nyuuzyou/wikis huggingface dataset:
Mittag-Leffler Expansion for Square of Secant Function

== Theorem ==
:$\ds \pi^2 \map {\sec^2} {\pi z} = 8 \sum_{n \mathop = 0}^\infty \frac {\paren {2 n + 1} + 4 z^2} {\paren {\paren {2 n + 1}^2 - 4 z^2}^2}$
where: 
:$z$ is a complex number that is not a half-integer
:$\sec$ denotes the secant function.
== Proof == 
== Sources == 
* : $\S 37$: Partial Fraction Expansions: $37.5$
Category:Mittag-Leffler Expansions
Category:Secant Function
Mittag-Leffler Expansion for Tangent Function

== Theorem == 
 
:$\ds \pi \map \tan {\pi z} = 8 \sum_{n \mathop = 0}^\infty \frac z {\paren {2 n + 1}^2 - 4 z^2}$
where: 
:$z \in \C$ is not a half-integer
:$\tan$ is the tangent function.
== Proof == 
== Sources == 
* : $\S 37$: Partial Fraction Expansions: $37.4$
* : $7.10$: Some Special Expansions
Category:Mittag-Leffler Expansions
Category:Tangent Function
Mixed Partial Derivative of Heaviside Step Function

== Theorem ==
Let $\tuple {x, y} \stackrel u {\longrightarrow} \map u {x, y}: \R^2 \to \R$ be the Heaviside step function.
Let $u := T_u$ be the distribution associated with $u$.
Let $\delta_{\tuple {0, 0} } \in \map {\DD'} {\R^2}$ be the Dirac delta distribution.
Then in the distributional sense:
:$\dfrac {\partial^2 u} {\partial x \partial y} = \delta_{\tuple {0, 0}}$
== Proof ==
Let $\phi \in \map \DD {\R^2}$ be a test function with support on $\openint 0 a^2 := \openint 0 a \times \openint 0 a$ where $\times$ is the Cartesian product and $a > 0$.
Then:
== Sources ==
* : Chapter $\S 6.2$: A glimpse of distribution theory. Test functions, distributions, and examples
Category:Heaviside Step Function
Category:Examples of Distributional Derivatives
Mixed Tensor/Examples/Rank 2

== Example of Mixed Tensor ==
A mixed tensor of rank $2$ is a set satisfying, for all $i$ and $j$:
:$\overline F^i_j = \ds \sum_r \sum s \dfrac {\partial \overline x^i} {\partial x^r} \dfrac {\partial x^s} {\partial \overline x^j} F^r_s$
== Sources ==
* 
* 
Category:Examples of Mixed Tensors
Modal Logic/Examples/S5

== Example of Modal Logic ==
The $\text {S5}$ system of modal logic contains the axioms and rules of inference of propositional logic together with the axioms:
In order to interpret $\text {S5}$, we need to state the condition under which a WFF of the form $\nec A$ is assigned the truth value true.
This is done via:
:$\nec A$ is true $A$ is true in all possible worlds.
This leads to:
:$\pos A$ is true $A$ is true in some possible world.
== Sources ==
* 
* 
Category:Examples of Modal Logic
Mode of Normal Distribution

== Theorem ==
Let $\mu$ be a real number. 
Let $\sigma$ be a strictly positive real number. 
Let $X \sim \Gaussian \mu {\sigma^2}$ where $\Gaussian \mu {\sigma^2}$ is the normal distribution with parameters $\mu$ and $\sigma^2$. 
Then the mode $M$ of $X$ is equal to $\mu$. 
== Proof == 
By the definition of the normal distribution, $X$ has probability density function: 
:$\map {f_X} x = \dfrac 1 {\sigma \sqrt {2 \pi} } \, \map \exp {-\dfrac {\paren {x - \mu}^2} {2 \sigma^2} }$
By the definition of a mode, $f_X$ attains its global maximum at $M$. 
From Exponential is Strictly Increasing, we have that $\exp x$ is strictly increasing for all real $x$.
Therefore, to maximise $f_X$, we must pick to $x$ to maximise: 
:$-\dfrac {\paren {x - \mu}^2} {2 \sigma^2}$
That is, to minimise:
:$\dfrac {\paren {x - \mu}^2} {2 \sigma^2}$
We have: 
:$\dfrac {\paren {x - \mu}^2} {2 \sigma^2} \ge 0$
with equality occurring at $x = \mu$.
Therefore, $\mu$ is a mode of $X$.
As this is the only value of $x$ for which equality occurs, we have that $\mu$ is the unique mode of $X$. 
Category:Modes
Category:Normal Distribution
Models for Propositional Logic

== Theorem ==
This page gathers together some useful results that can be used in the derivation of proofs by propositional tableau.
Let $\MM$ be a model for propositional logic, and let $\mathbf A$ and $\mathbf B$ be WFFs of propositional logic.
Then the following results hold.
The symbol $\models$ is used throughout to denote semantic consequence.
=== Double Negation ===
:$\MM \models \neg \neg \mathbf A$ $\MM \models \mathbf A$
This is the rule of Double Negation.
=== And ===
:$\MM \models \paren {\mathbf A \land \mathbf B}$ both $\MM \models \mathbf A$ and $\MM \models \mathbf B$
This follows by definition of Conjunction.
=== Not And ===
:$\MM \models \neg \paren {\mathbf A \land \mathbf B}$ either $\MM \models \neg \mathbf A$ or $\MM \models \neg \mathbf B$
This follows from De Morgan's Laws: Disjunction of Negations.
=== Or ===
:$\MM \models \paren {\mathbf A \lor \mathbf B}$ either $\MM \models \mathbf A$ or $\MM \models \mathbf B$
This follows by definition of Disjunction.
=== Not Or ===
:$\MM \models \neg \paren {\mathbf A \lor \mathbf B}$ $\MM \models \neg \mathbf A$ and $\MM \models \neg \mathbf B$
This follows from De Morgan's Laws: Conjunction of Negations.
=== Implies ===
:$\MM \models \paren {\mathbf A \implies \mathbf B}$ either $\MM \models \neg \mathbf A$ or $\MM \models \mathbf B$
This follows from Disjunction and Conditional.
=== Not Implies ===
:$\MM \models \neg \paren {\mathbf A \implies \mathbf B}$ $\MM \models \mathbf A$ and $\MM \models \neg \mathbf B$
This follows from Conjunction and Conditional.
=== Iff ===
:$\MM \models \paren {\mathbf A \iff \mathbf B}$ either:
::both $\MM \models \mathbf A$ and $\MM \models \mathbf B$
or:
::both $\MM \models \neg \mathbf A$ and $\MM \models \neg \mathbf B$
This follows by definition of biconditional.
=== Exclusive Or ===
:$\MM \models \mathbf A \oplus \mathbf B$ either:
::both $\MM \models \mathbf A$ and $\MM \models \neg \mathbf B$
or:
::both $\MM \models \neg \mathbf A$ and $\MM \models \mathbf B$
This follows by definition of exclusive or.
== Sources ==
* : $\S 1.5$: Semantics of Propositional Logic: Proposition $1.5.2$
Category:Propositional Tableaux
Modified Fort Space is Compact

== Theorem ==
Let $T = \struct {S, \tau_{a, b} }$ be a modified Fort space.
Then $T$ is a compact space.
== Proof ==
Let $\CC$ be an open cover of $T$.
Then $\exists U \in \CC: a \in U$.
Then by definition of modified Fort space, $U$ is cofinite.
That is, $S \setminus U$ is finite.
Then $S \setminus U$ is covered by a finite subcover of $\CC$.
Hence, by definition, $T$ is compact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $27$. Modified Fort Space: $1$
Category:Compact Spaces
Category:Modified Fort Space
Modified Fort Space is Scattered

== Theorem ==
Let $T = \struct {S, \tau_{a, b} }$ be a modified Fort space.
Then $T$ is scattered.
== Proof ==
We have that a modified Fort space is $T_1$.
We also have that a dense-in-itself subset of a $T_1$ space is infinite.
But from Isolated Points in Subsets of Modified Fort Space, we have that any subset of $T$ with more than two points has at least one isolated point.
So any dense-in-itself subset of $T$ must have an isolated point.
Hence the result, by definition of scattered space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $27$. Modified Fort Space: $6$
Category:Scattered Spaces
Category:Modified Fort Space
Modified Fort Space is Sequentially Compact

== Theorem ==
Let $T = \struct {S, \tau_{a, b} }$ be a modified Fort space.
Then $T$ is sequentially compact.
== Proof ==
Let $\sequence {x_n}_{n \mathop \in \N}$ be an infinite sequence in $T$.
Suppose $\sequence {x_n}$ takes an infinite number of distinct values in $S$.
Then there is an infinite subsequence $\sequence {x_{n_r} }_{r \mathop \in \N}$ with distinct terms.
Let $U$ be a neighborhood of $a$.
Then $S \setminus U$ is a finite set by definition.
Thus there exists $N \in \N$ such that $\forall r > N: x_{n_r} \in U$.
Thus $\sequence {x_{n_r} }$ converges to $a$.
Otherwise $\sequence {x_n}$ only takes a finite number of distinct values.
Then, since $\sequence {x_n}$ is infinite, there exists $x \in S$ such that:
:$\forall N \in \N: \exists n > N: x = x_n$
This implies that we can take a subsequence of $\sequence {x_n}$ which is constant, and which converges to that constant.
We can conclude then that, by definition, $T$ is a sequentially compact space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $27$. Modified Fort Space: $7$
Category:Sequentially Compact Spaces
Category:Modified Fort Space
Modified Fort Space is T1

== Theorem ==
Let $T = \struct {S, \tau_{a, b} }$ be a modified Fort space.
Then $T$ is a $T_1$ (Fréchet) space.
== Proof ==
Let $p, q \in S: p \ne q$.
There are two options:
:$(1): \quad$ At least one of $p$ and $q$ is not in $\set {a, b}$
:$(2): \quad$ Both $p$ and $q$ are in $\set {a, b}$.
$(1): \quad$ First let $p \notin \set {a, b}$ or $q \notin \set {a, b}$.
, let $p \notin \set {a, b}$.
From Clopen Points in Modified Fort Space both $\set p \in \tau_{a, b}$ and $S \setminus \set p \in \tau_{a, b}$.
So:
:$p \in \set p, q \notin \set p$
:$p \notin S \setminus \set p, q \in S \setminus \set p$
Similarly for $q \notin \set {a, b}$.
$(2): \quad$ Next, let $p, q \in \set {a, b}$.
Let $V_p \subseteq S$ such that $p \in V_p, q \notin V_p$ and $V_p$ is finite.
Then $S \setminus V_p$ is open in $T$ containing $q$ but not $p$.
Similarly, let $V_q \subseteq S$ such that $q \in V_q, p \notin V_q$ and $V_q$ is finite.
Then $S \setminus V_q$ is open in $T$ containing $p$ but not $q$.
Hence it can be seen that $T$ is a $T_1$ (Fréchet) space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $27$. Modified Fort Space: $2$
Category:T1 Spaces
Category:Modified Fort Space
Modified Fort Space is Totally Disconnected

== Theorem ==
Let $T = \struct {S, \tau_{a, b} }$ be a modified Fort space.
Then $T$ is totally disconnected.
== Proof ==
We have:
:Modified Fort Space is Scattered
:Modified Fort Space is $T_1$
We also have:
:Scattered $T_1$ Space is Totally Disconnected
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $27$. Modified Fort Space: $4$
Category:Totally Disconnected Spaces
Category:Modified Fort Space
Modified Fort Space is not Locally Connected

== Theorem ==
Let $T = \struct {S, \tau_{a, b} }$ be a modified Fort space.
Then $T$ is not locally connected.
== Proof ==
We have:
:Modified Fort Space is Totally Disconnected
:Totally Disconnected and Locally Connected Space is Discrete
We also have:
:Modified Fort Space is not $T_2$
:Discrete Space satisfies all Separation Properties
Hence modified Fort space is not the discrete space, and the result follows.
== Sources ==
* : Part $\text {II}$: Counterexamples: $27$. Modified Fort Space: $4$
Category:Locally Connected Spaces
Category:Modified Fort Space
Modified Fort Space is not T2

== Theorem ==
Let $T = \struct {S, \tau_{a, b} }$ be a modified Fort space.
Then $T$ is not a $T_2$ (Hausdorff) space.
== Proof ==
Consider $U, V \in \tau_{a, b}$ such that $a \in U, b \in V$.
We have that both $U$ and $V$ are cofinite.
So $U$ and $V$ must be infinite.
Suppose $U \cap V = \O$.
Then from Intersection with Complement is Empty iff Subset it follows that $U \subseteq \relcomp S V$ and so $U$ is finite.
But this contradicts the fact that $U$ is infinite.
So $U \cap V \ne \O$.
Hence the result by definition of $T_2$ (Hausdorff) space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $27$. Modified Fort Space: $3$
Category:Hausdorff Spaces
Category:Modified Fort Space
Modified Fort Space is not T3, T4 or T5

== Theorem ==
Let $T = \struct {S, \tau_{a, b} }$ be a modified Fort space.
Then $T$ is not a $T_3$ space, $T_4$ space or $T_5$ space.
== Proof ==
We have that a Modified Fort Space is $T_1$.
From $T_1$ Space is $T_0$ Space, $T$ is a $T_0$ space.
We then have that a Modified Fort Space is not $T_2$.
From Regular Space is $T_2$ Space, $T$ is not a regular space.
By definition, a regular space is a space that is both a $T_0$ space and a $T_3$ space.
But $T$ is a $T_0$ space and not a regular space.
So it follows that $T$ can not be a $T_3$ space.
Next we have that a Normal Space is a $T_3$ Space.
But as $T$ is not a $T_3$ space, $T$ can not be a normal space.
By definition, a normal space is a space that is both a $T_1$ space and a $T_4$ space.
But $T$ is a $T_1$ space and not a normal space.
So it follows that $T$ can not be a $T_4$ space.
Finally we have that a $T_5$ Space is a $T_4$ Space.
But as $T$ is not a $T_4$ space, $T$ can not be a $T_5$ space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $27$. Modified Fort Space: $3$
Category:T3 Spaces
Category:T4 Spaces
Category:T5 Spaces
Category:Modified Fort Space
Modified Fort Space is not Totally Separated

== Theorem ==
Let $T = \struct {S, \tau_{a, b} }$ be a modified Fort space.
Then $T$ is not totally separated.
== Proof ==
We have:
:Totally Separated Space is Completely Hausdorff and Urysohn
:Completely Hausdorff Space is Hausdorff Space
But we have:
:Modified Fort Space is not Hausdorff
The result follows from Modus Tollendo Tollens.
== Sources ==
* : Part $\text {II}$: Counterexamples: $27$. Modified Fort Space: $5$
Category:Totally Separated Spaces
Category:Modified Fort Space
Modified Fort Space is not Zero Dimensional

== Theorem ==
Let $T = \struct {S, \tau_{a, b} }$ be a modified Fort space.
Then $T$ is not zero dimensional.
== Proof ==
We have Modified Fort Space is not $T_3$.
We also have Zero Dimensional Space is $T_3$.
The result follows by Modus Tollendo Tollens.
== Sources ==
* : Part $\text {II}$: Counterexamples: $27$. Modified Fort Space: $5$
Category:Zero Dimensional Spaces
Category:Modified Fort Space
Modified Fort Topology is Topology

== Theorem ==
Let $T = \struct {S, \tau_{a, b} }$ be a modified Fort space.
Then $\tau_{a, b}$ is a topology on $S$.
== Proof ==
Recall the definition of a modified Fort space:
Let $S = N \cup \set {a, b}$ where $N$ is infinite, $a \ne b$ and $a, b \notin N$.
We have that $\O \subseteq N$ so $\O \in \tau_{a, b}$.
We have that $a, b \in S, S \setminus S = \O$ and $\O$ is trivially finite, so $S \in \tau_{a, b}$.
Now consider $A, B \in \tau_{a, b}$, and let $H = A \cap B$.
Let either $A \subseteq N$ or $B \subseteq N$.
From Intersection is Subset we have that $A \cap B \subseteq N$ and $A \cap B \subseteq B$.
Hence $A \cap B \subseteq N$ from Subset Relation is Transitive.
So by definition $A \cap B \in \tau_{a, b}$
Now suppose $A \cap \set {a, b} \ne \O$ and $B \cap \set {a, b} \ne \O$.
Then:
In order for $A$ and $B$ to be open sets we have that $N \setminus A$ and $N \setminus B$ are both finite.
Hence their union is also finite and so $N \setminus \paren {A \cap B}$ is finite.
So $H = A \cap B \in \tau_{a, b}$ as its complement is finite.
Now let $\UU \subseteq \tau_{a, b}$.
Then from De Morgan's Laws: Difference with Union:
:$\ds N \setminus \paren {\bigcup \UU} = \bigcap_{U \mathop \in \UU} \paren {N \setminus U}$
We have either of two options:
:$(1): \quad \forall U \in \UU: U \subseteq N$
in which case:
:$\ds \bigcup \UU \subseteq N$
Or:
:$(2): \quad \exists U \in \UU: N \setminus U$ is finite
in which case:
:$\ds \bigcap_{U \mathop \in \UU} \paren {N \setminus U}$ is finite, from Intersection is Subset.
So in either case:
:$\ds \bigcup \UU \in \tau_{a, b}$
== Sources ==
* : Part $\text {II}$: Counterexamples: $27$. Modified Fort Space
Category:Modified Fort Space
Modified Kaprekar Process on 4-Digit Number terminates in 2538/Historical Note

== Historical Note on Modified Kaprekar Process on 4-Digit Number terminates in 2538 ==
This result is reported by in his $1997$ work to appear in volume $22$ on page $34$ in an article authored by , but this has not been corroborated.
== Sources ==
* : $2538$
Category:Historical Notes
Module is Submodule of Itself

== Theorem ==
Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Then $\struct {G, +_G, \circ}_R$ is a submodule of itself.
== Proof ==
Follows directly from Group is Subgroup of Itself.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Example $27.1$
Category:Submodules
Module of All Mappings is Module

== Theorem ==
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $\struct {G, +_G, \circ}_R$ be an $R$-module.
Let $S$ be a set.
Let $\struct {G^S, +_G', \circ}_R$ be the module of all mappings from $S$ to $G$.
Then $\struct {G^S, +_G', \circ}_R$ is an $R$-module.
== Proof ==
To show that $\struct {G^S, +_G', \circ}_R$ is an $R$-module, we verify the following:
$\forall f, g, \in G^S, \forall \lambda, \mu \in R$:
:$(1): \quad \lambda \circ \paren {f +_G' g} = \paren {\lambda \circ f} +_G' \paren {\lambda \circ g}$
:$(2): \quad \paren {\lambda +_R \mu} \circ f = \paren {\lambda \circ f} +_G \paren {\mu \circ f}$
:$(3): \quad \paren {\lambda \times_R \mu} \circ f = \lambda \circ \paren {\mu \circ f}$
=== Criterion 1 ===
:$(1): \quad \lambda \circ \paren {f +_G' g} = \paren {\lambda \circ f} +_G' \paren {\lambda \circ g}$
Let $x \in S$.
Then:
Thus $(1)$ holds.
=== Criterion 2 ===
: $(2): \quad \paren {\lambda +_R \mu} \circ f = \paren {\lambda \circ f} +_G \paren {\mu \circ f}$
Let $x \in S$.
Thus $(2)$ holds.
=== Criterion 3 ===
:$(3): \quad \paren {\lambda \times_R \mu} \circ f = \lambda \circ \paren {\mu \circ f}$
Thus $(3)$ holds.
Hence the result.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Example $26.4$
Category:Examples of Modules
Module on Cartesian Product is Module

== Theorem ==
Let $\struct {R, +_R, \times_R}$ be a ring.
Let $n \in \N_{>0}$.
Let $\struct {R^n, +, \times}_R$ be the $R$-module $R^n$.
Then $\struct {R^n, +, \times}_R$ is an $R$-module.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Example $26.1$
Category:Examples of Modules
Category:Direct Products
Category:Module on Cartesian Product
Category:Module on Cartesian Product is Module
Module on Cartesian Product of Ring with Unity is Unitary Module

== Theorem ==
Let $\struct {R, +_R, \times_R}$ be a ring with unity.
Let $n \in \N_{>0}$.
Let $\struct {R^n, +, \times}_R$ be the $R$-module $R^n$.
Then $\struct {R^n, +, \times}_R$ is a unitary $R$-module.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Example $26.1$
Category:Examples of Unitary Modules
Category:Direct Products
Category:Module on Cartesian Product
Category:Module on Cartesian Product of Ring with Unity is Unitary Module
Modulo Addition/Cayley Table

== Cayley Table for Modulo Addition ==
The additive group of integers modulo $m$ can be described by showing its Cayley table.
=== Modulo 3 ===
=== Modulo 4 ===
=== Modulo 5 ===
=== Modulo 6 ===
Category:Examples of Cayley Tables
Category:Modulo Addition
Category:Examples of Additive Groups of Integers Modulo m
Modulo Addition/Cayley Table/Modulo 3

== Cayley Table for Addition Modulo $3$ ==
The additive group of integers modulo $3$ can be described by showing its Cayley table:
:$\begin {array} {r|rrr} \struct {\Z_3, +_3} & \eqclass 0 3 & \eqclass 1 3 & \eqclass 2 3 \\ \hline \eqclass 0 3 & \eqclass 0 3 & \eqclass 1 3 & \eqclass 2 3 \\ \eqclass 1 3 & \eqclass 1 3 & \eqclass 2 3 & \eqclass 0 0 \\ \eqclass 2 3 & \eqclass 2 3 & \eqclass 0 3 & \eqclass 1 3 \\ \end {array}$
It can also be presented:
:$\begin {array} {r|rrr} +_3 & 0 & 1 & 2 \\ \hline 0 & 0 & 1 & 2 \\ 1 & 1 & 2 & 0 \\ 2 & 2 & 0 & 1 \\ \end {array}$
== Sources ==
* : $\S 6.7$. Quotient groups: Example $126$
* : Chapter $\text{II}$: Groups: A Little Number Theory
* : Chapter $2$: Maps and relations on sets: Example $2.33$
Category:Examples of Cayley Tables
Category:Examples of Additive Groups of Integers Modulo m
Category:Cyclic Group of Order 3
Modulo Addition/Cayley Table/Modulo 4

== Cayley Table for Addition Modulo $4$ ==
The additive group of integers modulo $4$ can be described by showing its Cayley table:
$\quad \begin {array} {r|rrrr}
\struct {\Z_4, +_4} & \eqclass 0 4 & \eqclass 1 4 & \eqclass 2 4 & \eqclass 3 4 \\
\hline
\eqclass 0 4 & \eqclass 0 4 & \eqclass 1 4 & \eqclass 2 4 & \eqclass 3 4
\\
\eqclass 1 4 & \eqclass 1 4 & \eqclass 2 4 & \eqclass 3 4 & \eqclass 0 4 
\\
\eqclass 2 4 & \eqclass 2 4 & \eqclass 3 4 & \eqclass 0 4 & \eqclass 1 4
\\
\eqclass 3 4 & \eqclass 3 4 & \eqclass 0 4 & \eqclass 1 4 & \eqclass 2 4
\\
\end{array}$
It can also be presented:
$\quad \begin {array} {r|rrrr}
+_4 & 0 & 1 & 2 & 3 \\
\hline
0 & 0 & 1 & 2 & 3 
\\
1 & 1 & 2 & 3 & 0
\\
2 & 2 & 3 & 0 & 1
\\
3 & 3 & 0 & 1 & 2
\\
\end {array}$
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Example $6.2$
* : Chapter $1$: Integral Domains: $\S 6$. The Residue Classes
* : Chapter $2$: Maps and relations on sets: Exercise $5$
Category:Examples of Cayley Tables
Category:Cyclic Group of Order 4
Category:Examples of Additive Groups of Integers Modulo m
Modulo Addition/Cayley Table/Modulo 5

== Cayley Table for Modulo Addition ==
The additive group of integers modulo $m$ can be described by showing its Cayley table.
This one is for modulo $5$:
$\quad \begin {array} {r|rrrrr}
\struct {\Z_5, +_5} & \eqclass 0 5 & \eqclass 1 5 & \eqclass 2 5 & \eqclass 3 5 & \eqclass 4 5 \\
\hline
\eqclass 0 5 & \eqclass 0 5 & \eqclass 1 5 & \eqclass 2 5 & \eqclass 3 5 & \eqclass 4 5
\\
\eqclass 1 5 & \eqclass 1 5 & \eqclass 2 5 & \eqclass 3 5 & \eqclass 4 5 & \eqclass 0 5
\\
\eqclass 2 5 & \eqclass 2 5 & \eqclass 3 5 & \eqclass 4 5 & \eqclass 0 5 & \eqclass 1 5
\\
\eqclass 3 5 & \eqclass 3 5 & \eqclass 4 5 & \eqclass 0 5 & \eqclass 1 5 & \eqclass 2 5
\\
\eqclass 4 5 & \eqclass 4 5 & \eqclass 0 5 & \eqclass 1 5 & \eqclass 2 5 & \eqclass 3 5
\\
\end {array}$
which can also be presented:
$\quad \begin {array} {r|rrrrr}
+_5 & 0 & 1 & 2 & 3 & 4 \\
\hline
0 & 0 & 1 & 2 & 3 & 4 
\\
1 & 1 & 2 & 3 & 4 & 0
\\
2 & 2 & 3 & 4 & 0 & 1
\\
3 & 3 & 4 & 0 & 1 & 2
\\
4 & 4 & 0 & 1 & 2 & 3
\\
\end {array}$
== Sources ==
* : $\S 19$: Properties of $\Z_m$ as an algebraic system
* 
Category:Examples of Cayley Tables
Category:Examples of Additive Groups of Integers Modulo m
Modulo Addition/Cayley Table/Modulo 6

== Cayley Table for Modulo Addition ==
The additive group of integers modulo $m$ can be described by showing its Cayley table.
This one is for modulo $6$:
$\quad \begin{array}{r|rrrrrr}
\struct {\Z_6, +_6} & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 \\
\hline
\eqclass 0 6 & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6
\\
\eqclass 1 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 & \eqclass 0 6
\\
\eqclass 2 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 & \eqclass 0 6 & \eqclass 1 6
\\
\eqclass 3 6 & \eqclass 3 6 & \eqclass 4 6 & \eqclass 5 6 & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6
\\
\eqclass 4 6 & \eqclass 4 6 & \eqclass 5 6 & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6
\\
\eqclass 5 6 & \eqclass 5 6 & \eqclass 0 6 & \eqclass 1 6 & \eqclass 2 6 & \eqclass 3 6 & \eqclass 4 6
\\
\end{array}$
which can also be presented:
$\quad \begin{array}{r|rrrrrr}
+_6 & 0 & 1 & 2 & 3 & 4 & 5 \\
\hline
0 & 0 & 1 & 2 & 3 & 4 & 5 
\\
1 & 1 & 2 & 3 & 4 & 5 & 0
\\
2 & 2 & 3 & 4 & 5 & 0 & 1
\\
3 & 3 & 4 & 5 & 0 & 1 & 2
\\
4 & 4 & 5 & 0 & 1 & 2 & 3
\\
5 & 5 & 0 & 1 & 2 & 3 & 4
\\
\end{array}$
== Sources ==
* : Tables: $2$. Cyclic group of order $6$
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Example $2.3$
Category:Examples of Cayley Tables
Category:Groups of Order 6
Category:Examples of Additive Groups of Integers Modulo m
Modulo Addition has Identity

== Theorem ==
Let $m \in \Z$ be an integer.
Then addition modulo $m$ has an identity:
:$\forall \eqclass x m \in \Z_m: \eqclass x m +_m \eqclass 0 m = \eqclass x m = \eqclass 0 m +_m \eqclass x m$
That is:
:$\forall a \in \Z: a + 0 \equiv a \equiv 0 + a \pmod m$
== Proof ==
Thus $\eqclass 0 m$ is the identity for addition modulo $m$.
== Sources ==
* : Chapter $\text{II}$: Groups: A Little Number Theory
* : $\S 1$: Some examples of groups: Example $1.10$
* : Chapter $1$: Integral Domains: $\S 6$. The Residue Classes: Theorem $5$
* : $\S 19.1$: Properties of $\Z_m$ as an algebraic system
Category:Modulo Addition