nyuuzyou/wikis huggingface dataset:
Cauchy Sequence is Bounded/Normed Division Ring/Proof 2

== Theorem ==
== Proof ==
Let $d$ be the metric induced on $R$ be the norm $\norm {\,\cdot\,}$.
Let $\sequence {x_n} $ be a Cauchy sequence in $\struct {R, \norm {\,\cdot\,}}$.
By the definition of a Cauchy sequence in a normed division ring then $\sequence {x_n} $ is a Cauchy sequence in $\struct {R, d}$.
By Cauchy sequence is bounded in metric space then $\sequence {x_n} $ is a bounded sequence in $\struct {R, d}$.
By Sequence is Bounded in Norm iff Bounded in Metric then $\sequence {x_n} $ is a bounded sequence in $\struct {R, \norm {\,\cdot\,} }$.
== Sources ==
* : $\S 1.2$: Normed Fields, Exercise $11 \ (2)$
Category:Cauchy Sequence in Normed Division Ring is Bounded
Cauchy Sequence is Bounded/Normed Division Ring/Proof 3

== Theorem ==
== Proof ==
Let $\sequence {x_n} $ be a Cauchy sequence in $R$.
By Norm Sequence of Cauchy Sequence has Limit, $\sequence {\norm {x_n} }$ is a convergent sequence in $\R$.
By Convergent Real Sequence is Bounded, $\sequence {\norm {x_n} }$ is bounded.
That is:
:$\exists M \in \R_{\gt 0}: \forall n \in \N: \norm {x_n} = \size {\norm {x_n} } \le M$
Thus, by definition, $\sequence {x_n}$ is bounded.
== Sources ==
* : $\S 1.2$: Normed Fields, Exercise $11 \ (2)$
Category:Cauchy Sequence in Normed Division Ring is Bounded
Cauchy Sequence is Bounded/Normed Vector Space

== Theorem ==
Let $V = \struct {X, \norm {\,\cdot\,} }$ be a normed vector space.
Every Cauchy sequence in $X$ is bounded.
== Proof ==
Let $\sequence {x_n} $ be a Cauchy sequence in $V$.
Then by definition:
:$\forall \epsilon \in \R_{>0}: \exists N \in \N : \forall n, m \ge N: \norm {x_n - x_m} < \epsilon$
Let $N$ satisfy:
:$\forall n, m \ge N: \norm {x_n - x_m} < 1$
Let $m = N + 1 > N$.
Then $\forall n \ge N$:
Let $M = \max \set {\norm {x_1}, \norm {x_2}, \dots, \norm {x_N}, 1 + \norm {x_{N + 1} } }$.
Then:
:$\forall n < N: \norm {x_n} \le M$
:$\forall n \ge N: \norm {x_n} \le 1 + \norm {x_{N + 1} } \le M$
It follows by definition that $\sequence {x_n}$ is bounded.
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Normed Vector Spaces
Cauchy Sequence is Bounded/Real Numbers

== Theorem ==
Every Cauchy sequence in $\R$ is bounded.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 5$: Subsequences: Cauchy sequences: $\S 5.18$
Category:Real Analysis
Category:Cauchy Sequences
Category:Cauchy Sequence is Bounded
Cauchy Sequence is Bounded/Real Numbers/Proof 1

== Theorem ==
== Proof ==
Let $\sequence {a_n}$ be a Cauchy sequence in $\R$.
Then there exists $N \in \N$ such that:
:$\size {a_m - a_n} < 1$
for all $m, n \ge N$.
So for all $m \ge N$, we have:
So $\sequence {a_n}$ is bounded, as claimed.
Category:Cauchy Sequence is Bounded
Cauchy Sequence is Bounded/Real Numbers/Proof 2

== Theorem ==
== Proof ==
Let $\sequence {a_n}$ be a Cauchy sequence in $\R$.
Then there exists $N \in \N$ such that:
:$\size {a_m - a_n} < 1$
for all $m, n \ge N$.
Note that for $m \le N$:
Hence for all $m \in \N$:
:$\size {a_m} < \max \set {\size {a_1}, \size {a_2}, \dotsc, \size {a_N} } + 1$
Therefore $\sequence {a_n}$ is bounded, as claimed.
== Sources ==
* : $1$: Review of some real analysis: $\S 1.2$: Real Sequences: Theorem $1.2.9$
* : Appendix: $\S 18.4$: Subsequences
Category:Cauchy Sequence is Bounded
Cauchy Sequence of Subring iff Cauchy Sequence of Normed Division Ring

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\struct {S, \norm {\, \cdot \,}_S }$ be a normed division subring of $\struct {R, \norm {\, \cdot \,} }$.
Let $\sequence{x_n}$ be a sequence in $S$.
Then:
:$\sequence{x_n}$ is a Cauchy sequence in $\struct {S, \norm {\, \cdot \,}_S }$
:
:$\sequence{x_n}$ is a Cauchy sequence in $\struct {R, \norm {\, \cdot \,} }$
== Proof ==
The result follows immediately from:
:
:
Category:Normed Division Rings
Cauchy Sequence with Finite Elements Prepended is Cauchy Sequence

== Theorem ==
Let $\struct {R, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\sequence {x_n}$ be a sequence in $R$.
Let $N \in \N$
Let $\sequence {y_n}$ be the sequence defined by:
:$\forall n, y_n = x_{N + n}$
Let $\sequence {y_n}$ be a Cauchy sequence in $R$.
Then:
:$\sequence {x_n}$ is a Cauchy sequence in $R$.
== Proof ==
Given $\epsilon > 0$:
By the definition of a Cauchy sequence then:
:$\exists N': \forall n, m > N', \norm {y_n - y_m} < \epsilon$
Hence $\forall n, m > \paren {N' + N}$:
The result follows.
== Sources ==
Category:Cauchy Sequence in Normed Division Ring is Bounded
Cauchy Sequences form Ring with Unity

== Theorem ==
Let $\struct {R, +, \circ, \norm {\, \cdot \,} }$ be a normed division ring.
Let $\struct {R^\N, +, \circ}$ be the ring of sequences over $R$ with unity $\tuple {1, 1, 1, \dotsc}$.
Let $\CC \subset R^\N$ be the set of Cauchy sequences on $R$.
Then:
:$\struct {\CC, +, \circ}$ is a subring of $R^\N$ with unity $\tuple {1, 1, 1, \dotsc}$.
=== Corollary ===
== Proof ==
The Subring Test used to prove the result.
By Constant Rule for Cauchy sequences:
:the constant sequence $\tuple {1, 1, 1, \dotsc}$ is a Cauchy sequences.
Hence:
:$\CC \neq \O$
Let $\sequence {x_n}, \sequence {y_n} \in \CC$.
By definition of pointwise addition:
:$\sequence {x_n} + \paren {-\sequence {y_n}} = \sequence {x_n - y_n}$.
By Difference Rule for Normed Division Ring Sequences:
:$\sequence {x_n - y_n}$ is a Cauchy sequence.
Hence:
:$\sequence {x_n} + \paren {- \sequence {y_n}} \in CC$.
By definition of pointwise product:
:$\sequence {x_n} \circ \sequence {y_n} = \sequence {x_n \circ y_n}$.
By Product Rule for Normed Division Ring Sequences:
:the sequence $\sequence {x_n \circ y_n}$ is a Cauchy sequence.
Hence:
:$\sequence {x_n} \circ \sequence {y_n} \in \CC$.
By the Subring Test the result follows.
== Sources ==
* : $\S 3.2$: Completions: Proposition $3.2.5$
* : $\S 1.3$ Construction of the completion of a normed field
Category:Cauchy Sequences
Category:Normed Division Rings
Category:Cauchy Sequences in Normed Division Rings
Category:Rings of Sequences
Cauchy Sequences form Ring with Unity/Corollary

== Corollary to Cauchy Sequences form Ring with Unity ==
Let $\struct {F, +, \circ, \norm {\, \cdot \,} }$ be a valued field.
Let $\struct {F^\N, +, \circ}$ be the commutative ring of sequences over $F$ with unity $\tuple {1, 1, 1, \dotsc}$.
Let $\CC \subset F^\N$ be the set of Cauchy sequences on $F$.
Then:
:$\struct {\CC, +, \circ}$ is a commutative subring of $F^\N$ with unity $\tuple {1, 1, 1, \dotsc}$.
== Proof ==
The field $F$ is a commutative ring by definition.
From Structure Induced by Commutative Ring Operations is Commutative Ring, the ring of sequences over $F$ is a commutative ring.
Hence $\circ$ is commutative on $F^\N$.
By Cauchy Sequences form Ring with Unity, $\struct {\CC, +, \circ}$ is a subring of $F^\N$.
Hence by Restriction of Commutative Operation is Commutative the restriction of $\circ$ to $\CC$ is commutative.
== Sources ==
* : $\S 3.2$: Completions
Category:Cauchy Sequences in Normed Division Rings
Cauchy Sequences in Vector Spaces with Equivalent Norms Coincide

== Theorem ==
Let $M_a = \struct {X, \norm {\, \cdot \, }_a}$ and $M_b = \struct {X, \norm {\, \cdot \,}_b}$ be normed vector spaces.
Let $\sequence {x_n}_{n \mathop \in \N}$ be a Cauchy sequence in $M_a$.
Suppose, $\norm {\, \cdot \, }_a$ and $\norm {\, \cdot \,}_b$ are equivalent norms, that is:
:$\norm {\, \cdot \, }_a \sim \norm {\, \cdot \,}_b$
Then $\sequence {x_n}_{n \mathop \in \N}$ is also a Cauchy sequence in $M_b$. 
== Proof ==
We have that $\sequence {x_n}_{n \mathop \in \N}$ is a Cauchy sequence in $M_a$.
Then:
:$\forall \epsilon_a \in \R_{> 0} : \exists N \in \N : \forall n, m \in \N : n, m > N \implies \norm {x_n - x_m}_a < \epsilon_a$
By equivalence of norms:
:$\exists M \in \R_{> 0} : \norm {x_n - x_m}_b \le M \norm {x_n - x_m}_a < M \epsilon_a$
Let $\epsilon_b := M \epsilon_a$
Then:
:$\forall \epsilon_b \in \R_{> 0} : \exists N \in \N : \forall n \in \N : n, m > N \implies \norm {x_n - x_m}_b < \epsilon_b$
Therefore, $\sequence {x_n}_{n \mathop \in \N}$ is also a Cauchy sequence in $M_b$.
== Sources ==
* : $\S 1.4$: Normed and Banach spaces. Sequences in a normed space; Banach spaces
Category:Equivalence Relations
Category:Norm Theory
Category:Vector Spaces
Category:Cauchy Sequences
Cavalieri's Principle

== Theorem ==
Let two solid figures $S_1$ and $S_2$ have equal height.
Let sections made by planes parallel to their bases and at equal distances from the bases always have equal area.
Then the volumes of $S_1$ and $S_2$ are equal.
=== Extension ===
An extension of Cavalieri's Principle is as follows:
== Proof ==
Let $H$ be the common height of the two figures.
The volume of a solid figure is its Lebesgue Measure in $\R^3$.
Therefore:
In the same manner:
:$\ds \map V {S_2} = \int_{\closedint 0 H} \int_{\R^2} \paren {\chi_{S_2} }_x \rd \lambda^2 \rd \lambda$
The area of a plane figure is its Lebesgue Measure in $\R^2$.
The area of a section made in $S_1$ by the plane $x = x_0$ is:
Likewise:
:$\ds \map A {S_2 \cap P_{x \mathop = x_0} } = \int_{\R^2} \paren {\chi_{S_2} }_{x \mathop = x_0} \rd \lambda^2$
But by assumption:
:$\map A {S_1 \cap P_{x \mathop = x_0} } = \map A {S_2 \cap P_{x \mathop = x_0} }$
for every $x_0 \in \closedint 0 H$.
Therefore:
Thus we conclude that $\map V {S_1} = \map V {S_2}$
== Sources ==
* 
* : Chapter $\text {A}.14$: Cavalieri ()
* 
Category:Cavalieri's Principle
Category:Volume
Category:Solid Geometry
Cavalieri's Principle/Extension

== Theorem ==
Let two solid figures $S_1$ and $S_2$ have equal height.
Let the areas of the sections made by planes parallel to their bases and at equal distances from the bases always have the same ratio.
Then the volumes of $S_1$ and $S_2$ are in that same ratio.
== Proof ==
== Sources ==
* : Chapter $\text {A}.14$: Cavalieri ()
* 
* 
Category:Cavalieri's Principle
Category:Volume
Category:Solid Geometry
Cayley's Formula

== Theorem ==
The number of distinct labeled trees with $n$ nodes is $n^{n - 2}$.
== Proof ==
Follows directly from Bijection between Prüfer Sequences and Labeled Trees.
This shows that there is a bijection between the set of labeled trees with $n$ nodes and the set of all Prüfer sequences of the form:
:$\tuple {\mathbf a_1, \mathbf a_2, \ldots, \mathbf a_{n - 2} }$
where each of the $\mathbf a_i$'s is one of the integers $1, 2, \ldots, n$, allowing for repetition.
Since there are exactly $n$ possible values for each integer $\mathbf a_i$, the total number of such sequences is $\ds \prod_{i \mathop = 1}^{n - 2} n$.
The result follows from Equivalence of Mappings between Finite Sets of Same Cardinality.
== Examples ==
== Also known as ==
== Historical Note ==
== Sources ==
* 
* 
* 
Category:Cayley's Formula
Category:Graph Theory
Category:Combinatorics
Cayley's Formula/Examples/Order 4

== Examples of Use of Cayley's Formula ==
There are $16$ distinct labeled trees of order $4$.
== Proof ==
According to Cayley's Formula, there are $n^{n - 2}$ distinct labeled trees of order $n$.
The result follows by substituting $4$ for $n$.
== Sources ==
* 
* 
Category:Examples of Use of Cayley's Formula
Cayley's Formula/Historical Note

== Historical Note on Cayley's Formula ==
 first stated Cayley's Formula in his in $1889$.
However, his proof was unsatisfactory as he discussed only the case where $n = 6$, and his method cannot be generalized to larger $n$.
The proof given here, the work of , first appeared in $1918$.
== Sources ==
* 
* 
* 
Category:Historical Notes
Cayley's Representation Theorem

== Theorem ==
Let $S_n$ denote the symmetric group on $n$ letters.
Every finite group is isomorphic to a subgroup of $S_n$ for some $n \in \Z$.
=== General Case ===
== Proof 1 ==
== Proof 2 ==
== Also known as ==
== Also see ==
What this theorem tells us is that in order to study finite groups, it is necessary only to study subgroups of the symmetric groups on $n$ letters.
== Sources ==
* : Chapter $\text{II}$: Groups: Problem $\text{EE}$
* : Chapter $2$: The Symmetric Groups: $\S 77 \alpha$
* 
* 
* 
* 
Category:Cayley's Representation Theorem
Category:Symmetric Groups
Category:Group Theory
Category:Subgroups
Category:Representation Theorems
Cayley's Representation Theorem/Also known as

== Cayley's Representation Theorem: Also known as ==
Cayley's Representation Theorem is also known as the Representation Theorem for Groups.
Some present it as the Cayley Representation Theorem.
Some sources refer to it as just Cayley's Theorem, but as there is more than one result so named, it is better to use the more specific form.
== Sources ==
* 
* 
Category:Cayley's Representation Theorem
Cayley's Representation Theorem/General Case

== Theorem ==
Let $\struct {G, \cdot}$ be a group.
Then there exists a permutation group $P$ on some set $S$ such that:
:$G \cong P$
That is, such that $G$ is isomorphic to $P$.
== Proof ==
Let $G$ be a group and let $a \in G$.
Consider the left regular representation $\lambda_a: G \to G$ defined as:
:$\map {\lambda_a} x = a \cdot x$
From Regular Representations in Group are Permutations we have that $\lambda_a$ is a permutation.
Now let $b \in G$ and consider $\lambda_b: G \to G$ defined as:
:$\map {\lambda_b} x = b \cdot x$
From the Cancellation Laws it follows that $\lambda_a \ne \lambda_b \iff a \ne b$.
Let $H = \set {\lambda_x: x \in G}$.
Consider the mapping $\Phi: G \to H$ defined as:
:$\forall a \in G: \map \Phi a = \lambda_a$
From the above we have that $\Phi$ is a bijection.
Let $a, b \in G$.
From Composition of Regular Representations we have that:
:$\lambda_a \circ \lambda_b = \lambda_{a \cdot b}$
where $\circ$ denotes composition of mappings.
That is, $\Phi$ has the morphism property.
Thus $\Phi$ is seen to be a group isomorphism.
We also have that:
:$\paren {\lambda_a}^{-1} = \lambda_\paren {a^{-1} }$
because:
:$\lambda_a \circ \paren {\lambda_a}^{-1} = \lambda_\paren {a \cdot a^{-1} }$
Hence the set of left regular representations $\set {\lambda_x: x \in G}$ is a group which is isomorphic to $G$.
== Also known as ==
== Sources ==
* : $\S 1.6$: Theorem $1.6$
* : $\S 1.5$: Theorem
Category:Group Theory
Category:Cayley's Representation Theorem
Cayley's Representation Theorem/Proof 1

== Theorem ==
== Proof ==
Let $H = \set e$.
We can apply Permutation of Cosets to $H$ so that:
:$\mathbb S = G$
and:
:$\map \ker \theta = \set e$
The result follows by the First Isomorphism Theorem for Groups.
== Sources ==
* : Chapter $9$: Permutations: Corollary $9.24$
Category:Cayley's Representation Theorem
Cayley's Representation Theorem/Proof 2

== Theorem ==
== Proof ==
Let $G$ be any arbitrary finite group whose identity is $e_G$.
Let $S$ be the symmetric group on the elements of $G$, where $e_S$ is the identity of $S$.
For any $x \in G$, let $\lambda_x$ be the left regular representation of $G$ with respect to $x$.
From Regular Representations in Group are Permutations, $\forall x \in G: \lambda_x \in S$.
So, we can define a mapping $\theta: G \to S$ as:
:$\forall x \in G: \map \theta x = \lambda_x$
From Composition of Regular Representations, we have:
:$\forall x, y \in G: \lambda_x \circ \lambda_y = \lambda_{x y}$
where in this context $\circ$ denotes composition of mappings.
Thus by definition of $\theta$:
:$\map \theta x \circ \map \theta y = \map \theta {x y}$
demonstrating that $\theta$ is a homomorphism.
Having established that fact, we can now consider $\map \ker \theta$, where $\ker$ denotes the kernel of $\theta$.
Let $x \in G$.
We have that:
So $\map \ker \theta$ can contain no element other than $e_G$.
So, since clearly $e_G \in \map \ker \theta$, it follows that:
:$\map \ker \theta = \set {e_G}$
By Kernel is Trivial iff Monomorphism, it follows that $\theta$ is a monomorphism.
By Monomorphism Image is Isomorphic to Domain, we have that:
:$\theta \sqbrk G \cong \Img \theta$
that is, $\theta$ is isomorphic to its image.
Hence the result.
== Sources ==
* : $\S 7.4$. Kernel and image: Example $143$
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Theorem $8.7$
* : $\S 47.9$ Homomorphisms and their elementary properties
Category:Cayley's Representation Theorem
Cayley's Theorem

* Cayley's Representation Theorem:
::Every finite group is isomorphic to a subgroup of the symmetric group on $n$ letters for some $n \in \Z$.
* Cayley's Theorem (Graph Theory), also known as Cayley's Formula:
::The number of distinct labeled trees with $n$ nodes is $n^{n - 2}$.
Category:Named Theorems/Cayley
Cayley's Theorem (Category Theory)

== Theorem ==
Let $\mathbf C$ be a small category.
Denote with $\mathbf{Set}$ the category of sets.
Then there exists a category $\mathbf D$, subject to:
:$(1): \quad $ The objects of $\mathbf D$ are sets.
:$(2): \quad $ The morphisms of $\mathbf D$ are mappings.
:$(3): \quad \mathbf C \cong \mathbf D$, i.e. $\mathbf C$ and $\mathbf D$ are isomorphic.
That is, $\mathbf C$ is isomorphic to a subcategory of $\mathbf{Set}$.
== Proof ==
Define a functor $H: \mathbf C \to \mathbf{Set}$ by:
:$H C := \set {f \in \operatorname{mor} \mathbf C: \operatorname{cod} f = C}$
:$H f: H A \to H B, g \mapsto f \circ g$
for $f: A \to B$ a morphism of $\mathbf C$.
It is immediate by the definition of identity morphism that:
:$\map H {\operatorname{id}_A} = \operatorname{id}_{H A}$
For $f: A \to B$ and $g: B \to C$, observe:
Thus by Equality of Mappings:
:$\map H {g \circ f} = H g \circ H f$
It follows that $H$ is a functor.
It is clear that $H$ is injective on objects.
Suppose now that $H$ were not faithful.
Then there would be morphisms $g, h: A \to B$ of $\mathbf C$ such that $H g = H h$.
Since $\operatorname{id}_A \in H A$, this means in particular that:
:$g \circ \operatorname{id}_A = h \circ \operatorname{id}_A$
by Equality of Mappings.
But the definition of identity morphism then reduces this to $g = h$.
Hence, $H$ is faithful.
By Functor is Embedding iff Faithful and Injective on Objects, it follows that $H$ is an embedding.
Thus $\mathbf C$ is isomorphic to a subcategory of $\mathbf{Set}$.
== Also see ==
* Cayley's Representation Theorem
== Historical Note ==
== Sources ==
* : $\S 1.5$: Theorem $1.6$
Category:Category Theory
Cayley's Theorem (Category Theory)/Historical Note

== Historical Note on Cayley's Theorem (Category Theory) ==
Although did not prove this result, it is similar enough in both statement and proof to Cayley's Representation Theorem as to merit his name being appended to it.
As states it in :
:[Cayley's] theorem may be generalized to show that any category that is not "too big" can be represented as a [...] category of sets and functions.
:::: -- , p. 13
The contributor Lord_Farin subsequently was as audacious as to name the general result after as well.
== Sources ==
* : $\S 1.5$: Theorem $1.6$
Category:Historical Notes
Cayley-Dickson Construction forms Star-Algebra

== Theorem ==
Let $A = \struct {A_F, \oplus}$ be a $*$-algebra.
Let $A' = \struct {A'_F, \oplus'} = \struct {A, \oplus}^2$ be the algebra formed from $A$ by the Cayley-Dickson construction.
Then $A'$ is also a $*$-algebra.
== Proof ==
=== Bilinearity of $\oplus'$ ===
First we need to show that $\oplus'$ is bilinear.
$(1): \quad$ Let $\tuple {a_1, b_1}, \tuple {a_2, b_2}, \tuple {c, d} \in A'$.
Then:
Similarly (and equally tediously) we can show that:
:$\tuple {c, d} \oplus' \paren {\tuple {a_1, b_1} + \tuple {a_2, b_2}} = \paren {\tuple {c, d} \oplus' \tuple {a_1, b_1}} + \paren {\tuple {c, d} \oplus' \tuple {a_2, b_2}}$
$(2): \quad$ Let $\tuple {a, b}, \tuple {c, d} \in A'$ and $\alpha, \beta \in \R$.
Then:
Similarly:
:$\tuple {a, b} \oplus' \paren {\tuple {c, d} \beta} = \tuple {a, b} \oplus' \tuple {c, d} \beta$
So $\oplus'$ has been shown to be a bilinear mapping.
=== Conjugate Nature of $*'$ ===
We have that:
:$\forall \tuple {a, b} \in A': {\tuple {a, b}^*}' = \tuple {a^*, -b}$
So:
Finally:
thus proving that $*'$ is a conjugation on $A'$.
Hence the result.
== Sources ==
* : The Octonions (2002): 2.2 The Cayley-Dickson Construction
Category:Cayley-Dickson Construction
Category:Star-Algebras
Cayley-Dickson Construction from Commutative Associative Algebra is Associative

== Theorem ==
Let $A = \struct {A_F, \oplus}$ be a $*$-algebra.
Let $A' = \struct {A_F, \oplus'}$ be constructed from $A$ using the Cayley-Dickson construction.
Then:
:$A'$ is an associative algebra
:
:$A$ is both a commutative algebra and an associative algebra.
== Proof ==
Let the conjugation operator on $A$ be $*$.
Let $\tuple {a, b}, \tuple {c, d}, \tuple {e, f} \in A'$.
In order to streamline notation, let $\oplus$ and $\oplus'$ both be denoted by product notation:
:$a \oplus b =: a b$
:$x \oplus' y =: x y$
The context will make it clear which is meant.
Suppose $A$ is commutative and associative.
Then:
Now suppose $A'$ is an associative algebra.
By picking apart the above series of equations, it is clear that $A$ has to be both commutative and associative in order to assure the associativity of $A'$.
== Sources ==
* 2002: : The Octonions: $2.2$ The Cayley-Dickson Construction: Proposition $3$
Category:Cayley-Dickson Construction
Category:Associative Algebras
Category:Commutative Algebras
Cayley-Dickson Construction from Nicely Normed Algebra is Nicely Normed

== Theorem ==
Let $A = \struct {A_F, \oplus}$ be a $*$-algebra.
Let $A' = \struct {A_F, \oplus'}$ be constructed from $A$ using the Cayley-Dickson construction.
Then $A'$ is a nicely normed algebra $A$ is also a nicely normed algebra.
== Proof ==
Let the conjugation operator on $A$ be $*$.
Let $\tuple {a, b}, \tuple {c, d} \in A'$.
In order to streamline notation, let $\oplus$ and $\oplus'$ both be denoted by product notation:
:$a \oplus b =: a b$
:$x \oplus' y =: x y$
The context will make it clear which is meant.
Let $A$ be a nicely normed algebra.
Then:
As $A$ is itself nicely normed, we have that $a + a^*$ is real.
Hence it follows that $\tuple {a + a^*, 0}$ is real.
Next:
We have by definition of norm that $\norm a^2 + \norm b^2$ is real.
Hence it follows that $\tuple {\norm a^2 + \norm b^2, 0}$ is real.
It follows from reversing the argument that if $A'$ is not nicely normed then nor will $A$ be.
== Sources ==
* : The Octonions (2002): 2.2 The Cayley-Dickson Construction: Proposition $5$
Category:Cayley-Dickson Construction
Category:Nicely Normed Star-Algebras
Cayley-Dickson Construction from Real Star-Algebra is Commutative

== Theorem ==
Let $A = \struct {A_F, \oplus}$ be a $*$-algebra.
Let $A' = \struct {A_F, \oplus'}$ be constructed from $A$ using the Cayley-Dickson construction.
Then $A$ is a real star-algebra $A'$ is a commutative algebra.
== Proof ==
Let the conjugation operator on $A$ be $*$.
Let $\tuple {a, b}, \tuple {c, d} \in A'$.
Let $A$ be a real star-algebra.
So $A'$ is a commutative algebra.
Let $A'$ be a commutative algebra.
By picking apart the above equations, it is clear that for $A'$ to be a commutative algebra it is necessary for $A$ to be both real and commutative.
Hence the result.
== Sources ==
* : The Octonions (2002): 2.2 The Cayley-Dickson Construction: Proposition $2$
Category:Cayley-Dickson Construction
Category:Real Star-Algebras
Category:Commutative Algebras