nyuuzyou/wikis huggingface dataset:
Free Commutative Monoid is Commutative Monoid

== Theorem ==
The free commutative monoid on a set $\family {X_j: j \in J}$ is a commutative monoid.
== Proof ==
Let $M$ be the set of all monomials on the indexed set $\family {X_j: j \in J}$.
We are required to show that the following properties hold:
First note that using the multiindex notation described in the definition of monomials, for $r \in \N$, $m_i = \mathbf X^{k^i} \in M$, $i = 1, \ldots, r$, the product of the $m_i$ is given by:
:$m_1 \circ \cdots \circ m_r = \mathbf X^{k^1 + \cdots + k^r}$
Here the superscripts enumerate the multiindices, and do not indicate raising to a power.
So to show the closure, associativity and commutativity of monomials under $\circ$, it is sufficient to show the corresponding properties for multiindices under addition defined by:
:$\family {k^1 + k^2}_j := k^1_j + k^2_j$
In the following $k^1, k^2, k^3$ are multiindices, that is, families of non-negative integers indexed by $J$ such that only finitely many entries are non-zero.
=== Proof of Closure ===
Let $\family {k^1 + k^2}_j = k^1_j + k^2_j \ne 0$.
By definition of multiindex, at least one of $k^1_j$ and $k^2_j$ must be non-zero, and this can only be true for a finite number of entries.
Furthermore, since $k^1_j,\ k^2_j \ge 0$, we have $k^1_j + k^2_j \ge 0$.
Therefore $k^1 + k^2$ has finitely many non-zero entries, and these are all positive, and multiindices are closed under addition.
=== Proof of Associativity ===
Using associativity of integer addition, we have: 
:$\family {\paren {k^1 + k^2} + k^3}_j = \paren {k^1_j + k^2_j} + k^3_j = k^1_j + \paren {k^2_j + k^3_j} = \family {k^1 + \paren {k^2 + k^3} }_j$
So addition of multiindices is associative.
=== Proof of Commutativity ===
Using commutativity of integer addition, we have 
:$\family {k^1 + k^2}_j = k^1_j + k^2_j = k^2_j + k^1_j = \family {k^2 + k^1}_j$
So addition of multiindices is commutative.
=== Proof of Existence of Identity ===
Let $e_M$ be the multiindex such that $\family {e_M}_j = 0$ for all $j \in J$.
Then:
:$\family {e_M + k^1}_j = \family {e_M}_j + k^1_j = \family {k^1}_j$
so $e_M$ is an identity for the set of monomials.
Category:Examples of Commutative Monoids
Category:Polynomial Theory
Free Commutative Monoid on One Element is Isomorphic to Natural Numbers under Addition

== Theorem ==
Let $X = \set x$ be a singleton.
Let $M$ be the free commutative monoid on $X$.
Then $M$ is isomorphic to the additive monoid of natural numbers.
== Proof ==
By definition, the free commutative monoid $M$ on $\set x$ is:
:$M = \set {e, x, x^2, x^3, \ldots}$
where $e$ denotes the null sequence of elements of $X$.
Let $\phi$ denote the mapping from $M$ to $\N$ as:
:$\forall a \in M: \map \phi a = \begin{cases}
0 & : a = e \\
n & : a = x^n \end{cases}$
By definition of $\phi$:
:$\phi$ is injective: $\map \phi a = \map \phi b \implies a = b$
:$\phi$ is surjective: $\forall a \in \N: \exists b \in M: \map \phi b = a$
:$\phi$ is a monoid homomorphism: $\map \phi {a b} = \map \phi {a + b} = \map \phi a + \map \phi b$
Hence the result, by definition of isomorphism.
== Sources ==
* : $\S 3.1$: Monoids
Category:Natural Number Addition
Category:Free Monoids
Free Matroid is Matroid

== Theorem ==
Let $S$ be a finite set.
Let $\struct {S, \powerset S}$ be the free matroid of $S$.
Then $\struct {S, \powerset S}$ is a matroid.
== Proof ==
Let $S$ have cardinality $n$.
Let $\struct {S, \mathscr I_{n, n} }$ be the uniform matroid of rank $n$.
From Cardinality of Proper Subset of Finite Set, every subset of $S$ has cardinality less than or equal to $n$.
It follows that $\mathscr I_{n, n} = \powerset S$.
From Uniform Matroid is Matroid, then $\struct {S, \powerset S}$ Is a matroid.
== Sources ==
* Chapter $1.$ $\S 3.$ Examples of Matroids
Category:Matroid Theory
Free Module is Isomorphic to Free Module on Set

== Theorem ==
Let $M$ be a unitary $R$-module.
Let $\BB = \family {b_i}_{i \mathop \in I}$ be a family of elements of $M$.
Let $\Psi: R^{\paren I} \to M$ be the morphism given by Universal Property of Free Module on Set.
Then the following are equivalent:
:$\BB$ is a basis of $M$
:$\Psi$ is an isomorphism
== Proof ==
Follows directly from:
:Characterisation of Linearly Independent Set through Free Module Indexed by Set
:Characterisation of Spanning Set through Free Module Indexed by Set.
Category:Free Modules
Category:Module Theory
Free Module on Set is Free

== Theorem ==
Let $R$ be a ring with unity.
Let $I$ be a set.
Let $R^{\paren I}$ be the free $R$-module on $I$.
Then $R^{\paren I}$ is a free $R$-module.
== Proof ==
From Canonical Basis of Free Module on Set is Basis, $R^{\paren I}$ has a basis.
Category:Free Modules
Free Monoid is Unique

== Theorem ==
Let $S$ be a set.
Let $\struct {M, i}$ and $\struct {N, j}$ be free monoids over $S$.
Then there is a unique monoid isomorphism $f: M \to N$ such that:
:$\size f \circ i = j$
:$\size {f^{-1} } \circ j = i$
where $\size {\, \cdot \,}$ denotes the underlying set functor on $\mathbf{Mon}$.
== Proof ==
By the (categorial) definition of free monoid, we have the following commutative diagram:
$\quad\quad \begin{xy}
<-4em,4em>*{\mathbf{Mon} :},
<-4em,1em>*{\mathbf{Set} :},
<0em,4em>*+{N} = "N",
<4em,4em>*+{M} = "M",
<8em,4em>*+{N} = "N2",
"N";"M" **@{.} ?>*@{>} ?*!/_1em/{\bar i},
"M";"N2" **@{.} ?>*@{>} ?*!/_1em/{\bar j},
<0em,1em>*+{\left\vert{N}\right\vert} = "NN",
<4em,1em>*+{\left\vert{M}\right\vert} = "MM",
<8em,1em>*+{\left\vert{N}\right\vert} = "NN2",
<4em,-3em>*+{S} = "S",
"NN";"MM" **@{-} ?>*@{>} ?*!/_1em/{\left\vert{\bar i}\right\vert},
"MM";"NN2" **@{-} ?>*@{>} ?*!/_1em/{\left\vert{\bar j}\right\vert},
"S";"NN" **@{-} ?>*@{>} ?*!/_1em/{j},
"S";"MM" **@{-} ?>*@{>} ?<>(.7)*!/_.6em/{i},
"S";"NN2" **@{-} ?>*@{>} ?*!/^1em/{j},
\end{xy}$
The outer rim of these diagrams also needs to satisfy the universal property.
Since obviously $\operatorname{id}_N$ fits in place of $\bar j \circ \bar i$, we conclude these are equal.
Exchanging the rôles of $\struct {M, i}$ and $\struct {N, j}$ proves that $\bar i \circ \bar j = \operatorname{id}_M$ as well.
Hence $\bar i$ is the sought monoid isomorphism.
Its uniqueness follows from the universal property of the free monoid.
== Sources ==
* : $\S 1.7$
Category:Free Monoids
Category:Category of Monoids
Frege Set Theory is Logically Inconsistent

== Theorem ==
The system of axiomatic set theory that is Frege set theory is inconsistent.
== Proof ==
From Russell's Paradox, the leads to a contradiction.
Let $q$ be such a contradiction:
:$q = p \land \neg p$
for some statement $p$.
From the Rule of Explosion it then follows that every logical formula is a provable consequence of $q$.
Hence the result, by definition of inconsistent.
== Sources ==
* : Chapter $1$: General Background: $\S 8$ Russell's paradox
Category:Frege Set Theory
Frequency (Descriptive Statistics)/Examples/Arbitrary Example 1

== Example of Frequency in the context of Descriptive Statistics ==
Consider the sample:
:$2, 5, 3, 3, 3, 5, 3, 6, 2, 3, 9, 5$
The (absolute) frequency of the observation $3$ is $5$.
The (absolute) frequency of the observation $9$ is $1$.
== Sources ==
* 
* 
Category:Examples of Frequency (Descriptive Statistics)
Frequency Decreases when Mass Increases

== Theorem ==
=== Problem Definition ===
Then as the mass of $C$ increases, the frequency of the motion of $C$ decreases.
== Proof ==
From Position of Cart attached to Wall by Spring, the horizontal position of $C$ is given as:
:$(1): \quad x = C_1 \cos \alpha t + C_2 \sin \alpha t$
where:
:$C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$
:$\alpha = \sqrt {\dfrac k m}$
From the definition of frequency:
From Square Root is Strictly Increasing, the square root is strictly increasing on positive real numbers.
From Reciprocal Function is Strictly Decreasing, the reciprocal function is strictly decreasing on positive real numbers.
Hence, if the force constant of $S$ is constant (as is implicit in the statement of the problem), increasing $m$ causes a decrease in $\nu$.
== Sources ==
* : $\S 3.20$: Vibrations in Mechanical Systems
Category:Mechanics
Frequency Increases when Stiffness Increases

== Theorem ==
=== Problem Definition ===
Then as the force constant (or "stiffness") of $k$ increases, the frequency of the motion of $C$ likewise increases.
== Proof ==
From Position of Cart attached to Wall by Spring, the horizontal position of $C$ is given as:
:$(1): \quad x = C_1 \cos \alpha t + C_2 \sin \alpha t$
where:
:$C_1$ and $C_2$ depend upon the conditions of $C$ at time $t = 0$
:$\alpha = \sqrt {\dfrac k m}$
From the definition of frequency:
From Square Root is Strictly Increasing, the square root is strictly increasing on positive real numbers.
Hence, if the mass is constant (as is implicit in the statement of the problem), increasing $k$ causes an increase in $\nu$.
== Sources ==
* : $\S 3.20$: Vibrations in Mechanical Systems
Category:Mechanics
Frequency Table/Examples/Arbitrary Example 1

== Example of Frequency Table ==
Consider the observations:
:$2, 3, 5, 5, 5, 7, 9, 9, 9, 10$
The frequency table corresponding to the above is:
| $2$
| $3$
| $5$
| $7$
| $9$
| $10$
| $1$
| $1$
| $3$
| $1$
| $3$
| $1$
| $1$
| $2$
| $5$
| $6$
| $9$
| $10$
== Sources ==
* 
* 
Category:Examples of Frequency Tables
Frequency Table/Examples/Arbitrary Example 2

== Example of Frequency Table ==
For $200$ people, their exact height $x$ in centimetres is measured.
The observations are grouped in a cumulative frequency table as follows:
$\begin {array} {r|l|l} \hline
\text {Height $x$ (cm)} & \text {Count} & \text {Cumulative Count} \\
\hline
x < 165 & 6 & 6 \\
165 \le x < 170 & 39 & 45 \\
170 \le x < 175 & 93 & 138 \\
175 \le x < 180 & 44 & 182 \\
180 \le x < 185 & 15 & 197 \\
x \ge 185 & 3 & 200 \\
\hline \end {array}$
== Sources ==
* 
* 
Category:Examples of Grouped Data
Category:Examples of Frequency Tables
Frequency of Beats

== Theorem ==
Let $W_1$ and $W_2$ be harmonic waves whose frequencies are $f_1$ and $f_2$.
Let the superpositon of $W_1$ onto $W_2$ exhibit the phenomenon of beats.
The frequency $f_b$ of those beats is:
:$f_b = \size {f_2 - f_1}$
== Proof ==
Let $\omega_1$ and $\omega_2$ denote the angular frequency of $W_1$ and $W_2$ respectively.
Let us consider the harmonic waves that are $W_1$ and $W_2$ as they disturb the medium at $x = 0$.
, therefore, let $W_1$ and $W_2$ be be expressed as:
where:
Then:
Hence the superimposition of the two waves is equivalent to forming the product of the waves of frequency $\dfrac {\omega_1 + \omega_2} 2$ and $\dfrac {\size {\omega_1 - \omega_2} } 2$.
:
In the above diagram, there are $3$ waves superimposed:
:$(1) \quad \pm 2 \map \cos {\dfrac {\size {\omega_1 - \omega_2} t} 2}$ in red
:$(2) \quad 2 \map \sin {\dfrac {\paren {\omega_1 + \omega_2} t} 2}$ in green
:$(3) \quad 2 \map \sin {\dfrac {\paren {\omega_1 + \omega_2} t} 2} \map \cos {\dfrac {\size {\omega_1 - \omega_2} t} 2}$ in blue
Hence we see:
Hence the angular frequency $\omega_b$ of the slower component is seen to be:
:$\omega_b = \dfrac {\size {\omega_1 - \omega_2} } 2$
By definition of angular frequency:
Hence the frequency of the slower component is seen to be:
:$\dfrac {\size {f_1 - f_2} } 2$
Finally we note that the beats themselves occur at the peaks and troughs of the slower component.
That is, the frequency $f_b$ of those beats is twice $\dfrac {\size {f_1 - f_2} } 2$.
The result is apparent.
== Sources ==
* 
* 
Category:Beats
Frequency of Harmonic Wave

== Theorem ==
Let $\phi$ be a harmonic wave expressed as:
:$\forall x, t \in \R: \map \phi {x, t} = a \map \cos {\omega \paren {x - c t} }$
The frequency $\nu$ of $\phi$ can be expressed as:
:$\nu = \dfrac 1 \tau$
where $\tau$ is the period of $\phi$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text {I}$: The Equation of Wave Motion: $\S 3$: $(6)$
Category:Frequency of Harmonic Wave
Category:Frequency
Category:Harmonic Waves
Frequency of Harmonic Wave/Proof 2

== Theorem ==
== Proof ==
By definition, $\nu$ is the number of complete wavelengths of $\phi$ to pass an arbitrary point on the $x$-axis in unit time.
Let $x_0$ be that arbitrary point.
By definition, $\tau$ is the time taken for one complete wavelength of $\phi$ to pass $x_0$.
So after unit time, $\dfrac 1 \tau$ wavelengths of $\phi$ pass $x_0$.
Hence the result.
Category:Frequency of Harmonic Wave
Frequency of Periodic Wave

== Theorem ==
Let $\phi$ be a periodic wave expressed as:
:$\forall x, t \in \R: \map \phi {x, t} = \map f {x - c t}$
The frequency $\nu$ of $\phi$ can be expressed as:
:$\nu = \dfrac 1 \tau$
where $\tau$ is the period of $\phi$.
== Proof ==
By definition, $\nu$ is the number of complete wavelengths of $\phi$ to pass an arbitrary point on the $x$-axis in unit time.
Let $x_0$ be that arbitrary point.
By definition, $\tau$ is the time taken for one complete wavelength of $\phi$ to pass $x_0$.
So after unit time, $\dfrac 1 \tau$ wavelengths of $\phi$ pass $x_0$.
Hence the result.
== Examples ==
== Sources ==
* : Chapter $\text {I}$: The Equation of Wave Motion: $\S 3$
* 
* 
Category:Frequency of Periodic Wave
Category:Frequency
Category:Periodic Waves
Freshman's Dream

== Mistake ==
The freshman's dream is a name for the mistake:
:$\left({x + y}\right)^n = x^n + y^n$
where $n$ is a real number.
The correct result is given by the Binomial Theorem.
== Also known as ==
Some sources cite this as the child's binomial theorem.
== Also see ==
* Sophomore's Dream
Note the result given in Power of Sum Modulo Prime, which is:
:$\left({a + b}\right)^p \equiv a^p + b^p \pmod p$
where $p$ is a prime number.
This is also sometimes called the freshman's dream, but this one is in fact true.
Category:Fallacies and Mistakes
Category:Named Theorems
Fresnel Cosine Integral Function is Odd

== Theorem ==
:$\map {\operatorname C} {-x} = -\map {\operatorname C} x$
where: 
:$\operatorname C$ denotes the Fresnel cosine integral function
:$x$ is a real number.
== Proof ==
== Sources ==
* : $\S 35$: Fresnel Cosine Integral $\ds \map {\operatorname C} x = \sqrt {\frac 2 \pi} \int_0^x \cos u^2 \rd u$: $35.23$
Category:Fresnel Cosine Integral Function
Category:Examples of Odd Functions
Fresnel Cosine Integral Function of Zero

== Theorem ==
:$\map {\operatorname C} 0 = 0$
where $\operatorname C$ denotes the Fresnel cosine integral function.
== Proof ==
By Fresnel Cosine Integral Function is Odd, $\operatorname C$ is an odd function.
Therefore, by Odd Function of Zero is Zero:
:$\map {\operatorname C} 0 = 0$
== Sources ==
* : $\S 35$: Fresnel Cosine Integral $\ds \map {\operatorname C} x = \sqrt {\frac 2 \pi} \int_0^x \cos u^2 \rd u$: $35.23$
Category:Fresnel Cosine Integral Function
Fresnel Sine Integral Function is Odd

== Theorem ==
:$\map {\operatorname S} {-x} = -\map {\operatorname S} x$
where: 
:$\operatorname S$ denotes the Fresnel sine integral function
:$x$ is a real number.
== Proof ==
== Sources ==
* : $\S 35$: Fresnel Sine Integral $\ds \map {\operatorname S} x = \sqrt {\frac 2 \pi} \int_0^x \sin u^2 \rd u$: $35.20$
Category:Fresnel Sine Integral Function
Category:Examples of Odd Functions
Fresnel Sine Integral Function of Zero

== Theorem ==
:$\map {\operatorname S} 0 = 0$
where $\operatorname S$ denotes the Fresnel sine integral function.
== Proof ==
By Fresnel Sine Integral Function is Odd, $\operatorname S$ is an odd function.
Therefore, by Odd Function of Zero is Zero:
:$\map {\operatorname S} 0 = 0$
== Sources ==
* : $\S 35$: Fresnel Sine Integral $\ds \map {\operatorname S} x = \sqrt {\frac 2 \pi} \int_0^x \sin u^2 \rd u$: $35.20$
Category:Fresnel Sine Integral Function
Freyd-Mitchell Embedding Theorem

== Theorem ==
Let $\AA$ be a small abelian category.
Then there exists a ring with unity $R$ and a fully faithful and exact functor $F : \AA \to R \text{-} \mathbf{Mod}$ to the category of left $R$-modules.
== Proof ==
== Sources ==
* 
Category:Category Theory
Category:Homological Algebra
Friedrichs' Inequality

== Theorem ==
Let $G \subset \R^n$ be bounded domain.
Then for any $u \in \map {W^{1, 2}_0} G$:
:$\norm u_{\map {L^2} G} \le \map {\operatorname {diam} } G \norm {\nabla u}_{\map {L^2} G}$
where:
:$\map {\operatorname {diam} } G := \sup \limits_{x, y \mathop \in G} \size {x - y}$
== Proof ==
=== Smooth functions with compact support ===
Let $u \in \map {C_0^\infty} G$.
Put $\map u x = 0$ if $x = \tuple {x_1, x_2, \ldots, x_n} \notin G$.
Then:
:$u \in \map {C_0^\infty} {\R^n}$
Denote by $a,b$ the extremes of the last coordinate: $a = \inf \limits_{x \mathop \in G} x_n$ and $b = \sup \limits_{y \mathop \in G} x_n$
Then for any $x$:
:$\ds \size {\map u {x_1, x_2, \ldots, x_n} }^2 = \size {\int \limits_a^{x_n} \frac {\partial u} {\partial x_n} \tuple {x_1, x_2, \ldots, x_{n - 1}, t} \rd t}^2$
By Cauchy-Bunyakovsky-Schwarz Inequality:
Integrating this we get:
:$\ds \norm u_{\map {L^2} G} \le \map {\operatorname {diam} } G \int \limits_G \paren {\int \limits_a^b \size {\map {\nabla u} {x_1, \ldots, x_{m - 1}, t} }^2 \rd t} \rd x$
By Fubini's Theorem:
=== General case ===
Let now $u \in \map {W^{1, 2}_0} G$.
There exists a sequence $\ds \sequence {u_n}_{n \mathop = 1}^\infty \subset \map {C_0^\infty} G$ such that:
:$\norm {u - u_n}_{\map {W^{1, 2} } G} \to 0$
as $n \to \infty$.
Then:
:$\norm {u - u_n}_{\map {L^2} G} \to 0$
and:
:$\norm {\nabla u - \nabla u_n}_{\map {L^2} G} \to 0$
As $\size {\, \norm {u - u_n} \,} \le \norm {u - u_n}$, it follows that:
:$\norm {u_n}_{\map {L^2} G} \to \norm u_{\map {L^2} G}$
and:
:$\norm {\nabla u_n}_{\map {L^2} G} \to \norm {\nabla u}_{\map {L^2} G}$
Since the inequality is correct for all $u_n$, it is also correct for $u$.
Category:Sobolev Spaces
Friendship Theorem

== Theorem ==
Let there be a group of $6$ people.
The traditional setting is that these $6$ people are at a party.
Then (at least) one of the following $2$ statements is true:
:$(1): \quad$ At least $3$ of these $6$ people have all met each other before
:$(2): \quad$ At least $3$ of these $6$ people have never met each other before.
That is, either there exists a set of $3$ mutual acquaintances, or there exists a set of $3$ mutual strangers.
== Proof 1 ==
== Proof 2 ==
== Also known as ==
This theorem has many names and guises; has adopted the name the friendship theorem as it is short.
== Sources ==
* : $3$
Category:Ramsey Theory
Category:Classic Problems
Category:Graph Colorings
Category:Friendship Theorem
Category:Named Theorems
Friendship Theorem/Proof 2

== Theorem ==
== Proof ==
This can be expressed in graph theoretical language as:
Let the edges of $K_6$, the complete graph of $6$ vertices be edge-colored using $2$ colors.
Then there will be a triangle subgraph of $K_6$ whose edges all have the same color.
Let the colors be red and blue.
Choose any one vertex $P$.
There are $5$ edges incident to $P$.
They are each colored red or blue.
By the Pigeonhole Principle, at least $3$ of them must be of the same color.
, suppose that less than $3$ are red.
Then the others, $3$ or more, must be blue.
Let $A$, $B$, $C$ be the other vertices incident to these $3$ edges, all of which are blue.
There are $2$ cases:
:$(1): \quad$ Let any one of $AB$, $BC$ and $CA$ be blue.
Then that edge, together with the $2$ edges to $P$ forms a blue triangle.
:$(2): \quad$ Let none of $AB$, $BC$ and $CA$ be blue.
Then all three edges are red.
Thus $AB$, $BC$ and $CA$ form a red triangle.
Hence the result.
Category:Friendship Theorem
Frink's Metrization Theorem

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Then:
:$T$ is metrizable
:
:for all $s \in S$ there exists a countable neighborhood basis, denoted $\set {U_{s, n} : n \in \N}$, such that:
::$(1): \quad \forall s \in S, n \in \N : U_{s, n + 1} \subseteq U_{s, n}$
::$(2): \quad \forall s \in S, n \in \N : \exists m > n : \forall t \in S : U_{t, m} \cap U_{s ,m} \ne \O \implies U_{t, m} \subseteq U_{s, n}$
== Proof ==
== Sources ==
* : Chapter $7$: Metrizable Spaces: $\S23$: Metrization: Problem $23\text J$
Category:Metrization Theorems
Category:Neighborhood Bases
Category:Metrizable Topologies
Frobenius's Theorem

== Theorem ==
An algebraic associative real division algebra $A$ is isomorphic to $\R, \C$ or $\Bbb H$.
== Proof ==
Recall that an algebra $A$ is said to be quadratic if it is unital and the set $\set {1, x, x^2}$ is linearly dependent for every $x \in A$.
Thus, for every $x \in A$ there exist $\map t x, \map n x \in \R$ such that:
:$x^2 - \map t x x + \map n x = 0$
Obviously, $\map t x$ and $\map n x$ are uniquely determined if $x \notin \R$.
Suppose $x \notin \R$.
Then $x$ can be expressed as $x = a + b i$, with $a, b \in \R$ and $b \ne 0$.
Then:
:$x^2 = a^2 - b^2 + 2 a b i$
and:
:$x^2 - \map t x x + \map n x = a^2 - b^2 - a \map t x + \map n x + \paren {2 a b - b \map t x} i = 0$
It follows that both:
:$(1): \quad 2 a b - b \map t x = 0$
and:
:$(2): \quad a^2 - b^2 - a \map t x + \map n x = 0$
$(1)$ leads to:
:$\map t x = 2 a$
and $(2)$ leads to;
:$\map n x = a^2 + b^2$
Setting $\map t \lambda = 2 \lambda$ and $\map n \lambda = \lambda^2$ for $\lambda \in \R$, we can then consider $t$ and $n$ as maps from $A$ into $\R$.
(In this way $t$ becomes a linear functional).
We call $\map t x$ and $\map n x$ the trace and the norm of $x$ respectively.
From $x^2 - \paren {x + x^*} x + x^* x = 0$ we see that all algebras $\Bbb A_n$ are quadratic.
Further, every real division algebra $A$ that is algebraic and power-associative (this means that every subalgebra generated by one element is associative) is automatically quadratic.
Indeed, if $x \in A$ then there exists a nonzero polynomial $\map f X \in \R \sqbrk X$ such that $\map f x = 0$.
Writing $\map f X$ as the product of linear and quadratic polynomials in $\R \sqbrk X$ it follows that $\map p x = 0$ for some $\map p X \in \R \sqbrk X$ of degree $1$ or $2$.
In particular, algebraic alternative (and hence associative) real division algebras are quadratic.
Finally, if $A$ is a real unital algebra, that is, an algebra over $\R$ with unity $1$, then we shall follow a standard convention and identify $\R$ with $\R 1$.
Thus we shall write $\lambda$ for $\lambda 1$, where $\lambda \in \R$.
=== Lemma 1 ===
=== Lemma 2 ===
=== Lemma 3 ===
=== Lemma 4 ===
=== Lemma 5 ===
We have from above that $A$ is quadratic.
We may assume that $n = \dim A \ge 2$.
By Lemma 4 we can fix $i \in A$ such that $i^2 = -1$.
Thus, $A \cong \C$ if $n = 2$.
Let $n > 2$.
By Lemma 5:
:$\exists j \in A: j^2 = -1, i j = -j i$
Set $k = ij$.
It can immediately be checked that:
:$k^2 = -1$
:$k i = j = -i k$
:$j k = i = -k j$
:$\set {i, j, k}$ is a linearly independent set.
Therefore $A$ contains a subalgebra isomorphic to $\Bbb H$.
Finally, suppose $n > 4$.
By Lemma 5 there would exist $e \in A, e \ne 0$ such that:
:$(a): \quad e i = -i e$
:$(b): \quad e j = -j e$
:$(c): \quad e k = -k e$
However, from $(a)$ and $(b)$ it follows that $e i j = -i e j = i j e$.
Since $i j = k$, this contradicts $(c)$.
It follows that $n \le 4$, and so $\Bbb H$ is the highest order 
== Historical Note ==
== Sources ==
* 
* : Chapter $\text {B}.26$: Extensions of the Complex Number System. Algebras, Quaternions, and Lagrange's Four Squares Theorem
* 
* 
* 
Category:Frobenius's Theorem
Category:Abstract Algebra