\displaystyle \frac{\beta \gamma }{\alpha }}{x}_4\\ -\beta x_4& -\beta x_3\end{array}\\ \begin{array}{cc}\alpha x_3& \gamma x_4\\ \alpha x_4& \delta x_3\end{array}& \begin{array}{cc}\alpha x_1& \gamma x_2\\ \delta x_2& \alpha x_1\end{array}\end{array}\right)\left(\begin{array}{cc}\begin{array}{cc}\alpha y_1& {\displaystyle \frac{\gamma \delta }{\alpha }}{y}_2\\ \alpha y_2& \alpha y_1\end{array}& \begin{array}{cc}-{\displaystyle \frac{\beta \delta }{\alpha }}{y}_3& -{\displaystyle \frac{\beta \gamma }{\alpha }}{y}_4\\ -\beta y_4& -\beta y_3\end{array}\\ \begin{array}{cc}\alpha y_3& \gamma y_4\\ \alpha y_4& \delta y_3\end{array}& \begin{array}{cc}\alpha y_1& \gamma y_2\\ \delta y_2& \alpha y_1\end{array}\end{array}\right)=B,$$

where B is the resulting matrix. Let us calculate elements of the matrix B.

$$b_{11}={\alpha }^2{x}_1{y}_1+\gamma \delta x_2{y}_2-\beta \delta x_3{y}_3-\beta \gamma x_4{y}_4=\alpha z_1;$$

$$b_{12}=\gamma \delta x_1{y}_2+\gamma \delta x_2{y}_1-{\displaystyle \frac{\beta \gamma \delta }{\alpha }}{x}_3{y}_4-{\displaystyle \frac{\beta \gamma \delta }{\alpha }}{x}_4{y}_3={\displaystyle \frac{\gamma \delta }{\alpha }}{z}_2;$$

$$b_{13}=-\beta \delta x_1{y}_3-{\displaystyle \frac{\beta \gamma \delta }{\alpha }}{x}_2{y}_4-\beta \delta x_3{y}_1-{\displaystyle \frac{\beta \gamma \delta }{\alpha }}{x}_4{y}_2=-{\displaystyle \frac{\beta \delta }{\alpha }}{z}_3;$$

$$b_{14}=-\beta \gamma x_1{y}_4-{\displaystyle \frac{\beta \gamma \delta }{\alpha }}{x}_2{y}_3-{\displaystyle \frac{\beta \gamma \delta }{\alpha }}{x}_3{y}_2-\beta \gamma x_4{y}_1=-{\displaystyle \frac{\beta \gamma }{\alpha }}{z}_4;$$

$$b_{21}={\alpha }^2{x}_2{y}_1+{\alpha }^2{x}_1{y}_2-\alpha \beta x_4{y}_3-\alpha \beta x_3{y}_4=\alpha z_2;$$

$$b_{22}=\gamma \delta x_2{y}_2+{\alpha }^2{x}_1{y}_1-\beta \gamma x_4{y}_4-\beta \delta x_3{y}_3=\alpha z_1;$$

$$b_{23}=-\beta \delta x_2{y}_3-\alpha \beta x_1{y}_4-\alpha \beta x_4{y}_1-\beta \delta x_3{y}_2=-\beta z_4;$$

$$b_{24}=-\beta \gamma x_2{y}_4-\alpha \beta x_1{y}_3-\beta \gamma x_4{y}_2-\alpha \beta x_3{y}_1=-\beta z_3;$$

$$b_{31}={\alpha }^2{x}_3{y}_1+\alpha \gamma x_4{y}_2+{\alpha }^2{x}_1{y}_3+\alpha \gamma x_2{y}_4=\alpha z_3;$$

$$b_{32}=\gamma \delta x_3{y}_2+\alpha \gamma x_4{y}_1+\alpha \gamma x_1{y}_4+\gamma \delta x_2{y}_3=\gamma z_4;$$

$$b_{33}=-\beta \delta x_3{y}_3-\beta \gamma x_4{y}_4+{\alpha }^2{x}_1{y}_1+\gamma \delta x_2{y}_2=\alpha z_1;$$

$$b_{34}=-\beta \gamma x_3{y}_4-\beta \gamma x_4{y}_3+\alpha \gamma x_1{y}_2+\alpha \gamma x_2{y}_1=\gamma z_2;$$

$$b_{41}={\alpha }^2{x}_4{y}_1+\alpha \delta x_3{y}_2+\alpha \delta x_2{y}_3+{\alpha }^2{x}_1{y}_4=\alpha z_4;$$

$$b_{42}=\gamma \delta x_4{y}_2+\alpha \delta x_3{y}_1+\gamma \delta x_2{y}_4+\alpha \delta x_1{y}_3=\delta z_3;$$

$$b_{43}=-\beta \delta x_4{y}_3-\beta \delta x_3{y}_4+\alpha \delta x_2{y}_1+\alpha \delta x_1{y}_2=\delta z_2;$$

$$b_{44}=-\beta \gamma x_4{y}_4-\beta \delta x_3{y}_3-\gamma \delta x_2{y}_2+{\alpha }^2{x}_1{y}_1=\alpha z_1;$$

where $z=\left(z_1,z_2,z_3,z_4\right)=x·y$.

Let us prove property (4). $\gamma$(4), $x^{-1}={\displaystyle \frac{1}{\alpha ·{\left|x\right|}^4}}{x}^{*}$. Therefore,

$${\left(x^{-1}\right)}_1={\displaystyle \frac{1}{{\left|x\right|}^4}}{x}_1\left({\alpha }^2{x}_1^2-\gamma \delta x_2^2+\beta \delta x_3^2+\beta \gamma x_4^2\right)-{\displaystyle \frac{2\beta \gamma \delta }{\alpha {\left|x\right|}^4}}{x}_2{x}_3{x}_4,$$

$${\left(x^{-1}\right)}_2={\displaystyle \frac{1}{{\left|x\right|}^4}}{x}_2\left(-{\alpha }^2{x}_1^2+\gamma \delta x_2^2+\beta \delta x_3^2+\beta \gamma x_4^2\right)-{\displaystyle \frac{2\alpha \beta }{{\left|x\right|}^4}}{x}_1{x}_3{x}_4,$$

$${\left(x^{-1}\right)}_3={\displaystyle \frac{1}{{\left|x\right|}^4}}{x}_3\left(-{\alpha }^2{x}_1^2-\gamma \delta x_2^2-\beta \delta x_3^2+\beta \gamma x_4^2\right)+{\displaystyle \frac{2\alpha \gamma }{{\left|x\right|}^4}}{x}_1{x}_2{x}_4,$$

$${\left(x^{-1}\right)}_4={\displaystyle \frac{1}{{\left|x\right|}^4}}{x}_4\left(-{\alpha }^2{x}_1^2-\gamma \delta x_2^2+\beta \delta x_3^2-\beta \gamma x_4^2\right)+{\displaystyle \frac{2\alpha \delta }{{\left|x\right|}^4}}{x}_1{x}_2{x}_3,$$

and, consequently, we have

$$F\left(x^{-1}\right)=\left(\begin{array}{cc}\begin{array}{cc}\alpha {\left(x^{-1}\right)}_1& {\displaystyle \frac{\gamma \delta }{\alpha }}{\left(x^{-1}\right)}_2\\ \alpha {\left(x^{-1}\right)}_2& \alpha {\left(x^{-1}\right)}_1\end{array}& \begin{array}{cc}-{\displaystyle \frac{\beta \delta }{\alpha }}{\left(x^{-1}\right)}_3& -{\displaystyle \frac{\beta \gamma }{\alpha }}{\left(x^{-1}\right)}_4\\ -\beta {\left(x^{-1}\right)}_4& -\beta {\left(x^{-1}\right)}_3\end{array}\\ \begin{array}{cc}\alpha {\left(x^{-1}\right)}_3& \gamma {\left(x^{-1}\right)}_4\\ \alpha {\left(x^{-1}\right)}_4& \delta {\left(x^{-1}\right)}_3\end{array}& \begin{array}{cc}\alpha {\left(x^{-1}\right)}_1& \gamma {\left(x^{-1}\right)}_2\\ \delta {\left(x^{-1}\right)}_2& \alpha {\left(x^{-1}\right)}_1\end{array}\end{array}\right),$$

and

$$F\left(x\right)=\left(\begin{array}{cc}\begin{array}{cc}\alpha x_1& {\displaystyle \frac{\gamma \delta }{\alpha }}{x}_2\\ \alpha x_2& \alpha x_1\end{array}& \begin{array}{cc}-{\displaystyle \frac{\beta \delta }{\alpha }}{x}_3& -{\displaystyle \frac{\beta \gamma }{\alpha }}{x}_4\\ -\beta x_4& -\beta x_3\end{array}\\ \begin{array}{cc}\alpha x_3& \gamma x_4\\ \alpha x_4& \delta x_3\end{array}& \begin{array}{cc}\alpha x_1& \gamma x_2\\ \delta x_2& \alpha x_1\end{array}\end{array}\right).$$

Multiplying these matrices by each other, we find that $F\left(x^{-1}\right)·F\left(x\right)=E$, where E is the identity matrix that completes proof.

Let us prove property (5). By definition of the determinant we have

$$detF\left(x\right)=\alpha x_1\left|\begin{array}{ccc}\alpha x_1& -\beta x_4& -\beta x_3\\ \gamma x_4& \alpha x_1& \gamma x_2\\ \delta x_3& \delta x_2& \alpha x_1\end{array}\right|-{\displaystyle \frac{\gamma \delta }{\alpha }}{x}_2\left|\begin{array}{ccc}\alpha x_2& -\beta x_4& -\beta x_3\\ \alpha x_3& \alpha x_1& \gamma x_2\\ \alpha x_4& \delta x_2& \alpha x_1\end{array}\right|-.$$

$$-{\displaystyle \frac{\beta \delta }{\alpha }}{x}_3\left|\begin{array}{ccc}\alpha x_2& \alpha x_1& -\beta x_3\\ \alpha x_3& \gamma x_4& \gamma x_2\\ \alpha x_4& \delta x_3& \alpha x_1\end{array}\right|+{\displaystyle \frac{\beta \gamma }{\alpha }}{x}_4\left|\begin{array}{ccc}\alpha x_2& \alpha x_1& -\beta x_4\\ \alpha x_3& \gamma x_4& \alpha x_1\\ \alpha x_4& \delta x_3& \delta x_2\end{array}\right|.$$

Calculating determinants in the last equality, we make sure that

$$\left|\begin{array}{ccc}\alpha x_1& -\beta x_4& -\beta x_3\\ \gamma x_4& \alpha x_1& \gamma x_2\\ \delta x_3& \delta x_2& \alpha x_1\end{array}\right|=\alpha x_1^{*}, \left|\begin{array}{ccc}\alpha x_2& -\beta x_4& -\beta x_3\\ \alpha x_3& \alpha x_1& \gamma x_2\\ \alpha x_4& \delta x_2& \alpha x_1\end{array}\right|=-\alpha x_2^{*}, \left|\begin{array}{ccc}\alpha x_2& \alpha x_1& -\beta x_3\\ \alpha x_3& \gamma x_4& \gamma x_2\\ \alpha x_4& \delta x_3& \alpha x_1\end{array}\right|=\alpha x_3^{*},$$

$$\left|\begin{array}{ccc}\alpha x_2& \alpha x_1& -\beta x_4\\ \alpha x_3& \gamma x_4& \alpha x_1\\ \alpha x_4& \delta x_3& \delta x_2\end{array}\right|=-\alpha x_4^{*}.$$

Then $detF\left(x\right)={\alpha }^2{x}_1{x}_1^{*}+\gamma \delta x_2{x}_2^{*}-\beta \delta x_3{x}_3^{*}-\beta \gamma x_4{x}_4^{*}=\alpha ·{\left(x·x^{*}\right)}_1={\left|x\right|}^4$.

Let us prove property (6). $\det \left(F\left(x\right)\pm F\left(y\right)\right)=\det \left(F\left(x\pm y\right)\right)={\left|x\pm y\right|}^4$

Proofs of properties (7)–(9) are obvious.

Thus, there is the bijection between space of four-dimensional numbers and space of 4 × 4 matrices that maintains arithmetic operations, that is, the existing bijection is homomorphism.

Definition 4.

The set of characteristic numbers of the corresponding matrix $F\left(x\right)$ is called the spectrum of four-dimensional number $x$.

Let us work out characteristic equation for definition of the spectrum of four-dimensional number $x=\left(x_1,x_2,x_3,x_4\right)$ in space $M_2\left(\alpha ,\beta ,\gamma ,\delta \right)$:

$$\left|\begin{array}{cc}\begin{array}{cc}\alpha x_1-\mu & {\displaystyle \frac{\gamma \delta }{\alpha }}{x}_2\\ \alpha x_2& \alpha x_1-\mu \end{array}& \begin{array}{cc}-{\displaystyle \frac{\beta \delta }{\alpha }}{x}_3& -{\displaystyle \frac{\beta \gamma }{\alpha }}{x}_4\\ -\beta x_4& -\beta x_3\end{array}\\ \begin{array}{cc}\alpha x_3& \gamma x_4\\ \alpha x_4& \delta x_3\end{array}& \begin{array}{cc}\alpha x_1-\mu & \gamma x_2\\ \delta x_2& \alpha x_1-\mu \end{array}\end{array}\right|=0.$$

Let us calculate minors $M\left(a_{1i}\right)$ of elements of the first row of the matrix:

$$M\left(\alpha x_1-\mu \right)=\left(\alpha x_1-\mu \right)\left|\begin{array}{cc}\alpha x_1-\mu & \gamma x_2\\ \delta x_2& \alpha x_1-\mu \end{array}\right|+\beta x_4\left|\begin{array}{cc}\gamma x_4& \gamma x_2\\ \delta x_3& \alpha x_1-\mu \end{array}\right|-\beta x_3\left|\begin{array}{cc}\gamma x_4& \alpha x_1-\mu \\ \delta x_3& \delta x_2\end{array}\right|.$$

$$\begin{array}{cc}\hfill M\left(\alpha x_1-\mu \right)=& -{\mu }^3+3\alpha x_1{\mu }^2+\left(-3{\alpha }^2{x}_1^2+\gamma \delta x_2^2-\beta \delta x_3^2-\beta \gamma x_4^2\right)·\mu +{\alpha }^3{x}_1^3-\alpha \gamma \delta x_1{x}_2^2+\alpha \beta \delta x_1{x}_3^2\hfill \\ \hfill & +\alpha \beta \gamma x_1{x}_4^2-2\beta \gamma \delta x_2{x}_3{x}_4.\hfill \end{array}$$

$$M\left({\displaystyle \frac{\gamma \delta }{\alpha }}{x}_2\right)=\alpha x_2\left|\begin{array}{cc}\alpha x_1-\mu & \gamma x_2\\ \delta x_2& \alpha x_1-\mu \end{array}\right|+\beta x_4\left|\begin{array}{cc}\alpha x_3& \gamma x_2\\ \alpha x_4& \alpha x_1-\mu \end{array}\right|-\beta x_3\left|\begin{array}{cc}\alpha x_3& \alpha x_1-\mu \\ \alpha x_4& \delta x_2\end{array}\right|.$$

$$M\left({\displaystyle \frac{\gamma \delta }{\alpha }}{x}_2\right)=\alpha x_2{\mu }^2-2\alpha \left(\alpha x_1{x}_2+\beta x_3{x}_4\right)·\mu +\alpha \left({\alpha }^2{x}_1^2{x}_2-\gamma \delta x_2^3-\beta \delta x_2{x}_3^2-\beta \gamma x_2{x}_4^2+2\alpha \beta x_1{x}_3{x}_4\right).$$

$$M\left(-{\displaystyle \frac{\beta \delta }{\alpha }}{x}_3\right)=\alpha x_2\left|\begin{array}{cc}\gamma x_4& \gamma x_2\\ \delta x_3& \alpha x_1-\mu \end{array}\right|-\left(\alpha x_1-\mu \right)\left|\begin{array}{cc}\alpha x_3& \gamma x_2\\ \alpha x_4& \alpha x_1-\mu \end{array}\right|-\beta x_3\left|\begin{array}{cc}\alpha x_3& \gamma x_4\\ \alpha x_4& \delta x_3\end{array}\right|.$$

$$M\left(-{\displaystyle \frac{\beta \delta }{\alpha }}{x}_3\right)=-\alpha x_3{\mu }^2+2\alpha \left(\alpha x_1{x}_3-\gamma x_2{x}_4\right)·\mu +\alpha \left(2\alpha \gamma x_1{x}_2{x}_4-{\alpha }^2{x}_1^2{x}_3-\gamma \delta x_2^2{x}_3-\beta \delta x_3^3+\beta \gamma x_3{x}_4^2\right).$$

$$M\left(-{\displaystyle \frac{\beta \gamma }{\alpha }}{x}_4\right)=\alpha x_2\left|\begin{array}{cc}\gamma x_4& \alpha x_1-\mu \\ \delta x_3& \delta x_2\end{array}\right|-\left(\alpha x_1-\mu \right)\left|\begin{array}{cc}\alpha x_3& \alpha x_1-\mu \\ \alpha x_4& \delta x_2\end{array}\right|-\beta x_4\left|\begin{array}{cc}\alpha x_3& \gamma x_4\\ \alpha x_4& \delta x_3\end{array}\right|.$$

$$M\left(-{\displaystyle \frac{\beta \gamma }{\alpha }}{x}_4\right)=\alpha x_4{\mu }^2+2\alpha \left(\delta x_2{x}_3-\alpha x_1{x}_4\right)·\mu +\alpha \left({\alpha }^2{x}_1^2{x}_4+\gamma \delta x_2^2{x}_4-\beta \delta x_3^2{x}_4+\beta \gamma x_4^3-2\alpha \delta x_1{x}_2{x}_3\right).$$

Using these values of minors, we will calculate determinant of characteristic equation:

$$\begin{array}{cc}\hfill {\mu }^4-4\alpha x_1{\mu }^3+& 2\left(3{\alpha }^2{x}_1^2-\gamma \delta x_2^2+\beta \delta x_3^2+\beta \gamma x_4^2\right){\mu }^2\hfill \\ \hfill & +\left(-4{\alpha }^3{x}_1^3+4\alpha \gamma \delta x_1{x}_2^2-4\alpha \beta \delta x_1{x}_3^2-4\alpha \beta \gamma x_1{x}_4^2+8\beta \gamma \delta x_2{x}_3{x}_4\right)·\mu +{\alpha }^4{x}_1^4+{\gamma }^2{\delta }^2{x}_2^4\hfill \\ \hfill & +{\beta }^2{\delta }^2{x}_3^4+{\beta }^2{\gamma }^2{x}_4^4-2{\alpha }^2\gamma \delta x_1^2{x}_2^2+2{\alpha }^2\beta \delta x_1^2{x}_3^2+2{\alpha }^2\beta \gamma x_1^2{x}_4^2+2\beta \gamma {\delta }^2{x}_2^2{x}_3^2+2\beta {\gamma }^2\delta x_2^2{x}_4^2\hfill \\ \hfill & -2{\beta }^2\gamma \delta x_3^2{x}_4^2-8\alpha \beta \gamma \delta x_1{x}_2{x}_3{x}_4=0.\hfill \end{array}$$

(41)

Taking into account Equation (6), we rewrite Equation (41) as follows:

$$\begin{array}{cc}\hfill {\mu }^4-4\alpha x_1{\mu }^3+& 2\left(3{\alpha }^2{x}_1^2-\gamma \delta x_2^2+\beta \delta x_3^2+\beta \gamma x_4^2\right){\mu }^2\hfill \\ \hfill & +\left(-4{\alpha }^3{x}_1^3+4\alpha \gamma \delta x_1{x}_2^2-4\alpha \beta \delta x_1{x}_3^2-4\alpha \beta \gamma x_1{x}_4^2+8\beta \gamma \delta x_2{x}_3{x}_4\right)·\mu \hfill \\ \hfill & +\left[{\left(\alpha x_1-\sqrt{\gamma \delta }{x}_2\right)}^2+{\left(\sqrt{\beta \delta }{x}_3-\sqrt{\beta \gamma }{x}_4\right)}^2\right]·\left[{\left(\alpha x_1+\sqrt{\gamma \delta }{x}_2\right)}^2+{\left(\sqrt{\beta \delta }{x}_3+\sqrt{\beta \gamma }{x}_4\right)}^2\right]=0.\hfill \end{array}$$

Proceeding from the type of free term, we will consider the possibility of decomposition of the equation on the product of two square trinomials, namely, we will present the last equation in the form $\left({\mu }^2+t_1\mu +u_1\right)·\left({\mu }^2+t_2\mu +u_2\right)=0$, where

$$u_1={\left(\alpha x_1-\sqrt{\gamma \delta }{x}_2\right)}^2+{\left(\sqrt{\beta \delta }{x}_3-\sqrt{\beta \gamma }{x}_4\right)}^2{u}_2={\left(\alpha x_1+\sqrt{\gamma \delta }{x}_2\right)}^2+{\left(\sqrt{\beta \delta }{x}_3+\sqrt{\beta \gamma }{x}_4\right)}^2.$$

Expanding the brackets and equating coefficients of degrees $\mu$, we find for $t_1$ and $t_2$ the following system:

$$\{\begin{array}{c}{t}_1+t_2=-4\alpha x_1\\ t_1{t}_2+u_1+u_2=2\left(3{\alpha }^2{x}_1^2-\gamma \delta x_2^2+\beta \delta x_3^2+\beta \gamma x_4^2\right)\\ u_2{t}_1+u_1{t}_2=-4{\alpha }^3{x}_1^3+4\alpha \gamma \delta x_1{x}_2^2-4\alpha \beta \delta x_1{x}_3^2-4\alpha \beta \gamma x_1{x}_4^2+8\beta \gamma \delta x_2{x}_3{x}_4\end{array}.$$

This redefined system has a unique solution: $t_1=-2\left(\alpha x_1-\sqrt{\gamma \delta }{x}_2\right)$, $t_2=-2\left(\alpha x_1+\sqrt{\gamma \delta }{x}_2\right)$. Therefore, Equation (8) is equivalent to the following equation:

$$\left({\mu }^2-2\left(\alpha x_1-\sqrt{\gamma \delta }{x}_2\right)\mu +u_1\right)·\left({\mu }^2-2\left(\alpha x_1+\sqrt{\gamma \delta }{x}_2\right)\mu +u_2\right)=0$$

(42)

Equation (42) breaks up into two quadratic equations, and after solving them, four characteristic numbers of four-dimensional number $x$ are found:

$$\{\begin{array}{c}\begin{array}{c}{\mu }_1=\alpha x_1-\sqrt{\gamma \delta }{x}_2+\left(\sqrt{\beta \delta }{x}_3-\sqrt{\beta \gamma }{x}_4\right)i\\ {\mu }_2=\alpha x_1-\sqrt{\gamma \delta }{x}_2-\left(\sqrt{\beta \delta }{x}_3-\sqrt{\beta \gamma }{x}_4\right)i\end{array}\\ \begin{array}{c}{\mu }_3=\alpha x_1+\sqrt{\gamma \delta }{x}_2+\left(\sqrt{\beta \delta }{x}_3+\sqrt{\beta \gamma }{x}_4\right)i\\ {\mu }_4=\alpha x_1+\sqrt{\gamma \delta }{x}_2-\left(\sqrt{\beta \delta }{x}_3+\sqrt{\beta \gamma }{x}_4\right)i\end{array}\end{array}$$

(43)

Thus, the spectrum of four-dimensional number $x=\left(x_1,x_2,x_3,x_4\right)$ in space $M_2\left(\alpha ,\beta ,\gamma ,\delta \right)$ consists of four pairwise complex conjugate numbers of the form (43).

We will find in a similar way spectra of numbers in spaces $M_i\left(\alpha ,\beta ,\gamma ,\delta \right), i=3,4,5,6,7$. The spectrum of a number $x=\left(x_1,x_2,x_3,x_4\right)$ in space $M_3\left(\alpha ,\beta ,\gamma ,\delta \right)$ is

$$\{\begin{array}{c}\begin{array}{c}{\mu }_1=\alpha x_1-\sqrt{\beta \delta }{x}_3+\left(\sqrt{\gamma \delta }{x}_2-\sqrt{\beta \gamma }{x}_4\right)i\\ {\mu }_2=\alpha x_1-\sqrt{\beta \delta }{x}_3-\left(\sqrt{\gamma \delta }{x}_2-\sqrt{\beta \gamma }{x}_4\right)i\end{array}\\ \begin{array}{c}{\mu }_3=\alpha x_1+\sqrt{\beta \delta }{x}_3+\left(\sqrt{\gamma \delta }{x}_2+\sqrt{\beta \gamma }{x}_4\right)i\\ {\mu }_4=\alpha x_1+\sqrt{\beta \delta }{x}_3-\left(\sqrt{\gamma \delta }{x}_2+\sqrt{\beta \gamma }{x}_4\right)i\end{array}\end{array}$$

(44)

The spectrum of a number $x=\left(x_1,x_2,x_3,x_4\right)$ in space $M_4\left(\alpha ,\beta ,\gamma ,\delta \right)$ is

$$\{\begin{array}{c}\begin{array}{c}{\mu }_1=\alpha x_1-\sqrt{\beta \gamma }{x}_4+\left(\sqrt{\gamma \delta }{x}_2-\sqrt{\beta \delta }{x}_3\right)i\\ {\mu }_2=\alpha x_1-\sqrt{\beta \gamma }{x}_4-\left(\sqrt{\gamma \delta }{x}_2-\sqrt{\beta \delta }{x}_3\right)i\end{array}\\ \begin{array}{c}{\mu }_3=\alpha x_1+\sqrt{\beta \gamma }{x}_4+\left(\sqrt{\gamma \delta }{x}_2+\sqrt{\beta \delta }{x}_3\right)i\\ {\mu }_4=\alpha x_1+\sqrt{\beta \gamma }{x}_4-\left(\sqrt{\gamma \delta }{x}_2+\sqrt{\beta \delta }{x}_3\right)i\end{array}\end{array}$$

(45)

The spectrum of a number $x=\left(x_1,x_2,x_3,x_4\right)$ in space $M_5\left(\alpha ,\beta ,\gamma ,\delta \right)$ is

$$\{\begin{array}{c}\begin{array}{c}{\mu }_1=\alpha x_1-\sqrt{\beta \gamma }{x}_4+\left(\sqrt{\gamma \delta }{x}_2+\sqrt{\beta \delta }{x}_3\right)i\\ {\mu }_2=\alpha x_1-\sqrt{\beta \gamma }{x}_4-\left(\sqrt{\gamma \delta }{x}_2+\sqrt{\beta \delta }{x}_3\right)i\end{array}\\ \begin{array}{c}{\mu }_3=\alpha x_1+\sqrt{\beta \gamma }{x}_4+\left(\sqrt{\gamma \delta }{x}_2-\sqrt{\beta \delta }{x}_3\right)i\\ {\mu }_4=\alpha x_1+\sqrt{\beta \gamma }{x}_4-\left(\sqrt{\gamma \delta }{x}_2-\sqrt{\beta \delta }{x}_3\right)i\end{array}\end{array}$$

(46)

The spectrum of a number $x=\left(x_1,x_2,x_3,x_4\right)$ in space $M_6\left(\alpha ,\beta ,\gamma ,\delta \right)$ is

$$\{\begin{array}{c}\begin{array}{c}{\mu }_1=\alpha x_1-\sqrt{\beta \delta }{x}_3+\left(\sqrt{\gamma \delta }{x}_2+\sqrt{\beta \gamma }{x}_4\right)i\\ {\mu }_2=\alpha x_1-\sqrt{\beta \delta }{x}_3-\left(\sqrt{\gamma \delta }{x}_2+\sqrt{\beta \gamma }{x}_4\right)i\end{array}\\ \begin{array}{c}{\mu }_3=\alpha x_1+\sqrt{\beta \delta }{x}_3+\left(\sqrt{\gamma \delta }{x}_2-\sqrt{\beta \gamma }{x}_4\right)i\\ {\mu }_4=\alpha x_1+\sqrt{\beta \delta }{x}_3-\left(\sqrt{\gamma \delta }{x}_2-\sqrt{\beta \gamma }{x}_4\right)i\end{array}\end{array}$$

(47)

The spectrum of a number $x=\left(x_1,x_2,x_3,x_4\right)$ in space $M_7\left(\alpha ,\beta ,\gamma ,\delta \right)$ is

$$\{\begin{array}{c}\begin{array}{c}{\mu }_1=\alpha x_1-\sqrt{\gamma \delta }{x}_2+\left(\sqrt{\beta \delta }{x}_3+\sqrt{\beta \gamma }{x}_4\right)i\\ {\mu }_2=\alpha x_1-\sqrt{\gamma \delta }{x}_2-\left(\sqrt{\beta \delta }{x}_3+\sqrt{\beta \gamma }{x}_4\right)i\end{array}\\ \begin{array}{c}{\mu }_3=\alpha x_1+\sqrt{\gamma \delta }{x}_2+\left(\sqrt{\beta \delta }{x}_3-\sqrt{\beta \gamma }{x}_4\right)i\\ {\mu }_4=\alpha x_1+\sqrt{\gamma \delta }{x}_2-\left(\sqrt{\beta \delta }{x}_3-\sqrt{\beta \gamma }{x}_4\right)i\end{array}\end{array}$$

(48)

Let us denote the spectrum of a number $x$ by $\Lambda \left(x\right)$ and consider mapping $S:x\to \Lambda \left(x\right)$ in spaces $M_i\left(\alpha ,\beta ,\gamma ,\delta \right),i=2,3,4,5,6,7$.

Theorem 5.

The mapping S is one-to-one and onto, so it is a bijection.

Proof. 

Let us prove the theorem only for space $M_2\left(\alpha ,\beta ,\gamma ,\delta \right)$. For other spaces, the proof is similar. □

Letting $x\ne y$, we will show that then, $\Lambda \left(x\right)\ne \Lambda \left(y\right)$. Let us allow the opposite, which means that ${\mu }_i\left(x\right)={\mu }_i\left(y\right), i=1,2,3,4$. Therefore,

$$\alpha x_1-\sqrt{\gamma \delta }{x}_2+\left(\sqrt{\beta \delta }{x}_3-\sqrt{\beta \gamma }{x}_4\right)i=\alpha y_1-\sqrt{\gamma \delta }{y}_2+\left(\sqrt{\beta \delta }{y}_3-\sqrt{\beta \gamma }{y}_4\right)i,$$

$$\alpha x_1-\sqrt{\gamma \delta }{x}_2-\left(\sqrt{\beta \delta }{x}_3-\sqrt{\beta \gamma }{x}_4\right)i=\alpha y_1-\sqrt{\gamma \delta }{y}_2-\left(\sqrt{\beta \delta }{y}_3-\sqrt{\beta \gamma }{y}_4\right)i,$$

$$\alpha x_1+\sqrt{\gamma \delta }{x}_2+\left(\sqrt{\beta \delta }{x}_3+\sqrt{\beta \gamma }{x}_4\right)i=\alpha y_1+\sqrt{\gamma \delta }{y}_2+\left(\sqrt{\beta \delta }{y}_3+\sqrt{\beta \gamma }{y}_4\right)i,$$

$$\alpha x_1+\sqrt{\gamma \delta }{x}_2-\left(\sqrt{\beta \delta }{x}_3+\sqrt{\beta \gamma }{x}_4\right)i=\alpha y_1+\sqrt{\gamma \delta }{y}_2-\left(\sqrt{\beta \delta }{y}_3+\sqrt{\beta \gamma }{y}_4\right)i.$$

Moving the right parts to the left part and collecting similar terms, we obtain

$$(\alpha x_1-\alpha y_1)-\left(\sqrt{\gamma \delta }{x}_2-\sqrt{\gamma \delta }{y}_2\right)+\left(\left(\sqrt{\beta \delta }{x}_3-\sqrt{\beta \delta }{y}_3\right)-(\sqrt{\beta \gamma }{x}_4-\sqrt{\beta \delta }{y}_4)\right)i=0,$$

$$(\alpha x_1-\alpha y_1)-\left(\sqrt{\gamma \delta }{x}_2-\sqrt{\gamma \delta }{y}_2\right)-\left(\left(\sqrt{\beta \delta }{x}_3-\sqrt{\beta \delta }{y}_3\right)-(\sqrt{\beta \gamma }{x}_4-\sqrt{\beta \delta }{y}_4)\right)i=0,$$

$$(\alpha x_1-\alpha y_1)+\left(\sqrt{\gamma \delta }{x}_2-\sqrt{\gamma \delta }{y}_2\right)+\left(\left(\sqrt{\beta \delta }{x}_3-\sqrt{\beta \delta }{y}_3\right)+(\sqrt{\beta \gamma }{x}_4-\sqrt{\beta \delta }{y}_4)\right)i=0,$$

$$(\alpha x_1-\alpha y_1)+\left(\sqrt{\gamma \delta }{x}_2-\sqrt{\gamma \delta }{y}_2\right)-\left(\left(\sqrt{\beta \delta }{x}_3-\sqrt{\beta \delta }{y}_3\right)+(\sqrt{\beta \gamma }{x}_4-\sqrt{\beta \delta }{y}_4)\right)i=0.$$

From here, we obtain

$$\{\begin{array}{c}\begin{array}{c}\alpha x_1-\alpha y_1=0,\\ \sqrt{\gamma \delta }{x}_2-\sqrt{\gamma \delta }{y}_2=0,\end{array}\\ \begin{array}{c}\sqrt{\beta \delta }{x}_3-\sqrt{\beta \delta }{y}_3=0,\\ \sqrt{\beta \gamma }{x}_4-\sqrt{\beta \delta }{y}_4=0.\end{array}\end{array}$$

or $x-y=0$. We receive the contradiction with the condition $x\ne y$.

Moving back, we will show that one and only one four-dimensional number can correspond to any spectrum consisting of numbers of the form (43). Indeed, let ${\mu }_1=a+bi,{\mu }_2=a-bi,{\mu }_3=c+di,{\mu }_4=c-di$ be the spectrum of some four-dimensional number. It then follows from Formula (43) that

$$\{\begin{array}{c}\begin{array}{c}\alpha x_1={\displaystyle \frac{a+c}{2}}\\ \sqrt{\gamma \delta }{x}_2={\displaystyle \frac{c-a}{2}}\end{array}\\ \begin{array}{c}\sqrt{\beta \delta }{x}_3={\displaystyle \frac{b+d}{2}}\\ \sqrt{\beta \gamma }{x}_4={\displaystyle \frac{d-c}{2}}\end{array}\end{array}$$

Corollary 1.

The only number having zero spectrum is the number 0 = (0,0,0,0).

Corollary 2.

Spectra of basic numbers in space $M_2\left(\alpha ,\beta ,\gamma ,\delta \right)$ are $\Lambda \left(J_1\right)=\left(1,1,1,1\right)$, $\Lambda \left(J_2\right)={\displaystyle \frac{\sqrt{\gamma \delta }}{\alpha }}\left(-1,-1,1,1\right)$, $\Lambda \left(J_3\right)={\displaystyle \frac{\sqrt{\beta \delta }}{\alpha }}\left(i,-i,i,-i\right)$, $\Lambda \left(J_4\right)={\displaystyle \frac{\sqrt{\beta \gamma }}{\alpha }}\left(-i,i,i,-i\right)$. Spectra of basic numbers in other spaces are written out similarly.

Theorem 6.

For any four-dimensional number $x=\left(x_1,x_2,x_3,x_4\right)$, the equality holds:

$${\left|x\right|}^4={\mu }_1·{\mu }_2·{\mu }_3·{\mu }_4$$

(49)

Proof. 

Let us also carry out the proof of the theorem only for space $M_2\left(\alpha ,\beta ,\gamma ,\delta \right)$. For other spaces, the proof is absolutely similar. □

It follows from Equation (43) that ${\mu }_1·{\mu }_2={\left(\alpha x_1-\sqrt{\gamma \delta }{x}_2\right)}^2+{\left(\sqrt{\beta \delta }{x}_3-\sqrt{\beta \gamma }{x}_4\right)}^2{\mu }_3·{\mu }_4={\left(\alpha x_1+\sqrt{\gamma \delta }{x}_2\right)}^2+{\left(\sqrt{\beta \delta }{x}_3+\sqrt{\beta \gamma }{x}_4\right)}^2$. Therefore, it follows from Equality (6) that ${\mu }_1·{\mu }_2·{\mu }_3·{\mu }_4={\left|x\right|}^4$.

Corollary 3.

$\left|J_1\right|=1$, $\left|J_2\right|={\displaystyle \frac{\sqrt{\gamma \delta }}{\alpha }}$, $\left|J_3\right|={\displaystyle \frac{\sqrt{\beta \delta }}{\alpha }}$, $\left|J_4\right|={\displaystyle \frac{\sqrt{\beta \gamma }}{\alpha }}$ in any space $M_i\left(\alpha ,\beta ,\gamma ,\delta \right),i=2,3,4,5,6,7$.

Theorem 7.

In each of six anisotropic spaces of four-dimensional numbers, the following relations hold:

(1)

${\mu }_i\left(x\pm y\right)={\mu }_i\left(x\right)\pm {\mu }_i\left(y\right)$ for any $x\in R^4$, $y\in R^4$, $i=1,2,3,4$;

(2)

${\mu }_i\left(x·y\right)={\mu }_i\left(x\right)·{\mu }_i\left(y\right)$ for any $x\in R^4$, $y\in R^4$, $i=1,2,3,4$;

(3)

${\mu }_i\left(b·x\right)=b·{\mu }_i\left(x\right)$, for any $x\in R^4$, $b\in R^1$, $i=1,2,3,4$;

(4)

${\mu }_i\left(x^{-1}\right)={\left({\mu }_i\left(x\right)\right)}^{-1}$ for any nondegenerate, $i=1,2,3,4$,

(5)

${\mu }_i\left(x^n\right)={\mu }_i^n\left(x\right)$, for any $x\in R^4$, $i=1,2,3,4$, $n\in N$,

where ${\mu }_i\left(x\right)$is the i-th component of the spectrum of four-dimensional number $x$.

Proof. 

Let us study only for space $M_2\left(\alpha ,\beta ,\gamma ,\delta \right)$. □

Relations (1) are obvious. Let us prove the Relation (2). According to (5) and (43),

$$\begin{array}{cc}\hfill {\mu }_1\left(x·y\right)& =\left({\alpha }^2{x}_1{y}_1+\gamma \delta x_2{y}_2-\beta \delta x_3{y}_3-\beta \gamma x_4{y}_4-\alpha \sqrt{\gamma \delta }{x}_2{y}_1-\alpha \sqrt{\gamma \delta }{x}_1{y}_2+\beta \sqrt{\gamma \delta }{x}_4{y}_3+\beta \sqrt{\gamma \delta }{x}_3{y}_4\right)\hfill \\ \hfill & +(\alpha \sqrt{\beta \delta }{x}_3{y}_1+\gamma \sqrt{\beta \delta }{x}_4{y}_2+\alpha \sqrt{\beta \delta }{x}_1{y}_3+\gamma \sqrt{\beta \delta }{x}_2{y}_4-\alpha \sqrt{\beta \gamma }{x}_4{y}_1-\delta \sqrt{\beta \gamma }{x}_3{y}_2\hfill \\ \hfill & -\delta \sqrt{\beta \gamma }{x}_2{y}_3-\alpha \sqrt{\beta \gamma }{x}_1{y}_4)·i,\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\mu }_2\left(x·y\right)& =\left({\alpha }^2{x}_1{y}_1+\gamma \delta x_2{y}_2-\beta \delta x_3{y}_3-\beta \gamma x_4{y}_4-\alpha \sqrt{\gamma \delta }{x}_2{y}_1-\alpha \sqrt{\gamma \delta }{x}_1{y}_2+\beta \sqrt{\gamma \delta }{x}_4{y}_3+\beta \sqrt{\gamma \delta }{x}_3{y}_4\right)\hfill \\ \hfill & -(\alpha \sqrt{\beta \delta }{x}_3{y}_1+\gamma \sqrt{\beta \delta }{x}_4{y}_2+\alpha \sqrt{\beta \delta }{x}_1{y}_3+\gamma \sqrt{\beta \delta }{x}_2{y}_4-\alpha \sqrt{\beta \gamma }{x}_4{y}_1-\delta \sqrt{\beta \gamma }{x}_3{y}_2\hfill \\ \hfill & -\delta \sqrt{\beta \gamma }{x}_2{y}_3-\alpha \sqrt{\beta \gamma }{x}_1{y}_4)·i,\hfill \end{array}$$

$$\begin{gathered} {\mu }_3\left(x·y\right)=\left({\alpha }^2{x}_1{y}_1+\gamma \delta x_2{y}_2-\beta \delta x_3{y}_3-\beta \gamma x_4{y}_4+\alpha \sqrt{\gamma \delta }{x}_2{y}_1+\alpha \sqrt{\gamma \delta }{x}_1{y}_2-\beta \sqrt{\gamma \delta }{x}_4{y}_3-\beta \sqrt{\gamma \delta }{x}_3{y}_4\right)+ \\ \left(\alpha \sqrt{\beta \delta }{x}_3{y}_1+\gamma \sqrt{\beta \delta }{x}_4{y}_2+\alpha \sqrt{\beta \delta }{x}_1{y}_3+\gamma \sqrt{\beta \delta }{x}_2{y}_4+\alpha \sqrt{\beta \gamma }{x}_4{y}_1+\delta \sqrt{\beta \gamma }{x}_3{y}_2+\delta \sqrt{\beta \gamma }{x}_2{y}_3+\alpha \sqrt{\beta \gamma }{x}_1{y}_4\right)·i, \end{gathered}$$

$$\begin{array}{cc}\hfill {\mu }_4\left(x·y\right)& =\left({\alpha }^2{x}_1{y}_1+\gamma \delta x_2{y}_2-\beta \delta x_3{y}_3-\beta \gamma x_4{y}_4+\alpha \sqrt{\gamma \delta }{x}_2{y}_1+\alpha \sqrt{\gamma \delta }{x}_1{y}_2-\beta \sqrt{\gamma \delta }{x}_4{y}_3-\beta \sqrt{\gamma \delta }{x}_3{y}_4\right)\hfill \\ \hfill & -(\alpha \sqrt{\beta \delta }{x}_3{y}_1+\gamma \sqrt{\beta \delta }{x}_4{y}_2+\alpha \sqrt{\beta \delta }{x}_1{y}_3+\gamma \sqrt{\beta \delta }{x}_2{y}_4+\alpha \sqrt{\beta \gamma }{x}_4{y}_1+\delta \sqrt{\beta \gamma }{x}_3{y}_2\hfill \\ \hfill & +\delta \sqrt{\beta \gamma }{x}_2{y}_3+\alpha \sqrt{\beta \gamma }{x}_1{y}_4)·i.\hfill \end{array}$$

Similarly,

$${\mu }_1\left(x\right)=\left(\alpha x_1-\sqrt{\gamma \delta }{x}_2\right)+\left(\sqrt{\beta \delta }{x}_3-\sqrt{\beta \gamma }{x}_4\right)·i,$$

$${\mu }_2\left(x\right)=\left(\alpha x_1-\sqrt{\gamma \delta }{x}_2\right)-\left(\sqrt{\beta \delta }{x}_3-\sqrt{\beta \gamma }{x}_4\right)·i,$$

$${\mu }_3\left(x\right)=\left(\alpha x_1+\sqrt{\gamma \delta }{x}_2\right)+\left(\sqrt{\beta \delta }{x}_3+\sqrt{\beta \gamma }{x}_4\right)·i,$$

$${\mu }_4\left(x\right)=\left(\alpha x_1+\sqrt{\gamma \delta }{x}_2\right)-\left(\sqrt{\beta \delta }{x}_3+\sqrt{\beta \gamma }{x}_4\right)·i.$$

$${\mu }_1\left(y\right)=\left(\alpha y_1-\sqrt{\gamma \delta }{y}_2\right)+\left(\sqrt{\beta \delta }{y}_3-\sqrt{\beta \gamma }{y}_4\right)·i,$$

$${\mu }_2\left(y\right)=\left(\alpha y_1-\sqrt{\gamma \delta }{y}_2\right)-\left(\sqrt{\beta \delta }{y}_3-\sqrt{\beta \gamma }{y}_4\right)·i,$$

$${\mu }_3\left(y\right)=\left(\alpha y_1+\sqrt{\gamma \delta }{y}_2\right)+\left(\sqrt{\beta \delta }{y}_3+\sqrt{\beta \gamma }{y}_4\right)·i,$$

$${\mu }_4\left(y\right)=\left(\alpha y_1+\sqrt{\gamma \delta }{y}_2\right)-\left(\sqrt{\beta \delta }{y}_3+\sqrt{\beta \gamma }{y}_4\right)·i.$$

Then

$$\begin{array}{cc}\hfill {\mu }_1\left(x\right)·{\mu }_1\left(y\right)& =\left({\alpha }^2{x}_1{y}_1+\gamma \delta x_2{y}_2-\beta \delta x_3{y}_3-\beta \gamma x_4{y}_4-\alpha \sqrt{\gamma \delta }{x}_2{y}_1-\alpha \sqrt{\gamma \delta }{x}_1{y}_2+\beta \sqrt{\gamma \delta }{x}_4{y}_3+\beta \sqrt{\gamma \delta }{x}_3{y}_4\right)\hfill \\ \hfill & +(\alpha \sqrt{\beta \delta }{x}_3{y}_1+\gamma \sqrt{\beta \delta }{x}_4{y}_2+\alpha \sqrt{\beta \delta }{x}_1{y}_3+\gamma \sqrt{\beta \delta }{x}_2{y}_4-\alpha \sqrt{\beta \gamma }{x}_4{y}_1-\delta \sqrt{\beta \gamma }{x}_3{y}_2\hfill \\ \hfill & -\delta \sqrt{\beta \gamma }{x}_2{y}_3-\alpha \sqrt{\beta \gamma }{x}_1{y}_4)·i={\mu }_1\left(x·y\right),\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\mu }_2\left(x\right)·{\mu }_2\left(y\right)& =\left({\alpha }^2{x}_1{y}_1+\gamma \delta x_2{y}_2-\beta \delta x_3{y}_3-\beta \gamma x_4{y}_4-\alpha \sqrt{\gamma \delta }{x}_2{y}_1-\alpha \sqrt{\gamma \delta }{x}_1{y}_2+\beta \sqrt{\gamma \delta }{x}_4{y}_3+\beta \sqrt{\gamma \delta }{x}_3{y}_4\right)\hfill \\ \hfill & -(\alpha \sqrt{\beta \delta }{x}_3{y}_1+\gamma \sqrt{\beta \delta }{x}_4{y}_2+\alpha \sqrt{\beta \delta }{x}_1{y}_3+\gamma \sqrt{\beta \delta }{x}_2{y}_4-\alpha \sqrt{\beta \gamma }{x}_4{y}_1-\delta \sqrt{\beta \gamma }{x}_3{y}_2\hfill \\ \hfill & -\delta \sqrt{\beta \gamma }{x}_2{y}_3-\alpha \sqrt{\beta \gamma }{x}_1{y}_4)·i={\mu }_2\left(x·y\right),\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\mu }_3\left(x\right)·{\mu }_3\left(y\right)& =\left({\alpha }^2{x}_1{y}_1+\gamma \delta x_2{y}_2-\beta \delta x_3{y}_3-\beta \gamma x_4{y}_4+\alpha \sqrt{\gamma \delta }{x}_2{y}_1+\alpha \sqrt{\gamma \delta }{x}_1{y}_2-\beta \sqrt{\gamma \delta }{x}_4{y}_3-\beta \sqrt{\gamma \delta }{x}_3{y}_4\right)\hfill \\ \hfill & +(\alpha \sqrt{\beta \delta }{x}_3{y}_1+\gamma \sqrt{\beta \delta }{x}_4{y}_2+\alpha \sqrt{\beta \delta }{x}_1{y}_3+\gamma \sqrt{\beta \delta }{x}_2{y}_4+\alpha \sqrt{\beta \gamma }{x}_4{y}_1+\delta \sqrt{\beta \gamma }{x}_3{y}_2\hfill \\ \hfill & +\delta \sqrt{\beta \gamma }{x}_2{y}_3+\alpha \sqrt{\beta \gamma }{x}_1{y}_4)·i={\mu }_3\left(x·y\right),\hfill \end{array}$$

$$\begin{array}{cc}\hfill {\mu }_4\left(x\right)·{\mu }_4\left(y\right)& =\left({\alpha }^2{x}_1{y}_1+\gamma \delta x_2{y}_2-\beta \delta x_3{y}_3-\beta \gamma x_4{y}_4+\alpha \sqrt{\gamma \delta }{x}_2{y}_1+\alpha \sqrt{\gamma \delta }{x}_1{y}_2-\beta \sqrt{\gamma \delta }{x}_4{y}_3-\beta \sqrt{\gamma \delta }{x}_3{y}_4\right)\hfill \\ \hfill & -(\alpha \sqrt{\beta \delta }{x}_3{y}_1+\gamma \sqrt{\beta \delta }{x}_4{y}_2+\alpha \sqrt{\beta \delta }{x}_1{y}_3+\gamma \sqrt{\beta \delta }{x}_2{y}_4+\alpha \sqrt{\beta \gamma }{x}_4{y}_1+\delta \sqrt{\beta \gamma }{x}_3{y}_2\hfill \\ \hfill & +\delta \sqrt{\beta \gamma }{x}_2{y}_3+\alpha \sqrt{\beta \gamma }{x}_1{y}_4)·i={\mu }_4\left(x·y\right).\hfill \end{array}$$

The ratio (3) follows from the ratio (2). Let us prove the ratio (4). (2) follows from the ratio that ${\mu }_i\left(x·x^{-1}\right)={\mu }_i\left(x\right)·{\mu }_i\left(x^{-1}\right)$. On the other hand, $x·x^{-1}=J_1$. It follows from ratios (43) that ${\mu }_i\left(J_1\right)=1$ for all $i=1,2,3,4$. Therefore, ${\mu }_i\left(x^{-1}\right)={\displaystyle \frac{1}{{\mu }_i\left(x\right)}}$.

Relation (5) follows from Relation (2) as well.

The proven theorems show that there are three varying approaches for work with four-dimensional numbers: four-dimensional numbers, 4 × 4 matrices and four-dimensional imaginary numbers in the form of the spectrum. For carrying out arithmetic operations, these approaches are equivalent. For the solution of the algebraic equations, the most convenient is the spectra approach.

5. Application to the Solution of Systems of the Algebraic Equations

Let us consider the linear algebraic equation $ax=b$, where $a=\left(a_1,a_2,a_3,a_4\right)$ and $b=\left(b_1,b_2,b_3,b_4\right)$ are given four-dimensional numbers. Assuming, at first, that a is nondegenerate number, then this equation has as its only solution $x=b{a}^{-1}$. Of course, we consider this equation in one of the spaces $M_i\left(\alpha ,\beta ,\gamma ,\delta \right),i=2,3,4,5,6,7$. Let us find the solution of the equation in isotropic space $M_2$.

For finding of the inverse number $a^{-1}$, we will use Theorem 7, according to which we obtain ${\mu }_i\left(a^{-1}\right)={\left({\mu }_i\left(a\right)\right)}^{-1}$ for all i = 1, 2, 3, 4. Therefore, in isotropic space $M_2$, we have

$${\mu }_1^{-1}\left(a\right)={\displaystyle \frac{a_1-a_2}{{\left(a_1-a_2\right)}^2+{\left(a_3-a_4\right)}^2}}-{\displaystyle \frac{a_3-a_4}{{\left(a_1-a_2\right)}^2+{\left(a_3-a_4\right)}^2}}i={\left({\mu }_1\left(a\right)\right)}^{-1},$$

$${\mu }_2^{-1}\left(a\right)={\displaystyle \frac{a_1-a_2}{{\left(a_1-a_2\right)}^2+{\left(a_3-a_4\right)}^2}}+{\displaystyle \frac{a_3-a_4}{{\left(a_1-a_2\right)}^2+{\left(a_3-a_4\right)}^2}}i={\left({\mu }_2\left(a\right)\right)}^{-1},$$

$${\mu }_3^{-1}\left(a\right)={\displaystyle \frac{a_1+a_2}{{\left(a_1+a_2\right)}^2+{\left(a_3+a_4\right)}^2}}-{\displaystyle \frac{a_3+a_4}{{\left(a_1+a_2\right)}^2+{\left(a_3+a_4\right)}^2}}i={\left({\mu }_3\left(a\right)\right)}^{-1},$$

$${\mu }_4^{-1}\left(a\right)={\displaystyle \frac{a_1+a_2}{{\left(a_1+a_2\right)}^2+{\left(a_3+a_4\right)}^2}}+{\displaystyle \frac{a_3+a_4}{{\left(a_1+a_2\right)}^2+{\left(a_3+a_4\right)}^2}}i={\left({\mu }_4\left(a\right)\right)}^{-1}.$$

Knowing the spectrum, we will restore number $a^{-1}$:

$$\begin{array}{cc}\hfill a^{-1}& =({\displaystyle \frac{1}{2}}\left({\displaystyle \frac{a_1-a_2}{{\left(a_1-a_2\right)}^2+{\left(a_3-a_4\right)}^2}}+{\displaystyle \frac{a_1+a_2}{{\left(a_1+a_2\right)}^2+{\left(a_3+a_4\right)}^2}}\right),{\displaystyle \frac{1}{2}}(-{\displaystyle \frac{a_1-a_2}{{\left(a_1-a_2\right)}^2+{\left(a_3-a_4\right)}^2}}\hfill \\ \hfill & +{\displaystyle \frac{a_1+a_2}{{\left(a_1+a_2\right)}^2+{\left(a_3+a_4\right)}^2}}),{\displaystyle \frac{1}{2}}(-{\displaystyle \frac{a_3-a_4}{{\left(a_1-a_2\right)}^2+{\left(a_3-a_4\right)}^2}}\hfill \\ \hfill & -{\displaystyle \frac{a_3+a_4}{{\left(a_1+a_2\right)}^2+{\left(a_3+a_4\right)}^2}}),{\displaystyle \frac{1}{2}}\left({\displaystyle \frac{a_3-a_4}{{\left(a_1-a_2\right)}^2+{\left(a_3-a_4\right)}^2}}-{\displaystyle \frac{a_3+a_4}{{\left(a_1+a_2\right)}^2+{\left(a_3+a_4\right)}^2}}\right)).\hfill \end{array}$$

Next, the solution $x=\left(x_1,x_2,x_3,x_4\right)$ of linear equation $ax=b$ in the space $M_2$ is as follows:

$$\begin{array}{cc}\hfill x_1& ={\displaystyle \frac{b_1}{2}}\left({\displaystyle \frac{a_1-a_2}{{\left(a_1-a_2\right)}^2+{\left(a_3-a_4\right)}^2}}+{\displaystyle \frac{a_1+a_2}{{\left(a_1+a_2\right)}^2+{\left(a_3+a_4\right)}^2}}\right)\hfill \\ \hfill & -{\displaystyle \frac{b_2}{2}}\left({\displaystyle \frac{a_1-a_2}{{\left(a_1-a_2\right)}^2+{\left(a_3-a_4\right)}^2}}-{\displaystyle \frac{a_1+a_2}{{\left(a_1+a_2\right)}^2+{\left(a_3+a_4\right)}^2}}\right)\hfill \\ \hfill & +{\displaystyle \frac{b_3}{2}}\left({\displaystyle \frac{a_3-a_4}{{\left(a_1-a_2\right)}^2+{\left(a_3-a_4\right)}^2}}+{\displaystyle \frac{a_3+a_4}{{\left(a_1+a_2\right)}^2+{\left(a_3+a_4\right)}^2}}\right)\hfill \\ \hfill & -{\displaystyle \frac{b_4}{2}}\left({\displaystyle \frac{a_3-a_4}{{\left(a_1-a_2\right)}^2+{\left(a_3-a_4\right)}^2}}-{\displaystyle \frac{a_3+a_4}{{\left(a_1+a_2\right)}^2+{\left(a_3+a_4\right)}^2}}\right),\hfill \end{array}$$

$$\begin{array}{cc}\hfill x_2& ={\displaystyle \frac{b_2}{2}}\left({\displaystyle \frac{a_1-a_2}{{\left(a_1-a_2\right)}^2+{\left(a_3-a_4\right)}^2}}+{\displaystyle \frac{a_1+a_2}{{\left(a_1+a_2\right)}^2+{\left(a_3+a_4\right)}^2}}\right)\hfill \\ \hfill & -{\displaystyle \frac{b_1}{2}}\left({\displaystyle \frac{a_1-a_2}{{\left(a_1-a_2\right)}^2+{\left(a_3-a_4\right)}^2}}-{\displaystyle \frac{a_1+a_2}{{\left(a_1+a_2\right)}^2+{\left(a_3+a_4\right)}^2}}\right)\hfill \\ \hfill & +{\displaystyle \frac{b_4}{2}}\left({\displaystyle \frac{a_3-a_4}{{\left(a_1-a_2\right)}^2+{\left(a_3-a_4\right)}^2}}+{\displaystyle \frac{a_3+a_4}{{\left(a_1+a_2\right)}^2+{\left(a_3+a_4\right)}^2}}\right)\hfill \\ \hfill & -{\displaystyle \frac{b_3}{2}}\left({\displaystyle \frac{a_3-a_4}{{\left(a_1-a_2\right)}^2+{\left(a_3-a_4\right)}^2}}-{\displaystyle \frac{a_3+a_4}{{\left(a_1+a_2\right)}^2+{\left(a_3+a_4\right)}^2}}\right),\hfill \end{array}$$

$$\begin{array}{cc}\hfill x_3& ={\displaystyle \frac{b_3}{2}}\left({\displaystyle \frac{a_1-a_2}{{\left(a_1-a_2\right)}^2+{\left(a_3-a_4\right)}^2}}+{\displaystyle \frac{a_1+a_2}{{\left(a_1+a_2\right)}^2+{\left(a_3+a_4\right)}^2}}\right)\hfill \\ \hfill & -{\displaystyle \frac{b_4}{2}}\left({\displaystyle \frac{a_1-a_2}{{\left(a_1-a_2\right)}^2+{\left(a_3-a_4\right)}^2}}-{\displaystyle \frac{a_1+a_2}{{\left(a_1+a_2\right)}^2+{\left(a_3+a_4\right)}^2}}\right)\hfill \\ \hfill & -{\displaystyle \frac{b_1}{2}}\left({\displaystyle \frac{a_3-a_4}{{\left(a_1-a_2\right)}^2+{\left(a_3-a_4\right)}^2}}+{\displaystyle \frac{a_3+a_4}{{\left(a_1+a_2\right)}^2+{\left(a_3+a_4\right)}^2}}\right)\hfill \\ \hfill & +{\displaystyle \frac{b_2}{2}}\left({\displaystyle \frac{a_3-a_4}{{\left(a_1-a_2\right)}^2+{\left(a_3-a_4\right)}^2}}-{\displaystyle \frac{a_3+a_4}{{\left(a_1+a_2\right)}^2+{\left(a_3+a_4\right)}^2}}\right),\hfill \end{array}$$

$$\begin{array}{cc}\hfill x_4& ={\displaystyle \frac{b_4}{2}}\left({\displaystyle \frac{a_1-a_2}{{\left(a_1-a_2\right)}^2+{\left(a_3-a_4\right)}^2}}+{\displaystyle \frac{a_1+a_2}{{\left(a_1+a_2\right)}^2+{\left(a_3+a_4\right)}^2}}\right)\hfill \\ \hfill & -{\displaystyle \frac{b_3}{2}}\left({\displaystyle \frac{a_1-a_2}{{\left(a_1-a_2\right)}^2+{\left(a_3-a_4\right)}^2}}-{\displaystyle \frac{a_1+a_2}{{\left(a_1+a_2\right)}^2+{\left(a_3+a_4\right)}^2}}\right)\hfill \\ \hfill & -{\displaystyle \frac{b_2}{2}}\left({\displaystyle \frac{a_3-a_4}{{\left(a_1-a_2\right)}^2+{\left(a_3-a_4\right)}^2}}+{\displaystyle \frac{a_3+a_4}{{\left(a_1+a_2\right)}^2+{\left(a_3+a_4\right)}^2}}\right)\hfill \\ \hfill & +{\displaystyle \frac{b_1}{2}}\left({\displaystyle \frac{a_3-a_4}{{\left(a_1-a_2\right)}^2+{\left(a_3-a_4\right)}^2}}-{\displaystyle \frac{a_3+a_4}{{\left(a_1+a_2\right)}^2+{\left(a_3+a_4\right)}^2}}\right).\hfill \end{array}$$

We received the analog of formulas of Kramer for system

$$\{\begin{array}{c}\begin{array}{c}{a}_1{x}_1+a_2{x}_2-a_3{x}_3-a_4{x}_4=b_1\\ a_2{x}_1+a_1{x}_2-a_4{x}_3-a_3{x}_4=b_2\end{array}\\ \begin{array}{c}{a}_3{x}_1+a_4{x}_2+a_1{x}_3+a_2{x}_4=b_3\\ a_4{x}_1+a_3{x}_2+a_2{x}_3+a_1{x}_4=b_4\end{array}\end{array},$$

(50)

Which is equivalent to the four-dimensional equation $ax=b$ in isotropic space $M_2$.

Similar formulas can also be obtained for other four-dimensional spaces.

Let us assume now that a is a degenerate number. If b = (0,0,0,0), then, according to Theorem 2, considered linear equations have infinitely many solutions, namely, if a is a degenerate number of type II, then the solutions of the equation are all degenerate numbers of type I, and if a is a degenerate number of type I, then the solutions of the equation are all degenerate numbers of type II.

Let us assume now that a and b are degenerate numbers, and b ≠ (0,0,0,0). If a and b are degenerate numbers of different types, then according to the same Theorem 2, considered equation has no solution. If a and b are degenerate numbers of the same type, then the equation has an infinite number of solutions. Indeed, according to Theorem 7, we have

$${\mu }_i\left(a\right){\mu }_i\left(x\right)={\mu }_i\left(b\right), i=1,2,3,4.$$

Suppose that a and b are degenerate numbers of the first type, that is $a=\left(c_1,c_1,c_2,c_2\right)$, $b=\left(d_1,d_1,d_2,d_2\right)$. Then $\mu \left(a\right)=\left(0,0,2c_1+2c_{2}i,2c_1-2c_{2}i\right)$, $\mu \left(b\right)=\left(0,0,2d_1+2d_{2}i,2d_1-2d_{2}i\right)$, but $\mu \left(x\right)=\left(x_1-x_2+\left(x_3-x_4\right)i,x_1-x_2-\left(x_3-x_4\right)i,x_1+x_2+\left(x_3+x_4\right)i,x_1+x_2-\left(x_3+x_4\right)i\right)$. Substituting these expressions in the last equations, we have

$$\{\begin{array}{c}\left(c_1+c_{2}i\right)\left(x_1+x_2+\left(x_3+x_4\right)i\right)=d_1+d_{2}i\\ \left(c_1-c_{2}i\right)\left(x_1+x_2-\left(x_3+x_4\right)i\right)=d_1-d_{2}i\end{array}.$$

This system has an infinite number of solutions

$$x_1+x_2={\displaystyle \frac{c_1{d}_1+c_2{d}_2}{c_1^2+c_2^2}}, x_3+x_4={\displaystyle \frac{c_1{d}_2-c_2{d}_1}{c_1^2+c_2^2}}.$$

Solutions of linear equation are four-dimensional numbers $\left(x_1,-x_1+{\displaystyle \frac{c_1{d}_1+c_2{d}_2}{c_1^2+c_2^2}},x_3,-x_3+{\displaystyle \frac{c_1{d}_2-c_2{d}_1}{c_1^2+c_2^2}}\right)$, for any real $x_1$ and $x_2$.

Similar calculations can also be carried out in cases when a and b are degenerate numbers of the second type.

Thus, in four-dimensional spaces, linear equation $ax=b$ may have no solution, a unique solution, or an infinite set of solutions. We showed this fact in the example of isotropic space $M_2$. Check in other isotropic and anisotropic spaces does not cause difficulties.

Let us consider the quadratic equation $x^2=a$, where $a=\left(a_1,a_2,a_3,a_4\right)$ is a given four-dimensional number. As the example, we will consider this equation in isotropic space $M_2$. In this case, this equation can be rewritten by the definition of multiplication (6) in space $M_2$ in the following form:

$$\{\begin{array}{c}\begin{array}{c}{x}_1^2+x_2^2-x_3^2-x_4^2=a_1\\ 2x_1{x}_2-2x_3{x}_4=a_2\end{array}\\ \begin{array}{c}2x_1{x}_3+2x_2{x}_4=a_3\\ 2x_1{x}_4+2x_2{x}_3=a_4\end{array}\end{array}.$$

(51)

For the solution of this equation, we will use the spectra approach. Let us assume at first that a is a nondegenerate number. According to Theorem 7, ${\mu }_i\left(x^2\right)={\mu }_i^2\left(x\right)={\mu }_i\left(a\right), i=1,2,3,4$. By definition of the spectrum in space $M_2$

$$\begin{array}{c}\begin{array}{c}{\left(x_1-x_2+\left(x_3-x_4\right)i\right)}^2=a_1-a_2+\left(a_3-a_4\right)i\\ {\left(x_1-x_2-\left(x_3-x_4\right)i\right)}^2=a_1-a_2-\left(a_3-a_4\right)i\end{array}\\ \begin{array}{c}{\left(x_1+x_2+\left(x_3+x_4\right)i\right)}^2=a_1+a_2+\left(a_3+a_4\right)i\\ {\left(x_1+x_2-\left(x_3+x_4\right)i\right)}^2=a_1+a_2-\left(a_3+a_4\right)i\end{array}\end{array}$$

or

$$\begin{array}{c}{\left(x_1-x_2\right)}^2-{\left(x_3-x_4\right)}^2+2\left(x_1-x_2\right)\left(x_3-x_4\right)i=a_1-a_2+\left(a_3-a_4\right)i\\ {\left(x_1-x_2\right)}^2-{\left(x_3-x_4\right)}^2-2\left(x_1-x_2\right)\left(x_3-x_4\right)i=a_1-a_2-\left(a_3-a_4\right)i\\ {\left(x_1+x_2\right)}^2-{\left(x_3+x_4\right)}^2+2\left(x_1+x_2\right)\left(x_3+x_4\right)i=a_1+a_2+\left(a_3+a_4\right)i\\ {\left(x_1+x_2\right)}^2-{\left(x_3+x_4\right)}^2-2\left(x_1+x_2\right)\left(x_3+x_4\right)i=a_1+a_2-\left(a_3+a_4\right)i\end{array}$$

From here, equating the real and imaginary parts, we have

$$\begin{array}{c}\begin{array}{c}{\left(x_1-x_2\right)}^2-{\left(x_3-x_4\right)}^2=a_1-a_2\\ 2\left(x_1-x_2\right)\left(x_3-x_4\right)=a_3-a_4\end{array}\\ \begin{array}{c}{\left(x_1+x_2\right)}^2-{\left(x_3+x_4\right)}^2=a_1+a_2\\ 2\left(x_1+x_2\right)\left(x_3+x_4\right)=a_3+a_4\end{array}\end{array}.$$

(52)

This system has the following four solutions (if all the composed formulas given below make sense): $\left({\displaystyle \frac{b_{+}}{2\sqrt{2}}}+{\displaystyle \frac{b_{-}}{2\sqrt{2}}},{\displaystyle \frac{b_{+}}{2\sqrt{2}}}-{\displaystyle \frac{b_{-}}{2\sqrt{2}}},{\displaystyle \frac{a_3+a_4}{2\sqrt{2}{b}_{+}}}+{\displaystyle \frac{a_3-a_4}{2\sqrt{2}{b}_{-}}},{\displaystyle \frac{a_3+a_4}{2\sqrt{2}{b}_{+}}}-{\displaystyle \frac{a_3-a_4}{2\sqrt{2}{b}_{-}}}\right)$, $\left(-{\displaystyle \frac{b_{+}}{2\sqrt{2}}}+{\displaystyle \frac{b_{-}}{2\sqrt{2}}},-{\displaystyle \frac{b_{+}}{2\sqrt{2}}}-{\displaystyle \frac{b_{-}}{2\sqrt{2}}},-{\displaystyle \frac{a_3+a_4}{2\sqrt{2}{b}_{+}}}+{\displaystyle \frac{a_3-a_4}{2\sqrt{2}{b}_{-}}},-{\displaystyle \frac{a_3+a_4}{2\sqrt{2}{b}_{+}}}-{\displaystyle \frac{a_3-a_4}{2\sqrt{2}{b}_{-}}}\right)$, $\left({\displaystyle \frac{b_{+}}{2\sqrt{2}}}-{\displaystyle \frac{b_{-}}{2\sqrt{2}}},{\displaystyle \frac{b_{+}}{2\sqrt{2}}}+{\displaystyle \frac{b_{-}}{2\sqrt{2}}},{\displaystyle \frac{a_3+a_4}{2\sqrt{2}{b}_{+}}}-{\displaystyle \frac{a_3-a_4}{2\sqrt{2}{b}_{-}}},{\displaystyle \frac{a_3+a_4}{2\sqrt{2}{b}_{+}}}+{\displaystyle \frac{a_3-a_4}{2\sqrt{2}{b}_{-}}}\right)$, $\left(-{\displaystyle \frac{b_{+}}{2\sqrt{2}}}-{\displaystyle \frac{b_{-}}{2\sqrt{2}}},-{\displaystyle \frac{b_{+}}{2\sqrt{2}}}+{\displaystyle \frac{b_{-}}{2\sqrt{2}}},-{\displaystyle \frac{a_3+a_4}{2\sqrt{2}{b}_{+}}}-{\displaystyle \frac{a_3-a_4}{2\sqrt{2}{b}_{-}}},-{\displaystyle \frac{a_3+a_4}{2\sqrt{2}{b}_{+}}}+{\displaystyle \frac{a_3-a_4}{2\sqrt{2}{b}_{-}}}\right)$, where

$$b_{+}=\sqrt{a_1+a_2+\sqrt{{\left(a_1+a_2\right)}^2+{\left(a_3+a_4\right)}^2}}, b_{-}=\sqrt{a_1-a_2+\sqrt{{\left(a_1-a_2\right)}^2+{\left(a_3-a_4\right)}^2}}.$$

The obtained solutions are the square root from four-dimensional number $a$. In particular, in space $M_2$, solutions of the equation $x^2=J_1=\left(1,0,0,0\right)$ are four four-dimensional numbers (1,0,0,0), (0,1,0,0), (−1,0,0,0) and (0,−1,0,0). Solutions of the equation $x^2=J_3=\left(0,0,1,0\right)$ are the following numbers: $\left({\displaystyle \frac{1}{\sqrt{2}}},0,{\displaystyle \frac{1}{\sqrt{2}}},0\right),\left(-{\displaystyle \frac{1}{\sqrt{2}}},0,-{\displaystyle \frac{1}{\sqrt{2}}},0\right),\left(0,{\displaystyle \frac{1}{\sqrt{2}}},0,{\displaystyle \frac{1}{\sqrt{2}}}\right),\left(0,-{\displaystyle \frac{1}{\sqrt{2}}},0,-{\displaystyle \frac{1}{\sqrt{2}}}\right)$. If $a=J_2$, then the second terms of formulas for $x_3$ and $x_4$ are undefined. Solving the system (52) for this case, we will receive the following solutions: $\left({\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}},-{\displaystyle \frac{1}{2}}\right)$, $\left({\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}},-{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}}\right)$, $\left(-{\displaystyle \frac{1}{2}},-{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}},-{\displaystyle \frac{1}{2}}\right)$, $\left(-{\displaystyle \frac{1}{2}},-{\displaystyle \frac{1}{2}},-{\displaystyle \frac{1}{2}},{\displaystyle \frac{1}{2}}\right)$.

Thus, systems of type (51) have four roots.

Let us assume now that $a=\left(a_1,a_1,a_3,a_3\right)$ is nonzero degenerate number of the first type. Next, system (52) will take the following form:

$$\begin{array}{c}\begin{array}{c}{\left(x_1-x_2\right)}^2-{\left(x_3-x_4\right)}^2=0\\ 2\left(x_1-x_2\right)\left(x_3-x_4\right)=0\end{array}\\ \begin{array}{c}{\left(x_1+x_2\right)}^2-{\left(x_3+x_4\right)}^2=2a_1\\ 2\left(x_1+x_2\right)\left(x_3+x_4\right)=2a_3\end{array}\end{array}.$$

(53)

System (53) has the following two solutions: ${\displaystyle \frac{1}{2}}\left(b_{+},b_{+},{\displaystyle \frac{a_3}{b_{+}}},{\displaystyle \frac{a_3}{b_{+}}}\right)$ and $-{\displaystyle \frac{1}{2}}\left(b_{+},b_{+},{\displaystyle \frac{a_3}{b_{+}}},{\displaystyle \frac{a_3}{b_{+}}}\right)$, where $b_{+}=\sqrt{a_1+\sqrt{a_1^2+a_3^2}}$. Similarly, if $a=\left(a_1,-a_1,a_3,-a_3\right)$ is nonzero degenerate number of the second type, then the quadratic equation $x^2=a$ has the following two solutions: ${\displaystyle \frac{1}{2}}\left(b_{+},-b_{+},{\displaystyle \frac{a_3}{b_{+}}},-{\displaystyle \frac{a_3}{b_{+}}}\right)$ and ${\displaystyle \frac{1}{2}}\left(-b_{+},b_{+},-{\displaystyle \frac{a_3}{b_{+}}},{\displaystyle \frac{a_3}{b_{+}}}\right)$.

That is, if a is degenerate number, then the considered quadratic equation has two solutions that are also degenerate numbers of the same type.

Finally, if a = (0,0,0,0), then the considered equation has the only zero solution. Thus, we found all solutions of the equation $x^2=a$ in isotropic space $M_2$. In other spaces, everything is similar.

Moreover, in the four-dimensional space, any quadratic equation $a{x}^2+bx+c=0$ has four roots, which are defined by the formula

$$x={\displaystyle \frac{-b+\sqrt{b^2-4ac}}{2a}}$$

(54)

if a is nondegenerate number than can easily be made sure of, having substituted this expression in the equation. Thus, we can draw the following conclusion: if a and $b^2-4ac$ are nondegenerate numbers, then the quadratic equation $a{x}^2+bx+c=0$ has four roots, if a is nondegenerate number, and $b^2-4ac$ is the degenerate number, the quadratic equation has two roots, and if $b^2-4ac=0$, the quadratic equation has one root. Solutions of an equation are expressed by Formula (54).

Let us assume now that a is degenerate number. Next, we will consider various options of numbers b and c. If b is degenerate number of this kind, as in a, and c is degenerate number of other type, then the equation has no solution. Indeed, according to Theorem 2, $a{x}^2+bx$ is degenerate number of the same type, as in a, and c is degenerate number of other type. Let us give various options of coefficients of the equation for which the quadratic equation has no solution in the

Table 8. Options of coefficients of a, b, c.

No.	a Coefficient	b Coefficient	c Coefficient	Solution
1	Degenerate number of the I type	Degenerate number of the I type	Nondegenerate number	No solution
2	Degenerate number of the I type	Degenerate number of the I type	Degenerate number of the II type	No solution
3	Degenerate number of the I type	Degenerate number of the II type	Degenerate number of the I type	No solution
4	Degenerate number of the I type	Degenerate number of the II type	Degenerate number of the II type	No solution
5	Degenerate number of the II type	Degenerate number of the I type	Degenerate number of the I type	No solution
6	Degenerate number of the II type	Degenerate number of the I type	Degenerate number of the II type	No solution
7	Degenerate number of the II type	Degenerate number of the II type	Nondegenerate number	No solution
8	Degenerate number of the II type	Degenerate number of the II type	Degenerate number of the I type	No solution

.

In other cases when the coefficient of a is degenerate number, the quadratic equation has two solutions. Let us consider, for example, the case in which $a=\left(a_1,a_1,a_3,a_3\right)$ is degenerate number of the first type, $b=\left(b_1,b_2,b_3,b_4\right)$ is nondegenerate number, and $c=\left(c_1,-c_1,c_3,-c_3\right)$ is degenerate number of the second type, in isotropic space $M_2$. Having passed to the spectrum, we have

$${\mu }_i\left(a\right){\mu }_i^2\left(x\right)+{\mu }_i\left(b\right){\mu }_i\left(x\right)+{\mu }_i\left(c\right)=0, i=1,2,3,4.$$

This system will be transformed to the following form:

$$\begin{array}{c}\left(b_1-b_2\right)\left(x_1-x_2\right)-\left(b_3-b_4\right)\left(x_3-x_4\right)+2c_1=0\\ \left(b_3-b_4\right)\left(x_1-x_2\right)+\left(b_1-b_2\right)\left(x_3-x_4\right)+2c_3=0\\ 2a_1\left[{\left(x_1+x_2\right)}^2-{\left(x_3+x_4\right)}^2\right]-4a_3\left(x_1+x_2\right)\left(x_3+x_4\right)+\\ +\left(b_1+b_2\right)\left(x_1+x_2\right)-\left(b_3+b_4\right)\left(x_3+x_4\right)=0\\ 2a_3\left[{\left(x_1+x_2\right)}^2-{\left(x_3+x_4\right)}^2\right]+4a_1\left(x_1+x_2\right)\left(x_3+x_4\right)+\\ +\left(b_3+b_4\right)\left(x_1+x_2\right)+\left(b_1+b_2\right)\left(x_3+x_4\right)=0\end{array}.$$

(55)

Let us introduce the following denotations:

$$\begin{array}{c}\begin{array}{c}{y}_1=x_1-x_2\\ y_2=x_3-x_4\end{array}\\ \begin{array}{c}{y}_3=x_1+x_2\\ y_4=x_3+x_4\end{array}\end{array}.$$

Next, the system (55) will be rewritten as

$$\begin{array}{c}\begin{array}{c}\left(b_1-b_2\right)y_1-\left(b_3-b_4\right)y_2+2c_1=0\\ \left(b_3-b_4\right)y_1+\left(b_1-b_2\right)y_2+2c_3=0\end{array}\\ \begin{array}{c}\begin{array}{c}2a_1\left[y_3^2-y_4^2\right]-4a_3{y}_3{y}_4+\left(b_1+b_2\right)y_3-\left(b_3+b_4\right)y_4=0\\ 2a_3\left[y_3^2-y_4^2\right]+4a_1{y}_3{y}_4+\left(b_3+b_4\right)y_3+\left(b_1+b_2\right)y_4=0\end{array}\end{array}\end{array}.$$

(56)

From the first two equations, we find

$$y_1=-{\displaystyle \frac{2c_1\left(b_1-b_2\right)+2c_3\left(b_3-b_4\right)}{{\left(b_1-b_2\right)}^2+{\left(b_3-b_4\right)}^2}},$$

$$y_2={\displaystyle \frac{2c_1\left(b_3-b_4\right)-2c_3\left(b_1-b_2\right)}{{\left(b_1-b_2\right)}^2+{\left(b_3-b_4\right)}^2}}.$$

Multiply the third equation by $a_3$ and the fourth equation by $a_1$ and subtract the third equation from the fourth:

$$4\left(a_1^2+a_3^2\right)y_3{y}_4+\left[a_1\left(b_3+b_4\right)-a_3\left(b_1+b_2\right)\right]y_3+\left[a_1\left(b_1+b_2\right)+a_3\left(b_3+b_4\right)\right]y_4=0.$$

Further, we multiply the third equation by $a_1$ and the fourth equation by $a_3$ and add the following equations:

$$2\left(a_1^2+a_3^2\right)\left(y_3^2-y_4^2\right)+\left[a_1\left(b_1+b_2\right)+a_3\left(b_3+b_4\right)\right]y_3-\left[a_1\left(b_3+b_4\right)-a_3\left(b_1+b_2\right)\right]y_4=0.$$

Now, we divide both parts of the last equations by $2\left(a_1^2+a_3^2\right)$:

$$2y_3{y}_4+{\displaystyle \frac{\left[a_1\left(b_3+b_4\right)-a_3\left(b_1+b_2\right)\right]}{2\left(a_1^2+a_3^2\right)}}{y}_3+{\displaystyle \frac{\left[a_1\left(b_1+b_2\right)+a_3\left(b_3+b_4\right)\right]}{2\left(a_1^2+a_3^2\right)}}{y}_4=0.$$

$$y_3^2-y_4^2+{\displaystyle \frac{\left[a_1\left(b_1+b_2\right)+a_3\left(b_3+b_4\right)\right]}{2\left(a_1^2+a_3^2\right)}}{y}_3-{\displaystyle \frac{\left[a_1\left(b_3+b_4\right)-a_3\left(b_1+b_2\right)\right]}{2\left(a_1^2+a_3^2\right)}}{y}_4=0.$$

Let us introduce new variables:

$$z_3=y_3+{\displaystyle \frac{\left[a_1\left(b_1+b_2\right)+a_3\left(b_3+b_4\right)\right]}{4\left(a_1^2+a_3^2\right)}},$$

$$z_4=y_4+{\displaystyle \frac{\left[a_1\left(b_3+b_4\right)-a_3\left(b_1+b_2\right)\right]}{4\left(a_1^2+a_3^2\right)}}.$$

Subsequently, the last equations in new variables will take the following form:

$$z_3{z}_4={\displaystyle \frac{\left(a_1^2-a_3^2\right)\left(b_1+b_2\right)\left(b_3+b_4\right)-a_1{a}_3\left[{\left(b_1+b_2\right)}^2-{\left(b_3+b_4\right)}^2\right]}{16{\left(a_1^2+a_3^2\right)}^2}},$$

$$z_3^2-z_4^2={\displaystyle \frac{\left(a_1^2-a_3^2\right)\left[{\left(b_1+b_2\right)}^2-{\left(b_3+b_4\right)}^2\right]+4a_1{a}_3\left(b_1+b_2\right)\left(b_3+b_4\right)}{16{\left(a_1^2+a_3^2\right)}^2}}.$$

This system has two solutions:

$$z_3=\pm {\displaystyle \frac{\left[a_1\left(b_1+b_2\right)+a_3\left(b_3+b_4\right)\right]}{4\left(a_1^2+a_3^2\right)}},$$

$$z_4=\pm {\displaystyle \frac{\left[a_1\left(b_3+b_4\right)-a_3\left(b_1+b_2\right)\right]}{4\left(a_1^2+a_3^2\right)}}.$$

From here,

$$\left(y_3,y_4\right)=\left(0,0\right), \left(y_3,y_4\right)=\left(-{\displaystyle \frac{\left[a_1\left(b_1+b_2\right)+a_3\left(b_3+b_4\right)\right]}{2\left(a_1^2+a_3^2\right)}},-{\displaystyle \frac{\left[a_1\left(b_3+b_4\right)-a_3\left(b_1+b_2\right)\right]}{2\left(a_1^2+a_3^2\right)}}\right).$$

Consequently,

$$\left(x_1,x_2,x_3,x_4\right)=(-d_1,d_1,d_3,-d_4), \left(x_1,x_2,x_3,x_4\right)=(-d_1-d_2,d_1-d_2,d_3-d_4,-d_3-d_4),$$

where $d_1={\displaystyle \frac{c_1\left(b_1-b_2\right)+c_3\left(b_3-b_4\right)}{{\left(b_1-b_2\right)}^2+{\left(b_3-b_4\right)}^2}}$, $d_2={\displaystyle \frac{\left[a_1\left(b_1+b_2\right)+a_3\left(b_3+b_4\right)\right]}{4\left(a_1^2+a_3^2\right)}}$, $d_3={\displaystyle \frac{c_1\left(b_3-b_4\right)-c_3\left(b_1-b_2\right)}{{\left(b_1-b_2\right)}^2+{\left(b_3-b_4\right)}^2}}$, $d_4={\displaystyle \frac{\left[a_1\left(b_3+b_4\right)-a_3\left(b_1+b_2\right)\right]}{4\left(a_1^2+a_3^2\right)}}$.

Thus, we found two solutions of the quadratic equation in a case in which $a$ is degenerate number of the first type, $b$ is nondegenerate number, and $c$ is degenerate number of the second type, in isotropic space $M_2$. Other cases are investigated in the same way.

Let us note that in isotropic space $M_2$, the four-dimensional quadratic equation in components of four-dimensional number $x=\left(x_1,x_2,x_3,x_4\right)$ is written as

$$\begin{array}{c}{a}_1\left(x_1^2+x_2^2-x_3^2-x_4^2\right)+2a_2\left(x_1{x}_2-x_3{x}_4\right)-2a_3\left(x_1{x}_3+x_2{x}_4\right)-2a_4\left(x_1{x}_4+x_2{x}_3\right)+\\ +b_1{x}_1+b_2{x}_2-b_3{x}_3-b_4{x}_4+c_1=0\\ a_2\left(x_1^2+x_2^2-x_3^2-x_4^2\right)+2a_1\left(x_1{x}_2-x_3{x}_4\right)-2a_4\left(x_1{x}_3+x_2{x}_4\right)-2a_3\left(x_1{x}_4+x_2{x}_3\right)+\\ +b_2{x}_1+b_1{x}_2-b_4{x}_3-b_3{x}_4+c_2=0\\ a_3\left(x_1^2+x_2^2-x_3^2-x_4^2\right)+2a_4\left(x_1{x}_2-x_3{x}_4\right)+2a_1\left(x_1{x}_3+x_2{x}_4\right)+2a_2\left(x_1{x}_4+x_2{x}_3\right)+\\ +b_3{x}_1+b_4{x}_2+b_1{x}_3+b_2{x}_4+c_3=0\\ a_4\left(x_1^2+x_2^2-x_3^2-x_4^2\right)+2a_3\left(x_1{x}_2-x_3{x}_4\right)+2a_2\left(x_1{x}_3+x_2{x}_4\right)+2a_1\left(x_1{x}_4+x_2{x}_3\right)+\\ +b_4{x}_1+b_3{x}_2+b_2{x}_3+b_1{x}_4+c_4=0\end{array},$$

(57)

and in one-dimensional mathematics, there are no methods for solution of systems of (57) type. We have stated the simple method for finding all roots of such systems here.

Similarly to (57), systems can also be written in other four-dimensional spaces (anisotropic and isotropic), and all roots of such systems can be found.

6. Norms of Four-Dimensional Numbers

The symplectic module of four-dimensional number is not norm, as the whole space of degenerate numbers have the zero module. The possibility of determination of norm of four-dimensional numbers turns spaces $M_i\left(\alpha ,\beta ,\gamma ,\delta \right),i=2,3,4,5,6,7$ of four-dimensional numbers into normed spaces, which opens huge opportunities for expansion of results of one-dimensional mathematics in the four-dimensional case. Therefore, the concept of norm plays an important role in definition of topology of four-dimensional spaces and building of the four-dimensional calculus.

In this section, we define the concept of norm, and we investigate its properties.

Definition 5.

The number

$$\parallel x\parallel ={\displaystyle \frac{1}{4}}\left(\left|{\mu }_1\left(\mathrm{x}\right)\right|+\left|{\mu }_2\left(\mathrm{x}\right)\right|+\left|{\mu }_3\left(\mathrm{x}\right)\right|+\left|{\mu }_4\left(\mathrm{x}\right)\right|\right),$$

(58)

where $\left({\mu }_1\left(\mathrm{x}\right),{\mu }_2\left(\mathrm{x}\right),{\mu }_3\left(\mathrm{x}\right),{\mu }_4\left(\mathrm{x}\right)\right)$ is the spectrum of number $x$, is called spectral norm of four-dimensional number $x=\left(x_1,x_2,x_3,x_4\right)$.

This definition is universal in the sense that it does not depend on index of the four-dimensional space.

Let us present this definition for various spaces of four-dimensional numbers.

In space $M_2\left(\alpha ,\beta ,\gamma ,\delta \right)$, the Expression (58) is

$$\parallel x\parallel ={\displaystyle \frac{1}{2}}\left(\sqrt{{\left(\alpha x_1-\sqrt{\gamma \delta }{x}_2\right)}^2+{\left(\sqrt{\beta \delta }{x}_3-\sqrt{\beta \gamma }{x}_4\right)}^2}+\sqrt{{\left(\alpha x_1+\sqrt{\gamma \delta }{x}_2\right)}^2+{\left(\sqrt{\beta \delta }{x}_3+\sqrt{\beta \gamma }{x}_4\right)}^2}\right).$$

(59)

Similarly, the norm in space $M_3\left(\alpha ,\beta ,\gamma ,\delta \right)$ has the form

$$\parallel x\parallel ={\displaystyle \frac{1}{2}}\left(\sqrt{{\left(\alpha x_1-\sqrt{\beta \delta }{x}_3\right)}^2+{\left(\sqrt{\gamma \delta }{x}_2-\sqrt{\beta \gamma }{x}_4\right)}^2}+\sqrt{{\left(\alpha x_1+\sqrt{\beta \delta }{x}_3\right)}^2+{\left(\sqrt{\gamma \delta }{x}_2+\sqrt{\beta \gamma }{x}_4\right)}^2}\right).$$

(60)

In space $M_4\left(\alpha ,\beta ,\gamma ,\delta \right)$ the norm (58) is

$$\parallel x\parallel ={\displaystyle \frac{1}{2}}\left(\sqrt{{\left(\alpha x_1-\sqrt{\beta \gamma }{x}_4\right)}^2+{\left(\sqrt{\gamma \delta }{x}_2-\sqrt{\beta \delta }{x}_3\right)}^2}+\sqrt{{\left(\alpha x_1+\sqrt{\beta \gamma }{x}_4\right)}^2+{\left(\sqrt{\gamma \delta }{x}_2+\sqrt{\beta \delta }{x}_3\right)}^2}\right).$$

(61)

Further, in spaces $M_5\left(\alpha ,\beta ,\gamma ,\delta \right)$, $M_6\left(\alpha ,\beta ,\gamma ,\delta \right)$ and $M_7\left(\alpha ,\beta ,\gamma ,\delta \right)$, norms have the forms (62), (63) and (64), respectively:

$$\parallel x\parallel ={\displaystyle \frac{1}{2}}\left(\sqrt{{\left(\alpha x_1-\sqrt{\beta \gamma }{x}_4\right)}^2+{\left(\sqrt{\gamma \delta }{x}_2+\sqrt{\beta \delta }{x}_3\right)}^2}+\sqrt{{\left(\alpha x_1+\sqrt{\beta \gamma }{x}_4\right)}^2+{\left(\sqrt{\gamma \delta }{x}_2-\sqrt{\beta \delta }{x}_3\right)}^2}\right).$$

(62)

$$\parallel x\parallel ={\displaystyle \frac{1}{2}}\left(\sqrt{{\left(\alpha x_1-\sqrt{\beta \delta }{x}_3\right)}^2+{\left(\sqrt{\gamma \delta }{x}_2+\sqrt{\beta \gamma }{x}_4\right)}^2}+\sqrt{{\left(\alpha x_1+\sqrt{\beta \delta }{x}_3\right)}^2+{\left(\sqrt{\gamma \delta }{x}_2-\sqrt{\beta \gamma }{x}_4\right)}^2}\right).$$

(63)

$$\parallel x\parallel ={\displaystyle \frac{1}{2}}\left(\sqrt{{\left(\alpha x_1-\sqrt{\gamma \delta }{x}_2\right)}^2+{\left(\sqrt{\beta \delta }{x}_3+\sqrt{\beta \gamma }{x}_4\right)}^2}+\sqrt{{\left(\alpha x_1+\sqrt{\gamma \delta }{x}_2\right)}^2+{\left(\sqrt{\beta \delta }{x}_3-\sqrt{\beta \gamma }{x}_4\right)}^2}\right).$$

(64)

Theorem 8.

The spectral norm (58) is norm in spaces $M_i\left(\alpha ,\beta ,\gamma ,\delta \right),i=2,3,\dots ,7$.

Proof. 

For the proof, it is necessary to prove that Expression (58) satisfies the following conditions:

(1)

$\parallel x\parallel \ge 0$, $\parallel x\parallel =0$ if and only if $x=\left(0,0,0,0\right)$.

(2)

$\parallel a·x\parallel =\left|a\right|\parallel x\parallel$ for any real number $a\in R$.

(3)

$\parallel x+y\parallel \le \parallel x\parallel +\parallel y\parallel$ for any four-dimensional numbers $x$ and $y$ (triangle inequality).

□

The first two conditions are obvious. Let us prove triangle inequality, following [8]. For this purpose, we will consider two separate inequalities:

$${\left(L_1{M}_2-L_2{M}_1\right)}^2\ge 0,$$

$${\left(N_1{K}_2-N_2{K}_1\right)}^2\ge 0,$$

where $L_1$, $L_2$, $M_1$, $M_2$, $N_1$, $N_2$, $K_1$, $K_2$ are arbitrary real numbers.

Expanding the brackets, and after algebraic transformations, we receive

$$L_1{L}_2{M}_1{M}_2\le {L_1}^2{M_2}^2+{L_2}^2{M_1}^2$$

$$2N_1{N}_2{K}_1{K}_2\le {N_1}^2{K_2}^2+{N_2}^2{K_1}^2$$

Adding to both parts of the first inequality ${\left(L_1{L}_2\right)}^2+{\left(M_1{M}_2\right)}^2$ and the second inequality ${\left(N_1{N}_2\right)}^2+{\left(K_1{K}_2\right)}^2$, and taking the root, we write

$$L_1{L}_2+M_1{M}_2\le \sqrt{\left({L_1}^2+{M_1}^2\right)\left({L_2}^2+{M_2}^2\right)}$$

$$N_1{N}_2+K_1{K}_2\le \sqrt{\left({N_1}^2+{K_1}^2\right)\left({N_2}^2+{K_2}^2\right)}$$

Let us double and add to both parts of the first and second inequalities ${L_1}^2+{L_2}^2+{M_1}^2+{M_2}^2$ and ${N_1}^2+{N_2}^2+{K_1}^2+{K_2}^2$, respectively, and take the root:

$$\sqrt{{\left(L_1+L_2\right)}^2+{\left(M_1+M_2\right)}^2}\le \sqrt{{L_1}^2+{M_1}^2}+\sqrt{{L_2}^2+{M_2}^2}$$

$$\sqrt{{\left(N_1+N_2\right)}^2+{\left(K_1+K_2\right)}^2}\le \sqrt{{N_1}^2+{K_1}^2}+\sqrt{{N_2}^2+{K_2}^2}$$

Therefore,

$$\begin{gathered} {\displaystyle \frac{1}{2}}\sqrt{{\left(L_1+L_2\right)}^2+{\left(M_1+M_2\right)}^2}+{\displaystyle \frac{1}{2}}\sqrt{{\left(N_1+N_2\right)}^2+{\left(K_1+K_2\right)}^2}\le \\ {\displaystyle \frac{1}{2}}\sqrt{{L_1}^2+{M_1}^2}+{\displaystyle \frac{1}{2}}\sqrt{{N_1}^2+{K_1}^2}+{\displaystyle \frac{1}{2}}\sqrt{{L_2}^2+{M_2}^2}+{\displaystyle \frac{1}{2}}\sqrt{{N_2}^2+{K_2}^2}. \end{gathered}$$

(65)

Now, depending on the space in which we work, we will enter the corresponding changes of variables. For example, for space $M_2\left(\alpha ,\beta ,\gamma ,\delta \right)$, we will enter the following changes of variables:

$$L_1=\alpha x_1-\sqrt{\gamma \delta }{x}_2,L_2=\alpha y_1-\sqrt{\gamma \delta }{y}_2,$$

$$M_1=\sqrt{\beta \delta }{x}_3-\sqrt{\beta \gamma }{x}_4,M_2=\sqrt{\beta \delta }{y}_3-\sqrt{\beta \gamma }{y}_4,$$

$$N_1=\alpha x_1+\sqrt{\gamma \delta }{x}_2,N_2=\alpha y_1+\sqrt{\gamma \delta }{y}_2,$$

$$K_1=\sqrt{\beta \delta }{x}_3-\sqrt{\beta \gamma }{x}_4,K_2=\sqrt{\beta \delta }{y}_3-\sqrt{\beta \gamma }{y}_4.$$

Next, inequality (65) turns into inequality $\parallel x+y\parallel \le \parallel x\parallel +\parallel y\parallel$.

Definition 6.

The number

$${\parallel x\parallel }_E=\sqrt{{\alpha }^2{x}_1^2+\gamma \delta x_2^2+\beta \delta x_3^2+\beta \gamma x_4^2}$$

(66)

is called Euclidean norm of four-dimensional number $x=\left(x_1,x_2,x_3,x_4\right)$ in space $M_i\left(\alpha ,\beta ,\gamma ,\delta \right),i=2,3,\dots ,7.$

Let us note that there are the following relations between spectral and Euclidean norms:

$$\begin{array}{cc}\hfill M_2\left(\alpha ,\beta ,\gamma ,\delta \right):\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}& \hfill \\ \hfill & \parallel x\parallel ={\displaystyle \frac{1}{2}}\left(\sqrt{{\parallel x\parallel }_E^2-2\sqrt{\gamma \delta }\left(\alpha x_1{x}_2+\beta x_3{x}_4\right)}+\sqrt{{\parallel x\parallel }_E^2+2\sqrt{\gamma \delta }\left(\alpha x_1{x}_2+\beta x_3{x}_4\right)}\right).\hfill \end{array}$$

(67)

$$\begin{array}{cc}\hfill M_3\left(\alpha ,\beta ,\gamma ,\delta \right):\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}& \hfill \\ \hfill & \parallel x\parallel ={\displaystyle \frac{1}{2}}\left(\sqrt{{\parallel x\parallel }_E^2-2\sqrt{\beta \delta }\left(\alpha x_1{x}_3+\gamma x_2{x}_4\right)}+\sqrt{{\parallel x\parallel }_E^2+2\sqrt{\beta \delta }\left(\alpha x_1{x}_3+\gamma x_2{x}_4\right)}\right).\hfill \end{array}$$

(68)

$$\begin{array}{cc}\hfill M_4\left(\alpha ,\beta ,\gamma ,\delta \right):\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}& \hfill \\ \hfill & \parallel x\parallel ={\displaystyle \frac{1}{2}}\left(\sqrt{{\parallel x\parallel }_E^2-2\sqrt{\beta \gamma }\left(\alpha x_1{x}_4+\delta x_2{x}_3\right)}+\sqrt{{\parallel x\parallel }_E^2+2\sqrt{\beta \gamma }\left(\alpha x_1{x}_4+\delta x_2{x}_3\right)}\right).\hfill \end{array}$$

(69)

$$\begin{array}{cc}\hfill M_5\left(\alpha ,\beta ,\gamma ,\delta \right):\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}& \hfill \\ \hfill & \parallel x\parallel ={\displaystyle \frac{1}{2}}\left(\sqrt{{\parallel x\parallel }_E^2-2\sqrt{\beta \gamma }\left(\alpha x_1{x}_4-\delta x_2{x}_3\right)}+\sqrt{{\parallel x\parallel }_E^2+2\sqrt{\beta \gamma }\left(\alpha x_1{x}_4-\delta x_2{x}_3\right)}\right).\hfill \end{array}$$

(70)

$$\begin{array}{cc}\hfill M_6\left(\alpha ,\beta ,\gamma ,\delta \right):\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}& \hfill \\ \hfill & \parallel x\parallel ={\displaystyle \frac{1}{2}}\left(\sqrt{{\parallel x\parallel }_E^2-2\sqrt{\beta \delta }\left(\alpha x_1{x}_3-\gamma x_2{x}_4\right)}+\sqrt{{\parallel x\parallel }_E^2+2\sqrt{\beta \delta }\left(\alpha x_1{x}_3-\gamma x_2{x}_4\right)}\right).\hfill \end{array}$$

(71)

$$\begin{array}{cc}\hfill M_7\left(\alpha ,\beta ,\gamma ,\delta \right):\hspace{1em}\hspace{1em}\hspace{1em}\hspace{1em}& \hfill \\ \hfill & \parallel x\parallel ={\displaystyle \frac{1}{2}}\left(\sqrt{{\parallel x\parallel }_E^2-2\sqrt{\gamma \delta }\left(\_x_1{x}_2-\beta x_3{x}_4\right)}+\sqrt{{\parallel x\parallel }_E^2+2\sqrt{\gamma \delta }\left(\alpha x_1{x}_2-\beta x_3{x}_4\right)}\right).\hfill \end{array}$$

(72)

Theorem 9.

Spectral and Euclidean norms are equivalent in spaces $M_i\left(\alpha ,\beta ,\gamma ,\delta \right),i=2,3,\dots ,7$, namely

$${\displaystyle \frac{1}{\sqrt{2}}}{\parallel x\parallel }_E\le \parallel x\parallel \le {\parallel x\parallel }_E.$$

(73)

Proof. 

Let us carry this out following [9,10]. Consider inequality

$$0\le \sqrt{L^2-M^2}\le L$$

where $0\le \left|\mathrm{M}\right|\le L$. Let us add to both parts of inequality L and multiply all parts of inequality by 2:

$$2L\le L+M+2\sqrt{L^2-M^2}+L-M\le 4L.$$

From here, after simple transformations, we obtain

$${\displaystyle \frac{1}{\sqrt{2}}}\sqrt{L}\le {\displaystyle \frac{1}{2}}\left(\sqrt{L+M}+\sqrt{L-M}\right)\le \sqrt{L}$$

Out of these inequalities and out of Equalities (67)–(72), (73) is easily brought for all spaces $M_i\left(\alpha ,\beta ,\gamma ,\delta \right),i=2,3,\dots ,7$. □

Let us denote as $N_0$ the set of all points in the four-dimensional space for which $\parallel x\parallel ={\parallel x\parallel }_E$.

It follows from Relations (67)–(72) that in spaces $M_i\left(\alpha ,\beta ,\gamma ,\delta \right),i=2,3,\dots ,7$, the coordinates of points from the set $N_0$ satisfy to the following relations:

In $M_2\left(\alpha ,\beta ,\gamma ,\delta \right)$:$\alpha x_1{x}_2+\beta x_3{x}_4=0$.

In $M_3\left(\alpha ,\beta ,\gamma ,\delta \right)$: $\alpha x_1{x}_3+\gamma x_2{x}_4=0$.

In $M_4\left(\alpha ,\beta ,\gamma ,\delta \right)$:$\alpha x_1{x}_4+\delta x_2{x}_3=0$.

In $M_5\left(\alpha ,\beta ,\gamma ,\delta \right)$:$\alpha x_1{x}_4-\delta x_2{x}_3=0$.

In $M_6\left(\alpha ,\beta ,\gamma ,\delta \right)$: $\alpha x_1{x}_3-\gamma x_2{x}_4=0$.

In $M_7\left(\alpha ,\beta ,\gamma ,\delta \right)$: $\alpha x_1{x}_2-\beta x_3{x}_4=0$.

Theorem 10.

In spaces $M_i\left(\alpha ,\beta ,\gamma ,\delta \right),i=2,3,\dots ,7$, the set $N_0$ is closed with respect to multiplication and the following equalities hold:

(1)

For any $x,y\in N_0,\parallel xy\parallel =\parallel x\parallel \parallel y\parallel$, $\parallel xy{\parallel }_E={\parallel x\parallel }_E{\parallel y\parallel }_E$.

(2)

For any $x,y\in O_I, \parallel x\parallel ={\displaystyle \frac{1}{\sqrt{2}}}{\parallel x\parallel }_E$, $\parallel xy\parallel =2\parallel x\parallel \parallel y\parallel$, $\parallel xy{\parallel }_E=\sqrt{2}{\parallel x\parallel }_E{\parallel y\parallel }_E$.

(3)

For any $x,y\in O_{II}, \parallel x\parallel ={\displaystyle \frac{1}{\sqrt{2}}}{\parallel x\parallel }_E$, $\parallel xy\parallel =2\parallel x\parallel \parallel y\parallel$, $\parallel xy{\parallel }_E=\sqrt{2}{\parallel x\parallel }_E{\parallel y\parallel }_E$.

Proof. 

We will carry out the proof for space $M_2\left(\alpha ,\beta ,\gamma ,\delta \right)$. For other spaces the proof is carried out in the same way. Let $x,y\in N_0$ and $z=xy$. Next, αz₁z₂ + βz₃z₄ = (αx₁x₂ + βx₃x₄)(${\alpha }^2{y}_1^2+\gamma \delta y_2^2+\beta \delta y_2^3+\beta \gamma y_2^4$) + (αy₁y₂ + βy₃y₄)(${\alpha }^2{x}_1^2+\gamma \delta x_2^2+\beta \delta x_2^3+\beta \gamma x_2^4$) = 0, that is, $z\in N_0$. This means that in space $M_2\left(\alpha ,\beta ,\gamma ,\delta \right)$, the set $N_0$ is closed with respect to multiplication. □

Let us prove equality $\parallel xy\parallel =\parallel x\parallel \parallel y\parallel$. Represent $\parallel xy\parallel$ in the form

$$\parallel xy\parallel ={\displaystyle \frac{1}{2}}\left(\sqrt{A^2+B^2}+\sqrt{C^2+D^2}\right),$$

where

$$A=\left(\alpha x_1-\sqrt{\gamma \delta }{x}_2\right)\left(\alpha y_1-\sqrt{\gamma \delta }{y}_2\right)-\left(\sqrt{\beta \delta }{x}_3-\sqrt{\beta \gamma }{x}_4\right)\left(\sqrt{\beta \delta }{y}_3-\sqrt{\beta \gamma }{y}_4\right),$$

$$B=\left(\sqrt{\beta \delta }{x}_3-\sqrt{\beta \gamma }{x}_4\right)\left(\alpha y_1-\sqrt{\gamma \delta }{y}_2\right)+\left(\alpha x_1-\sqrt{\gamma \delta }{x}_2\right)\left(\sqrt{\beta \delta }{y}_3-\sqrt{\beta \gamma }{y}_4\right),$$

$$C=\left(\alpha x_1+\sqrt{\gamma \delta }{x}_2\right)\left(\alpha y_1+\sqrt{\gamma \delta }{y}_2\right)-\left(\sqrt{\beta \delta }{x}_3+\sqrt{\beta \gamma }{x}_4\right)\left(\sqrt{\beta \delta }{y}_3+\sqrt{\beta \gamma }{y}_4\right),$$

$$D=\left(\sqrt{\beta \delta }{x}_3+\sqrt{\beta \gamma }{x}_4\right)\left(\alpha y_1+\sqrt{\gamma \delta }{y}_2\right)+\left(\alpha x_1+\sqrt{\gamma \delta }{x}_2\right)\left(\sqrt{\beta \delta }{y}_3+\sqrt{\beta \gamma }{y}_4\right).$$

Then, removing the brackets and considering that $x,y\in N_0$, we have $\sqrt{A^2+B^2}={\parallel x\parallel }_E{\parallel y\parallel }_E$ and $\sqrt{C^2+D^2}={\parallel x\parallel }_E{\parallel y\parallel }_E$. Next, it follows from definition of the set $N_0$ that $\parallel xy\parallel =\parallel x\parallel \parallel y\parallel$.

Equality $\parallel xy{\parallel }_E={\parallel x\parallel }_E{\parallel y\parallel }_E$ follows from the proven equality.

Now, prove equality $\parallel x\parallel ={\displaystyle \frac{1}{\sqrt{2}}}{\parallel x\parallel }_E$ for degenerate numbers of the first type. Let $x=\left(c_1,{\displaystyle \frac{\alpha }{\sqrt{\gamma \delta }}}{c}_1,c_2,\sqrt{{\displaystyle \frac{\delta }{\gamma }}}{c}_2\right)$ and $y=\left(d_1,{\displaystyle \frac{\alpha }{\sqrt{\gamma \delta }}}{d}_1,d_2,\sqrt{{\displaystyle \frac{\delta }{\gamma }}}{d}_2\right)$ be degenerate numbers of the first type. Then $\parallel x\parallel =\sqrt{{\alpha }^2{c}_1^2+\beta \delta c_2^2}$, ${\parallel x\parallel }_E=\sqrt{2}\sqrt{{\alpha }^2{c}_1^2+\beta \delta c_2^2}$. Similarly, $\parallel y\parallel =\sqrt{{\alpha }^2{d}_1^2+\beta \delta d_2^2}$, ${\parallel y\parallel }_E=\sqrt{2}\sqrt{{\alpha }^2{d}_1^2+\beta \delta d_2^2}$. $xy=\left(2\alpha c_1{d}_1-2{\displaystyle \frac{\beta \delta }{\alpha }}{c}_2{d}_2,{\displaystyle \frac{\alpha }{\sqrt{\gamma \delta }}}\left(2\alpha c_1{d}_1-2{\displaystyle \frac{\beta \delta }{\alpha }}{c}_2{d}_2\right),2\alpha c_1{d}_2+2\alpha c_2{d}_1,\sqrt{{\displaystyle \frac{\delta }{\gamma }}}\left(2\alpha c_1{d}_2+2\alpha c_2{d}_1\right)\right)$. Consequently,

$$\parallel xy\parallel =2\sqrt{{\alpha }^4{c}_1^2{d}_1^2+{\alpha }^2\beta \delta c_1^2{d}_2^2+{\alpha }^2\beta \delta c_2^2{d}_1^2+{\beta }^2{\delta }^2{c}_2^2{d}_2^2}=2\parallel x\parallel \parallel y\parallel ,$$

$$\parallel xy{\parallel }_E=2\sqrt{2}\sqrt{{\alpha }^4{c}_1^2{d}_1^2+{\alpha }^2\beta \delta c_1^2{d}_2^2+{\alpha }^2\beta \delta c_2^2{d}_1^2+{\beta }^2{\delta }^2{c}_2^2{d}_2^2}=\sqrt{2}{\parallel x\parallel }_E{\parallel y\parallel }_E.$$

For degenerate numbers of the second type, the approval of the theorem is proven similarly.

Thus, in the subspace $N_0$, the spectral norm accepts the greatest values coinciding with Euclidean norm, and in the subspace of degenerate numbers, its value is less than values of Euclidean norm.

7. Discussion

In this work, various anisotropic, including isotropic, spaces of four-dimensional numbers with associative and commutative multiplication are constructed. At the same time, in these spaces, according to the Frobenius [5] theorem, there are so-called “zero divisors”. But the good news is that the general form of zero divisors is described explicitly, which makes it possible “to fight” against them effectively.

In this article, the algebra in four-dimensional numerical spaces is described, and an important feature of the constructed algebras is that they are natural generalization one-dimensional and two-dimensional (complex) algebras, which favorably distinguishes them from the theory of quaternions. The given common solutions of linear and quadratic equations allowed us to solve explicitly the system of four square equations with four unknowns. It was shown that such systems generally have four roots. Generally, modern mathematics has no effective methods for the solution of such systems.

Most importantly, spectral and Euclidean norms are introduced that turn four-dimensional spaces into normed spaces. Further, it is possible to define and investigate the convergence of the sequences of four-dimensional numbers and to consider series of four-dimensional numbers. It is possible to enter various topologies and to define the functions of four-dimensional numbers.