nyuuzyou/wikis huggingface dataset:
Internal Direct Product Theorem/General Result/Proof 1

== Theorem ==
== Proof ==
By definition, $G$ is the internal group direct product of $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ the mapping:
:$\ds C: \prod_{k \mathop = 1}^n H_k \to G: \map C {h_1, \ldots, h_n} = \prod_{k \mathop = 1}^n h_k$
is a group isomorphism from the cartesian product $\struct {H_1, \circ {\restriction_{H_1} } } \times \cdots \times \struct {H_n, \circ {\restriction_{H_n} } }$ onto $\struct {G, \circ}$. 
=== Necessary Condition ===
Let $G$ be the internal group direct product of $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$.
:$(1): \quad$ From Subgroup Product is Internal Group Direct Product iff Surjective, $G = H_1 H_2 \cdots H_n$.
:$(2): \quad$ From Internal Group Direct Product is Injective: General Result, $\sequence {H_k}_{1 \mathop \le k \mathop \le n}$ is a sequence of independent subgroups.
:$(3): \quad$ From Internal Group Direct Product Isomorphism:
:::: $\forall k \in \set {1, 2, \ldots, n}: H_k \lhd G$
=== Sufficient Condition ===
Now suppose the three conditions hold.
:$(1): \quad$ From Subgroup Product is Internal Group Direct Product iff Surjective, $C$ is surjective.
:$(2): \quad$ From Internal Group Direct Product is Injective: General Result, $C$ is injective.
:$(3): \quad$ From Internal Group Direct Product of Normal Subgroups, $C$ is a group homomorphism.
Putting these together, we see that $C$ is a bijective homomorphism, and therefore an isomorphism.
So by definition, $G$ is the internal group direct product of $\sequence {H_k} _{1 \mathop \le k \mathop \le n}$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Theorem $18.14$
Category:Internal Direct Product Theorem
Internal Direct Product Theorem/General Result/Proof 2

== Theorem ==
== Proof ==
It is to be shown that:
by definition of Internal Group Direct Product.
Condition $(3)$ already gives that $H_i$ is normal for all $i \in \set {1, 2, \ldots, n}$.
Condition $(1)$ gives us that each element $g$ of $G$ can be expressed in the form:
:$g = h_1 h_2 \dotsm h_n$ with $h_i \in H_i$ for all $i \in \set {1, 2, \ldots, n}$.
It is now shown that this expression is unique.
Suppose that:
:$g = h_1 h_2 \dotsm h_n = k_1 k_2 \dotsm k_n$
where $h_i, k_i \in H_i$ for all $i \in \set {1, 2, \ldots, n}$ and $h_j \ne k_j$ for at least one $j$.
Let $j$ be the largest integer such that $h_j \ne k_j$, so that $h_i = k_i$ for $i > j$.
Cancelling $h_i$ for $i > j$ gives:
:$h_i h_2 \dotsm h_j = k_1 k_2 \dotsm k_j$
and so:
:$h_j {k_j}^{-1} = \paren {h_1 h_2 \dotsm h_{j - 1} }^{-1} \paren {k_1 k_2 \dotsm k_{j - 1} } \in \paren {H_1 H_2 \dotsm H_{j - 1} } \cap G_j$
But by condition $(2)$:
:$\paren {H_1 H_2 \dotsm H_{j - 1} } \cap G_j = \set e$
by definition of independent subgroups.
Thus $h_j = k_j$, which contradicts our assertion that $h_j \ne k_j$.
Hence the decomposition is unique.
== Sources ==
* : Chapter $13$: Direct products: Proposition $13.5$
Category:Internal Direct Product Theorem
Internal Direct Product Theorem/Proof 1

== Theorem ==
== Proof ==
From Conditions for Internal Group Direct Product it is sufficient to show that if:
:$\quad G = H \circ K$
and:
:$\quad H \cap K = \set e$
then:
:$\struct {H, \circ {\restriction_H} }$ and $\struct {H_2, \circ {\restriction_K} }$ are both normal subgroups of $\struct {G, \circ}$
:$\forall \tuple {h, k} \in H \times K: h \circ k = k \circ h$
=== Sufficient Condition ===
Let $\struct {H, \circ {\restriction_H} }$ and $\struct {H_2, \circ {\restriction_K} }$ both be normal subgroups of $\struct {G, \circ}$.
Let $x \in H$ and $y \in K$.
Then:
$x$ and $y$ are arbitrary, so:
:$\forall \tuple {h, k} \in H \times K: h \circ k = k \circ h$
=== Necessary Condition ===
Let:
:$\forall \tuple {h, k} \in H \times K: h \circ k = k \circ h$
Let $z = G$.
We have:
Then we have:
Similarly:
Thus, by definition, $H$ and $K$ are both $\struct {H, \circ {\restriction_H} }$ and $\struct {H_2, \circ {\restriction_K} }$ are both normal subgroups of $\struct {G, \circ}$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Theorem $13.5$
Category:Internal Direct Product Theorem
Internal Direct Product Theorem/Proof 2

== Theorem ==
== Proof ==
=== Sufficient Condition ===
Let $\phi: H \times K \to G$ be the mapping defined as:
:$\forall \tuple {h, k} \in H \times K: \map \phi {h, k} = h \circ k$
Let $\phi$ be an isomorphism.
:$(1): \quad$ From Codomain of Internal Direct Isomorphism is Subset Product of Factors, $G = H \circ K$.
:$(2): \quad$ From Internal Group Direct Product is Injective, $H$ and $K$ are independent subgroups of $G$.
:$(3): \quad$ From Internal Group Direct Product Isomorphism, $H$ and $K$ are normal subgroups of $G$.
=== Necessary Condition ===
Let $\phi: H \times K \to G$ be the mapping defined as:
:$\forall \tuple {h, k} \in H \times K: \map \phi {h, k} = h \circ k$
Suppose the three conditions hold.
:$(1): \quad$ From $G = H \circ K$, $\phi$ is surjective.
:$(2): \quad$ From Internal Group Direct Product is Injective, $\phi$ is injective.
:$(3): \quad$ From Internal Group Direct Product of Normal Subgroups, $\phi$ is a group homomorphism.
Putting these together, we see that $\phi$ is a bijective homomorphism, and therefore an isomorphism.
So by definition, $G$ is the internal group direct product of $H$ and $K$.
Category:Internal Direct Product Theorem
Internal Group Direct Product/Examples/C2 x C3

== Example of Internal Group Direct Product ==
The direct product of the cyclic groups $C_2$ and $C_3$ is isomorphic to the cyclic groups $C_6$.
Hence it is seen to be an internal group direct product.
== Proof ==
Let $x$ be a generator for $C_6$.
Let $G_1$ be the subgroup of $C_6$ generated by $x^2$.
Let $G_2$ be the subgroup of $C_6$ generated by $x^3$.
From Subgroup of Abelian Group is Normal, both $G_1$ and $G_2$ are normal in $G$.
Then we have:
and so the conditions are fulfilled for $G_1 \times G_2$ to be an internal group direct product.
== Sources ==
* : Chapter $13$: Direct products: Direct products: Example $13.4$
Category:Examples of Internal Group Direct Products
Internal Group Direct Product/Examples/D4

== Example of Internal Group Direct Product ==
Consider the dihedral group $D_4$, which is the symmetry group of the square.
Suppose $D_4$ is the internal group direct product of two subgroups.
Then those two subgroups are $\set e$ and $D_4$ itself, where $e$ is the identity element of $D_4$.
== Proof ==
Let $H$ and $K$ be subgroups of $D_4$ such that $D_4$ is the internal group direct product of $H$ and $K$.
Then from the Internal Direct Product Theorem $H$ and $K$ must be:
:normal subgroups
:whose intersection is $\set e$
:whose subset product is $D_4$.
Let $D_4$ be expressed in its group presentation:
Pairs of subgroups of $D_4$ whose intersection is $\set e$ and whose subset product is $D_4$ are:
:$(1): \quad \set e$ and $D_4$
:$(2): \quad \set {e, b}$ and $\set {e, a, a^4, a^3}$
:$(3): \quad \set {e, b a}$ and $\set {e, a^2, b, b a^2}$
:$(3): \quad \set {e, b a^2}$ and $\set {e, a^2, b, b a^3}$
:$(5): \quad \set {e, b a^3}$ and $\set {e, a^2, b, b a^2}$
From Normal Subgroups of Dihedral Group, none of $\set {e, b}$, $\set {e, b a}$, $\set {e, b a^2}$ or $\set {e, b a^3}$ are normal.
Hence the result.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.7$
Category:Examples of Internal Group Direct Products
Category:Symmetry Group of Square
Internal Group Direct Product/Non-Examples/Pointwise Addition on Continuous Real Functions on Unit Interval

== Example of Structure which is not Internal Group Direct Product ==
Let $J \subseteq \R$ denote the closed unit interval $\closedint 0 1$.
Let $\map {\mathscr C} J$ denote the set of all continuous real functions from $J$ to $\R$.
Let $G = \struct {\map {\mathscr C} J, +}$ be the group formed on $\map {\mathscr C} J$ by pointwise addition.
Let $\struct {H, +}$ and $\struct {K, +}$ be the algebraic substructures of $G$ such that:
:$H := \set {f \in G: \forall x \in J: \map f x \ge 0}$
:$K := \set {f \in G: \forall x \in J: \map f x \le 0}$
Then, while $G$, $H$ and $K$ fulfil the $3$ conditions of Conditions for Internal Group Direct Product, $G$ is not the internal group direct product of $H$ and $K$.
== Proof ==
From Pointwise Addition on Continuous Real Functions on Closed Unit Interval forms Group we have that $G$ is indeed a group.
Recall from Conditions for Internal Group Direct Product:
$G$ is the internal group direct product of $H$ and $K$ :
:$(1): \quad \forall h \in H, k \in K: h + k = k + h$
:$(2): \quad G = H + K$
:$(3): \quad H \cap K = \set e$
where:
:$e$ is the identity element of $G$
:$H$ and $K$ are subgroups of $G$.
From Pointwise Addition on Real-Valued Functions is Commutative:
:$\forall h \in H, h \in K: h + k = k + h$
and so condition $(1)$ is fulfilled.
Condition $(3)$ is satisfied by $f_0$, defined as:
:$\forall x \in J: \map {f_0} x = 0$
Now let $f \in G$.
Let $h: J \to \R$ be defined as:
:$\forall x \in J: \map h x = \begin{cases} \map f x & : \map f x > 0 \\ 0 & : \map f x \le 0 \end{cases}$
We have that $h \in H$ by definition.
Let $k: J \to \R$ be defined as:
:$\forall x \in J: \map k x = \begin{cases} \map f x & : \map f x < 0 \\ 0 & : \map f x \ge 0 \end{cases}$
We have that $k \in K$ by definition.
Then we have:
:$\forall x \in J: \map f x = \begin{cases} \map h x & : \map h x > 0 \\ \map k x & : \map k x \le 0 \end{cases}$
Hence:
:$f = h + k$
and such $h$ and $k$ are unique for any given $f$.
Hence we have that:
:$G = H + K$
and so condition $(2)$ is fulfilled.
But we have that:
:$\forall h \in H: -h \in K$
and:
:$\forall k \in K: -k \in H$
and so elements of $H$ and $K$ do not have inverses in those sets.
Hence neither $H$ nor $K$ is a group.
Hence neither $H$ nor $K$ is a subgroup of $G$.
Thus $G$ is not the internal group direct product of $H$ and $K$.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Exercise $13.17 \ \text {(b)}$
Category:Examples of Internal Group Direct Products
Internal Group Direct Product Commutativity

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $H$ and $K$ be subgroups of $G$.
Let $\struct {G, \circ}$ be the internal group direct product of $H$ and $K$.
Then:
:$\forall h \in H, k \in K: h \circ k = k \circ h$
=== General Result ===
== Proof 1 ==
== Proof 2 ==
Category:Internal Group Direct Products
Category:Examples of Commutative Operations
Category:Internal Group Direct Product Commutativity
Internal Group Direct Product Commutativity/General Result

== Theorem ==
Let $\struct {G, \circ}$ be the internal group direct product of $H_1, H_2, \ldots, H_n$.
Let $h_i$ and $h_j$ be elements of $H_i$ and $H_j$ respectively, $i \ne j$.
Then $h_i \circ h_j = h_j \circ h_i$.
== Proof ==
Let $g = h_i \circ h_j \circ h_i^{-1} \circ h_j^{-1}$.
From the Internal Direct Product Theorem: General Result, $H_i$ and $H_j$ are normal in $G$.
Hence $h_i \circ h_j \circ h_i^{-1} \in H_j$ and thus $g \in H_j$.
Similarly, $g \in H_i$ and thus $g \in H_i \cap H_j$.
But:
== Sources ==
* : Chapter $13$: Direct products: Corollary $13.6$
Category:Internal Group Direct Product Commutativity
Internal Group Direct Product Commutativity/Proof 1

== Theorem ==
== Proof ==
Let $G$ be the internal group direct product of $H$ and $K$.
Then by definition the mapping:
:$\phi: H \times K \to G: \map \phi {h, k} = h \circ k$
is a (group) isomorphism from the (external) direct product $\struct {H, \circ \restriction_H} \times \struct {K, \circ \restriction_K}$ onto $\struct {G, \circ}$.
Let the symbol $\circ$ also be used for the operation induced on $H \times K$ by $\circ \restriction_H$ and $\circ \restriction_K$.
Let $h \in H, k \in K$.
Then:
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Theorem $13.4$
Category:Internal Group Direct Product Commutativity
Internal Group Direct Product Commutativity/Proof 2

== Theorem ==
== Proof ==
Let $\sqbrk {x, y}$ denote the commutator of $x, y \in G$:
:$\sqbrk {x, y} := x^{-1} y^{-1} x y$
We have that:
Let $h \in H$, $k \in K$.
We have:
and:
Thus:
:$\sqbrk {h, k} \in H \cap K$
But as $H \cap K = \set e$, it follows that:
:$\sqbrk {h, k} = e$
It follows from Commutator is Identity iff Elements Commute that:
:$h k = k h$
and the result follows.
== Sources ==
* : Chapter $8$: Homomorphisms, Normal Subgroups and Quotient Groups: Exercise $15$
Category:Internal Group Direct Product Commutativity
Internal Group Direct Product Isomorphism

== Theorem ==
Let $G$ be a group.
Let $H_1, H_2$ be subgroups of $G$.
Let $\phi: H_1 \times H_2 \to G$ be the mapping defined by $\map \phi {h_1, h_2} := h_1 h_2$.
If $\phi$ is a (group) isomorphism, then both $H_1$ and $H_2$ are normal subgroups of $G$.
== Proof ==
$\phi$ is an isomorphism, so in particular a (group) homomorphism.
Thus by Induced Group Product is Homomorphism iff Commutative, every element of $H_1$ commutes with every element of $H_2$.
Now suppose $a \in G$.
As $\phi$ is an isomorphism, it follows that $\phi$ is surjective.
Thus by Subgroup Product is Internal Group Direct Product iff Surjective:
:$\exists h_1 \in H_1, h_2 \in H_2: a = h_1 h_2$
Now any element of $a H_1 a^{-1}$ is in the form $a h a^{-1}$ for some $h \in H_1$.
Thus:
Thus $a H_1 a^{-1} \subseteq H_1$, and $H_1$ is therefore a normal subgroup of $G$.
Similarly, any element of $a H_2 a^{-1}$ is in the form $a h a^{-1}$ for some $h \in H_2$.
Thus:
Thus $a H_2 a^{-1} \subseteq H_2$ and $H_2$ is normal as well.
Category:Internal Group Direct Products
Category:Group Isomorphisms
Category:Normal Subgroups
Internal Group Direct Product is Injective

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $H_1, H_2$ be subgroups of $G$.
Let $\phi: H_1 \times H_2 \to G$ be a mapping defined by:
:$\map \phi {h_1, h_2} = h_1 h_2$
Then $\phi$ is injective :
:$H_1 \cap H_2 = \set e$
=== General Result ===
== Proof ==
=== Necessary Condition ===
Let $\phi$ be an injection.
Let $\map \phi {h_1, h_2} = \map \phi {k_1, k_2}$.
As $\phi$ is injective, this means that:
:$\tuple {h_1, h_2} = \tuple {k_1, k_2}$
and thus:
:$h_1 = k_1, h_2 = k_2$
From the definition of $\phi$, this means:
:$h_1 h_2 = k_1 k_2$
Thus, each element of $G$ that can be expressed as a product of the form $h_1 h_2$ can be thus expressed uniquely.
Now, suppose $h \in H_1 \cap H_2$.
We have:
Thus we see that:
Thus:
:$H_1 \cap H_2 = \set e$
=== Sufficient Condition ===
Let $H_1 \cap H_2 = \set e$.
Let:
:$\map \phi {h_1, h_2} = \map \phi {k_1, k_2}$
Then:
:$h_1 h_2 = k_1 k_2: h_1, k_1 \in H_1, h_2, k_2 \in H_2$
Thus:
:$k_1^{-1} h_1 = k_2 h_2^{-1}$
But:
:$k_1^{-1} h_1 \in H_1$ and $k_2 h_2^{-1} \in H_2$
As they are equal, we have:
:$k_1^{-1} h_1 = k_2 h_2^{-1} \in H_1 \cap H_2 = \set e$
It follows that:
:$h_1 = k_1, h_2 = k_2$
and thus:
:$\tuple {h_1, h_2} = \tuple {k_1, k_2}$
Thus $\phi$ is injective and the result follows.
Category:Internal Group Direct Products
Category:Injections
Internal Group Direct Product of Normal Subgroups

== Theorem ==
Let $G$ be a group whose identity is $e$.
Let $H_1$ and $H_2$ be normal subgroups of $G$ such that $H_1 \cap H_2 = \set e$.
Let $\phi: H_1 \times H_2 \to G$ be a mapping defined by $\map \phi {h_1, h_2} = h_1 h_2$.
Then $\phi$ is a (group) homomorphism.
== Proof ==
Let $H_1$ and $H_2$ be normal subgroups of $G$.
Let $h_1 \in H_1, h_2 \in H_2$.
Consider $x \in G: x = h_1 h_2 h_1^{-1} h_2^{-1}$.
As $H_2$ is normal, we have $h_1 h_2 h_1^{-1} \in H_2$ and thus $x \in H_2$.
Similarly, we can show that $x \in H_1$ and so $x \in H_1 \cap H_2$ and thus $x = e$.
From Product of Commuting Elements with Inverses, $h_1 h_2 h_1^{-1} h_2^{-1} = e$ $h_1$ and $h_2$ commute.
Thus $h_1 h_2 = h_2 h_1$. 
As $h_1$ and $h_2$ are arbitrary elements of $H_1$ and $H_2$, it follows that every element of $H_1$ commutes with every element of $H_2$.
Thus from Induced Group Product is Homomorphism iff Commutative, $\phi$ is a homomorphism.
Category:Internal Group Direct Products
Category:Normal Subgroups
Category:Group Homomorphisms
Internal and External Group Direct Products are Isomorphic

== Theorem ==
Let $G$ be a group whose identity is $e$.
Then $G$ is the (external) group direct product of $G_1, G_2, \ldots, G_n$ $G$ is the internal group direct product of $N_1, N_2, \ldots, N_n$ such that:
:$\forall i \in \N_n: N_i \cong G_i$
where:
:$\cong$ denotes (group) isomorphism
:$\N_n$ denotes $\set {1, 2, \ldots, n}$
== Proof ==
=== Necessary Condition ===
Let $G$ be the external direct product of groups $G_1, G_2, \ldots, G_n$.
For all $i \in \N_n$, let $N_i$ be defined as the set:
:$N_i = \set e \times \cdots \times \set e \times G_i \times \set e \times \cdots \times \set e$
of elements which have entry $e$ everywhere except possibly in the $i$th co-ordinate.
It remains to be checked that:
:$(1): \quad N_i$ is isomorphic to $G_i$
:$(2): \quad N_i$ is a normal subgroup of $G$
:$(3): \quad$ Every element of $G$ has a unique expression:
:$\tuple {g_1, \ldots, g_n} = \tuple {g_1, e, \ldots, e} \tuple {e, g_2, e \ldots e} \ldots \tuple {e, \ldots, g_n}$
as a product of elements of $N_1, \ldots, N_n$.
==== $N_i$ is isomorphic to $G_i$ ====
==== $N_i$ is a normal subgroup of $G$ ====
==== Every element of $G$ has a unique expression ====
=== Sufficient Condition ===
Now let $G$ be the internal group direct product of $N_1, N_2, \ldots, N_n$.
We define a mapping $\theta: G \to N_1 \times N_2 \times \cdots \times N_n$ by:
:$\map \theta {g_1 g_2 \ldots g_n} = \tuple {g_1, g_2, \ldots, g_n}$
The fact that $\theta$ is a bijection follows from the definitions.
It remains to be shown that $\theta$ is a homomorphism.
The proof proceeds by induction.
For all $n \in \Z_{\ge 2}$, let $\map P n$ be the proposition:
:$\map \theta {g_1 g_2 \ldots g_n} \map \theta {h_1 h_2 \ldots h_n} = \map \theta {g_1 g_2 \ldots g_n h_1 h_2 \ldots h_n}$
==== Basis for the Induction ====
$\map P 2$ is the case:
Thus $\map P 2$ is seen to hold.
This is the basis for the induction.
==== Induction Hypothesis ====
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\map \theta {g_1 g_2 \ldots g_k} \map \theta {h_1 h_2 \ldots h_k} = \map \theta {g_1 g_2 \ldots g_k h_1 h_2 \ldots h_k}$
from which it is to be shown that:
:$\map \theta {g_1 g_2 \ldots g_{k + 1} } \map \theta {h_1 h_2 \ldots h_{k + 1} } = \map \theta {g_1 g_2 \ldots g_{k + 1} h_1 h_2 \ldots h_{k + 1} }$
==== Induction Step ====
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{>1}: \map \theta {g_1 g_2 \ldots g_n} \map \theta {h_1 h_2 \ldots h_n} = \map \theta {g_1 g_2 \ldots g_n h_1 h_2 \ldots h_n}$
So $\theta$ is a homomorphism.
== Sources ==
* : Introduction: Notation
* : Chapter $13$: Direct products: Proposition $13.7$
Category:Group Direct Products
Category:Internal Group Direct Products
Category:Group Isomorphisms
Interpretation/Examples/Propositional Calculus

== Example of an Interpretation ==
The propositional calculus $\CC$ has a domain $\set {\T, \F}$ representing true and false.
The semantic rules of $\CC$ assign to each WFF of $\CC$ one or other of $\T$ and $\F$.
The connectives of $\CC$ are assumed in this context to have a fixed meaning.
== Sources ==
* 
* 
Category:Examples of Interpretations
Intersecting Chord Theorem for Conic Sections

== Theorem ==
Consider a right circular cone $\CC$ with opening angle $2 \alpha$ whose apex is at $O$.
Consider a slicing plane $\PP$, not passing through $O$, at an angle $\beta$ to the axis of $\CC$.
Let the plane $OAA'$ through the axis of $\CC$ perpendicular to $\PP$ intersect $\PP$ in the line $AA'$.
Let $P$ be an arbitrary point on the intersection of $\PP$ with $\CC$.
Let $PM$ be constructed perpendicular to $AA'$.
Then:
:$PM^2 = k \cdot AM \cdot MA'$
where $k$ is the constant:
:$k = \dfrac {\map \sin {\beta + \alpha} \map \sin {\beta - \alpha} } {\cos \alpha}$
== Proof ==
== Sources ==
* : Chapter $\text {IV}$. The Ellipse: $1 \text b$.
Category:Conic Sections
Intersecting Chords Theorem

== Theorem ==
Let $AC$ and $BD$ both be chords of the same circle.
Let $AC$ and $BD$ intersect at $E$.
Then $AE \cdot EC = DE \cdot EB$.
:If in a circle two straight lines cut one another, the rectangle contained by the segments of the one is equal to the rectangle contained by the segments of the other.
== Proof 1 ==
== Proof 2 ==
== Also known as ==
== Also see ==
* Power of a Point Theorem
== Sources ==
* $(6)$
* $(6)$
* 
Category:Intersecting Chords Theorem
Category:Circles
Category:Named Theorems
Intersecting Chords Theorem/Proof 1

== Theorem ==
== Proof ==
:
Let $AC$ and $BD$ be intersecting chords of circle $ABCD$.
Let the point of intersection be $E$.
If $E$ is the center of $ABCD$ the solution is trivial, as $AE = EC = BE = ED$ and so $AE \cdot EC = BE \cdot ED$.
Otherwise, let $F$ be the center of $ABCD$.
Let $FG$ be drawn perpendicular to $AC$, and $FH$ be drawn perpendicular to $BD$.
From Conditions for Diameter to be Perpendicular Bisector, $G$ bisects $AC$ and $H$ bisects $BD$.
So $AG = GC$ and $BH = HD$.
From Difference of Two Squares we have that $AE \cdot EC + EG^2 = GC^2$.
Let us add $GF^2$ to these.
So $AE \cdot EC + EG^2 + GF^2 = GC^2 + GF^2$.
But from Pythagoras's Theorem we have that:
:$GC^2 + GF^2 = CF^2$
:$EG^2 + GF^2 = EF^2$
So:
:$AE \cdot EC + EF^2 = CF^2$
Using the same construction, we have that:
:$DE \cdot EB + EF^2 = BF^2$
But $BF = CF$ as both are the radius of the circle $ABCD$.
That gives us:
:$AE \cdot EC + EF^2 = DE \cdot EB + EF^2$
It follows that $AE \cdot EC = DE \cdot EB$
== Sources ==
* : Book $\text{III}$. Propositions
Category:Intersecting Chords Theorem
Intersecting Circles have Different Centers

== Theorem ==
== Proof ==
:
Let $ABC$ and $BDCG$ be circles which cut one another at $B$ and $C$.
Suppose they had the same center $E$.
Join $EC$ and let $EG$ be drawn at random through $F$.
As $E$ is the center of $ABC$, by , we have that $EC = EF$.
Similarly, as $E$ is also the center of $BDCG$, we have that $EC = EG$.
But they are clearly unequal by the method of construction.
So from this contradiction, the two circles can not have the same center.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Circles
Intersection Complement of Set with Itself is Complement

== Theorem ==
Let $A$ and $B$ be subsets of a universal set $\Bbb U$.
Let $\uparrow$ denote the operation on $A$ and $B$ defined as:
:$\paren {A \uparrow B} \iff \paren {\relcomp {\Bbb U} {A \cap B} }$
where $\relcomp {\Bbb U} A$ denotes the complement of $A$ in $\Bbb U$.
Then:
:$A \uparrow A = \relcomp {\Bbb U} A$
== Proof ==
== Sources ==
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: Exercise $16$
Category:Set Intersection
Category:Set Complement
Intersection Complex Measure is Complex Measure

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space.
Let $\mu$ be a complex measure on $\struct {X, \Sigma}$. 
Let $F \in \Sigma$.
Let $\mu_F$ be the intersection complex measure of $\mu$ by $F$.
Then $\mu_F$ is a complex measure.
== Proof ==
Since $\mu$ is a complex measure, we have: 
:$\map \mu E \in \C$
for each $E \in \Sigma$.
So, in particular: 
:$\map \mu {E \cap F} \in \C$
for all $E \in \Sigma$.
That is: 
:$\map {\mu_F} E \in \C$ 
for all $E \in \Sigma$. 
We verify the two conditions required of a complex measure.
We have:
Now let $\sequence {S_n}_{n \mathop \in \N}$ be a sequence of pairwise disjoint sets in $\Sigma$.
Then, we have: 
So $\mu$ is a complex measure.
Category:Complex Measures
Category:Intersection Complex Measures