Complex number: Difference between revisions

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In this context the complex numbers have been called the '''binarions'''.<ref>Kevin McCrimmon (2004) ''A Taste of Jordan Algebras'', pp 64, Universitext, Springer {{isbn|0-387-95447-3}} {{mr|id=2014924}}</ref>
 
Just as by applying the construction to reals the property of [[ordered field|ordering]] is lost, properties familiar from real and complex numbers vanish with each extension. The [[quaternion]]s lose commutativity, that is, {{math|''x''·''y'' ≠ ''y''·''x''}} for some quaternions {{math|''x'', ''y'',}}, and the multiplication of [[octonion]]s, additionally to not being commutative, fails to be associative: {{math|(''x''·''y'')·''z'' ≠ ''x''·(''y''·''z'')}} for some octonions {{math|''x'', ''y'', ''z''.}}.
 
Reals, complex numbers, quaternions and octonions are all [[normed division algebra]]s over '''R'''. By [[Hurwitz's theorem (normed division algebras)|Hurwitz's theorem]] they are the only ones; the [[sedenion]]s, the next step in the Cayley–Dickson construction, fail to have this structure.
 
The Cayley–Dickson construction is closely related to the [[regular representation]] of '''C''', thought of as an '''R'''-[[Algebra (ring theory)|algebra]] (an '''R'''-vector space with a multiplication), with respect to the basis {{math|(1, ''i'')}}. This means the following: the '''R'''-linear map
:<math>\begin{align}
\mathbb{C} \rightarrow \mathbb{C}, \\
z \mapsto wz</math>
\end{align}</math>
for some fixed complex number {{mvar|w}} can be represented by a {{math|2 × 2}} matrix (once a basis has been chosen). With respect to the basis {{math|(1, ''i'')}}, this matrix is
:<math>\begin{pmatrix}
\begin{pmatrix}
\operatorname{Re}(w) & -\operatorname{Im}(w) \\
\operatorname{Im}(w) & \;\; \operatorname{Re}(w)
\end{pmatrix},</math>
</math>,
that is, the one mentioned in the section on matrix representation of complex numbers above. While this is a [[linear representation]] of '''C''' in the [[2 × 2 real matrices]], it is not the only one. Any matrix
:<math>J = \begin{pmatrix}p & q \\ r & -p \end{pmatrix}, \quad p^2 + qr + 1 = 0</math>