nyuuzyou/wikis huggingface dataset:
Associated Legendre Function of the First Kind/Examples/3, 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:
:$\map { {P_3}^1} x = \dfrac 3 2 \paren {5 x^2 - 1} \paren {1 - x^2}^{1 / 2}$
== Proof ==
== Sources ==
* : $\S 26$: Associated Legendre Functions: Associated Legendre Functions of the First Kind: $26.8$
* : $\S 28$: Legendre and Associated Legendre Functions: Associated Legendre Functions of the First Kind: $28.56.$
Category:Examples of Associated Legendre Functions of the First Kind
Associated Legendre Function of the First Kind/Examples/3, 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:
:$\map { {P_3}^2} x = 15 x \paren {1 - x^2}$
== Proof ==
== Sources ==
* : $\S 26$: Associated Legendre Functions: Associated Legendre Functions of the First Kind: $26.9$
* : $\S 28$: Legendre and Associated Legendre Functions: Associated Legendre Functions of the First Kind: $28.57.$
Category:Examples of Associated Legendre Functions of the First Kind
Associated Legendre Function of the First Kind/Examples/3, 3

== 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:
:$\map { {P_3}^3} x = 15 \paren {1 - x^2}^{3 / 2}$
== Proof ==
== Sources ==
* : $\S 26$: Associated Legendre Functions: Associated Legendre Functions of the First Kind: $26.10$
* : $\S 28$: Legendre and Associated Legendre Functions: Associated Legendre Functions of the First Kind: $28.58.$
Category:Examples of Associated Legendre Functions of the First Kind
Associated Legendre Function of the First Kind/Examples/m gt n

== Example of Associated Legendre Function of the First Kind ==
Let $\map { {P_n}^m} x$ denote an associated Legendre function of the first kind.
Let $m > n$.
Then:
:$\map { {P_n}^m} x = 0$
== Proof ==
From Derivative Form of Associated Legendre Function of the First Kind:
:$\map { {P_n}^m} x = \dfrac {\paren {1 - x^2}^{m / 2} } {2^n n!} \dfrac {\d^{m + n} } {\d x^{m + n} } \paren {x^2 - 1}^n$
We have that:
:$\paren {x^2 - 1}^n = x^{2 n} + \map Q x$
where $\map Q x$ is a polynomial of degree $2 n - 2$.
Next we note that $m > n \implies m + n > 2 n > 2 n - 2$.
Hence:
Hence:
:$\dfrac {\paren {1 - x^2}^{m / 2} } {2^n n!} \dfrac {\d^{m + n} } {\d x^{m + n} } \paren {x^2 - 1}^n = 0$
and the result follows.
== Sources ==
* : $\S 26$: Associated Legendre Functions: Associated Legendre Functions of the First Kind: $26.4$
* : $\S 28$: Legendre and Associated Legendre Functions: Associated Legendre Functions of the First Kind: $28.52.$
Category:Examples of Associated Legendre Functions of the First Kind
Associated Legendre Function of the First Kind/Examples/n, 0

== Example of Associated Legendre Function of the First Kind ==
:$\map { {P_n}^0} x = \map {P_n} x$
where:
:$\map { {P_n}^0} x$ denotes the associated Legendre function of the first kind
:$\map {P_n} x$ denotes the Legendre polynomial.
== Proof ==
From the definition of the associated Legendre function of the first kind:
:$\map { {P_n}^m} x = \paren {1 - x^2}^{m / 2} \dfrac {\d^m} {\d x^m} \map {P_n} x$
where $\map {P_n} x$ is the Legendre polynomial of order $n$.
When $m = 0$ we have:
== Sources ==
* : $\S 26$: Associated Legendre Functions: Associated Legendre Functions of the First Kind: $26.3$
* : $\S 28$: Legendre and Associated Legendre Functions: Associated Legendre Functions of the First Kind: $28.51.$
Category:Examples of Associated Legendre Functions of the First Kind
Associatehood is Equivalence Relation

== Theorem ==
Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$ and whose unity is $1_D$.
Let $\sim$ be the relation defined on $D$ as:
:$\forall x, y \in D: x \sim y$ $x$ is an associate of $y$
Then $\sim$ is an equivalence relation.
== Proof ==
Checking in turn each of the criteria for equivalence:
=== Reflexive ===
Clearly $x \divides x$ as $x = 1_D \circ x$, so $x \sim x$.
Thus $\sim$ is reflexive.
=== Symmetric ===
By the definition:
:$x \sim y \iff x \divides y \land y \divides x \iff y \sim x$
Thus $\sim$ is symmetric.
=== Transitive ===
Thus $\sim$ is transitive.
All criteria are fulfilled for $\sim$ to be an equivalence relation.
Hence the result.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm: Exercise $24.18 \ \text{(a)}$
* : $\S 62.2$ Factorization in an integral domain
Category:Integral Domains
Category:Examples of Equivalence Relations
Category:Associates
Associates in Ring of Polynomial Forms over Field

== Theorem ==
Let $F \sqbrk X$ be the ring of polynomial forms over the field $F$.
Let $\map d X$ and $\map {d'} X$ be polynomial forms in $F \sqbrk X$.
Then $\map d X$ is an associate of $\map {d'} X$ $\map d X = c \cdot \map {d'} X$ for some $c \in F, c \ne 0_F$.
Hence any two polynomials in $F \sqbrk X$ have a unique monic GCD.
== Proof ==
From the definition of associate, there exist $\map e X$ and $\map {e'} X$ \in $F \sqbrk X$ such that:
:$\map d X = \map e X \cdot \map {d'} X$
:$\map {d'} X = \map {e'} X \cdot \map d X$
From Field is Integral Domain, $F$ is an integral domain.
From Degree of Product of Polynomials over Integral Domain it follows that necessarily $\deg e = \deg e' = 0$, as $F$ has no proper zero divisors.
Thus for some $c, c' \in F$, it must be that $\map e X = c$ and $\map {e'} X = c'$.
From the two equations above it follows that $c \cdot c' = 1_F$, where $1_F$ is the unity of $F$.
Hence, it follows that $c \ne 0_F$.
The result follows.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 28$. Highest Common Factor: Example $55$
Category:Factorization
Category:Polynomial Theory
Category:Euclidean Domains
Category:Associates
Associative Algebra has Multiplicative Inverses iff Unitary Division Algebra

== Theorem ==
Let $\struct {A_R, \oplus}$ be an associative algebra over the ring $A_R$.
Then:
:$\struct {A_R, \oplus}$ has a unique multiplicative inverse for every non-zero $a \in A_R$
:
:$\struct {A_R, \oplus}$ is a unitary division algebra.
== Proof ==
Let $A = \struct {A_R, \oplus}$ be a unitary division algebra whose zero is $\mathbf 0_A$.
From the unitary nature of $A$ we have that $\oplus$ has a unit $1$ such that:
:$\forall a \in A_R, a \ne \mathbf 0_A: a \oplus 1 = a = 1 \oplus a$
From the division algebra nature of $A$ we have that $R$ is a field and that:
: $\forall a, b \in A_R, b \ne \mathbf 0_A: \exists_1 x \in A_R, y \in A_R: b \oplus x = a = y \oplus b$
As $1 \in A_R$ it follows that the above holds if $1$ is substituted for $a$:
:$\exists_1 x \in A_R, y \in A_R: b \oplus x = 1 = y \oplus b$
Hence:
So if we let $b^{-1} := x = y$ in the above, we have:
:$\exists_1 b^{-1} \in A_R: 1 = b \oplus b^{-1} = 1 = b^{-1} \oplus b$
That is, $b^{-1}$ is a unique multiplicative inverse of $b$.
So every non-zero element of $A_R$ has a unique multiplicative inverse in $A_R$.
Now suppose that $A_R$ has a unique multiplicative inverse in $A_R$.
It follows directly that, by definition, $\struct {A_R, \oplus}$ has to be a unitary algebra or there is no $1$ for $a \oplus a^{-1}$ to equal.
First we prove existence, i.e we show that:
: $\forall a, b \in A_R, b \ne \mathbf 0_A: \exists x \in A_R, y \in A_R: b \oplus x = a = y \oplus b$
Take $a \oplus b^{-1}$ where $b^{-1}$ is the unique multiplicative inverse of $b$.
Then:
Similarly, we show that $\exists y \in A_R: a = y \oplus b$.
Now we prove uniqueness.
Now suppose $\exists x_1, x_2 \in A_R: a = x_1 \oplus b, a = x_2 \oplus b$.
Then:
In exactly the same way, $a = x_2 \oplus b \implies a \oplus b^{-1} = x_2$.
And so $x_1 = x_2$ thus proving uniqueness.
In a similar way we prove that 
:$\forall a, b \in A_R, b \ne \mathbf 0_A: \exists_1 y \in A_R: a = y \oplus b$
== Sources ==
* 2002: : The Octonions: 1.1 Preliminaries
Category:Associative Algebras
Category:Multiplicative Inverses
Category:Unitary Division Algebras
Associative Commutative Idempotent Operation is Self-Distributive

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure, such that:
:$(1): \quad \circ$ is associative
:$(2): \quad \circ$ is commutative
:$(3): \quad \circ$ is idempotent.
Then $\circ$ is self-distributive.
That is:
:$\forall a, b, c \in S: \paren {a \circ b} \circ \paren {a \circ c} = a \circ b \circ c = \paren {a \circ c} \circ \paren {b \circ c}$
== Proof ==
Category:Abstract Algebra
Category:Commutativity
Category:Associativity
Category:Idempotence
Category:Self-Distributive Operations
Associative Law of Addition

== Theorem ==
Let $\mathbb F$ be one of the standard number sets: $\N, \Z, \Q, \R$ and $\C$.
Then:
:$\forall x, y, z \in \mathbb F: x + \paren {y + z} = \paren {x + y} + z$
That is, the operation of addition on the standard number sets is associative.
 
=== Natural Number Addition is Associative ===
 
=== Integer Addition is Associative ===
 
=== Rational Addition is Associative ===
 
=== Real Addition is Associative ===
 
=== Complex Addition is Associative ===
== Also see ==
* Associative Law of Multiplication
* Commutative Law of Addition
* Commutative Law of Multiplication
== Sources ==
* : Chapter $\text I$: Definitions. Elements of Vector Algebra: $1$. Scalar and Vector Quantities
* : $\S 1.1$. Number Systems: $\text{II}.$
* : $\S 4.2$. Commutative and associative operations: Example $64$
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: $\S 2$: Example $2.1$
* : Part $\text I$: Prologue: Chapter $1$: Basic Properties of Numbers: $(\text P 1)$
* 
* 
* 
* 
Category:Associative Law of Addition
Category:Numbers
Category:Addition
Category:Examples of Associative Operations‎
Associative Law of Multiplication

== Theorem ==
Let $\mathbb F$ be one of the standard number sets: $\N, \Z, \Q, \R$ and $\C$.
Then:
:$\forall x, y, z \in \mathbb F: x \times \paren {y \times z} = \paren {x \times y} \times z$
That is, the operation of multiplication on the standard number sets is associative.
 
=== Natural Number Multiplication is Associative ===
 
=== Integer Multiplication is Associative ===
 
=== Rational Multiplication is Associative ===
 
=== Real Multiplication is Associative ===
 
=== Complex Multiplication is Associative ===
=== Euclid's Statement ===
== Euclid's Proof ==
== Also see ==
* Associative Law of Addition
* Commutative Law of Multiplication
* Commutative Law of Addition
== Sources ==
* : Chapter $\text I$: Definitions. Elements of Vector Algebra: $1$. Scalar and Vector Quantities
* : $\S 1.1$. Number Systems: $\text{V}.$
* : Chapter $\text {I}$: The Group Concept: $\S 1$: Introduction: $(1.2)$
* : $\S 4.2$. Commutative and associative operations: Example $64$
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Example $2.1$
* : Part $\text I$: Prologue: Chapter $1$: Basic Properties of Numbers: $(\text P 5)$
* 
* 
Category:Associative Law of Multiplication
Category:Numbers
Category:Examples of Associative Operations‎
Category:Multiplication
Associative Law of Multiplication/Euclid's Proof

== Theorem ==
== Proof ==
Let a first magnitude $A$ be the same multiple of a second $B$ that a third $C$ is of a fourth $D$.
Let equimultiples $EF, GH$ be taken of $A, C$.
We need to show that $EF$ is the same multiple of $B$ that $GH$ is of $D$.
We have that $EF$ is the same multiple of $A$ that $GH$ is of $C$.
Therefore as many magnitudes as there are in $EF$ equal to $A$, so many also are there in $GH$ equal to $C$.
Let $EF$ be divided into the magnitudes $EK, KF$ equal to $A$, and $GH$ into the magnitudes $GL, LH$ equal to $C$.
Then the multitude of the magnitudes $EK, KF$ will be equal to the multitude of the magnitudes $GL, LH$.
:
We have that $A$ is the same multiple of $B$ that $C$ is of $D$, while $EK = A$ and $GL = C$.
So $EK$ is the same multiple of $B$ that $GL$ is of $D$.
For the same reason, $KF$ is the same multiple of $B$ that $LH$ is of $D$.
So we have that:
:a first magnitude $EK$ is the same multiple of a second $B$ that a third $GL$ is of a fourth $D$
:a fifth $KF$ is also of the same multiple of the second $B$ that a sixth $LH$ is of the fourth $D$.
Therefore the sum of the first and fifth, $EF$, is also the same multiple of the second $B$ that the sum of the third and sixth, $GH$ is of the fourth $D$.
== Sources ==
* : Book $\text{V}$. Propositions
Category:Associative Law of Multiplication
Associative Law of Multiplication/Euclid's Statement

== Theorem ==
That is, if:
:$n a, n b$ are equimultiples of $a, b$
and if:
:$m \cdot n a, m \cdot nb$ are equimultiples of $n a, n b$
then:
:$m \cdot n a$ is the same multiple of $a$ that $m \cdot n b$ is of $b$
This can also be expressed as:
:$m \cdot n a = m n \cdot a$
Category:Associative Law of Multiplication
Associative Operation/Examples/Associative/x circ a circ y

== Example of Associative Operation ==
Let $\struct {S, \circ}$ be an algebraic structure where $\circ$ is an associative operation.
Let $a \in S$ be an arbitrary element of $S$.
Let $*$ be the operation defined on $S$ by:
:$\forall x, y \in S: x * y := x \circ a \circ y$
Then $*$ is associative on $S$.
== Proof ==
Let $x, y, z \in S$ be arbitrary.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.15$
Category:Examples of Associative Operations
Associative Operation/Examples/Non-Associative/Arbitrary Order 3 Structure

== Example of Non-Associative Operations ==
Consider the algebraic structure of order $3$ defined by the Cayley table:
:$\begin{array}{c|cccc}
\circ & a & b & c \\
\hline
a & b & c & b \\
b & b & a & c \\
c & a & c & c \\
\end{array}$
demonstrating non-associativity.
Also note that $a \circ b \ne b \circ a$, so $\circ$ is non-commutative as well.
== Sources ==
* : $\S 4.2$. Commutative and associative operations: Example $66$
Category:Examples of Associative Operations
Associative Operation/Examples/Non-Associative/xy+1

== Example of Non-Associative Operations ==
Let $\R$ denote the set of real numbers.
Let $\circ$ denote the operation on $\R$ defined as:
:$\forall x, y \in \R: x \circ y := x y + 1$
Then $\circ$ is not an associative operation, despite being commutative.
== Proof ==
Let $x, y, z \in \R$.
We have:
But unless $x = z$ it is not the case that $\paren {x y + 1} z + 1 = x y z + x + 1$.
However, $x y + 1 = y x + 1$ and so $\circ$ is commutative.
== Sources ==
* : Chapter $4$. Groups: Exercise $2$
Category:Examples of Associative Operations
Associative and Anticommutative

== Theorem ==
Let $\circ$ be a binary operation on a set $S$.
Let $\circ$ be both associative and anticommutative.
Then:
:$\forall x, y, z \in S: x \circ y \circ z = x \circ z$
== Proof ==
Let $\circ$ be both associative and anticommutative.
Then from Associative Idempotent Anticommutative:
:$\forall x, z \in S: x \circ z \circ x = x$
and $\circ$ is idempotent.
Consider $x \circ y \circ z \circ x \circ z$.
We have:
Also:
Hence the result.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.17 \ \text{(c)}$
Category:Anticommutativity
Category:Associativity
Associativity of Group Direct Product

== Theorem ==
The group direct product $G \times \paren {H \times K}$ is (group) isomorphic to $\paren {G \times H} \times K$.
== Proof ==
Let $G, H, K$ be groups.
The mapping $\theta: G \times \paren {H \times K} \to \paren {G \times H} \times K$ defined as:
:$\forall g \in G, h \in H, k \in K: \map \theta {\tuple {g, \tuple {h, k} } } = \tuple {\tuple {g, h}, k}$
is shown to be a group isomorphism, as follows:
=== Injective ===
Let $\map \theta {\tuple {g_1, \tuple {h_1, k_1} } } = \map \theta {\tuple {g_2, \tuple {h_2, k_2} } }$.
By the definition of $\theta$:
:$\tuple {\tuple {g_1, h_1}, k_1} = \tuple {\tuple {g_2, h_2}, k_2}$
By Equality of Ordered Pairs:
:$\tuple {g_1, h_1} = \tuple {g_2, h_2}$
:$k_1 = k_2$
and consequently:
:$g_1 = g_2, h_1 = h_2$
Thus:
:$\tuple {g_1, \tuple {h_1, k_1} } = \tuple {g_2, \tuple {h_2, k_2} }$
and so $\theta$ is injective.
=== Surjective ===
If $\tuple {\tuple {g, h}, k} \in \paren {G \times H} \times K$, then $g \in G, h \in H, k \in K$.
Thus:
:$\tuple {g, \tuple {h, k} } \in G \times \paren {H \times K}$
and:
:$\map \theta {\tuple {g, \tuple {h, k} } } = \tuple {\tuple {g, h}, k}$
and so $\theta$ is surjective.
Hence, by definition, $\theta$ is bijective.
=== Group Homomorphism ===
Now let $\tuple {g_1, \tuple {h_1, k_1} }, \tuple {g_2, \tuple {h_2, k_2} } \in G \times \tuple {H \times K}$.
Then:
Hence, $\theta$ is a (group) homomorphism.
Hence, by definition, $\theta$ is a group isomorphism.
Category:Group Direct Products
Category:Associativity
Associativity of Hadamard Product

== Theorem ==
Let $\struct {S, \cdot}$ be an algebraic structure.
Let $\map {\MM_S} {m, n}$ be a $m \times n$ matrix space over $S$.
For $\mathbf A, \mathbf B \in \map {\MM_S} {m, n}$, let $\mathbf A \circ \mathbf B$ be defined as the Hadamard product of $\mathbf A$ and $\mathbf B$.
The operation $\circ$ is associative on $\map {\MM_S} {m, n}$ $\cdot$ is associative on $\struct {S, \cdot}$.
== Proof ==
=== Necessary Condition ===
Let the operation $\cdot$ be associative on $\struct {S, \cdot}$.
Let $\mathbf A = \sqbrk a_{m n}$, $\mathbf B = \sqbrk b_{m n}$ and $\mathbf C = \sqbrk c_{m n}$ be elements of the $m \times n$ matrix space over $S$.
Then:
That is, $\circ$ is associative on $\map {\MM_S} {m, n}$.
=== Sufficient Condition ===
Suppose $\struct {S, \cdot}$ is such that $\cdot$ is not associative.
Then there exists $a$, $b$ and $c$ in $S$ such that:
:$a \cdot \paren {b \cdot c} \ne \paren {a \cdot b} \cdot c$
Let $\mathbf A$, $\mathbf B$ and $\mathbf C$ be elements of $\map {\MM_S} {m, n}$ such that:
:$a_{i j} = a$, $b_{i j} = b$, $c_{i j} = c$
where:
:$a_{i j}$ is the $\tuple {i, j}$th element of $\mathbf A$
:$b_{i j}$ is the $\tuple {i, j}$th element of $\mathbf B$
:$c_{i j}$ is the $\tuple {i, j}$th element of $\mathbf C$
Then:
:$a_{i j} \cdot \paren {b_{i j} \cdot c_{i j} } \ne \paren {a_{i j} \cdot b_{i j} } \cdot c_{i j}$
 
That is:
:$\paren {\mathbf A \circ \mathbf B} \circ \mathbf C \ne \mathbf A \circ \paren {\mathbf B \circ \mathbf C}$
because they differ (at least) at indices $\tuple {i, j}$.
That is, $\circ$ is not associative on $\map {\MM_S} {m, n}$.
== Also see ==
* Closure of Hadamard Product
* Commutativity of Hadamard Product
Category:Hadamard Product
Category:Examples of Associative Operations‎
Associativity of Operation in Group Direct Product

== Theorem ==
Let $\struct {G \times H, \circ}$ be the group direct product of the two groups $\struct {G, \circ_1}$ and $\struct {H, \circ_2}$.
Then the operation $\circ$ in $\struct {G \times T, \circ}$ is associative.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $\text{II}$: Groups: Direct Products
Category:Group Direct Products
Category:Associativity of Operation in Group Direct Product
Associativity on Four Elements

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Let $a, b, c, d \in S$.
Then:
:$a \circ b \circ c \circ d$
gives a unique answer no matter how the elements are associated.
== Proof ==
As $\struct {S, \circ}$ is a semigroup:
: it is closed
: $\circ$ is associative
From Parenthesization of Word of $4$ Elements, there are exactly $5$ different ways of inserting brackets in the expression $a \circ b \circ c \circ d$.
As $\circ$ is associative, we have that:
:$\forall s_1, s_2, s_3 \in S: \paren {s_1 \circ s_2} \circ s_3 = s_1 \circ \paren {s_2 \circ s_3}$
As $\struct {S, \circ}$ is closed, we know that all products of elements from $\set {a, b, c, d}$ are in $S$, and are likewise bound by the associativity of $S$.
So:
== Also see ==
* General Associativity Theorem, which demonstrates this rule for any number of elements.
== Sources ==
* : Exercise $2.1: \ 11$
* : $\S 4.2$. Commutative and associative operations
* : Part $\text I$: Prologue: Chapter $1$: Basic Properties of Numbers
:::in the context of addition
* : Chapter $2$: The Definition of Group Structure: $\S 27 \alpha$
* : $\S 28$. Associativity and commutativity
* : Chapter $3$: Elementary consequences of the definitions
Category:Associativity
Associativity on Indexing Set

== Theorem ==
Let $\struct {S, \circ}$ be a commutative semigroup.
Let $\family {x_\alpha}_{\alpha \mathop \in A}$ be a family of terms of $S$ indexed by a finite non-empty set $A$.
Let $\family {B_k}_{1 \mathop \le k \mathop \le n}$ be a family of distinct subsets of $A$ forming a partition of $A$.
Then:
:$\ds \prod_{k \mathop = 1}^n \paren {\prod_{a \mathop \in B_k} x_\alpha} = \prod_{\alpha \mathop \in A} x_\alpha$
== Proof ==
For each $k \in \N_{>0}$, let $\card {B_k} = p_k$.
Let:
:$r_0 = 0$
:$\ds \forall k \in \N_{>0}: r_k = \sum_{j \mathop = 1}^k {p_j}$
and:
:$p = r_n$
Then:
:$r_k - r_{k - 1} = p_k$
So, by Isomorphism to Closed Interval, both integer intervals $\closedint 1 {p_k}$ and $\closedint {r_{k - 1} + 1} {r_k}$ have $p_k$ elements.
By Unique Isomorphism between Equivalent Finite Totally Ordered Sets, there is a unique isomorphism $\tau_k: \closedint 1 {p_k} \to \closedint {r_{k - 1} + 1} {r_k}$ as both are totally ordered.
The orderings on both of these are those induced by the ordering on $\N$.
It is clear that $\tau_k$ is defined as:
:$\forall j \in \closedint 1 {p_k}: \map {\tau_k} j = r_k + j$
For each $k \in \closedint 1 n$, let $\rho_k: \closedint 1 {p_k} \to B_k$ be a bijection.
By Strictly Increasing Sequence induces Partition, the mapping $\sigma: \closedint 1 p \to A$ defined by:
:$\ds \forall j \in \closedint {r_{k - 1} + 1} {r_k}: \forall k \in \closedint 1 n: \map \sigma j = \map {\rho_k} {\map {\tau_k^{-1} } j}$
is a bijection.
Let $\forall j \in \closedint 1 p: y_j = x_{\map \sigma j}$.
By definition:
:$\ds \prod_{\alpha \mathop \in A} {x_\alpha} = \prod_{j \mathop = 1}^p {x_{\map \sigma j} } = \prod_{j \mathop = 1}^p {y_j}$
Also:
:$\ds \forall k \in \closedint 1 n: \prod_{\alpha \mathop \in B_k} {x_\alpha} = \prod_{i \mathop = 1}^{p^k} {x_{\map {\rho_k} i} }$
Also by definition:
:$\ds \prod_{j \mathop = r_{k - 1} + 1}^{r_k} y_j = \prod_{i \mathop = 1}^{p_k} y_{\map {\tau_k} i} = \prod_{i \mathop = 1}^{p_k} x_{\map \sigma {\map {\tau_k} i} } = \prod_{i \mathop = 1}^{p_k} x_{\map {\rho_k} i}$
So by the General Associativity Theorem:
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Theorem $18.9$
Category:Indexed Families
Category:Associativity
Associator of Associative Algebra is Zero

== Theorem ==
Let $\struct {A_R, \oplus}$ be an associative algebra.
Let $\sqbrk {a, b, c}$ denote the associator of $a, b, c \in A_R$.
Then:
:$\forall a, b, c \in A_R: \sqbrk {a, b, c} = \mathbf 0_R$
== Proof ==
== Sources ==
* 2002: : The Octonions: 1.1 Preliminaries
Category:Associators
Category:Associative Algebras
Astroid is Envelope of Line Segment whose End Points lie on Coordinate Axes

== Theorem ==
Let $\LL$ be a line segment of length $a$.
Let the endpoints of $\LL$ be constrained to lie on the coordinate axes of a Cartesian plane: one on the $x$-axis and one on the $y$-axis
The envelope of $\LL$ forms an astroid whose cusps lie on the points at which the coordinate axes intersect a circle of radius $a$ whose center is at the origin.
== Proof ==
== Sources ==
* 
Category:Astroids
Asymmetric Relation/Examples/Greater Than

== Example of Asymmetric Relation ==
The relation $\RR$ defined on the set of real numbers $\R$:
:$\forall x, y \in \R: x \mathop \RR y \iff x > y$
where:
:$>$ denotes the conventional greater than relation
is an asymmetric relation.
== Sources ==
* 
* 
Category:Examples of Asymmetric Relations
Asymmetric Relation is Antireflexive

== Theorem ==
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a relation on $S$.
Let $\RR$ be asymmetric.
Then $\RR$ is also antireflexive.
== Proof ==
Let $\RR$ be asymmetric.
Then, by definition:
:$\tuple {x, y} \in \RR \implies \tuple {y, x} \notin \RR$
 $\tuple {x, x} \in \RR$.
Then:
Thus $\RR$ is antireflexive.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $5$ Properties of Relations: $163$
* : Chapter $1$: Preliminaries: $1.5$ Relations: Properties of Relations
Category:Asymmetric Relations
Category:Antireflexive Relations