nyuuzyou/wikis huggingface dataset:
Complement of Set of Rational Pairs in Real Euclidean Plane is Arc-Connected/Mistake

== Source Work ==
:
:Part $\text {II}$: Counterexamples
::Section $33$: Special Subsets of the Plane
:::Item $2$
== Mistake ==
:Let $A$ be the subset of $\R^2 = \R \times \R$ consisting of all point with at least one irrational coordinate, and let $A$ have the induced topology. $A$ is arc-connected since a point $\tuple {x_1, y_1}$ with two irrational coordinates may be joined by an arc to any point $\tuple {a, b} \in A$ either $a$ or $b$ is irrational, say $a$. Then the union of the lines $x = a, y = y_1$ is an arc-connected subset of $X$ connecting $\tuple {x_1, y_1}$ to $\tuple {a, b}$. ...
== Correction ==
There is a minor typo here: that last sentence should say:
:Then the union of the lines $x = x_1, y = y_1$ is an arc-connected subset of $A$ connecting $\tuple {x_1, y_1}$ to $\tuple {a, b}$.
== Also see ==
* Complement of Set of Rational Pairs in Real Euclidean Plane is Arc-Connected
== Sources ==
* : Part $\text {II}$: Counterexamples: $33$. Special Subsets of the Plane: $2$
Category:Mistakes/Counterexamples in Topology
Complement of Singleton Closure is Meet-Irreducible

== Theorem ==
Let $\struct {S, \tau}$ be a topological space.
Let $x \in S$.
Then:
:$S \setminus \set x^-$ is a meet-irreducible open set
where $\set x^-$ denotes the closure of $\set x$
== Proof ==
From Characterization of Meet-Irreducible Open Set:
$S \setminus \set x^-$ is a meet-irreducible open set 
:$\forall U, V \in \tau : \paren {U \cap V \subseteq S \setminus \set x^- \implies U \subseteq S \setminus \set x^- \text { or } V \subseteq S \setminus \set x^-}$
Let:
:$U, V \in \tau: U \cap V \subseteq S \setminus \set x^-$
By Set is Subset of its Topological Closure:
:$x \in \set x^-$
By definition of set difference:
:$x \notin S \setminus \set x^-$
By definition of subset:
:$x \notin U \cap V$
By definition of set intersection:
:either $x \notin U$ or $x \notin V$
, suppose $x \notin U$.
By definition of set complement:
:$x \in S \setminus U$
We have $S \setminus U$ is a closed set by definition.
From Closure of Subset of Closed Set of Topological Space is Subset:
:$\set x^- \subseteq S \setminus U$
From Subset of Set Difference iff Disjoint Set:
:$U \subseteq S \setminus \set x^-$
The result follows.
== Sources ==
* : Chapter $1$: Spaces and Lattices of Open Sets, $\S 1$ Sober spaces, Definition $1.1$
Category:Meet-Irreducible Open Sets
Category:Set Closures
Complement of Strict Total Ordering

== Theorem ==
Let $\struct {S, \prec}$ be a relational structure such that $\prec$ is a strict total ordering.
Then the complement of $\prec$ is a weak total ordering.
== Proof ==
We need to show that $\nprec$ is a weak total ordering.
First we check in turn each of the criteria for an ordering:
=== Reflexivity ===
By Complement of Reflexive Relation, it follows directly that $\nprec$ is reflexive.
=== Transitivity ===
Let $a \nprec b$ and $b \nprec c$.
Now, suppose that $a \prec c$.
As $b \nprec c$, from the Trichotomy Law, one of the following is the case:
:$b = c$ and so $a \prec b$;
:$c \prec b$ and $a \prec b$ from transitivity of $\prec$.
Both are an immediate contradiction.
So $a \nprec c$ and so $\nprec$ is transitive.
=== Antisymmetry ===
Suppose $a \nprec b$.
Then from the Trichotomy Law, either $a = b$ or $b \prec a$.
Similarly, if $b \nprec a$ we have that either $a = b$ or $a \prec b$.
As $\prec$ is asymmetric, $b \prec a \iff \neg a \prec b$.
So if $a \nprec b$ and $b \nprec a$, it follows that $a = b$.
So $\nprec$ is antisymmetric.
The fact that $\nprec$ is connected follows directly from the Trichotomy Law.
Category:Total Orderings
Complement of Subset with Property (S) is Closed under Directed Suprema

== Theorem ==
Let $L = \struct {S, \preceq}$ be an up-complete ordered set.
Let $X$ be a subset of $S$ with property (S).
Then $\relcomp S X$ is closed under directed suprema.
== Proof ==
Let $D$ be a directed subset of $S$ such that:
:$D \subseteq \relcomp S X$
:$\sup D \notin \relcomp S X$
By definition of relative complement:
:$\sup D \in X$
By definition of property (S):
:$\exists y \in D: \forall x \in D: y \preceq x \implies x \in X$
By definition of reflexivity:
:$y \in X$
By definitions of intersection and non-empty:
:$D \cap X \ne \O$
Thus this by Empty Intersection iff Subset of Complement contradicts:
:$D \subseteq \relcomp S X$
== Sources ==
* 
Category:Order Theory
Complement of Symmetric Relation

== Theorem ==
Let $\RR \subseteq S \times S$ be a relation.
Then $\RR$ is symmetric its complement $\relcomp {S \times S} \RR \subseteq S \times S$ is also symmetric.
== Proof ==
Let $\RR \subseteq S \times S$ be symmetric.
Then from Symmetry of Relations is Symmetric:
:$\tuple {x, y} \in \RR \iff \tuple {y, x} \in \RR$
 $\relcomp {S \times S} \RR \subseteq S \times S$ is not symmetric.
Then:
:$\exists \tuple {x, y} \in \relcomp {S \times S} \RR: \tuple {y, x} \notin \relcomp {S \times S} \RR$
But then by definition of complement of $\RR$:
:$\tuple {y, x} \in \RR$
As $\RR$ is symmetric it follows that:
:$\tuple {x, y} \in \RR$
This contradicts the premise that $\tuple {x, y} \in \relcomp {S \times S} \RR$.
Hence by Proof by Contradiction it follows that $\relcomp {S \times S} \RR$ is symmetric.
The converse follows similarly.
Category:Symmetric Relations
Category:Relative Complement
Complement of Top/Boolean Algebra

== Theorem ==
Let $\struct {S, \vee, \wedge, \neg}$ be a Boolean algebra.
Then:
:$\neg \top = \bot$
== Proof ==
Since $\top$ is the identity for $\wedge$, the second condition for $\neg \top$:
:$\top \wedge \neg \top = \bot$
implies that $\neg \top = \bot$ is the only possibility.
Since $\bot$ is the identity for $\vee$, it follows that:
:$\top \vee \bot = \top$
and we conclude that:
:$\neg \top = \bot$
as desired.
== Also see ==
* Complement of Bottom (Boolean Algebras)
== Sources ==
* : $\S 2$: Exercise $2$
Category:Boolean Algebras
Complement of Top/Bounded Lattice

== Theorem ==
Let $\struct {S, \vee, \wedge, \preceq}$ be a bounded lattice.
Then the top $\top$ has a unique complement, namely $\bot$, bottom.
== Proof ==
By Dual Pairs (Order Theory), $\top$ is dual to $\bot$.
The result follows from the Duality Principle and Complement of Bottom.
== Also see ==
* Complement of Bottom
Category:Bounded Lattices
Complement of Universal Set is Empty Set

== Theorem ==
The complement of the universal set is the empty set:
:$\map \complement {\mathbb U} = \O$
== Proof ==
Substitute $\mathbb U$ for $S$ in $\relcomp S S = \O$ from Relative Complement with Self is Empty Set.
== Also see ==
* Complement of Empty Set is Universal Set
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{B i}$
* : Appendix $\text{A}.1$: Sets and Subsets
* : $\S 6.6$: Subsets
* : $\S 2$
* : $\text {(viii)}$
Category:Set Complement
Category:Universal Set
Category:Empty Set
Complement of Upper Closure of Element is Open in Lower Topology

== Theorem ==
Let $T = \struct {S, \preceq, \tau}$ be a relational structure with lower topology.
Let $x \in S$.
Then $\relcomp S {x^\succeq}$ is open and $x^\succeq$ is closed.
== Proof ==
Define $B := \set {\relcomp S {y^\succeq}: y \in S}$
By definition of lower topology:
:$B$ is sub-basis of $T$.
By definition of sub-basis:
:$B \subseteq \tau$
By definition of $B$:
:$\relcomp S {x^\succeq} \in B$
Thus by definition of subset:
:$\relcomp S {x^\succeq} \in \tau$
Thus by definition:
:$x^\succeq$ is closed.
== Sources ==
* 
Category:Topological Order Theory
Complement of Upper Section is Lower Section

== Theorem ==
Let $\struct {S, \preceq}$ be an ordered set.
Let $L$ be an upper section.
Then $S \setminus L$ is a lower section.
== Proof ==
This follows mutatis mutandis from the proof of Complement of Lower Section is Upper Section.
== Also see ==
* Complement of Lower Section is Upper Section
== Sources ==
* 
Category:Upper Sections
Category:Lower Sections
Complement of Vertical Section of Set is Vertical Section of Complement

== Theorem ==
Let $X$ and $Y$ be sets.
Let $E \subseteq X \times Y$. 
Let $x \in X$. 
Then: 
:$\paren {\paren {X \times Y} \setminus E}_x = Y \setminus E_x$
where:
:$\paren {\paren {X \times Y} \setminus E}_x$ is the $x$-vertical section of the set difference $\paren {X \times Y} \setminus E$
:$E_x$ is the $x$-vertical section of $E$. 
 
== Proof ==
Note that from the definition of set difference, we have that: 
:$y \in Y \setminus E_x$
: 
:$y \in Y$ and $y \not \in E_x$.
That is, from the definition of the $x$-vertical section: 
:$y \in Y$ and $\tuple {x, y} \not \in E$.
This is equivalent to:
:$\tuple {x, y} \in \paren {X \times Y} \setminus E$ 
From the definition of the $x$-vertical section, this is then equivalent to: 
:$y \in \paren {\paren {X \times Y} \setminus E}_x$
So we have:
:$y \in Y \setminus E_x$ $y \in \paren {\paren {X \times Y} \setminus E}_x$.
So:
:$\paren {\paren {X \times Y} \setminus E}_x = Y \setminus E_x$
Category:Vertical Section of Sets
Complementary Error Function of Zero

== Theorem ==
:$\map \erfc 0 = 1$
where $\erfc$ denotes the complementary error function.
== Proof ==
== Sources ==
* : $\S 35$: Complementary Error Function $\ds \map \erfc x = 1 - \map \erf x = \frac 2 {\sqrt \pi} \int_x^\infty e^{-u^2} \rd u$: $35.6$
Category:Complementary Error Function
Complementary Idempotent is Idempotent

== Theorem ==
Let $\HH$ be a Hilbert space.
Let $I$ be an identity operator on $\HH$.
Let $A$ be an idempotent operator.
Then the complementary idempotent $I - A$ is also idempotent.
== Proof ==
That is, $I - A$ is idempotent.
== Also see ==
* Complementary Idempotent of Complementary Idempotent is Idempotent
== Sources ==
* $\text{II}.3.2 (a)$
Category:Linear Transformations on Hilbert Spaces
Complementary Idempotent of Complementary Idempotent is Idempotent

== Theorem ==
Let $\HH$ be a Hilbert space.
Let $I$ be an identity operator on $\HH$.
Let $A$ be an idempotent operator.
Let $B$ be the complementary idempotent of A.
Then the complementary idempotent of $B$ is $A$.
== Proof ==
From Complementary Idempotent is Idempotent the complementary idempotent of $B$ is well-defined.
Let $C$ be the complementary idempotent of $B$.
We have:
Category:Linear Transformations on Hilbert Spaces
Complementary Projection is Complementary Idempotent

== Theorem ==
Let $\HH$ be a Hilbert space.
Let $A$ be a projection.
Let $B$ be the complementary projection of A.
Then $B$ is the complementary idempotent of $A$.
== Proof ==
By the definition of projection, $A$ is an idempotent operator.
The result follows immediately from the definitions of:
* complementary projection
* complementary idempotent
where the constructions of the complementary projection and the complementary idempotent from $A$ are identical.
Category:Linear Transformations on Hilbert Spaces
Complementary Projection is Projection

== Theorem ==
Let $\HH$ be a Hilbert space.
Let $A$ be a projection.
Then the complementary projection $I - A$ is also a projection.
== Proof ==
By Characterization of Projections, $A$ is Hermitian.
Then $\paren {I - A}^* = I^* - A^* = I - A$ from Adjoining is Linear.
So $I - A$ is also Hermitian.
From Complementary Idempotent is Idempotent, $I - A$ is idempotent.
Hence, applying Characterization of Projections, $I - A$ is a projection.
== Also ==
* Complementary Projection of Complementary Projection is Projection 
Category:Linear Transformations on Hilbert Spaces
Complementary Projection of Complementary Projection is Projection

== Theorem ==
Let $\HH$ be a Hilbert space.
Let $I$ be an identity operator on $\HH$.
Let $A$ be a projection.
Let $B$ be the complementary projection of A.
Then the complementary projection of $B$ is $A$.
== Proof ==
From Complementary Projection is Projection the complementary projection of $B$ is well-defined.
Let $C$ be the complementary projection of $B$.
We have:
Category:Linear Transformations on Hilbert Spaces
Complements of Components of Two-Component Partition form Partition

== Theorem ==
Let $S$ be a set with at least two elements.
Let $A, B \subseteq S$.
Let $\complement_S$ denote the complement relative to $S$.
$A \mid B$ is a partition of $S$ $\relcomp S A \mid \relcomp S B$ is a partition of $S$.
== Proof ==
=== Necessary Condition ===
Let $A \mid B$ be a partition of $S$.
That is, by definition:
:$(1): \quad A \cap B = \O$
:$(2): \quad A \cup B = S$
:$(3): \quad A, B \ne \O$
Thus:
, now suppose that $\relcomp S A = \O$.
Thus by Rule of Transposition:
:$B \ne \O \implies \relcomp S A \ne \O$
Mutatis mutandis:
:$A \ne \O \implies \relcomp S B \ne \O$
by hypothesis:
:$A, B \ne \O$
and so:
:$\relcomp S A, \relcomp S B \ne \O$
Thus all three conditions are satisfied for $\relcomp S A \mid \relcomp S B$ to be a partition of $S$.
=== Sufficient Condition ===
Let $\relcomp S A \mid \relcomp S B$ be a partition of $S$.
By the necessary condition, it follows that $\relcomp S {\relcomp S A} \mid \relcomp S {\relcomp S B}$ is a partition of $S$.
From Relative Complement of Relative Complement:
:$\relcomp S {\relcomp S A} = A$ and $\relcomp S {\relcomp S B} = B$
and so $A \mid B$ is a partition of $S$.
Category:Relative Complement
Category:Set Partitions
Complements of Parallelograms are Equal

== Theorem ==
== Proof ==
:
Let $ABCD$ be a parallelogram, and let $AC$ be a diameter.
Let $EKHA$ and $FKGC$ be parallelograms about $AC$.
Let $BEKG$ and $DFKH$ be the complements of $EKHA$ and $FKGC$.
From Opposite Sides and Angles of Parallelogram are Equal:
:$\triangle ABC = \triangle ACD$
Also from Opposite Sides and Angles of Parallelogram are Equal:
:$\triangle AEK = \triangle AHK$
:$\triangle KFC = \triangle KGC$
So from Common Notion 2:
:$\triangle AEK + \triangle KGC = \triangle AHK + \triangle KFC$
But the whole of $\triangle ABC$ equals the whole of $\triangle ACD$.
So when $\triangle AEK + \triangle KGC$ and $\triangle AHK + \triangle KFC$ are subtracted from $\triangle ABC$ and $\triangle ACD$ respectively, the complement $BEKG$ which remains is equal in area to $DFKH$, from Common Notion 3.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Parallelograms
Complete Archimedean Valued Field is Real or Complex Numbers

== Theorem ==
Let $\struct{k, \norm{\,\cdot\,}}$ be a complete valued field with Archimedean norm $\norm{\,\cdot\,}$.
Then either:
:$k$ is isomorphic to the real numbers $\R$ and $\norm{\,\cdot\,}$ is equivalent to the absolute value $\size{\,\cdot\,}$ on $\R$
or
:$k$ is isomorphic to the complex numbers $\C$ and $\norm{\,\cdot\,}$ is equivalent to the complex modulus $\size{\,\cdot\,}$ on $\C$
== Proof ==
== Also see ==
This result is sometimes called Ostrowski's theorem. 
In the same paper, published in $1918$, also proved that every non-trivial norm on the rational numbers is equivalent to either the $p$-adic Norm for some prime $p$
or the absolute value.
It is this latter result that is more often referred to as Ostrowski's Theorem.
 
== Sources ==
* : $\S 1$ Introduction, Theorem $1.1$
Category:Field Theory
Category:Norm Theory
Complete Bipartite Graphs which are Complete Graphs

== Theorem ==
Let $K_{m, n}$ be a complete bipartite graph.
:$K_{0, 0}$ is the complete graph $K_0$
:$K_{0, 1}$ and $K_{1, 0}$ are the complete graph $K_1$
:$K_{1, 1}$ is the complete graph $K_2$
and no other complete bipartite graphs are complete.
== Proof ==
$K_{0, 0}$ is the null graph by Null Graph is Complete Bipartite Graph.
Then by Null Graph is Complete Graph, $K_{0, 0}$ is the complete graph $K_0$.
$K_{0, 1}$ and $K_{1, 0}$ consist of one vertex and no edges.
Then by Complete Graph of Order 1 is Edgeless, $K_{0, 1}$ and $K_{1, 0}$ are both the complete graph $K_1$.
That $K_{1, 1}$ is the complete graph $K_2$ can be determined by inspection:
:
Suppose either $m > 1$ or $n > 1$.
Then one of the partite sets contains more than $1$ vertex.
By the nature of a bipartite graph, $2$ such vertices are not joined by an edge.
Hence by definition such a $K_{m, n}$ is not a complete graph.
Category:Complete Bipartite Graphs
Category:Complete Graphs
Complete Bipartite Graphs which are Cycle Graphs

== Theorem ==
Let $K_{m, n}$ be a complete bipartite graph.
:$K_{2, 2}$ is the cycle graph $C_4$
and no other complete bipartite graphs are cycle graphs.
== Proof ==
From Cycle Graph is 2-Regular, a cycle graph is $2$-regular.
From Complete Bipartite Graphs which are Regular, the only $2$-regular complete bipartite graph is $K_{2, 2}$.
Hence the result.
Category:Complete Bipartite Graphs
Category:Cycle Graphs
Complete Bipartite Graphs which are Path Graphs

== Theorem ==
Let $K_{m, n}$ be a complete bipartite graph.
:$K_{0, 0}$ is the path graph $P_0$
:$K_{0, 1}$ and $K_{1, 0}$ are the path graph $P_1$
:$K_{1, 1}$ is the path graph $P_2$
:$K_{1, 2}$ and $K_{2, 1}$ are the path graphs $P_3$
and no other complete bipartite graphs are path graphs.
== Proof ==
$K_{0, 0}$, $K_{0, 1}$ and $K_{1, 0}$ have no edges.
Hence they are path graphs vacuously.
That $K_{1, 1}$ is the path graph $P_2$ can be determined by inspection:
:
Similarly, that $K_{1, 2}$ and $K_{2, 1}$ are the path graph $P_3$ can be determined by inspection:
:
We have that Path Graph is Tree.
From Complete Bipartite Graphs which are Trees, a path graph must be of the form $K_{1, n}$ or $K_{n, 1}$ for some $n$.
But for $n \ge 3$ there exists one vertex which is of degree greater than $2$.
Hence from Degree of Vertex in Path Graph is Less than 3, $K_{1, n}$ or $K_{n, 1}$ is not a path graph for $n \ge 3$.
Hence the result.
Category:Complete Bipartite Graphs
Category:Path Graphs
Complete Bipartite Graphs which are Regular

== Theorem ==
Let $G = \struct {A \mid B, E} =: K_{m, n}$ be the complete bipartite graph with $m$ vertices in $A$ and $n$ vertices in $B$.
Then:
:$K_{n, n}$ is $n$-regular for all $n$
:$K_{n, 0}$ and $K_{0, n}$ are $0$-regular for all $n$
and no other complete bipartite graphs are regular.
== Proof ==
First consider $K_{n, 0}$ and $K_{0, n}$ for $n \in \N$.
From Condition for Complete Bipartite Graph to be Edgeless, every vertex in $K_{n, 0}$ and $K_{0, n}$ has degree $0$.
From Graph is 0-Regular iff Edgeless, $K_{n, 0}$ and $K_{0, n}$ are $0$-regular.
Now consider $K_{m, n}$ where $m, n > 0$.
By definition of complete bipartite graph:
:every vertex in $A$ is adjacent to every vertex in $B$
and so:
:every vertex in $B$ is adjacent to every vertex in $A$.
Hence:
:every vertex in $A$ has degree $m$
:every vertex in $B$ has degree $n$
So when $m, n > 0$, every vertex in $K_{m, n}$ has the same degree $m = n$.
Hence the result.
Category:Complete Bipartite Graphs
Category:Regular Graphs
Complete Bipartite Graphs which are Trees

== Theorem ==
Let $G = \struct {A \mid B, E} =: K_{m, n}$ be the complete bipartite graph with $m$ vertices in $A$ and $n$ vertices in $B$.
Then:
:$K_{0, 0}$ is a tree
:$K_{1, n}$ and $K_{n, 1}$ is a tree for all $n$
and no other complete bipartite graphs are trees.
== Proof ==
We note from Null Graph is Complete Bipartite Graph and Null Graph is Tree that $K_{0, 0}$ is a complete bipartite graph that is also a tree.
Next we note that the order of $K_{1, n}$ is $n + 1$.
Indeed, there is $1$ vertex in $A$ and $n$ vertices in $B$, for a total of $n + 1$.
Then we note that the size of $K_{1, n}$ is $n$.
Indeed, each vertex in $B$ has exactly $1$ edge to the $1$ vertex in $A$.
By the definition of complete bipartite graph, that accounts for all edges in $G$.
From Finite Connected Simple Graph is Tree iff Size is One Less than Order it follows that $K_{1, n}$ is a tree.
Similarly:
:the order of $K_{n, 1}$ is $n + 1$
:the size of $K_{n, 1}$ is $n$
and again, $K_{n, 1}$ is a tree
Now consider the complete bipartite graph $G = \struct {A \mid B, E} = K_{0, n}$ where $n > 1$.
Let $b_1, b_2 \in B$.
As $K_{0, n}$ is bipartite, $b_1$ and $b_2$ can be adjacent to vertices in $A$ only.
But $A = \O$ and so $b_1$ and $b_2$ are isolated.
Hence $b_1$ and $b_2$ are not connected.
Hence by definition $G$ is not a connected graph
So by definition $K_{0, n}$ is not a tree.
By it follows that $K_{m, 0}$ is also not a tree for $m > 1$.
Now consider the complete bipartite graph $K_{m, n}$ where $m, n > 1$.
We recall that $K_{m, n} := G = \struct {A \mid B, E}$ such that every vertex in $A$ is adjacent to every vertex in $B$.
As $m, n > 1$ we have that:
Hence we can find a cycle in $G$:
:$\tuple {a_1, b_1, a_2, b_2, a_1}$
As a tree is defined as a (connected) graph with no cycles, it follows that such a $K_{m, n}$ is not a tree.
Hence the result.
Category:Complete Bipartite Graphs
Category:Tree Theory
Complete Boolean Lattice is a Frame

== Theorem ==
Let $\struct{B, \vee, \wedge, \neg, \preceq}$ be a complete Boolean lattice.
Then:
:$\struct{B, \vee, \wedge, \preceq}$ is a frame.
== Proof ==
From Boolean Lattice is Heyting Lattice:
:$\struct{B, \vee, \wedge, \preceq}$ is a complete Heyting lattice.
From Characterization of Locale:
:$\struct{B, \vee, \wedge, \preceq}$ is a frame.
Category:Boolean Lattices
Category:Frames
Complete Boolean Lattice is a Locale

== Theorem ==
Let $\struct{B, \vee, \wedge, \neg, \preceq}$ be a complete Boolean lattice.
Then:
:$\struct{B, \vee, \wedge, \preceq}$ is a locale.
== Proof ==
From Boolean Lattice is Heyting Lattice:
:$\struct{B, \vee, \wedge, \preceq}$ is a complete Heyting lattice.
From Characterization of Locale:
:$\struct{B, \vee, \wedge, \preceq}$ is a locale.
Category:Boolean Lattices
Category:Locales
Complete Elliptic Integral of the First Kind as Power Series

== Theorem ==
The complete elliptic integral of the first kind:
:$\ds \map K k = \int_0^{\pi / 2} \frac {\rd \phi} {\sqrt {1 - k^2 \sin^2 \phi} } = \int_0^1 \frac {\rd v} {\sqrt {\paren {1 - v^2} \paren {1 - k^2 v^2} } }$
can be expressed as the power series:
== Proof ==
From Reduction Formula for Integral of Power of Sine, $\forall i \in \N$:
Hence:
== Sources ==
* : $34.2$: Complete Elliptic Integral of the First Kind
Category:Complete Elliptic Integral of the First Kind
Complete Elliptic Integral of the Second Kind as Power Series

== Theorem ==
The complete elliptic integral of the second kind:
:$\ds \map E k = \int_0^{\pi / 2} \sqrt {1 - k^2 \sin^2 \phi} \, \rd \phi = \int_0^1 \dfrac {\sqrt {1 - k^2 v^2} } {\sqrt {1 - v^2}} \, \rd v$
can be expressed as the power series:
== Proof ==
From Reduction Formula for Integral of Power of Sine, $\forall i \in \N$:
Hence:
== Sources ==
* : $34.4$: Complete Elliptic Integral of the Second Kind
Category:Complete Elliptic Integral of the Second Kind
Complete Factorizations of Proper Element in Principal Ideal Domain are Equivalent

== Theorem ==
Let $\struct {D, +, \circ}$ be a principal ideal domain.
Let $x \in D$ be a proper element of $D$.
Let there be two complete factorizations of $x$:
:$x = u_y \circ y_1 \circ y_2 \circ \cdots \circ y_m = F_1$
:$x = u_z \circ z_1 \circ z_2 \circ \cdots \circ z_n = F_2$
Then $F_1$ and $F_2$ are equivalent.
== Proof ==
#
== Sources ==
* : $\S 62$. Factorization in an integral domain
Category:Factorization
Category:Principal Ideal Domains
Complete Graph K5 is not Planar

== Theorem ==
The complete graph $K_5$ is not planar.
== Proof ==
Recall the definition of planar graph:
First we note that the complete graph $K_4$ is planar by demonstrating its planarity:
:
Now we attempt to create $K_5$ by building it from its subgraph $K_4$.
Each of the faces of $K_4$ is in the same form as each of the others.
That is, each face consists of a triangle of $3$ vertices which is not incident to the $4$th vertex.
So exactly where in the plane we place the $5$th vertex does not matter.
 let us place vertex $E$ outside the triangle $ABC$ as shown below:
:
Edges $AE$, $BE$ and $CE$ can be constructed without crossing an existing edge of $K_4$, as indicated by dotted black lines.
However, it is not possible to construct edge $DE$ without crossing one of $AB$, $AC$ or $BC$.
This is shown in dotted $\color{red} {\text {red} }$.
Hence the result.
== Sources ==
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Category:Planar Graphs
Category:Complete Graphs