On Number System

除了複數﹐目前知道的有四元數﹐八元數﹐十六元數。R3是沒有的。clifford algebraist 將這些數叫做 hypercomplex number (若果有的話)
先說說 hypercomplex number 的性質
The general form of a hypercomplex number is simply:




(1)




a0 + a1 i1 + a2 i2 + ... + an in
where n is a fixed integer, and a0, a1, a2, ... an are arbitrary real numbers and i0, i1, i2, ... i3 are symbols such that

a0 + a1 i1 + a2 i2 + ... + an in = b0 + b1 i1 + b2 i2 + ... + bn in
if and only if

a0 = b0, a1 = b1, ..., an = bn

另外
Each multiplication of two bases ia and ib is necessarily a member of the set of hypercomplex numbers being defined. In other words, given two real integers (from 1 to n) a and b, and real numbers p0 through pn, we can define a multiplication table such that

ia ib = p0 + p1 i1 + p2 i2 + ... + pn in
Therefore, for an nth order hypercomplex number, n*n*(n+1) number of such constants must be defined to determine the form of the algebra.
其它必須條件
The product of a real component a, viewed as a hypercomplex number, with any other general hypercomplex number is established via:

(a + 0 i1 + ... + 0 in) (b0 + b1 i1 + ... bn in) = a b0 + a b1 i1 + ... + a bn in
and

(b0 + b1 i1 + ... bn in) (a + 0 i1 + ... + 0 in) = a b0 + a b1 i1 + ... + a bn in

If a and b are real components and u and v are hypercomplex numbers, then

(a u) (b v) = (ab) (uv)

Left and right distributive laws hold:

u (v + w) = uv + uw
(v + w) u = vu + wu

証明R3不存在

令a+bi+cj包含複數子集，所以ii=-1等複數性質應保留=>jj=-1那麼，ij與ji=？

例如ij=1=>iij=i=>i=-j 但這樣就不需要j啦﹐這是一個矛盾。