nyuuzyou/wikis huggingface dataset:
Complex Multiplication is Associative

== Theorem ==
The operation of multiplication on the set of complex numbers $\C$ is associative:
:$\forall z_1, z_2, z_3 \in \C: z_1 \paren {z_2 z_3} = \paren {z_1 z_2} z_3$
== Proof ==
From the definition of complex numbers, we define the following:
where $x_1, x_2, x_3, y_1, y_2, y_3 \in \R$.
Thus:
== Examples ==
== Sources ==
* : Chapter $\text I$: Functions of a Complex Variable: $\S 1$. Complex Numbers: $\text {(iii)}$ The fundamental operations $\text {(b)}$
* : Chapter $\text{II}$: Groups: Subgroups
* : $1$: Complex Numbers: Axiomatic Foundations of the Complex Number System: $5$
* : $1$: Complex Numbers: Supplementary Problems: Axiomatic Foundations of Complex Numbers: $77 \ \text{(b)}$
* : $1$ The complex plane: Complex numbers $\S 1.2$ The algebraic structure of the complex numbers
Category:Complex Multiplication is Associative
Category:Associative Law of Multiplication
Complex Multiplication is Associative/Examples

== Examples of Use of Complex Multiplication is Associative ==
=== Example: $\paren {2 - i} \paren {\paren {-3 + 2 i} \paren {5 - 4 i} } = \paren {\paren {2 - i} \paren {-3 + 2 i} } \paren {5 - 4 i}$ ===
Category:Examples of Use of Complex Multiplication is Associative
Complex Multiplication is Associative/Examples/(2 - i) ((-3 + 2i) (5 - 4i))

== Examples of Use of Complex Multiplication is Associative ==
==== Example: $\paren {2 - i} \paren {\paren {-3 + 2 i} \paren {5 - 4 i} }$ ====
==== Example: $\paren {\paren {2 - i} \paren {-3 + 2 i} } \paren {5 - 4 i}$ ====
As can be seen:
:$\paren {2 - i} \paren {\paren {-3 + 2 i} \paren {5 - 4 i} } = \paren {\paren {2 - i} \paren {-3 + 2 i} } \paren {5 - 4 i}$
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Fundamental Operations with Complex Numbers: $1$
Category:Examples of Use of Complex Multiplication is Associative
Complex Multiplication is Closed

== Theorem ==
The set of complex numbers $\C$ is closed under multiplication:
:$\forall z, w \in \C: z \times w \in \C$
== Proof from Informal Definition ==
== Proof from Formal Definition ==
== Sources ==
* : Chapter $1$: Integral Domains: $\S 2$. Operations: Example $1$
Category:Complex Multiplication
Category:Algebraic Closure
Category:Complex Multiplication is Closed
Complex Multiplication is Closed/Proof 1

== Theorem ==
== Proof ==
From the informal definition of complex numbers, we define the following:
:$z = x_1 + i y_1$
:$w = x_2 + i y_2$
where $i = \sqrt {-1}$ and $x_1, x_2, y_1, y_2 \in \R$.
Then from the definition of complex multiplication:
:$z w = \paren {x_1 x_2 - y_1 y_2} + i \paren {x_1 y_2 + x_2 y_1}$
From Real Numbers form Field:
:$x_1 x_2 - y_1 y_2 \in \R$
and:
:$x_1 y_2 + x_2 y_1 \in \R$
Hence the result.
Category:Complex Multiplication is Closed
Complex Multiplication is Closed/Proof 2

== Theorem ==
== Proof ==
From the formal definition of complex numbers, we define the following:
:$z = \tuple {x_1, y_1}$
:$w = \tuple {x_2, y_2}$
Then from the definition of complex multiplication:
:$z w = \tuple {x_1 x_2 - y_1 y_2, x_1 y_2 + x_2 y_1}$
From Real Numbers form Field:
:$x_1 x_2 - y_1 y_2 \in \R$
and:
:$x_1 y_2 + x_2 y_1 \in \R$
Hence the result.
Category:Complex Multiplication is Closed
Complex Multiplication is Commutative

== Theorem ==
The operation of multiplication on the set of complex numbers $\C$ is commutative:
:$\forall z_1, z_2 \in \C: z_1 z_2 = z_2 z_1$
== Proof ==
From the definition of complex numbers, we define the following:
where $x_1, x_2, y_1, y_2 \in \R$.
Then:
== Examples ==
== Also see ==
* Complex Multiplication is Associative
== Sources ==
* : Chapter $\text I$: Functions of a Complex Variable: $\S 1$. Complex Numbers: $\text {(iii)}$ The fundamental operations $\text {(b)}$
* : $1$: Complex Numbers: Axiomatic Foundations of the Complex Number System: $4$
* : $1$: Complex Numbers: Supplementary Problems: Axiomatic Foundations of Complex Numbers: $76 \ \text{(b)}$
* : $1$ The complex plane: Complex numbers $\S 1.2$ The algebraic structure of the complex numbers
Category:Complex Multiplication
Category:Examples of Commutative Operations
Category:Commutative Law of Multiplication
Complex Multiplication is Commutative/Examples/(2 - 3i) (4 + 2i)

== Examples of Use of Complex Multiplication is Commutative ==
==== Example: $\paren {2 - 3 i} \paren {4 + 2 i}$ ====
==== Example: $\paren {4 + 2 i} \paren {2 - 3 i}$ ====
As can be seen:
:$\paren {2 - 3 i} \paren {4 + 2 i} = \paren {4 + 2 i} \paren {2 - 3 i}$
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Fundamental Operations with Complex Numbers: $1$
Category:Examples of Use of Complex Multiplication is Commutative
Complex Natural Logarithm/Examples/-1

== Examples of Complex Natural Logarithm ==
:$\map \ln {-1} = \paren {2 k + 1} \pi i$
for all $k \in \Z$.
== Proof ==
== Sources ==
* : $\S 4$. Elementary Functions of a Complex Variable: Exercise $6 \ \text{(i)}$
Category:Examples of Complex Natural Logarithms
Complex Natural Logarithm/Examples/-2

== Examples of Complex Natural Logarithm ==
:$\ln \paren {-2} = \ln 2 + \paren {2 k + 1} \pi i$
for all $k \in \Z$.
== Proof ==
== Sources ==
* : $\S 4.6$. The Logarithm: Examples: $\text {(i)}$
Category:Examples of Complex Natural Logarithms
Complex Natural Logarithm/Examples/1 - i tan alpha

== Examples of Complex Natural Logarithm ==
:$\ln \paren {1 - i \tan \alpha} = \ln \sec \alpha + i \paren {-\alpha + 2 k \pi}$
for all $k \in \Z$.
== Proof ==
== Sources ==
* : $\S 4$. Elementary Functions of a Complex Variable: Exercise $6 \ \text{(ii)}$
Category:Examples of Complex Natural Logarithms
Complex Natural Logarithm/Examples/i

== Example of Complex Natural Logarithm ==
:$\ln \paren i = \paren {4 k + 1} \dfrac {\pi i} 2$
for all $k \in \Z$.
== Proof ==
== Sources ==
* : $\S 4.6$. The Logarithm: Examples: $\text {(ii)}$
Category:Examples of Complex Natural Logarithms
Complex Number equals Conjugate iff Wholly Real

== Theorem ==
Let $z \in \C$ be a complex number.
Let $\overline z$ be the complex conjugate of $z$.
Then $z = \overline z$ $z$ is wholly real.
== Proof ==
Let $z = x + i y$.
Then:
Hence by definition, $z$ is wholly real.
Now suppose $z$ is wholly real.
Then:
== Sources ==
* : $\S 1.2$. The Algebraic Theory
Category:Complex Conjugates
Complex Number equals Negative of Conjugate iff Wholly Imaginary

== Theorem ==
Let $z \in \C$ be a complex number.
Let $\overline z$ be the complex conjugate of $z$.
Then $\overline z = -z$ $z$ is wholly imaginary.
== Proof ==
Let $z = x + i y$.
Then:
Hence by definition, $z$ is wholly imaginary.
Now suppose $z$ is wholly imaginary.
Then:
== Sources ==
* : $\S 1.2$. The Algebraic Theory
Category:Complex Conjugates
Complex Number is Algebraic over Real Numbers

== Theorem ==
Let $z \in \C$ be a complex number.
Then $z$ is algebraic over $\R$.
== Proof ==
Let $z = a + i b$.
Let $\overline z = a - i b$ be the complex conjugate of $z$.
From Product of Complex Number with Conjugate:
:$z \overline z = a^2 + b^2$
From Sum of Complex Number with Conjugate:
:$z + \overline z = 2 a$
Thus from Viète's Formulas, both $z$ and $\overline z$ are roots of the polynomial:
:$X^2 - 2 a X + \paren {a^2 + b^2}$
Hence the result by definition of algebraic over $\R$.
== Sources ==
* : Chapter $8$: Field Extensions: $\S 38$. Simple Algebraic Extensions: Example $75$
Category:Algebraic Numbers
Complex Numbers are Parallel iff Cross Product is Zero

== Theorem ==
Let $z_1$ and $z_2$ be complex numbers in vector form such that $z_1 \ne 0$ and $z_2 \ne 0$.
Then $z_1$ and $z_2$ are parallel :
:$z_1 \times z_2 = 0$
where $z_1 \times z_2$ denotes the complex cross product of $z_1$ with $z_2$.
== Proof ==
By definition of complex cross product:
:$z_1 \times z_2 = \cmod {z_1} \, \cmod {z_2} \sin \theta$
:$\cmod {z_1}$ denotes the complex modulus of $z_1$
:$\theta$ denotes the angle from $z_1$ to $z_2$, measured in the positive direction.
=== Necessary Condition ===
Let $z_1$ and $z_2$ be parallel.
Then either $\theta = 0^\circ$ or $\theta = 180^\circ$.
Either way, from Sine of Zero is Zero or Sine of Straight Angle:
:$\sin \theta = 0$
and so:
:$\cmod {z_1} \, \cmod {z_2} \sin \theta = 0$
Hence by definition:
:$z_1 \times z_2 = 0$
=== Sufficient Condition ===
Let $z_1 \times z_2 = 0$.
Then by definition:
:$\cmod {z_1} \, \cmod {z_2} \sin \theta = 0$
As neither $z_1 = 0$ or $z_2 = 0$ it follows that $\sin \theta = 0$.
From Sine of Multiple of Pi it follows that either:
:$\theta = 0^\circ$
:$\theta = 180^\circ$
or:
:$\theta$ is either of the above plus a full circle.
That is, $z_1$ and $z_2$ are parallel.
== Sources ==
* : $1$: Complex Numbers: Dot and Cross Product: $2.$
Category:Geometry of Complex Plane
Category:Complex Cross Product
Complex Numbers are Perpendicular iff Dot Product is Zero

== Theorem ==
Let $z_1$ and $z_2$ be complex numbers in vector form such that $z_1 \ne 0$ and $z_2 \ne 0$.
Then $z_1$ and $z_2$ are perpendicular :
:$z_1 \circ z_2 = 0$
where $z_1 \circ z_2$ denotes the complex dot product of $z_1$ with $z_2$.
== Proof ==
By definition of complex dot product:
:$z_1 \circ z_2 = \cmod {z_1} \, \cmod {z_2} \cos \theta$
:$\cmod {z_1}$ denotes the complex modulus of $z_1$
:$\theta$ denotes the angle from $z_1$ to $z_2$, measured in the positive direction.
=== Necessary Condition ===
Let $z_1$ and $z_2$ be perpendicular.
Then either $\theta = 90^\circ$ or $\theta = 270^\circ$.
Either way, from Cosine of Right Angle or Cosine of Three Right Angles:
:$\cos \theta = 0$
and so:
:$\cmod {z_1} \, \cmod {z_2} \cos \theta = 0$
Hence by definition:
:$z_1 \circ z_2 = 0$
=== Sufficient Condition ===
Let $z_1 \circ z_2 = 0$.
Then by definition:
:$\cmod {z_1} \, \cmod {z_2} \cos \theta = 0$
As neither $z_1 = 0$ or $z_2 = 0$ it follows that $\cos \theta = 0$.
From Cosine of Half-Integer Multiple of Pi it follows that either:
:$\theta = 90^\circ$
:$\theta = 270^\circ$
or:
:$\theta$ is either of the above plus a full circle.
That is, $z_1$ and $z_2$ are perpendicular.
== Sources ==
* : $1$: Complex Numbers: Dot and Cross Product: $1.$
Category:Geometry of Complex Plane
Category:Complex Dot Product
Complex Numbers are Uncountable

== Theorem ==
The set of complex numbers $\C$ is uncountably infinite. 
== Proof ==
For all $r \in \R$, we have $r = r + 0 i \in C$.
Thus the set of real numbers $\R$ can be considered a subset of $\C$.
As the Real Numbers are Uncountable, it follows from Sufficient Conditions for Uncountability, proposition $(1)$, that $\C$ is uncountably infinite.
Category:Complex Numbers
Category:Set Theory
Complex Numbers as External Direct Product

== Theorem ==
Let $\struct {\C_{\ne 0}, \times}$ be the group of non-zero complex numbers under multiplication.
Let $\struct {\R_{> 0}, \times}$ be the group of positive real numbers under multiplication.
Let $\struct {K, \times}$ be the circle group.
Then:
:$\struct {\C_{\ne 0}, \times} \cong \struct {\R_{> 0}, \times} \times \struct {K, \times}$
== Proof ==
Let $\phi: \C_{\ne 0} \to \R_{> 0} \times K$ be the mapping:
:$\map \phi {r e^{i \theta} } = \paren {r, e^{i \theta} }$
$\forall \tuple {a, b} \in \R_{> 0} \times K:\exists z = a \times b \in \C$ such that:
:$\map \phi z = \tuple {a, b}$
by Complex Multiplication is Closed and $\R \subset \C$.
So $\phi$ is surjective.
To prove $\phi$ is injective, let $\map \phi {r_1 e^{i \theta_1} } = \map \phi {r_2 e^{i \theta_2} }$.
So $\phi$ is injective, thus bijective.
Also:
So $\phi$ is a group homomorphism.
Since it is bijective, it is a group isomorphism.
Category:Complex Numbers
Category:External Direct Products
Complex Numbers as Quotient Ring of Real Polynomial

== Theorem ==
Let $\C$ be the set of complex numbers.
Let $P \sqbrk x$ be the set of polynomials over real numbers, where the coefficients of the polynomials are real.
Let $\ideal {x^2 + 1} = \set {\map Q x \paren {x^2 + 1}: \map Q x \in P \sqbrk x}$ be the ideal generated by $x^2 + 1$ in $P \sqbrk x$.
Let $D = P \sqbrk x / \ideal {x^2 + 1}$ be the quotient of $P \sqbrk x$ modulo $\ideal {x^2 + 1}$.
Then:
:$\struct {\C, +, \times} \cong \struct {D, +, \times}$
== Proof ==
By Division Algorithm of Polynomial, any set in $D$ has an element in the form $a + b x$.
Define $\phi: D \to \C$ as a mapping:
:$\map \phi {\eqclass {a + b x} {x^2 + 1} } = a + b i$
We have that:
:$\forall z = a + b i \in \C : \exists \eqclass {a + b x} {x^2 + 1} \in D$
such that:
:$\map \phi {\eqclass {a + b x} {x^2 + 1} } = a + b i = z$
So $\phi$ is a surjection.
To prove that it is a injection, we let:
:$\map \phi {\eqclass {a + b x} {x^2 + 1} } = \map \phi {\eqclass {c + d x} {x^2 + 1} }$
So:
So $\phi$ is an injection and thus a bijection.
It remains to show that $\phi$ is a homomorphism for the operation $+$ and $\times$.
Thus $\phi$ has been demonstrated to be a bijective ring homomorphism and thus by definition a ring isomorphism.
Category:Complex Numbers
Category:Quotient Rings
Complex Numbers cannot be Extended to Algebra in Three Dimensions with Real Scalars

== Theorem ==
It is not possible to extend the complex numbers to an algebra of $3$ dimensions with real scalars.
== Proof ==
 that $\set {1, i, j}$ forms a basis for an algebra of $3$ dimensions with real scalars.
Let $1$ and $i$ have their usual properties as they do as complex numbers:
:$\forall a: 1 a = a 1 = a$
:$i \cdot i = -1$
Then:
:$i j = a_1 + a_2 i + a_3 j$
for some $a_1, a_2, a_3 \in \R$.
Multiplying through by $i$:
:$(1): \quad i \paren {i j} = \paren {i i} j = -j$
and:
But this implies that ${a_3}^2 = -1$, which contradicts our supposition that $a_3 \in \R$.
Hence the result by Proof by Contradiction.
== Sources ==
* : Chapter $\text {B}.26$: Extensions of the Complex Number System. Algebras, Quaternions, and Lagrange's Four Squares Theorem
Category:Algebras
Complex Numbers cannot be Ordered Compatibly with Ring Structure

== Theorem ==
Let $\struct {\C, +, \times}$ be the field of complex numbers.
There exists no total ordering on $\struct {\C, +, \times}$ which is compatible with the structure of $\struct {\C, +, \times}$.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also presented as ==
Some sources word this result in the following form:
:As applied to complex numbers, the phrases "greater than" or "less than" have no meaning. Inequalities can only occur in relations between the moduli of complex numbers.
This tells only part of the story.
By Zermelo's Well-Ordering Theorem, it is possible to apply an ordering of some kind upon $\C$.
Unfortunately, as is proven here, such an ordering would not actually be compatible with the structure of $\struct {\C, +, \times}$, and so of little practical use.
== Historical Note ==
== Sources ==
* : Chapter $\text I$: Functions of a Complex Variable: $\S 1$. Complex Numbers: $\text {(iii)}$ The fundamental operations
* : $\S 1.2$. The Algebraic Theory
* : $-1$ and $i$
* : $1$ The complex plane: Complex numbers $\S 1.4$ Inequalities
* : $-1$ and $i$
Category:Complex Numbers
Category:Complex Numbers cannot be Ordered Compatibly with Ring Structure
Complex Numbers cannot be Ordered Compatibly with Ring Structure/Proof 1

== Theorem ==
== Proof ==
 there exists a relation $\preceq$ on $\C$ which is ordering compatible with the ring structure of $\C$.
That is:
:$(1): \quad z \ne 0 \implies 0 \prec z \lor z \prec 0$, but not both
:$(2): \quad 0 \prec z_1, z_2 \implies 0 \prec z_1 z_2 \land 0 \prec z_1 + z_2$
By Totally Ordered Ring Zero Precedes Element or its Inverse, $(1)$ can be replaced with:
:$(1'): \quad z \ne 0 \implies 0 \prec z \lor 0 \prec -z$, but not both.
As $i \ne 0$, it follows that:
:$0 \prec i$ or $0 \prec -i$
Suppose $0 \prec i$.
Then:
Otherwise, suppose $0 \prec \paren {-i}$.
Then:
Thus by Proof by Cases:
:$0 \prec -1$
Thus it follows that:
Thus both:
:$0 \prec -1$
and:
:$0 \prec 1$
This contradicts hypothesis $(1')$:
:$(1): \quad z \ne 0 \implies 0 \prec z \lor 0 \prec -z$, but not both
Hence, by Proof by Contradiction, there can be no such ordering.
== Historical Note ==
Category:Complex Numbers cannot be Ordered Compatibly with Ring Structure
Complex Numbers cannot be Ordered Compatibly with Ring Structure/Proof 2

== Theorem ==
== Proof ==
 such a total ordering $\preceq$ exists.
By the definition of a total ordering, $\preceq$ is connected.
That is:
:$0 \preceq i \lor i \preceq 0$
Using Proof by Cases, we will prove that:
:$0 \preceq -1$
;Case 1
Assume that $0 \preceq i$.
By definition of an ordering compatible with the ring structure of $\C$:
:$\forall x, y \in \C: \left({0 \preceq x \land 0 \preceq y}\right) \implies 0 \preceq x \times y$
Substituting $x = i$ and $y = i$ gives:
:$0 \preceq i \times i$
Simplifying:
:$0 \preceq -1$
which is the result required.
;Case 2
Assume that $i \preceq 0$.
By definition of compatibility with addition:
:$\forall x, y, z \in \C: x \preceq y \implies \paren {x + z} \preceq \paren {y + z}$
Substituting $x = i$, $y = 0$, $z = -i$ gives:
:$i + \paren {-i} \preceq 0 + \paren {-i}$
Simplifying:
:$0 \preceq -i$
By definition of an ordering compatible with the ring structure of $\C$:
:$\forall x, y \in \C: \paren {0 \preceq x \land 0 \preceq y} \implies 0 \preceq x \times y$
Substituting $x = -i$ and $y = -i$ gives:
:$0 \preceq \paren {-i} \times \paren {-i}$
Simplifying:
:$0 \preceq -1$
This has been demonstrated to follow from both cases, and so by Proof by Cases:
:$0 \preceq -1$
By definition of an ordering compatible with the ring structure of $\C$:
:$\forall x, y \in \C: \paren {0 \preceq x \land 0 \preceq y} \implies 0 \preceq x \times y$
Substituting $x = -1$ and $y = -1$:
:$0 \preceq \paren {-1} \times \paren {-1}$
Simplifying:
:$0 \preceq 1$
By definition of compatibility with addition:
:$\forall x, y, z \in \C: x \preceq y \implies \paren {x + z} \preceq \paren {y + z}$
Substituting $x = 0$, $y = 1$, $z = -1$ gives:
:$0 + \paren {-1} \preceq 1 + \paren {-1}$
Simplifying:
:$-1 \preceq 0$
From the definition of ordering:
:$\forall a, b \in \C: \paren {a \preceq b \land b \preceq a} \implies a = b$
Substituting $a = -1$ and $b = 0$ gives:
:$-1 = 0$
which is a contradiction.
Hence, from Proof by Contradiction, there can be no such ordering.
== Historical Note ==
Category:Complex Numbers cannot be Ordered Compatibly with Ring Structure
Complex Numbers cannot be Ordered Compatibly with Ring Structure/Proof 3

== Theorem ==
== Proof ==
From Complex Numbers form Integral Domain, $\struct {\C, +, \times}$ is an integral domain.
 that $\struct {\C, +, \times}$ can be ordered.
Thus, by definition, it possesses a (strict) positivity property $P$.
Then from Strict Positivity Property induces Total Ordering, let $\le$ be the total ordering induced by $P$.
From Unity of Ordered Integral Domain is Strictly Positive:
:$1$ is strictly positive.
Thus by strict positivity, axiom $3$:
:$-1$ is not strictly positive.
Consider the element $i \in \C$.
By definition of strict positivity, axiom $3$, either:
:$i$ is strictly positive
or:
:$-i$ is strictly positive.
Suppose $i$ is strictly positive.
Then by Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive:
:$i^2 = -1$ is strictly positive.
Similarly, suppose $-i$ is strictly positive.
Then by Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive:
:$\paren {-i}^2 = -1$ is strictly positive.
In both cases we have that $-1$ is strictly positive.
But it has already been established that $-1$ is not strictly positive.
Hence, by Proof by Contradiction, there can be no such ordering.
== Historical Note ==
== Sources ==
* : Chapter $2$: Ordered and Well-Ordered Integral Domains: $\S 7$. Order: Example $12$
Category:Complex Numbers cannot be Ordered Compatibly with Ring Structure
Complex Numbers form Algebra

== Theorem ==
The set of complex numbers $\C$ forms an algebra over the field of real numbers.
This algebra is:
:$(1): \quad$ An associative algebra.
:$(2): \quad$ A commutative algebra.
:$(3): \quad$ A normed division algebra.
:$(4): \quad$ A nicely normed $*$-algebra.
However, $\C$ is not a real $*$-algebra.
== Proof ==
The complex numbers $\C$ are formed by the Cayley-Dickson Construction from the real numbers $\R$.
From Real Numbers form Algebra, we have that $\R$ forms:
:$(1): \quad$ An associative algebra.
:$(2): \quad$ A commutative algebra.
:$(3): \quad$ A normed division algebra.
:$(4): \quad$ A nicely normed $*$-algebra whose $*$ operator is the identity mapping.
:$(5): \quad$ A real $*$-algebra.
From Cayley-Dickson Construction forms Star-Algebra, $\C$ is a $*$-algebra.
From Cayley-Dickson Construction from Nicely Normed Algebra is Nicely Normed, $\C$ is a nicely normed $*$-algebra.
From Cayley-Dickson Construction from Real Star-Algebra is Commutative, $\C$ is a commutative algebra.
From Cayley-Dickson Construction from Commutative Associative Algebra is Associative, $\C$ is an associative algebra.
However, from Algebra from Cayley-Dickson Construction is not Real Star-Algebra, $\C$ is not a real $*$-algebra.
=== Proof of Normed Division Algebra ===
Consider the element $\tuple {1, 0}$ of $\R^2$.
We have:
As $\times$ has already been shown to be commutative, it follows that:
:$\tuple {1, 0} \times \tuple {x_1, x_2} = \tuple {x_1, x_2}$.
So $\tuple {1, 0} \in \R^2$ functions as a unit.
That is, $\struct {\R^2, \times}$ is a unitary algebra.
We define a norm on $\struct {\R^2, \times}$ by:
:$\forall \mathbf a = \tuple {a_1, a_2} \in \R^2: \norm {\mathbf a} = \sqrt { {a_1}^2 + {a_2}^2}$
This is a norm because:
:$(1): \quad \forall \mathbf x \in \R^2: \norm {\mathbf x} = 0 \iff \mathbf x = \mathbf 0$
:$(2): \quad \forall \mathbf x \in \R^2: \norm {\lambda \mathbf x} = \size \lambda \norm x$
:$(3): \quad \forall x, y, z \in \R: \norm {x - y} \le \norm {x - z} + \norm {z - y}$
It also follows that:
and so $\struct {\R^2, \times}$ is a normed division algebra.
== Sources ==
* : Chapter $\text {B}.26$: Extensions of the Complex Number System. Algebras, Quaternions, and Lagrange's Four Squares Theorem
Category:Complex Numbers
Category:Algebras
Complex Numbers form Field

== Theorem ==
Consider the algebraic structure $\struct {\C, +, \times}$, where:
:$\C$ is the set of all complex numbers
:$+$ is the operation of complex addition
:$\times$ is the operation of complex multiplication
Then $\struct {\C, +, \times}$ forms a field.
== Proof ==
From Complex Numbers under Addition form Infinite Abelian Group, we have that $\struct {\C, +}$ forms an abelian group.
From Non-Zero Complex Numbers under Multiplication form Infinite Abelian Group, we have that $\struct {\C_{\ne 0}, \times}$ forms an abelian group.
Finally, we have that Complex Multiplication Distributes over Addition.
Thus all the criteria are fulfilled, and $\struct {\C, +, \times}$ is a field.
== Also see ==
* Definition:Field of Complex Numbers
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 1$. Rings and Fields: Example $2$
* : Chapter $4$: Fields: $\S 15$. Examples of Fields: Example $16$
* : Chapter $1$: Rings - Definitions and Examples: $2$: Some examples of rings: Ring Example $4$
* : Chapter $3$: Field Theory: Definition and Examples of Field Structure: $\S 88$
* : Chapter $1$: Axioms for the Real Numbers: $2$. Fields: Example $5$
* : $\S 19$: Properties of $\Z_m$ as an algebraic system
* : $\S 55$. Special types of ring and ring elements: $(4)$
* : $1$: Complex Numbers: Axiomatic Foundations of the Complex Number System
* : $\S 2.4$: The rational numbers and some finite fields
* : $1$ The complex plane: Complex numbers $\S 1.2$ The algebraic structure of the complex numbers
* 
* 
Category:Complex Numbers
Category:Examples of Fields
Complex Numbers form Integral Domain

== Theorem ==
The set of complex numbers $\C$ forms an integral domain under addition and multiplication: $\struct {\C, +, \times}$.
== Proof ==
Recall that Complex Numbers form Field.
The result then follows directly from Field is Integral Domain.
== Sources ==
* : Chapter $1$: Integral Domains: $\S 3$. Definition of an Integral Domain
Category:Examples of Integral Domains
Category:Complex Numbers