nyuuzyou/wikis huggingface dataset:
Anisotropic Vector Gives Composition of Bilinear Space

== Theorem ==
Let $\mathbb K$ be a field.
Let $\struct {V, f}$ be a bilinear space over $\mathbb K$.
Let $v \in V$ be anisotropic.
Let $\sequence v$ be its span.
Let $v^\perp$ be its orthogonal complement.
Then $\struct {V, f}$ is the internal orthogonal sum of $\sequence v$ and $v^\perp$:
:$V = \sequence v \oplus v^\perp$
== Proof ==
Category:Bilinear Forms (Linear Algebra)
Annihilator is Submodule of Algebraic Dual

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $G$ be a module over $R$.
Let $M$ be a submodule of $G$.
Let $G^*$ be the algebraic dual of $G$.
Then the annihilator $M^\circ$ of $M$ is a submodule of $G^*$.
=== Corollary ===
== Proof ==
By definition, $M^\circ$ is a subset of $G^*$.
Recall that by definition of algebraic dual, the elements of $G^*$ are linear transformations from $G$ to the $R$-module $R$.
By Submodule Test, it remains to be shown that:
Indeed:
and:
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations
Category:Submodules
Category:Annihilators
Category:Algebraic Duals
Category:Annihilator is Submodule of Algebraic Dual
Annihilator is Submodule of Algebraic Dual/Corollary

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $G$ be a module over $R$.
Let $M$ be a submodule of $G$.
Let $G^*$ be the algebraic dual of $G$.
Let $N$ be a submodule of $G^*$.
Let $G^{**}$ be the algebraic dual of $G^*$.
Then the annihilator $N^\circ$ of $N$ is a submodule of $G^{**}$.
== Proof ==
Follows directly as an example of Annihilator is Submodule of Algebraic Dual.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations
Category:Annihilator is Submodule of Algebraic Dual
Annihilator of Annihilator on Algebraic Dual of Subspace is Image under Evaluation Isomorphism

== Theorem ==
Let $G$ be an $n$-dimensional vector space over a field.
Let $G^*$ be the algebraic dual of $G$.
Let $G^{**}$ be the algebraic dual of $G^*$.
Let $J: G \to G^{**}$ be the evaluation isomorphism.
Let $M$ be an $m$-dimensional subspace of $G$.
Let $M^\circ$ be the annihilator of $M$.
Then:
:$M^{\circ \circ} = J \sqbrk M$
where $J \sqbrk M$ denotes the image of $M$ under $J$.
== Proof ==
Let Dimension of Annihilator on Algebraic Dual be applied to $M^\circ$ instead of $M$.
It is seen that the annihilator $M^{\circ \circ}$ of $M^\circ$ has dimension $n - \paren {n - m} = m$.
But clearly:
:$J \sqbrk M \subseteq M^{\circ \circ}$.
As $J$ is an isomorphism, $J \sqbrk M$ has dimension $m$.
So by Dimension of Proper Subspace is Less Than its Superspace:
:$J \sqbrk M = M^{\circ \circ}$.
As a consequence:
:$\map {J^\gets} {M^{\circ \circ} } = M$
Hence the result:
:$M^{\circ \circ} = J \sqbrk M$
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 28$. Linear Transformations: Theorem $28.10$
Category:Annihilators
Category:Vector Subspaces
Annihilator of Image of Bounded Linear Transformation is Kernel of Dual Operator

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ and $Y$ be normed vector spaces over $\GF$.
Let $T : X \to Y$ be a bounded linear transformation.
Let $X^\ast$ and $Y^\ast$ be the normed duals of $X$ and $Y$ respectively.
Let $T^\ast : Y^\ast \to X^\ast$ be the dual operator of $T$. 
Let $T \sqbrk X^\bot$ be the annihilator of $T \sqbrk X$.
Then:
:$T \sqbrk X^\bot = \map \ker {T^\ast}$ 
== Proof ==
We have: 
:$f \in \map \ker {T^\ast}$
:
:$\map {\paren {T^\ast f} } x = \map f {T x} = 0$ for each $x \in X$. 
That is, :
:$\map f y = 0$ for all $y \in T \sqbrk X$
To conclude, we have $f \in \map \ker {T^\ast}$ :
:$f \in T \sqbrk X^\bot$
== Sources ==
* : Exercise $2.39 \text { (i)}$
Category:Annihilators of Subspaces of Banach Spaces
Category:Dual Operators
Annihilator of Ring Always Contains Zero

== Theorem ==
Let $\struct {R, +, \times}$ be a ring.
Let $\map {\mathrm {Ann} } R$ be the annihilator of $R$.
Then $0 \in \map {\mathrm {Ann} } R$.
== Proof ==
We have by definition of integral multiple that:
:$\forall r \in R: 0 \cdot r = 0_R$
where $0_R$ is the zero of $R$.
Hence the result by definition of annihilator.
Category:Ring Theory
Annihilator of Subspace of Banach Space as Intersection of Kernels

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ be a Banach space over $\GF$.
Let $M$ be a vector subspace of $X$. 
Let $X^\ast$ be the normed dual space of $X$. 
Let $M^\bot$ be the annihilator of $M$. 
Then: 
:$\ds M^\bot = \bigcap_{x \mathop \in M} \map \ker {x^\wedge}$
where $x^\wedge$ denotes the evaluation linear transformation evaluated at $x$. 
== Proof ==
We have: 
Category:Annihilators of Subspaces of Banach Spaces
Annihilator of Subspace of Banach Space is Subspace of Normed Dual

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ be a Banach space over $\GF$.
Let $M$ be a vector subspace of $X$. 
Let $X^\ast$ be the normed dual space of $X$. 
Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$.
Let $M^\bot$ be the annihilator of $M$. 
Then $M^\bot$ is a linear subspace of $X^\ast$. 
== Proof ==
For each $x \in X$, we have: 
:$\map { {\mathbf 0}_{X^\ast} } x = 0$ for each $x \in X$.
In particular:
:$\map { {\mathbf 0}_{X^\ast} } x = 0$ for each $x \in M$
so that: 
:${\mathbf 0}_{X^\ast} \in M^\bot$
So we have $M^\bot \ne \O$.
From One-Step Vector Subspace Test, it is enough to show that for each $f, g \in M^\bot$ and $\lambda \in \GF$ we have: 
:$f + \lambda g \in M^\bot$
Let $f, g \in M^\bot$ and $\lambda \in \GF$. 
Then $\map f x = 0$ and $\map g x = 0$.
So $0 = \map f x + \lambda \map g x = \map {\paren {f + \lambda g} } x$.
So we have $f + \lambda g \in M^\bot$.
From One-Step Vector Subspace Test, $M^\bot$ is a linear subspace of $X^\ast$.
Category:Annihilators of Subspaces of Banach Spaces
Annihilator of Subspace of Banach Space is Weak-* Closed

== Theorem ==
Let $X$ be a Banach space.
Let $M$ be a vector subspace of $X$. 
Let $X^\ast$ be the normed dual space of $X$. 
Let $w^\ast$ be the weak-$\ast$ topology on $X^\ast$.
Let $M^\bot$ be the annihilator of $M$. 
Then $M^\bot$ is closed in $\struct {X^\ast, w^\ast}$. 
== Proof 1 ==
== Proof 2 ==
Category:Annihilator of Subspace of Banach Space is Weak-* Closed
Category:Annihilators of Subspaces of Banach Spaces
Category:Weak-* Topologies
Annihilator of Subspace of Banach Space is Weak-* Closed/Proof 1

== Theorem ==
== Proof ==
From Set is Closed iff Equals Topological Closure, we aim to show:
:$M^\bot = \map {\cl_{w^\ast} } {M^\bot}$
From Set is Subset of its Topological Closure, we have: 
:$M^\bot \subseteq \map {\cl_{w^\ast} } {M^\bot}$
Now let: 
:$f \in \map {\cl_{w^\ast} } {M^\bot}$
From Point in Set Closure iff Limit of Net, there exists a directed set $\struct {\Lambda, \preceq}$ and a net $\family {f_\lambda}_{\lambda \in \Lambda}$ in $M^\bot$ converging to $f$ in $\struct {X^\ast, w^\ast}$.
That is, from Characterization of Convergent Net in Weak-* Topology: 
:for each $x \in X$ the net $\family {\map {f_\lambda} x}_{\lambda \in \Lambda}$ converges to $\map f x$ in $\GF$.
Since $f_\lambda \in M^\bot$ for each $\lambda \in \Lambda$, we have: 
:$\map {f_\lambda} x = 0$ for each $x \in M$ and $\lambda \in \Lambda$.
From Constant Net is Convergent, we obtain:
:the net $\family {\map {f_\lambda} x}_{\lambda \in \Lambda}$ converges to $\map f x$ and $0$ in $\GF$. 
From Metric Space is Hausdorff, $\GF$ is Hausdorff. 
Hence from Characterization of Hausdorff Property in terms of Nets, we obtain $\map f x = 0$ for each $x \in M$.
So:
:$f \in M^\bot$
So we obtain: 
:$\map {\cl_{w^\ast} } {M^\bot} \subseteq M^\bot$
and hence:
:$M^\bot = \map {\cl_{w^\ast} } {M^\bot}$
From Set is Closed iff Equals Topological Closure, we have that $M^\bot$ is closed in $\struct {X^\ast, w^\ast}$.
Category:Annihilator of Subspace of Banach Space is Weak-* Closed
Annihilator of Subspace of Banach Space is Weak-* Closed/Proof 2

== Theorem ==
== Proof ==
From Annihilator of Subspace of Banach Space as Intersection of Kernels, we have:
:$\ds M^\bot = \bigcap_{x \in M} \map \ker {x^\wedge}$ 
From Characterization of Continuity of Linear Functional in Weak-* Topology:
:the linear functional $x^\wedge : \struct {X^\ast, w^\ast} \to \GF$ is continuous.
From Characterization of Continuous Linear Functionals on Topological Vector Space, $\map \ker {x^\wedge}$ is closed in $\struct {X^\ast, w^\ast}$.
So $M^\bot$ is the intersection of closed sets in $\struct {X^\ast, w^\ast}$, and hence is closed itself.
Category:Annihilator of Subspace of Banach Space is Weak-* Closed
Annihilator of Subspace of Banach Space is Zero iff Subspace is Everywhere Dense

== Theorem ==
Let $\GF \in \set {\R, \C}$. 
Let $X$ be a Banach space over $\GF$.
Let $M$ be a vector subspace of $X$. 
Let $X^\ast$ be the normed dual space of $X$. 
Let $M^\bot$ be the annihilator of $M$. 
Then: 
:$M^\bot = \set { {\mathbf 0}_{X^\ast} }$
:
:$M$ is everywhere dense in $X$. 
== Proof ==
=== Necessary Condition ===
Suppose that:
:$M$ is not everywhere dense in $X$.
From Closure of Linear Subspace of Topological Vector Space is Linear Subspace, $\map \cl M$ is a closed linear subspace of $X$.
From Set is Closed iff Equals Topological Closure, we have $X \ne \map \cl M$.
So $\map \cl M$ is a proper closed linear subspace of $X$.
From Existence of Distance Functional, there exists $f \in X^\ast \setminus \set { {\mathbf 0}_{X^\ast} }$ such that: 
:$\map f x = 0$ for all $x \in M$
so that $f \in M^\bot$.
So $M^\bot \ne \set { {\mathbf 0}_{X^\ast} }$
From Proof by Contraposition, we have: 
:if $M^\bot = \set { {\mathbf 0}_{X^\ast} }$ then $M$ is everywhere dense in $X$.
=== Sufficient Condition ===
Suppose that: 
:$M$ is everywhere dense in $X$. 
Let $f \in M^\bot$. 
Then for all $x \in M$ we have $\map f x = 0$. 
From Metric Space is Hausdorff, $\GF$ is Hausdorff.
Since $M$ is everywhere dense, we have:
:$\map f x = 0$ for all $x \in X$.
So if $f \in M^\bot$, then $f = {\mathbf 0}_{X^\ast}$. 
Further, we have ${\mathbf 0}_{X^\ast} \in M^\bot$. 
Hence we obtain: 
:$M^\bot = \set { {\mathbf 0}_{X^\ast} }$
Category:Annihilators of Subspaces of Banach Spaces
Anning's Theorem

== Theorem ==
In any base greater than $1$, the fraction:
:$\dfrac {101 \, 010 \, 101} {110 \, 010 \, 011}$
has the property that if the two $1$'s in the center of the numerator and the denominator are replaced by the same odd number of $1$'s, the value of the fraction remains the same.
For example:
:$\dfrac {101 \, 010 \, 101} {110 \, 010 \, 011} = \dfrac {1 \, 010 \, 111 \, 110 \, 101} {1 \, 100 \, 111 \, 110 \, 011} = \dfrac {9091} {9901}$ (in base $10$).
== Proof ==
Let $b$ be the base in question.
Let $F = \dfrac {101 \, 010 \, 101} {110 \, 010 \, 011}$.
Then:
:$F = \dfrac {1 + b^2 + b^4 + b^6 + b^8} {1 + b + b^4 + b^7 + b^8}$
It is necessary to prove that for all $k \in \Z_{>0}$:
:$F = \dfrac {1 + b^2 + b^4 + b^5 + \cdots + b^{2 k + 2} + b^{2 k + 4} + b^{2 k + 6} } {1 + b + b^4 + b^5 + \cdots + b^{2 k + 2} + b^{2 k + 5} + b^{2 k + 6} }$
This is done by:
: multiplying the numerator of one by the denominator of the other
and then:
: multiplying the denominator of one by the numerator of the other
and checking that they are equal.
Thus we proceed:
and:
Equality can be demonstrated.
== Historical Note ==
== Sources ==
* 
* : No. $208$
* : $101,010,101 / 110,010,011$
Category:Number Theory
Anning's Theorem/Historical Note

== Historical Note on Anning's Theorem ==
Anning's Theorem appears as a problem in The $1970$ publication by , who cites as its source.
Confusingly, later he refers to P. Anning in the context of the same result. It can be assumed that this is a mistake.
The name Anning's Theorem has been coined by as a convenient way to refer to this result, whose description would otherwise be too unwieldy for convenience.
== Sources ==
* : $101,010,101 / 110,010,011$
Category:Historical Notes
Anomalous Cancellation/Examples/143 185 over 17 018 560

== Theorem ==
The fraction $\dfrac {143 \, 185} {17 \, 018 \, 560}$ exhibits the phenomenon of anomalous cancellation:
:$\dfrac {143 \, 185} {17 \, 018 \, 560} = \dfrac {1435} {170 \, 560}$
as can be seen by deleting the $18$ from both numerator and denominator.
This is part of a longer pattern:
:$\dfrac {1435} {170 \, 560} = \dfrac {143 \, 185} {17 \, 018 \, 560} = \dfrac {14 \, 318 \, 185} {1 \, 701 \, 818 \, 560} = \cdots$
== Proof ==
Let $q = \dfrac r s = \dfrac {1431818 \cdots 185} {1701818 \cdots 18560}$.
Let the number of digits in $r$ be $n + 2$.
Then the number of digits in $s$ is $n + 4$.
By inspection, it is seen that $n$ is even.
Then it is seen that:
and similarly:
Hence:
== Sources ==
* : $16 / 64$
* : $16 / 64$
Category:Anomalous Cancellation
Anomalous Cancellation/Examples/3544 over 7531

== Theorem ==
The fraction $\dfrac {3544} {7531}$ exhibits the phenomenon of anomalous cancellation:
:$\dfrac {3544} {7531} = \dfrac {344} {731}$
as can be seen by deleting the $5$ from both numerator and denominator.
This is part of a longer pattern:
:$\dfrac {344} {731} = \dfrac {3544} {7531} = \dfrac {35544} {75531} = \cdots$
== Proof ==
== Sources ==
* : $16 /64$
* : $16/64$
Category:Anomalous Cancellation
Anomalous Cancellation/Variants/37 + 13 over 37 + 24

== Theorem ==
:$\dfrac {37 + 13} {37 + 24} = \dfrac {37^3 + 13^3} {37^3 + 24^3}$
== Proof ==
Hence:
:$\dfrac {37^3 + 13^3} {37^3 + 24^3} = \dfrac {1057 \paren {37 + 13} } {1057 \paren {37 + 24} } = \dfrac {37 + 13} {37 + 24}$
== Sources ==
* : $16 / 64$
* : $16 / 64$
Category:Anomalous Cancellation
Anomalous Cancellation/Variants/3 + 25 + 38 over 7 + 20 + 39

== Theorem ==
:$\dfrac {3 + 25 + 38} {7 + 20 + 39} = \dfrac {3^4 + 25^4 + 38^4} {7^4 + 20^4 + 39^4}$
== Proof ==
== Sources ==
* : $16 / 64$
* : $16 / 64$
Category:Anomalous Cancellation
Anomalous Cancellation on 2-Digit Numbers

== Theorem ==
There are exactly four anomalously cancelling proper fractions having two-digit numerator and denominator when expressed in base $10$ notation:
== Proof ==
Let $\dfrac {\sqbrk {a x} } {\sqbrk {x b} }$ be an anomalously cancelling proper fraction.
Then we have:
From $(3)$:
Therefore $\dfrac {9 a b} {10 a - b}$ must be an integer.
 $10 a - b$ is divisible by a prime $p$ such that $p > 9$.
We have that:
:$\paren {10 a - b} \divides 9 a b$
and so:
:$p \divides 9 a b$
where $\divides$ denotes divisibility.
By Euclid's Lemma for Prime Divisors, that means one of the elements of $\set {9, a, b}$ must be divisible by $p$.
Hence from Absolute Value of Integer is not less than Divisors, either $a$ or $b$ must be greater than $9$.
This contradicts $(1)$ above.
Hence by Proof by Contradiction it cannot be the case that $10 a - b$ is divisible by a prime greater than $9$.
We continue:
It remains for us to check whether $x = \dfrac {9 a b} {10 a - b}$ is an integer for $\dfrac b {10 - b} \le a < b$, when $b$ ranges from $2$ to $9$.
We note first that the following pairs $\tuple {a, b}$ are such that $\dfrac b {10 - b} > a$:
Then we eliminate the pairs $\tuple {a, b}$ where $10 a - b$ has a prime factor greater than $9$:
Thus it remains to check whether $\dfrac {9 a b} {10 a - b}$ is an integer for the following pairs $\tuple {a, b}$:
:$\tuple {1, 2}$, $\tuple {1, 3}$, $\tuple {2, 3}$, $\tuple {1, 4}$, $\tuple {2, 4}$, $\tuple {1, 5}$, $\tuple {2, 5}$, $\tuple {3, 5}$, $\tuple {4, 5}$, $\tuple {2, 6}$, $\tuple {3, 6}$, $\tuple {4, 8}$, $\tuple {5, 8}$
So the only solutions for $\tuple {a, x, b}$ are:
:$\tuple {1, 6, 4}$, $\tuple {1, 9, 5}$, $\tuple {2, 6, 5}$ and $\tuple {4, 9, 8}$
These correspond to the fractions:
:$\dfrac {16} {64}$, $\dfrac {19} {95}$, $\dfrac {26} {65}$ and $\dfrac {49} {98}$
== Historical Note ==
== Sources ==
* : $16 /64$
* : $16/64$
Category:Anomalous Cancellation
Anomalous Cancellation on 2-Digit Numbers/Examples/16 over 64

== Example of Anomalous Cancellation on 2-Digit Numbers ==
The fraction $\dfrac {16} {64}$ exhibits the phenomenon of anomalous cancellation:
:$\dfrac {16} {64} = \dfrac 1 4$
as can be seen by deleting the $6$ from both numerator and denominator.
This is part of a longer pattern:
:$\dfrac 1 4 = \dfrac {16} {64} = \dfrac {166} {664} = \dfrac {1666} {6664} = \cdots$
== Proof ==
== Sources ==
* : $16 /64$
* : $16/64$
Category:Anomalous Cancellation
Anomalous Cancellation on 2-Digit Numbers/Examples/19 over 95

== Example of Anomalous Cancellation on 2-Digit Numbers ==
The fraction $\dfrac {19} {95}$ exhibits the phenomenon of anomalous cancellation:
:$\dfrac {19} {95} = \dfrac 1 5$
as can be seen by deleting the $9$ from both numerator and denominator.
This is part of a longer pattern:
:$\dfrac 1 5 = \dfrac {19} {95} = \dfrac {199} {995} = \dfrac {1999} {9995} = \cdots$
== Proof ==
Formally written, we have to show that:
:$\ds \paren {\paren {\sum_{i \mathop = 0}^{n - 1} 9 \times 10^i} + 10^n} \Big / \paren {5 + \paren {\sum_{i \mathop = 1}^n 9 \times 10^i} } = \frac 1 5$ for integers $n > 1$.
So:
== Sources ==
* : $16 /64$
* : $16/64$
Category:Anomalous Cancellation
Anomalous Cancellation on 2-Digit Numbers/Examples/26 over 65

== Example of Anomalous Cancellation on 2-Digit Numbers ==
The fraction $\dfrac {26} {65}$ exhibits the phenomenon of anomalous cancellation:
:$\dfrac {26} {65} = \dfrac 2 5$
as can be seen by deleting the $6$ from both numerator and denominator.
This is part of a longer pattern:
:$\dfrac 2 5 = \dfrac {26} {65} = \dfrac {266} {665} = \dfrac {2666} {6665} = \cdots$
== Proof ==
== Sources ==
* : $16 / 64$
* : $16 / 64$
Category:Anomalous Cancellation
Anomalous Cancellation on 2-Digit Numbers/Examples/49 over 98

== Example of Anomalous Cancellation on 2-Digit Numbers ==
The fraction $\dfrac {49} {98}$ exhibits the phenomenon of anomalous cancellation:
:$\dfrac {49} {98} = \dfrac 4 8$
as can be seen by deleting the $9$ from both numerator and denominator.
This is part of a longer pattern:
:$\dfrac 4 8 = \dfrac {49} {98} = \dfrac {499} {998} = \dfrac {4999} {9998} = \cdots$
== Proof ==
== Sources ==
* : $16 /64$
* : $16/64$
Category:Anomalous Cancellation
Antiassociative Operation has no Idempotent Elements

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\circ$ be antiassociative on $S$.
Then no element of $S$ is idempotent under $ \circ$.
That is:
:$\forall x \in S: x \circ x \ne x$
== Proof ==
 $a \in S$ such that $a$ is idempotent under $\circ$.
That is:
:$a \circ a = a$
Then:
:$\paren {a \circ a} \circ a = a \circ a$ 
and 
:$a \circ \paren {a \circ a} = a \circ a$
This contradicts our assumption that $\circ$ is antiassociative on $S$.
Category:Antiassociative Structures
Antiassociative Operation is not Commutative

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\circ$ be antiassociative on $S$.
Then $\circ$ is not commutative on $S$.
== Proof ==
We will show there are two elements in $S$ that do not commute.
Let $a \in S$.
From Antiassociative Operation has no Idempotent Elements:
:$a \circ a \ne a$
So for some $b \in S$: 
:$a \circ a = b$
Then:
:$\paren {a \circ a} \circ a = b \circ a$
and:
:$a \circ \paren {a \circ a} = a \circ b$
From our assumption, $\circ$ is antiassociative.
So:
:$\paren {a \circ a} \circ a \ne a \circ \paren {a \circ a}$
Hence for some $a, b \in S$:
:$a \circ b \ne b \circ a$
Category:Antiassociative Structures
Antiassociative Structure of Finite Order

== Theorem ==
Let $n \in \N$ such that $n > 2$.
Then there exists an algebraic structure $\struct {S, \circ}$ of order $n$ such that $\circ$ is antiassociative on $S$.
== Proof ==
Let $S = \set {a_0, \ldots, a_{n - 1} }$.
Let $\circ$ be a binary operation defined on $S$ such that:
:$\forall x \in S: x \circ a_i = a_{\paren {i + 1} \pmod n}$
Then $\forall j, k, m \in \closedint 0 {n - 1}$:
and:
As $n \ge 2$ it follows that:
:$m + 1 \ne m + 2 \pmod n$
and so $\forall j, k, m \in \closedint 0 {n - 1}$:
:$\paren {a_j \circ a_k} \circ a_m \ne a_j \circ \paren {a_k \circ a_m}$
Hence the result.
Category:Antiassociative Structures
Anticommutativity of External Direct Product

== Theorem ==
Let $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ be algebraic structures where $S$ and $T$ both have at least two distinct elements.
Let $\struct {S \times T, \circ}$ be their external direct product.
Then $\struct {S \times T, \circ}$ is anticommutative at least one of $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ is.
== Proof ==
=== Sufficient Condition ===
Suppose $\struct {S \times T, \circ}$ is anticommutative and neither $\struct {S, \circ_1}$ nor $\struct {T, \circ_2}$ is.
Then for some distinct $a, b \in S$:
:$a \circ_1 b = b \circ_1 a$
Similarly, for some distinct $c, d \in T$:
:$c \circ_2 d = d \circ_2 c$
Then we compute by definition of $\circ$ in the external direct product:
:$\tuple {a, c} \circ \tuple {b, d} = \tuple {a \circ_1 b, c \circ_2 d}$
:$\tuple {b, d} \circ \tuple {a, c} = \tuple {b \circ_1 a, d \circ_2 c}$
So: 
:$\tuple {a, c} \circ \tuple {b, d} = \tuple {b, d} \circ \tuple {a, c}$
which contradicts our assumption that $\struct {S \times T, \circ}$ was anticommutative.
Hence at least one of $\struct {S, \circ_1}$ and $\struct {T, \circ_2}$ is anticommutative.
=== Necessary Condition ===
Suppose $\struct {S \times T, \circ}$ is not anticommutative. 
Then for some $\tuple {a, b}, \tuple {c, d} \in S \times T$:
:$\tuple {a, b} \circ \tuple {c, d} = \tuple {c, d} \circ \tuple {a, b}$
which by the definition of $\circ$ implies:
:$\tuple {a \circ_1 c, b \circ_2 d} = \tuple {c \circ_1 a, d \circ_2 b}$
Therefore we conclude:
:$a \circ_1 c = c \circ_1 a$
:$b \circ_2 d = d \circ_2 b$
Hence, neither $\struct {S, \circ_1}$ nor $\struct {T, \circ_2}$ can be anticommutative.
Hence if one of $\struct {S, \circ_1}$ or $\struct {T, \circ_2}$ is anticommutative, then $\struct {S \times T, \circ}$ must be as well.
Category:Anticommutativity
Category:External Direct Products
Antilexicographic Order/Examples/Unit Square with Open Side

== Example of Antilexicographic Order ==
Consider the antilexicographic product of the real intervals $\hointr 0 1$ and $\closedint 0 1$ under the usual ordering:
:$\struct {T, \preccurlyeq_a} := \struct {\hointr 0 1, \le} \otimes^a \struct {\closedint 0 1, \le}$
$\struct {T, \preccurlyeq_a}$ has one minimal element:
:$\tuple {0, 0}$
which is also the smallest element: of $\struct {T, \preccurlyeq_l}$.
$\struct {T, \preccurlyeq_a}$ has no greatest element and no maximal elements.
== Proof ==
Recall the definition of the antilexicographic order:
::$\tuple {a, b} \preccurlyeq_a \tuple {c, d} \iff \tuple {c \prec_2 d} \lor \paren {c = d \land a \preccurlyeq_1 b}$
Let $\tuple {a, b} \in T$ be a minimal element of $T$.
:$\forall \tuple {x, y} \in T: \tuple {x, y} \preccurlyeq_l \tuple {a, b} \implies \tuple {x, y} = \tuple {a, b}$
Suppose $a \ne 0$.
Then:
:$\tuple {0, b} \preccurlyeq_a \tuple {a, b}$ but $\tuple {0, b} \ne \tuple {a, b}$
Similarly, suppose $b \ne 0$.
Then:
:$\tuple {a, 0} \preccurlyeq_a \tuple {a, b}$ but $\tuple {a, 0} \ne \tuple {a, b}$
Hence if $\tuple {a, b}$ is a minimal element of $T$ it follows that:
:$\tuple {a, b} = \tuple {0, 0}$
Next we note that:
:$\forall \tuple {c, d} \in T: 0 \le c, 0 \le d$
and so:
:$\forall \tuple {c, d} \in T: \tuple {0, 0} \preccurlyeq_a \tuple {c, d}$
Hence by definition $\tuple {0, 0}$ is a smallest element of $T$.
By Smallest Element is Unique, $\tuple {0, 0}$ is then the smallest element of $T$.
 $T$ has a maximal element $\tuple {u, v}$.
Then we have:
:$\tuple {u, v} \preccurlyeq_a \tuple {x, y} \implies \tuple {u, v} = \tuple {x, y}$
and in particular:
:$\tuple {u, v} \preccurlyeq_a \tuple {x, v} \implies \tuple {u, v} = \tuple {x, v}$
That is:
:$u \le x \implies u = x$
But we have that:
:$u \in \hointr 0 1$
and so:
:$u < 1$
So take:
:$x = \dfrac u 1 + \dfrac 1 1 = \dfrac {u + 1} 2$
Hence from Mediant is Between:
:$u < x$
and so it is not the case that $u \le x \implies u = x$.
Hence it is not the case that:
:$\tuple {u, v} \preccurlyeq_a \tuple {x, v} \implies \tuple {u, v} = \tuple {x, v}$
and so $\tuple {u, v}$ is not a maximal element of $T$.
So by Proof by Contradiction it follows that there is no maximal element of $T$.
 $\tuple {u, v}$ is a greatest element of $T$.
From Greatest Element is Maximal, $\tuple {u, v}$ would have to be a maximal element of $T$.
But from above $T$ has no maximal element.
Hence by Proof by Contradiction it follows that there is no greatest element of $T$.
Hence the result.
== Sources ==
* : Part $1$: Not Entirely Naive Set Theory: Chapter $2$: Partial Order Relations: Exercise $31$
Category:Examples of Order Products