nyuuzyou/wikis huggingface dataset:
Argument Principle

== Theorem == 
Let $\gamma$ be a closed contour.
Let $D$ be the region enclosed by $\gamma$. 
Let $f$ be a function meromorphic on $D$. 
Let $f$ be holomorphic with no zeroes on $\gamma$.
Let $N$ denote the number of zeroes of $f$ in $D$, counted up to multiplicity.
Let $P$ denote the number of poles of $f$ in $D$, counted up to order.
Then:
 
:$\ds N - P = \frac 1 {2\pi i} \oint_\gamma \frac {\map {f'} z} {\map f z} \rd z$
== Proof == 
Let $n_1, n_2, n_3 \ldots n_N$ denote the zeroes of $f$, and $p_1, p_2, p_3 \ldots p_P$ denote its poles. 
Then, there exists a non-zero holomorphic function $g$ such that:
:$\ds \map f x = \frac {\prod_{k \mathop = 1}^N \paren {z - n_k} } {\prod_{j \mathop = 1}^P \paren {z - p_j} } \map g z$
Taking the logarithmic derivative:
Then: 
By our construction of $g$, $\frac {g'} g$ is holomorphic on $D$ so, by the Cauchy Integral Theorem, the integral is equal to zero.
We therefore have:
:$\ds \oint_\gamma \frac {\map {f'} z} {\map f z} \rd z = 2\pi i \paren {N - P}$
Hence:
:$\ds N - P = \frac 1 {2\pi i} \oint_\gamma \frac {\map {f'} z} {\map f z} \rd z$
Category:Complex Analysis
Argument of Complex Conjugate equals Argument of Reciprocal

== Theorem ==
Let $z \in \C$ be a complex number.
Then:
:$\arg {\overline z} = \arg \dfrac 1 z$
where:
:$\arg$ denotes the argument of a complex number
:$\overline z$ denotes the complex conjugate of $z$.
== Proof ==
Let $z$ be expressed in polar form:
:$z := r \paren {\cos \theta + i \sin \theta}$
Then:
The result follows by definition of the argument of a complex number
== Also see ==
* Argument of Complex Conjugate equals Negative of Argument
== Sources ==
* : $\S 2$. Geometrical Representations
Category:Complex Conjugates
Argument of Complex Conjugate equals Negative of Argument

== Theorem ==
Let $z \in \C$ be a complex number.
Then:
:$\arg {\overline z} = -\arg z$
where:
:$\arg$ denotes the argument of a complex number
:$\overline z$ denotes the complex conjugate of $z$.
== Proof ==
Let $z$ be expressed in polar form:
:$z := r \paren {\cos \theta + i \sin \theta}$
Then:
The result follows by definition of the argument of a complex number
== Also see ==
* Argument of Complex Conjugate equals Argument of Reciprocal
== Sources ==
* : $\S 2$. Geometrical Representations
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Complex Conjugate of $z$: $3.7.9$
Category:Complex Conjugates
Argument of Complex Number/Examples

== Examples of Argument of Complex Number ==
=== Example: $\arg 3$ ===
=== Example: $\map \arg {-3}$ ===
=== Example: $\map \arg {1 + i}$ ===
=== Example: $\map \arg {-1 - i}$ ===
=== Example: $\map \arg {2 i}$ ===
=== Example: $\map \arg {-i}$ ===
Category:Examples of Arguments of Complex Numbers
Argument of Complex Number/Examples/-1-i

== Example of Argument of Complex Number ==
:$\map \arg {-1 - i} = -\dfrac {3 \pi} 4$
== Proof ==
We have that:
Hence:
Hence:
:$\map \arg {-1 - i} = -\dfrac {3 \pi} 4$
== Sources ==
* : $\S 2$. Geometrical Representations: Example $\text{(iv)}$
Category:Examples of Arguments of Complex Numbers
Argument of Complex Number/Examples/-1-i/Mistake

== Source Work ==
:
:Chapter $2$: Geometrical Representations
::Example $\text{(iv)}$
This mistake can be seen in the $1960$ edition as published by Routledge & Kegan Paul.
== Mistake ==
:$\map \arg {-1 - i} = -\dfrac \pi 4$
The correct result is:
:$\map \arg {-1 - i} = -\dfrac {3 \pi} 4$
== Sources ==
* : $\S 2$. Geometrical Representations: Example $\text{(iv)}$
Category:Mistakes/Complex Numbers/Walter Ledermann
Argument of Complex Number/Examples/-3

== Example of Argument of Complex Number ==
:$\map \arg {-3} = \pi$
== Proof ==
We have that:
:$-3 = -3 + 0 i$
and so:
Hence:
Hence:
:$\map \arg {-3} = \pi$
== Sources ==
* : $\S 2$. Geometrical Representations: Example $\text{(ii)}$
Category:Examples of Arguments of Complex Numbers
Argument of Complex Number/Examples/-i

== Example of Argument of Complex Number ==
:$\map \arg {-i} = -\dfrac \pi 2$
== Proof ==
We have that:
Hence:
Hence:
:$\map \arg {-i} = -\dfrac \pi 2$
== Sources ==
* : $\S 2$. Geometrical Representations: Example $\text{(vi)}$
Category:Examples of Arguments of Complex Numbers
Argument of Complex Number/Examples/1+i

== Example of Argument of Complex Number ==
:$\map \arg {1 + i} = \dfrac \pi 4$
== Proof ==
We have that:
Hence:
Hence:
:$\map \arg {1 + i} = \dfrac \pi 4$
== Sources ==
* : $\S 2$. Geometrical Representations: Example $\text{(iii)}$
Category:Examples of Arguments of Complex Numbers
Argument of Complex Number/Examples/2i

== Example of Argument of Complex Number ==
:$\map \arg {2 i} = \dfrac \pi 2$
== Proof ==
We have that:
Hence:
Hence:
:$\map \arg {2 i} = \dfrac \pi 2$
== Sources ==
* : $\S 2$. Geometrical Representations: Example $\text{(v)}$
Category:Examples of Arguments of Complex Numbers
Argument of Complex Number/Examples/3

== Example of Argument of Complex Number ==
:$\arg 3 = 0$
== Proof ==
We have that:
:$3 = 3 + 0 i$
and so:
Hence:
Hence:
:$\map \arg 3 = 0$
== Sources ==
* : $\S 2$. Geometrical Representations: Example $\text{(i)}$
Category:Examples of Arguments of Complex Numbers
Argument of Exponential is Imaginary Part plus Multiple of 2 Pi

== Theorem ==
Let $z \in \C$ be a complex number.
Let $\exp z$ denote the complex exponential of $z$.
Let $\arg z$ denote the argument of $z$.
Then:
:$\map \arg {\exp z} = \set {\Im z + 2 k \pi: k \in \Z}$
where $\Im z$ denotes the imaginary part of $z$.
== Proof ==
Let $z = x + i y$.
Let $\theta \in \map \arg {\exp z}$.
We have:
== Sources ==
* : $\S 4.5$. The Functions $e^z$, $\cos z$, $\sin z$
Category:Argument of Complex Number
Category:Exponential Function
Argument of Negative Real Number is Pi

== Theorem ==
Let $x \in \R_{>0}$ be a positive real number.
Then:
:$\arg \paren {-x} = \pi$
where $\arg$ denotes the argument of a complex number.
== Proof ==
We have that:
:$-x = -x + 0 i$
and so:
Hence:
Hence:
: $\arg \paren {-x} = \pi$
Category:Examples of Arguments of Complex Numbers
Argument of Product equals Sum of Arguments

== Theorem ==
Let $z_1, z_2 \in \C$ be complex numbers.
Let $\arg$ be the argument operator.
Then:
:$\map \arg {z_1 z_2} = \map \arg {z_1} + \map \arg {z_2} + 2 k \pi$
where $k$ can be $0$, $1$ or $-1$.
== Proof ==
Let $\theta_1 = \map \arg {z_1}, \theta_2 = \map \arg {z_2}$.
Then the polar forms of $z_1, z_2$ are:
By the definition of complex multiplication, factoring $\cmod {z_1} \cmod {z_2}$ from all terms, we have:
:$z_1 z_2 = \cmod {z_1} \cmod {z_2} \paren {\paren {\cos \theta_1 \cos \theta_2 - \sin \theta_1 \sin \theta_2} + i \paren {\cos \theta_1 \sin \theta_2 + \sin \theta_1 \cos \theta_2} }$
Using Sine of Sum and Cosine of Sum, we have:
:$z_1 z_2 = \cmod {z_1} \cmod {z_2} \paren {\map \cos {\theta_1 + \theta_2} + i \, \map \sin {\theta_1 + \theta_2} }$
The theorem follows from the definition of $\map \arg z$, which says that $\map \arg {z_1 z_2}$ satisfies the equations:
which in turn means that:
There are $3$ possibilities for the size of $\theta_1 + \theta_2$:
:$(1): \quad \theta_1 + \theta_2 > \pi$
Then:
:$-\pi < \theta_1 + \theta_2 - 2 \pi \le \pi$
and we have:
and so $\theta_1 + \theta_2 - 2 \pi$ is the argument of $z_1 z_2$ within its principal range.
:$(2): \quad \theta_1 + \theta_2 \le -\pi$
Then:
:$-\pi < \theta_1 + \theta_2 + 2 \pi \le \pi$
and we have:
and so $\theta_1 + \theta_2 + 2 \pi$ is within the principal range of $z_1 z_2$.
:$(3): \quad -\pi < \theta_1 + \theta_2 \le \pi$
Then $\theta_1 + \theta_2$ is already within the principal range of $z_1 z_2$.
Therefore:
:$\map \arg {z_1 z_2} = \theta_1 + \theta_2 = \map \arg {z_1} + \map \arg {z_2} + 2 k \pi$
where $k$ can be $0$, $1$ or $-1$.
== Sources ==
* : $\S 2$. Geometrical Representations: $(2.13)$
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Multiplication and Division: $3.7.12$
* : $1$: Complex Numbers: Solved Problems: De Moivre's Theorem: $27 \ \text{(a)}$
Category:Argument of Complex Number
Category:Complex Multiplication
Argument of Quotient equals Difference of Arguments

== Theorem ==
Let $z_1$ and $z_2$ be complex numbers.
Then:
:$\map \arg {\dfrac {z_1} {z_2} } = \map \arg {z_1} - \map \arg {z_1} + 2 k \pi$
where:
:$\arg$ denotes the argument of a complex number
:$k$ can be $0$, $1$ or $-1$.
== Proof ==
Let $z_1$ and $z_2$ be expressed in polar form.
:$z_1 = \polar {r_1, \theta_1}$
:$z_2 = \polar {r_2, \theta_2}$
From Division of Complex Numbers in Polar Form:
:$\dfrac {z_1} {z_2} = \dfrac {r_1} {r_2} \paren {\map \cos {\theta_1 - \theta_2} + i \, \map \sin {\theta_1 - \theta_2} }$
By the definition of argument:
:$\map \arg {z_1} = \theta_1$
:$\map \arg {z_2} = \theta_2$
:$\map \arg {\dfrac {z_1} {z_2} } = \theta_1 - \theta_2$
There are $3$ possibilities for the size of $\theta_1 + \theta_2$:
:$(1): \quad \theta_1 - \theta_2 > \pi$
Then:
:$-\pi < \theta_1 - \theta_2 - 2 \pi \le \pi$
and we have:
and so $\theta_1 + \theta_2 - 2 \pi$ is the argument of $\dfrac {z_1} {z_2}$ within its principal range.
:$(2): \quad \theta_1 - \theta_2 \le -\pi$
Then:
:$-\pi < \theta_1 - \theta_2 + 2 \pi \le \pi$
and we have:
and so $\theta_1 - \theta_2 + 2 \pi$ is within the principal range of $\dfrac {z_1} {z_2}$.
:$(3): \quad -\pi < \theta_1 + \theta_2 \le \pi$
Then $\theta_1 - \theta_2$ is already within the principal range of $\dfrac {z_1} {z_2}$.
== Sources ==
* : $\S 2$. Geometrical Representations
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Multiplication and Division: $3.7.15$
* : $1$: Complex Numbers: Solved Problems: De Moivre's Theorem: $27 \ \text{(b)}$
Category:Complex Division
Category:Argument of Complex Number
Argument of x to the n Equals n Times The Argument

== Theorem ==
Let $z$ be a complex number.
Then:
:$\forall n \in \N_{>0}: \map \arg {z^n} = n \map \arg z$
== Proof ==
For $n = 1$
:$\map \arg {z^1} = 1 \cdot \map \arg z$
Assuming the result is true for $n = k$, we have:
Argumentum ad Hominem/Examples

== Examples of Argumentum ad Hominem ==
Examples of argumentum ad hominem:
:"I'm not even going to bother to read that proof you just posted up - if it's by you then it's bound to be rubbish."
:" Don't believe everything you read in there, those reporters are all crackpots."
:"No butcher can be expected to argue honestly about the meat trade."
All of the above may or may not be true statements, but all are ultimately fallacious.
== Sources ==
* 
Category:Examples of Argumentum ad Hominem
Arithmetic Average of Second Chebyshev Function/Lemma 1

== Lemma ==
Let $x \ge 1$ be a real number.
Then: 
:$\paren {x + 1} \map \ln {x + 1} - x \ln x \le 2 \map \ln {x + 1}$
== Proof ==
Define a function $f : \hointr 1 \infty \to \R$ by: 
:$\map f x = \paren {x + 1} \map \ln {x + 1} - x \ln x - 2 \map \ln {x + 1}$
for each $x \in \hointr 1 \infty$.
Note that: 
:$\map f 1 = 2 \ln 2 - 1 \ln 1 - 2 \ln 2 = 0$
So it suffices to show that $f$ is decreasing for $x \ge 1$, then we will have $\map f x < 0$ for $x \ge 1$.
Note that $f$ is differentiable and: 
We can now see that $\map {f'} x \le 0$ for $x \ge 1$. 
So, from Real Function with Negative Derivative is Decreasing:
:$f$ is decreasing.
So, for $x \ge 1$, we have: 
:$\paren {x + 1} \map \ln {x + 1} - x \ln x - 2 \map \ln {x + 1} \le \map f 1 = 0$
That is: 
:$\paren {x + 1} \map \ln {x + 1} - x \ln x \le 2 \map \ln {x + 1}$
Category:Arithmetic Average of Second Chebyshev Function
Arithmetic Average of Second Chebyshev Function/Lemma 2

== Lemma ==
Let $x \ge 3$ be a real number.
Then: 
:$-2 \map \ln {x + 1} \le \paren {x - 1} \map \ln {x - 1} - x \ln x$
== Proof ==
Define a function $g : \openint 1 \infty \to \R$ by:
:$\map g x = \paren {x - 1} \map \ln {x - 1} - x \ln x + 2 \map \ln {x + 1}$
We have that $g$ is differentiable with: 
We can see that $\map {g'} x \ge 0$ for $x \ge 3$. 
So from Real Function with Positive Derivative is Increasing:
:$g$ is increasing. 
So for $x \ge 3$ we have: 
so for $x \ge 3$ we have: 
Category:Arithmetic Average of Second Chebyshev Function
Arithmetic Mean is Weighted Mean with Equal Weights

== Theorem ==
Let $S = \sequence {x_1, x_2, \ldots, x_n}$ be a set of real numbers.
Let $W$ be a weight function to be applied to the terms of $S$ such that:
:$\forall x \in S: \map W x = c$
where $c$ is a real constant.
Let $\bar x$ denote the weighted mean of $S$ $W$.
Then:
:$\bar x = A_n$
where $A_n$ denotes the arithmetic mean of $S$.
== Proof ==
Category:Weighted Means
Category:Arithmetic Mean
Arithmetic Mean of two Real Numbers is Between them

== Theorem ==
Let $a, b \in \R_{\ne 0}$ be real numbers such that $a < b$.
Let $\map A {a, b}$ denote the arithmetic mean of $a$ and $b$.
Then:
:$a < \map A {a, b} < b$
== Proof ==
By definition of arithmetic mean:
:$\map A {a, b} := \dfrac {a + b} 2$
Thus:
and:
Hence the result.
Category:Arithmetic Mean
Arithmetic Sequence of 16 Primes

== Theorem ==
The $16$ integers in arithmetic sequence defined as:
:$2\,236\,133\,941 + 223\,092\,870 n$
are prime for $n = 0, 1, \ldots, 15$.
== Proof ==
First we note that:
:$2\,236\,133\,941 - 223\,092\,870 = 2\,013\,041\,071 = 53 \times 89 \times 426\,763$
and so this arithmetic sequence of primes does not extend to $n < 0$.
But note that $2\,236\,133\,941 + 16 \times 223\,092\,870 = 5\,805\,619\,861 = 79 \times 73\,488\,859$ and so is not prime.
== Sources ==
* : $2,236,133,941$
Category:Prime Numbers
Category:Arithmetic Sequences
Arithmetic Sequence of 4 Terms with 3 Distinct Prime Factors

== Theorem ==
The arithmetic sequence:
:$30, 66, 102, 138$
is the smallest of $4$ terms which consists entirely of positive integers each with $3$ distinct prime factors.
== Proof ==
We demonstrate that this is indeed an arithmetic sequence:
demonstrating the common difference of $36$.
Then we note:
It is understood that smallest arithmetic sequence means:
:integer sequence of length $4$ whose terms are in arithmetic sequence such that their sum is the smallest of all such sequences.
To demonstrate that this sequence is indeed the smallest, according to this definition, we list all positive integers less than $138$, each with $3$ distinct prime factors.
They are:
One cannot pick any other $4$ numbers from the list to form an arithmetic sequence.
Hence any further such arithmetic sequence will contain larger terms whose sum will be consequently larger.
== Sources ==
* : $30$
Category:Arithmetic Sequences
Arithmetic iff Way Below Relation is Multiplicative in Algebraic Lattice

== Theorem ==
Let $L = \struct {S, \vee, \wedge, \preceq}$ be a bounded below algebraic lattice.
Then:
:$L$ is arithmetic
:
:$\ll$ is a multiplicative relation
where $\ll$ denotes the way below relation of $L$.
== Proof ==
=== Sufficient Condition ===
Let $L$ be arithmetic.
Let $a, x, y \in S$ such that
:$a \ll x$ and $a \ll y$
By Algebraic iff Continuous and For Every Way Below Exists Compact Between:
:$\exists c \in \map K L: a \preceq c \preceq x$
and
:$\exists k \in \map K L: a \preceq k \preceq y$
where $\map K L$ denotes the compact subset of $L$.
By Meet Semilattice is Ordered Structure:
:$c \wedge k \preceq x \wedge y$
By definition of arithmetic:
:$\map K L$ is meet closed.
By definition of meet closed:
:$c \wedge k \in \map K L$
By definition of compact subset:
:$c \wedge k$ is compact.
By definition of compact:
:$c \wedge k \ll c \wedge k$
By Meet is Idempotent:
:$a \wedge a = a$
By Meet Semilattice is Ordered Structure:
:$a \preceq c \wedge k$
Thus by Preceding and Way Below implies Way Below
:$a \ll x \wedge y$
=== Necessary Condition ===
Let $\ll$ be a multiplicative relation.
Thus $L$ is algebraic.
It remains to prove that
:$\map K L$ is meet closed.
Let $x, y \in \map K L$.
By definition of compact subset:
:$x$ and $y$ are compact.
By definition of compact:
:$x \ll x$ and $y \ll y$
By Way Below Relation is Auxiliary Relation:
:$\ll$ is auxiliary relation.
By Multiplicative Auxiliary Relation iff Congruent:
:$x \wedge y \ll x \wedge y$
By definition:
:$x \wedge y$ is compact.
Thus by definition of compact subset:
:$x \wedge y \in \map K L$
== Sources ==
* 
* 
Category:Way Below Relation
Category:Continuous Lattices
Arnauld's Paradox

== Paradox ==
Suppose negative numbers exist.
Then:
:$\dfrac {-1} 1 = \dfrac 1 {-1}$
Thus the ratio of a smaller quantity to a larger quantity equals the ratio of the same larger quantity to the same smaller quantity.
== Resolution ==
== Sources ==
* : $-1$ and $i$
* : $-1$ and $i$
Category:Paradoxes
Arrange 3 Knives to Rest on Tips of Handles

== Problem ==
How can you arrange $3$ knives to balance while supported only by their handles?
=== Variant ===
== Solution ==
:
== Historical Note ==
== Sources ==
* 
* : Henry van Etten: $112$
Category:Physics
Category:Classic Problems
Category:Arrange 3 Knives to Rest on Tips of Handles
Arrange 3 Knives to Rest on Tips of Handles/Historical Note

== Historical Note on Arrange 3 Knives to Rest on Tips of Handles ==
This problem appears for the first time in of $1624$ by .
In its English translation, it is worded to arrange three knives so that they:
:hang in the air without being supported by anything but themselves.
== Sources ==
* : Henry van Etten: $112$
Category:Historical Notes
Arrange 3 Knives to Rest on Tips of Handles/Variant

== Problem ==
Arrange $3$ knives so they can support a drinking glass, while in the space between $3$ other glasses placed on the table, another glass can be placed underneath that first glass.
== Solution ==
From the original solution (for which the author of this page had to use beer bottles instead of his wife's precious glassware):
:
It can be seen that a bottle can be balanced on the platform formed by the three knife blades, if sufficient care is taken to ensure symmetry.
It is also noted that the three blades are higher off the surface of the table than the tops of the three supporting bottles, and so a further bottle can easily be placed underneath them:
:
== Historical Note ==
== Sources ==
* : Henry van Etten: $112$
Category:Arrange 3 Knives to Rest on Tips of Handles
Arrange 3 Knives to Rest on Tips of Handles/Variant/Historical Note

== Historical Note on Arrange 3 Knives to Rest on Tips of Handles: Variant ==
This problem is a slightly more challenging variant on the original Arrange 3 Knives to Rest on Tips of Handles as published by in his $1624$ work , and dates from Victorian times.
== Sources ==
* : Henry van Etten: $112$
Category:Historical Notes
Arrange 3 Knives to Rest on Tips of Handles/Variant/Mistake

== Source Work ==
: 
:The Puzzles:
::Henry van Etten
:::$112$
== Mistake ==
:Variants in Victorian puzzle books demanded how three knives might be used to support a drinking glass, in the ample space between three other drinking glasses placed on the table with more than enough space for a fourth glass to be placed on the table between them.
== Correction ==
If you count carefully:
:one glass to be balanced
:three glasses forming the ample space
you notice that the glass to be placed on the table between them is actually the fifth glass.
== Sources ==
* : Henry van Etten: $112$
Category:Mistakes/Curious and Interesting Puzzles
Array/Examples/Arbitrary 1

== Example of Array ==
An example of a rectangular array:
:$\begin {bmatrix} \alpha_{1 1} & \alpha_{1 2} & \cdots & \alpha_{1 n} \\
\alpha_{2 1} & \alpha_{2 2} & \cdots & \alpha_{2 n} \\
\vdots & \vdots & \ddots & \vdots \\
\alpha_{m 1} & \alpha_{m 2} & \cdots & \alpha_{m n} \end {bmatrix}$
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices
Category:Arrays
Arrow Paradox

== Paradox ==
At any particular instant, a moving arrow is either at rest or in motion.
If the instant is indivisible, the arrow cannot move, because if it did, the instant would immediately be divided.
But time is made up of instants.
As the arrow cannot move in any one instant, it cannot move in time.
Hence the arrow remains at rest.
== Resolution ==
This is an example of how an incorrect model of reality (time is made up of instants) leads to incorrect predictions (movement is impossible).
== Historical Note ==
== Sources ==
* : Chapter $\text{II}$: Modern Minds in Ancient Bodies
* 
* 
* 
Category:Zeno's Paradoxes
Category:Antinomies
Art Gallery Problem

== Classic Problem ==
Consider an whose shape is a (simple) polygon.
The Art Gallery Problem asks:
:What is the minimum number of security guards needed, positioned on the perimeter of the , so that every point within the can be seen by at least one of the guards?
That is, for a given (simple) polygon $P$:
:What is the smallest set $S$ of points on the perimeter of $P$ such that for every point $p \in P$ there exists some $q \in S$ such that there exists a straight line segment $p q$ lying wholly inside $P$?
Category:Classic Problems
Category:Plane Geometry
Artin's Theorem on Alternative Algebras

== Theorem ==
Let $A = \struct {A_R, \oplus}$ be an algebra over the ring $R$ such that $A$ is not a boolean algebra.
Then $A$ is alternative :
:$\forall a, b \in A: \paren {a \oplus a} \oplus b = a \oplus \paren {a \oplus b}$
:$\forall a, b \in A: \paren {b \oplus a} \oplus a = b \oplus \paren {a \oplus a}$
== Alternative Definition ==
Some sources take this as a definition of an alternative algebra and from it deduce that $A$ is alternative :
:for all $a, b \in A_R$, the subalgebra generated by $\set {a, b}$ is an associative algebra.
However, the latter statement is how an alternative algebra is defined on this site, and from it we deduce the statements given in the exposition of this theorem.
The approaches are clearly equivalent.
== Proof ==
When $A$ is 
So suppose that:
:$\forall a, b \in A: \paren {a \oplus a} \oplus b = a \oplus \paren {a \oplus b}$
:$\forall a, b \in A: \paren {b \oplus a} \oplus a = b \oplus \paren {a \oplus a}$
Then:
:$\sqbrk {a, a, b} = 0$
:$\sqbrk {b, a, a} = 0$
where $\sqbrk {a, a, b}$ denotes the associator of $a, b \in A_R$.
Now let us compute, using the linearity of $\oplus$ and the two suppositions:
== Sources ==
* : The Octonions (2002): 1.1 Preliminaries
* 
Category:Alternative Algebras
Associated Legendre Function of the First Kind/Examples

== Examples of Associated Legendre Functions of the First Kind ==
=== Example: ${P_n}^0$ ===
=== Example: ${P_n}^m$ where $m > n$ ===
=== Example: ${P_1}^1$ ===
=== Example: ${P_2}^1$ ===
=== Example: ${P_2}^2$ ===
=== Example: ${P_3}^1$ ===
=== Example: ${P_3}^2$ ===
=== Example: ${P_3}^3$ ===
Category:Examples of Associated Legendre Functions of the First Kind
Associated Legendre Function of the First Kind/Examples/1, 1

== Example of Associated Legendre Function of the First Kind ==
Let $\map { {P_n}^m} x$ denote an associated Legendre function of the first kind.
Then:
:$\map { {P_1}^1} x = \paren {1 - x^2}^{1/2} = \sqrt {1 - x^2}$
== Proof ==
Then we have:
Hence:
== Sources ==
* : $\S 26$: Associated Legendre Functions: Associated Legendre Functions of the First Kind: $26.5$
* : $\S 28$: Legendre and Associated Legendre Functions: Associated Legendre Functions of the First Kind: $28.53.$
Category:Examples of Associated Legendre Functions of the First Kind
Associated Legendre Function of the First Kind/Examples/2, 1

== Example of Associated Legendre Function of the First Kind ==
Let $\map { {P_n}^m} x$ denote an associated Legendre function of the first kind.
Then:
:$\map { {P_2}^1} x = 3 x \paren {1 - x^2}^{1/2} = 3 x \sqrt {1 - x^2}$
== Proof ==
== Sources ==
* : $\S 26$: Associated Legendre Functions: Associated Legendre Functions of the First Kind: $26.6$
* : $\S 28$: Legendre and Associated Legendre Functions: Associated Legendre Functions of the First Kind: $28.54.$
Category:Examples of Associated Legendre Functions of the First Kind
Associated Legendre Function of the First Kind/Examples/2, 2

== Example of Associated Legendre Function of the First Kind ==
Let $\map { {P_n}^m} x$ denote an associated Legendre function of the first kind.
Then:
:$\map { {P_2}^2} x = 3 \paren {1 - x^2}$
== Proof ==
== Sources ==
* : $\S 26$: Associated Legendre Functions: Associated Legendre Functions of the First Kind: $26.7$
* : $\S 28$: Legendre and Associated Legendre Functions: Associated Legendre Functions of the First Kind: $28.55.$
Category:Examples of Associated Legendre Functions of the First Kind