# Associative Composition Algebra/Binarions

The division binarions C is the field of complex numbers:
    $z=x+yi,\ \ i^{2}=-1,\ \ x,y\in R,\quad z^{*}=x-yi.$
Several academic journals and many university texts are dedicated to the function theory of C, for example the Wikibook Complex Analysis.

## Division binarions

With $z=x+iy,\ \ zz^{*}=x^{2}+y^{2}=N(z)$ is the square of the Euclidean distance from 0 to z. Furthermore, the vectors from 0 to w and z in C are perpendicular, z ⊥ w, when $zw^{*}+wz^{*}=0.$ These features make C an ideal vehicle for display of topics in Euclidean geometry.
Proposition: The diagonals of a w:rhombus are perpendicular.
    proof: The four binarions $\{0,z,w,z+w\},\ \ zz^{*}=ww^{*}$ form a rhombus. One diagonal is z + w and the other is parallel to z – w. They are perpendicular because
    $(z+w)(z-w)^{*}+(z-w)(z+w)^{*}=0.$.
Proposition: A Euclidean plane isometry is either a translation z → z + t or a rotation such as $z\rightarrow \omega z,\quad \omega \omega ^{*}=1.$
Note that a rotation about p is obtained by arithmetic with $z\mapsto \omega (z-p)+p=\omega z+p(1-\omega )$
where the last expression shows the mapping equivalent to rotation at 0 and a translation. Therefore, given direct isometry $z\mapsto \omega z+a,$ one can solve $p(1-\omega )=a$ to obtain $p=a/(1-\omega )$ as the center for an equivalent rotation, provided that $\omega \neq 1$, that is, provided the direct isometry is not a pure translation.

### Möbius transformations

Möbius transformations act on the projective line over division binarions. The points on this line use projective coordinates: (a,b)~(c,d) if there is a non-zero u such that ua=c and ub=d. This binary relation ~ on pairs of division binarions is an equivalence relation, where an equivalence class is written [a:b] for any pair (a,b) in the class. No point in the projective line corresponds to (0,0).
As a projective linear transformation, a Möbius transformation may be written
    $[z,w]{\begin{pmatrix}a&c\\b&d\end{pmatrix}}=[az+bw,\ cz+dw].$
The point [z,0] = [1,0] corresponds to the point at infinity with respect to the rest of C, which is represented by [z,w]=[zw$^{−1}$, 1].
Exercise: Show that the Möbius transformation takes infinity to a/c.
For other points, let w = 1 so
    $[z,1]{\begin{pmatrix}a&c\\b&d\end{pmatrix}}=[az+b,\ cz+d]=\left[{\frac {az+b}{cz+d}},1\right]$ when z ≠ −d/c.
To avoid singular transformations, ad −bc is taken to be non-zero. Particular transformations coming under the Möbius umbrella include:
* Case 1: b=c=0, d=1. a>1 magnification, 0<a<1 contraction, a=−1 reflection in 0, aa*=1 rotation
* Case 2: c=0, a=d=1, b=t translation of division binarion plane by t in C
* Case 3: a=d=0, b=c=1 multiplicative inversion of C extended to 0 and infinity.
Note that the rotation z to uz leaves [0,1] and [1,0] fixed. Furthermore, any two distinct points p and q can be placed to these polar opposites by
    $[z,1]{\begin{pmatrix}1&1\\-q&-p\end{pmatrix}}=[z-q,\ z-p].$
The image of a third point r cannot be [1,1] since p ≠ q, but r can be moved there: Let $w={\frac {r-p}{r-q}}.$ Then
    $[r,1]{\begin{pmatrix}w&1\\-wq&-p\end{pmatrix}}=[rw-wq,\ r-p]=[1,1].$
Thus the constructed transformation maps p,q,r to infinity, 0, 1 respectively. Applied to a fourth division binarion z, the image is the cross ratio [z,p,q,r].
Exercises:
* Construct the transformation taking −i to [1,0], 0 to [i,1] and i to [1,1]. What is the image of the unit disk zz* < 1 under this transformation?
* Two points in the projective line are fixed with a rotation. Show that a Möbius transformation that leaves three points fixed must be the identity mapping.
* If μ is a circle or line and g is a Möbius transformation, then μg is a circle or line.
* If the image of z under the cross ratio transformation of p, q, and r is a real number, then the four points z, p, q, and r lie on a circle or line.

## Hamilton

Nearly two centuries ago (in 1834 and 1837) William Rowan Hamilton wrote on algebraic couples as he explored the formation of division binarions from real numbers. See his essays Theory of Conjugate Functions or Algebraic Couples edited by David R. Wilkins. Hamilton wrote the product of a couple as
    $(a_{1},\ a_{2})(b_{1},\ b_{2})=(a_{1}b_{1}-a_{2}b_{2},\ a_{1}b_{2}+a_{2}b_{1}).$
This is equation 37, from Transactions of the Royal Irish Academy, volume 17, page 93 of Wilkins text.
This approach to establishing the field of division binarions ("complex numbers") was taken by w:Raymond Wilder in 1965 in Introduction to the Foundations of Mathematics, second edition, page 62. The same author, in 1981, identified the use of "ordered pairs (a,b) of real numbers, a and b along with rules for operating with such pairs" as a forced origin of a new concept, by Hamillton, in his Mathematics as a Cultural System (page 33).

## Euclid contradicted

The parallel postulate of Euclid has been phrased as a unique parallel existing off a given line and passing through a given point. The geometry of w:Lobachevski provides an infinite number of lines through a point, not a given line, and parallel to it. In the illustration, consider the given blue arc, and the pink arc illustrating an arc not intersecting the blue arc.
The division binarions enable a model of a hyperbolic plane in the unit disk. The geodesics in this model are circular arcs that meet the unit circle at right angles. The motions of the model are Mobius transformations that preserve the disk. These transformations are represented by homographies with $uu^{*}-vv^{*}=1.$
Indeed, for a point [z : 1] in the projective line over division binarions, the action of SU(1,1) is given by
    ${\bigl\;z:\;1\;{\bigr ]}\,{\begin{pmatrix}u&v\\v^{*}&u^{*}\end{pmatrix}}=[\;u\,z+v^{*}:\,v\,z+u^{*}\;]\,=\,\left[\;{\frac {uz+v^{*}}{vz+u^{*}}}:\,1\;\right]$
since in projective coordinates $(\;u\,z+v^{*}:\;v\,z+u^{*}\;)\thicksim \left(\;{\frac {\,u\,z+v^{*}\,}{v\,z+u^{*}}}:\;1\;\right)~.$
Writing $\;suv+{\overline {suv}}=2\,\Re {\mathord {\bigl}\,suv\,{\bigr )}\;,$ division-binarion arithmetic shows
    ${\bigl |}u\,z+v^{*}{\bigr |}^{2}=S+z\,z^{*}\quad {\text{ and }}\quad {\bigl |}v\,z+u^{*}{\bigr |}^{2}=S+1~,$
where $~S=v\,v^{*}\left(z\,z^{*}+1\right)+2\,\Re {\mathord {\bigl}\,uvz\,{\bigr )}~.$ Therefore, $~z\,z^{*}<1\implies {\bigl |}uz+v^{*}{\bigr |}<{\bigl |}\,v\,z+u^{*}\,{\bigr |}~$ so that their ratio lies in the open disk.
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