3
$\begingroup$

In this post user William Ryman asked what would happen if we try to build "complex numbers" with shapes other than circle or hyperbola in the role of a "unit circle".

Here I proposed three shapes that could work. The common principle behind them being that if the unit curve is defined as $r=r(\phi)$, an arbitrary point, corresponding to a 2-dimensional number on the plane $z=(a,b)$ is characterized by angle $\alpha(z)=\arctan(b/a)$, magnitude $M(z)=\frac{\sqrt{a^2+b^2}}{r(\alpha(z))}$ and argument $\operatorname{arg}(z)=\int_0^{\alpha(z)} r(\phi)^2 d\phi$, twice the area of a sector between the radius-vector and $x$ axis.

The addition of numbers is defined element-wise as $(a_1,b_1)+(a_2,b_2)=(a_1+a_2,b_1+b_2)$.

The multiplication is defined in such a way that the arguments are added and magnitudes are multiplied: $\operatorname{arg}(uv)=\operatorname{arg}(u)+\operatorname{arg}(v)$ and $M(uv)=M(u)M(v)$.

These definitions make addition and multiplication commutative and associative.

So, I decided to consider the number system based on the following equation for unit curve: $r=|\cos\phi|$ (which is two tangent circles). This function is reciprocal to the function defining dual numbers, so I called the system "anti-dual numbers".

enter image description here

The expressions for modulus and argument of a number $z=(a,b)$ thus would be:

$M(z)=\frac{a^2+b^2}{a}$

$\arg z=\frac{1}{2} \left(\frac{a b}{a^2+b^2}+\arctan \left(b/a\right)\right)$

These expressions are valid for the first quarter of the plane, in other quarters we should account that negative modulus corresponds to a shift of argument by $\pi/2$ (not by $\pi$ as in complex numbers!), that's why we have to add the functions arg and mod which are intended to represent the canonical form.

The expression for the angle of direction of radius-vector as a function of argument is from this post by Tyma Gaidash:

$\phi (z)=\arcsin\sqrt{I_{\frac{4 \arg z}{\pi }}^{-1}\left(\frac{1}{2},\frac{3}{2}\right)}$

This expression involves inverse beta regularized function.

The code below for Mathematica system provides functions for determining argument and modulus of a number $(a,b)$, determining Cartesian coordinates based on modulus and argument as well as a function that multiplies two numbers given in Cartesian coordinates.

arg[a_, b_] := 1/2 ((a b)/(a^2 + b^2) + ArcTan[b/a])
mod[a_, b_] := (a^2 + b^2)/a
argc[a_, b_] := arg[a, b] + Pi/2 HeavisideTheta[-mod[a, b]]
modc[a_, b_] := Abs[mod[a, b]] 
\[Phi][A_] := 
 ArcSin[Sqrt[InverseBetaRegularized[4 A/Pi, 1/2, 3/2]]] // FullSimplify
angle[A_] := 
 Piecewise[{{\[Phi][A], 0 <= A < Pi/4}, {\[Phi][A - Pi/4] + Pi/2, 
    Pi/4 < A <= Pi/2}, {-\[Phi][-A], -Pi/4 < A < 
     0}, {-\[Phi][-A + Pi/4] - Pi/2, -Pi/2 < A < -Pi/4}}]
X[m_, A_] := m Cos[angle[A]] Abs[Cos[angle[A]]]
Y[m_, A_] := m Sin[angle[A]] Abs[Cos[angle[A]]]
Multiply[{a1_, b1_}, {a2_, b2_}] := {X[mod[a1, b1] mod[a2, b2], 
   arg[a1, b1] + arg[a2, b2]], 
  Y[mod[a1, b1] mod[a2, b2], arg[a1, b1] + arg[a2, b2]]}

Example:

a := -1; b := -1
arg[a, b]
mod[a, b]

Output:

1/2 (1/2 + Pi/4) - Pi/2
2

Multiplication:

Multiply[{1, 1}, {1, 1}] // N

Output:

{-1.10363, 1.78788}

The following code makes plot of multiplication of two points:

p1 := {1, 1}
p2 := {-1, -1}
p := Multiply[p1, p2] // N
ContourPlot[(x^2 + y^2)^2 == x^2, {x, -4, 4}, {y, -4, 4}, 
 Axes -> True, AxesLabel -> Automatic, Frame -> None, 
 Epilog -> {Gray, Point[p1], Blue, Point[p2]}, Red, Point[p]]

enter image description here


That said, I wonder, what algebraic and analytic properties this system has? It seems to be a 2-dimensional hypercomplex commutative numbering system that is not isomorphic to complex, split-complex and dual numbers.

One interesting feature of this system is existence of divisors of infinity because $(0,1)(0,1)=\infty$ (multiplication by divisors of infinity cannot be handled by the provided code though). This makes the system not closed under multiplication unless an improper element $\infty$ is attached. On the other hand, the system has no zero divisors. The system is not distributive, though, despite being commutative and associative.

What else can be said about the system? Can analysis be built on these numbers?

$\endgroup$
3
  • 1
    $\begingroup$ Maybe you could start by listing precisely for which pairs of inputs multiplication is defined, and for which it is not. $\endgroup$ Jun 7, 2022 at 3:34
  • $\begingroup$ @TorstenSchoeneberg the code works for all pairs except those involving elements laying on the vertical ($y$) axis. In that case it gives an infinity because those elements have infinite modulus. But this does not mean the multiplication cannot be specifically defined in this case as well. It just should be handled separately. $\endgroup$
    – Anixx
    Jun 7, 2022 at 3:37
  • 5
    $\begingroup$ Sure, e.g. define $(a,b) \cdot (0,y) = \begin{cases} (0, a^5\cdot \ln(b^2+y^2+0.3) \text{ if } a \ge b \\(-12\pi y, -2.1\pi y) \text{ if } a < b \text{ and } a\neq 8 \\ (17, 17) \text{ else}\end{cases}$. My point: If you want the multiplication to be defined everywhere so that it makes the thing into a two-dimensional $\mathbb R$-algebra, it will be isomorphic to either $\mathbb C$ or $\mathbb R \times \mathbb R$ or $\mathbb R[x]/x^2$. Since it seems like you don't want that, you have to say what axioms of algebras you want to loosen, and how. Otherwise, my above proposal is as good as any. $\endgroup$ Jun 7, 2022 at 16:57

2 Answers 2

2
$\begingroup$

Your unit-norm curve passes through zero. This means the norm is nor well-defined at zero, which then implies the norm is not well defined on some path through zero (which turns out to be the vertical axis). Lack of closure under multiplication arises from this condition.

Closure under multiplication required choosing a unit-norm curve that avoids zero.

$\endgroup$
3
  • $\begingroup$ It passes through zero intentionally so to have divisors of infinity (elements with infinite modulus). Conversely, if unit-norm has asymptotes at infinity, this leads to zero divisors (elements with modulus zero). $\endgroup$
    – Anixx
    Jun 23, 2022 at 23:36
  • $\begingroup$ So, indeed, to be closed under multiplication this needs some compactification. $\endgroup$
    – Anixx
    Jun 23, 2022 at 23:38
  • $\begingroup$ The aim was twofold 1.Investigate number systems with unit-norm being an unusual curve (in this case, inverse of two vertical lines of dual numbers in polar coordinates). 2.Investigate the possibility of a number system with divisors of infinity. $\endgroup$
    – Anixx
    Jun 24, 2022 at 0:02
1
$\begingroup$

$\DeclareMathOperator{arccsc}{arccsc}$ The power equation for $(a_1+b_1)^n=a_n+b_n\omega$ is $$a_n=\frac{\sqrt{(a_1^2+b_1^2)(a_{n-1}^2+b_{n-1}^2)}}{r(\alpha(\theta_1))\cdot r(\alpha(\theta_{n-1}))},\quad b_n=\sin(\alpha(z_{n-1}))+\sin(\alpha(z_{1})).$$ This is an idea of the equation, and I may have made some errors along the way. I used $z^n=M(a_1,b_1)M(b_{n-1},a_{n-1})+\arg(b_{1},a_{1})+\arg(b_{n-1},a_{n-1}).$ Expanding the general $M(a,b)$s gives you the formula for $a_n$, and using the general arg() instead of the specified equation, then using the FTC gives you $$\int_0^{\alpha(z_{n-1})}{|\cos(x_n)|dx_n}=\int_0^{\pi/2}{\cos(x_n)dx_n}+\int_{\pi/2}^{\alpha(z_{n-1})}{\cos(x_n)dx_n}\\ =\sin(\alpha(z_{n-1}))+\sin(\alpha(z_{1})).$$ The exponential and log can be found by using their taylor series', then use the exponential and logarithm definitions of other functions to define them. $$\frac{1}{z}=e^{-\ln(x)},\;\; \sinh(z)=\frac{e^z+e^{-z}}{2},\;\; \arcsin(z)=−i\ln{(iz+\sqrt{1−z^2})},\\ \arccsc(z)=\arcsin{\left(\frac{1}{z}\right)} \ldots$$ For arcsin() and other ln() based functions, use $\ln{\big((a+bi)(c+d\omega)\big)}=\ln(a+bi)+\ln(c+d\omega).$ I realized just now that $b_n\le2$ when $n\gt1,n\in\mathbb{N}$ For this specific pair, simplifying the original equation gives you $$\frac{a_{n-1}^2+b_{n-1}^2}{a_{n-1}}*\frac{a_{1}^2+b_{1}^2}{a_{1}}+\omega(\frac{b_{n-1}}{a_{n-1}^2+b_{n-1}^2}+\frac{b_{1}}{a_{1}^2+b_{1}^2})$$

$\endgroup$
11
  • $\begingroup$ Actually, it should be just $\arctan$, not atan2, verified. $\endgroup$
    – Anixx
    Dec 13, 2022 at 15:59
  • $\begingroup$ What I am interested in, is whether we can define addition on this set so that the multiplication is distributive. element-wise addition does not satisfy this, unfortunately. $\endgroup$
    – Anixx
    Dec 13, 2022 at 16:02
  • $\begingroup$ @Anixx we could maybe inverse the PEMDAS (Parentheses, Exponents, Multiply, Divide, Add, Subtract) operations in the set to be PEASMD? I do not know much about rings, sets, and groups, but changing the order of solving the equation may work? Also, I meant to change arctan2 to $\alpha(z)$ in all of my equations, but forgot about it… oops! $\endgroup$
    – ItsErtHere
    Dec 13, 2022 at 16:51
  • $\begingroup$ I replaced the arctan2’s. Thank you Anixx. Also, this post and the one you linked have given me some inspiration to study more of these numbers! $\endgroup$
    – ItsErtHere
    Dec 13, 2022 at 17:10
  • 1
    $\begingroup$ I think, logarithm should be something like $\ln [a,b]=[\ln \frac{a^2+b^2}{a},\frac{1}{2} \left(\frac{a b}{a^2+b^2}+\arctan \left(b/a\right)\right)]$ or, maybe, $\ln[a,b]=[1,0]\cdot \ln \frac{a^2+b^2}{a}+[0,1]\cdot \frac{1}{2} \left(\frac{a b}{a^2+b^2}+\arctan \left(b/a\right)\right)$, I am not sure, because real part of logarithm should be logarithm of modulus and non-real part should be argument... $\endgroup$
    – Anixx
    Dec 14, 2022 at 17:39

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .