1550251515-Classical_Complex_Analysis__Gonzalez_

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72 Chapter^1


Obviously, the commutative law for multiplication does not hold, since
ij = -ji = k. To show that the quaternion system Q is a noncommutative
field, we first note that Q is clearly a ring with a unit, namely, (1, 0, 0, 0) =
1, and neutral or zero element (0, 0, 0, 0) = 0. Hence it will suffice to
show that every nonzero element x E Q has a multiplicative inverse; i.e.,


corresponding to every x i= 0 there is an element, denoted x-^1 , such that


xx-^1 = x-^1 x = 1. To see this, we begin by defining the modulus Jxl and


the conjugate x of the quaternion x = ( x 1 , x2, xa, x 4 ) by the formulas


Jxl = J xi + x~ + x~ + x~
X = X1 - X2i - Xaj - X4k
Then we have, by (1.19-7),
XX - = XX - = X1^2 + X2^2 + Xa^2 + X4^2 = I X 12

Hence if x '/= O, we have Jxl '/= 0, and by choosing x-^1 = x/lxl^2 we get


xx-^1 = x-^1 x = -xx = 1
lxl2
F. G. Frobenius [10] has shown that the only hypercomplex systems over
the reals which are fields are the system of the reals, that of the complex
numbers, and that of the quaternions. The first two are commutative fields;
the third is not. R. Arens [1] has proved that a normed algebra in which
IJa,Bll = llall · ll,811 is either the reals, the complexes, or the 'quaternions.
Makoto Itoh [15] has shown that the quaternion system may be considered
as a binary hypercomplex system over the complex field.
In 1845, A. Cayley [4] introduced a real linear algebra with eight units

e 0 = 1, ei, ... , e 7 , the elements of which are called octaves or octonions

and have the form

Z = xo -1-x1e1 -1-x2e2 + · · · + x7e7


where Xm E ~ (m = O, 1, ... , 7). The multiplication table for the units


ei, ... , e1 is


ei e2 ea e4 es e6 e1

ei -1 ea -e2 es -e4 -e7 e6


e2 -ea -1 ei e6 e1 -e4 -es


ea e2 -e1 -1 e1 -e6 es -e4


e4 -es -e6 -e7 -1 ei ez ea


es e4 -e7 e6 -e1 -1 -ea ez


e6 e1 e4 -es -e2 ea -1 -e1


e1 -e5 es e4 -ea -e2 ei -1

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