nyuuzyou/wikis huggingface dataset:
Non-Trivial Commutative Division Ring is Field

== Theorem ==
Let $\struct {R, +, \circ}$ be a non-trivial division ring such that $\circ$ is commutative.
Then $\struct {R, +, \circ}$ is a field.
Similarly, let $\struct {F, +, \circ}$ be a field.
Then $\struct {F, +, \circ}$ is a non-trivial division ring such that $\circ$ is commutative.
== Proof ==
Suppose $\struct {R, +, \circ}$ is a non-trivial division ring such that $\circ$ is commutative.
By definition $\struct {R, +}$ is an abelian group.
Thus are satisfied.
Also by definition, $\struct {R, \circ}$ is a semigroup such that $\circ$ is commutative.
Thus to are satisfied.
As $\struct {R, +, \circ}$ is a ring with unity, is satisfied.
 is satisfied by dint of $\struct {R, +, \circ}$ being a ring.
Finally note that by definition of division ring, is satisfied.
Thus all the field axioms are satisfied, and so $\struct {R, +, \circ}$ is a field.
Suppose $\struct {F, +, \circ}$ is a field.
Then by definition $\struct {F, +, \circ}$ is a non-trivial division ring such that $\circ$ is commutative.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $23$. The Field of Rational Numbers
Category:Field Theory
Category:Division Rings
Non-Trivial Connected Set in T1 Space is Dense-in-itself

== Theorem ==
Let $T = \struct {S, \tau}$ be a $T_1$ (Fréchet) topological space.
Let $H \subseteq S$ be connected in $T$.
If $H$ has more than one element, then $H$ is dense-in-itself.
== Proof ==
 $H$ is not dense-in-itself.
Then $\exists x \in H$ such that $x$ is isolated in $H$.
That is, $\exists U \in \tau: U \cap H = \set x$.
Since $H$ has more than one element, we can find $y \in H$ with $y \ne x$.
Since $T$ is $T_1$:
:$\forall y \in H: y \ne x \implies \paren {\exists V_y \in \tau: y \in V_y, x \notin V_y}$
Define $V = \ds \bigcup_{y \mathop \in H \\ y \mathop \ne x} V_y$.
Then $\forall z \in H$:
:$z = x \implies z \in U$
:$z \ne x \implies z \in V_z \subseteq V$
Showing that $H \subseteq U \cup V$.
This also shows that both $U \cap H$ and $V \cap H$ are non-empty.
Since $x \notin V_y$ for any $y \in H$, 
we have $x \notin V$. 
This implies $\set x \cap V = \O$
So $H \cap U \cap V = \O$.
Thus we have:
:$H \subseteq U \cup V$
:$H \cap U \cap V = \O$
:$U \cap H \ne \O$
:$V \cap H \ne \O$
showing that $H$ is disconnected.
This is a contradiction.
Hence by Proof by Contradiction it must be the case that $H$ is dense-in-itself. 
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness: Disconnectedness
Category:T1 Spaces
Category:Denseness
Category:Connected Sets
Non-Trivial Discrete Space is not Arc-Connected

== Corollary to Non-Trivial Discrete Space is not Connected ==
Let $T = \struct {S, \tau}$ be a non-trivial discrete topological space.
$T$ is not arc-connected.
== Proof ==
 $T$ is arc-connected.
From:
:Arc-Connected Space is Path-Connected
:Path-Connected Space is Connected
we have that $T$ is connected.
But this directly contradicts Non-Trivial Discrete Space is not Connected.
The result follows from Proof by Contradiction.
== Sources ==
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $10$
Category:Discrete Topology
Category:Arc-Connected Spaces
Category:Non-Trivial Discrete Space is not Connected
Non-Trivial Discrete Space is not Connected

== Theorem ==
Let $T = \struct {S, \tau}$ be a non-trivial discrete topological space.
Then $T$ is not connected.
Thus also:
=== Corollary 1 ===
=== Corollary 2 ===
=== Corollary 3 ===
=== Corollary 4 ===
== Proof ==
Let $T = \struct {S, \tau}$ be a non-trivial discrete space.
Let $a \in S$.
Let $A = \set a$ and $B = \relcomp S {\set a}$, where $\relcomp S {\set a}$ is the complement of $A$ in $S$.
As $T$ is not trivial, $B \ne \O$.
Then from Set in Discrete Topology is Clopen, $A$ and $B$ are both open.
So $A \mid B$ is a separation of $S$.
It follows, by definition, that $T$ is not connected.
== Sources ==
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $10$
Category:Discrete Topology
Category:Connected Spaces
Category:Non-Trivial Discrete Space is not Connected
Non-Trivial Discrete Space is not Irreducible

== Corollary to Non-Trivial Discrete Space is not Connected ==
Let $T = \left({S, \tau}\right)$ be a non-trivial discrete topological space.
$T$ is not irreducible.
== Proof ==
 $T$ is irreducible.
From Irreducible Space is Connected, we have that $T$ is connected.
But this directly contradicts Non-Trivial Discrete Space is not Connected.
The result follows from Proof by Contradiction.
Category:Discrete Topology
Category:Irreducible Spaces
Category:Non-Trivial Discrete Space is not Connected
Non-Trivial Discrete Space is not Path-Connected

== Corollary to Non-Trivial Discrete Space is not Connected ==
Let $T = \struct {S, \tau}$ be a non-trivial discrete topological space.
$T$ is not path-connected.
== Proof ==
 $T$ is path-connected.
From Path-Connected Space is Connected, we have that $T$ is connected.
But this directly contradicts Non-Trivial Discrete Space is not Connected.
The result follows from Proof by Contradiction.
== Sources ==
* : Part $\text {II}$: Counterexamples: $1 \text { - } 3$. Discrete Topology: $10$
Category:Discrete Topology
Category:Path-Connected Spaces
Category:Non-Trivial Discrete Space is not Connected
Non-Trivial Discrete Space is not Ultraconnected

== Corollary to Non-Trivial Discrete Space is not Connected ==
Let $T = \struct {S, \tau}$ be a non-trivial discrete topological space.
$T$ is not ultraconnected.
== Proof ==
 $T$ is ultraconnected.
From Ultraconnected Space is Connected, we have that $T$ is connected.
But this directly contradicts Non-Trivial Discrete Space is not Connected.
The result follows from Proof by Contradiction.
Category:Discrete Topology
Category:Ultraconnected Spaces
Category:Non-Trivial Discrete Space is not Connected
Non-Trivial Event is Union of Simple Events

== Theorem ==
Let $\EE$ be an experiment.
Let $e$ be an event in $\EE$ such that $e \ne \O$.
That is, such that $e$ is non-trivial.
Then $e$ can be expressed as the union of a set of simple events in $\EE$.
== Proof ==
By definition of event, $e$ is a subset of the sample space $\Omega$ of $\EE$.
By hypothesis:
:$e \ne \O$
and so:
:$\exists s \in \Omega: s \in e$
Let $S$ be the set defined as:
:$S = \set {\set s: s \in e}$
By Union is Smallest Superset: Set of Sets it follows that:
:$\ds \bigcup S \subseteq e$
Let $x \in e$.
Then by Singleton of Element is Subset:
:$\set x \subseteq e$
and by definition of $S$ it follows that:
:$\set x \in S$
and so by definition of set union:
:$x \in \ds \bigcup S$
Thus we have:
:$e \subseteq \ds \bigcup S$
The result follows by definition of set equality.
== Sources ==
* : Chapter $2$: Probability and Discrete Sample Spaces: $2.2$ Sample spaces and events
Category:Events
Non-Trivial Excluded Point Topology is not T1

== Theorem ==
Let $T = \struct {S, \tau_{\bar p} }$ be a excluded point space such that $S$ is not a singleton.
Then $T$ is not a $T_1$ (Fréchet) space.
== Proof ==
Follows directly from:
:Excluded Point Topology is Open Extension Topology of Discrete Topology
:Open Extension Topology is not $T_1$
== Sources ==
* : Part $\text {II}$: Counterexamples: $13 \text { - } 15$. Excluded Point Topology: $2$
Category:T1 Spaces
Category:Excluded Point Topology
Non-Trivial Group has Non-Trivial Cyclic Subgroup

== Theorem ==
Let $G$ be a group whose identity element is $e$.
Let $g \in G$.
If $g$ has infinite order, then $\gen g$ is an infinite cyclic group.
If $\order g = n$, then $\gen g$ is a cyclic group with $n$ elements.
Thus, every group which is non-trivial has at least one cyclic subgroup which is also non-trivial.
In the case that $G$ is itself cyclic, that cyclic subgroup may of course be itself.
== Proof ==
=== Infinite Order ===
Suppose that $g$ has infinite order.
We have that $\gen g$ consists of all possible powers of $g$.
So $\gen g$ can contain a finite number of elements only if some of these were equal.
Then we would have:
:$\exists i, j \in \Z, i < j: g^i = g^j$
and so:
:$g^{j - i} = g^{i - i} = g^0 = e$
which would mean that $g$ was of finite order.
This contradiction leads to the conclusion that $\gen g$ must be an infinite cyclic group.
=== Finite Order ===
If $\order g = n$, we have from Equal Powers of Finite Order Element that there are exactly $n$ different elements of $G$ of the form $g^i$.
Hence $\gen g$ is a cyclic group with $n$ elements.
== Sources ==
* : Chapter $4$: Subgroups: Proposition $4.11$
Category:Cyclic Groups
Non-Trivial Particular Point Topology is not T1

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space such that $S$ is not a singleton.
Then $T$ is not a $T_1$ (Fréchet) space.
== Proof ==
Follows directly from:
:Particular Point Topology is Closed Extension Topology of Discrete Topology
:Closed Extension Topology is not $T_1$.
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $4$
Category:T1 Spaces
Category:Particular Point Topology
Non-Trivial Particular Point Topology is not T3

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space such that $S$ is not a singleton.
Then $T$ is not a $T_3$ space.
== Proof ==
We have that there are at least two distinct elements of $S$.
So, consider $x, p \in S: x \ne p$.
Then $X = \set x$ is closed in $T$ and $p \notin X$.
Suppose $U \in \tau_p$ is an open set in $T$ such that $X \subseteq U$.
We have that $\set p \in \tau_p$ such that $p \in \set p$.
But as $p \in U, p \in \set p$ we have that $U \cap \set p \ne \O$.
So $T$ is not a $T_3$ space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $4$
Category:T3 Spaces
Category:Particular Point Topology
Non-Trivial Particular Point Topology is not T4/Mistake

== Source Work ==
:
:Part $\text {II}$: Counterexamples
::Section $8 \text { - } 10$: Particular Point Topology
:::Item $4$
== Mistake ==
:Every particular point topology is $T_0$, but since there are no disjoint open sets, none of the higher separation axioms are satisfied unless $X$ has only one point.
In the above, $X$ is a particular point space.
However, this is not true for the $T_4$ axiom.
The Sierpiński space is a particular point topology with exactly two points.
But the Sierpiński Space is T4.
== Also see ==
* Particular Point Topology with Three Points is not T4
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $4$
Category:Mistakes/Counterexamples in Topology
Non-Trivial Ultraconnected Space is not T1

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space which is ultraconnected.
If $S$ has more than one element, then $T$ is not a $T_1$ (Fréchet) space.
That is, if $T$ is a $T_1$ (Fréchet) space with more than one element, it is not ultraconnected.
== Proof ==
$T = \struct {S, \tau}$ be ultraconnected.
Thus by definition:
:$(1): \quad \forall x, y \in S: \set x^- \cap \set y^- \ne \O$
Let $a, b \in S$ such that $a \ne b$.
 $T$ is a $T_1$ (Fréchet) space.
By definition of $T_1$ Space, $\set a$ and $\set b$ are closed.
From Closed Set Equals its Closure we have that $\set a^- = \set a$ and $\set b^- = \set b$.
It immediately follows that:
:$\set a^- \cap \set b^- = \O$
But that contradicts $(1)$ above. 
The result follows by Proof by Contradiction.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $4$: Connectedness
Category:T1 Spaces
Category:Ultraconnected Spaces
Non-Unity Variant of Echelon Matrix/Examples

== Examples of Non-Unity Echelon Matrices ==
=== Arbitrary Matrix 1 ===
=== Arbitrary Matrix $2$ ===
=== Arbitrary Matrix $3$ ===
=== Arbitrary Matrix $4$ ===
=== Arbitrary Matrix $5$ ===
=== Arbitrary Matrix $6$ ===
=== Arbitrary Matrix $7$ ===
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.3$
Category:Examples of Echelon Matrices
Non-Unity Variant of Echelon Matrix/Examples/Arbitrary Matrix 1

== Examples of Non-Unity Echelon Matrices ==
:$\begin {bmatrix}
0 & 1 & 2 & 3 \\
0 & 0 & 0 & 1 \\
0 & 0 & 0 & 0 \\
\end {bmatrix}$
is an echelon matrix.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.3 \ \text {(a)}$
Category:Examples of Echelon Matrices
Non-Unity Variant of Echelon Matrix/Examples/Arbitrary Matrix 2

== Examples of Non-Unity Echelon Matrices ==
:$\begin {bmatrix}
1 & 0 & 1 & 2 & 3 \\
2 & 0 & 0 & 0 & 1 \\
0 & 1 & 0 & 0 & 0 \\
\end {bmatrix}$
is not an echelon matrix, because the leading coefficient of row $2$ is in the same column as that of the row above it.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.3 \ \text {(b)}$
Category:Examples of Echelon Matrices
Non-Unity Variant of Echelon Matrix/Examples/Arbitrary Matrix 3

== Examples of Non-Unity Echelon Matrices ==
:$\begin {bmatrix}
1 & 1 & 1 & 1 \\
0 & 2 & 2 & 2 \\
0 & 0 & 3 & 4 \\
\end {bmatrix}$
is a non-unity echelon matrix.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.3 \ \text {(c)}$
Category:Examples of Echelon Matrices
Non-Unity Variant of Echelon Matrix/Examples/Arbitrary Matrix 4

== Examples of Non-Unity Echelon Matrices ==
:$\begin {bmatrix}
1 & 1 & 2 & 3 \\
0 & 0 & 1 & 1 \\
0 & 0 & 3 & 3 \\
\end {bmatrix}$
is not an echelon matrix, because the leading coefficient of row $3$ is in the same column as that of the row above it.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.3 \ \text {(d)}$
Category:Examples of Echelon Matrices
Non-Unity Variant of Echelon Matrix/Examples/Arbitrary Matrix 5

== Examples of Non-Unity Echelon Matrices ==
:$\begin {bmatrix}
1 & 1 & 1 & 1 & 1 \\
0 & 2 & 2 & 2 & 2 \\
0 & 0 & 0 & 3 & 3 \\
\end {bmatrix}$
is a non-unity echelon matrix.
== Sources ==
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.5$ Row and column operations: Exercise $1.3 \ \text {(e)}$
Category:Examples of Echelon Matrices
Non-Unity Variant of Echelon Matrix/Examples/Arbitrary Matrix 6

== Examples of Non-Unity Echelon Matrices ==
:$\begin {bmatrix}
1 & -1 & 2 & 3 \\
0 & 2 & 1 & 0 \\
0 & 0 & 0 & 3 \\
0 & 0 & 0 & 0 \\
\end {bmatrix}$
is a non-unity echelon matrix.
== Sources ==
* 
Category:Examples of Echelon Matrices
Non-Unity Variant of Echelon Matrix/Examples/Arbitrary Matrix 7

== Examples of Non-Unity Echelon Matrices ==
:$\begin {bmatrix}
2 & 1 \\
0 & 2 \\
0 & 0 \\
\end {bmatrix}$
is a non-unity echelon matrix.
== Sources ==
* 
Category:Examples of Echelon Matrices
Non-Zero-Sum Game/Examples/Naval Warfare

== Example of Non-Zero-Sum Game ==
The loss of a particular ship may cause a setback of considerably greater value to the enemy than the cost of the materiel used by the attacker to destroy it.
Hence naval warfare can often be considered as a non-zero-sum game.
== Sources ==
* : Chapter $\text{I}$: An Outline of the Theory of Games: $2$
Category:Examples of Non-Zero-Sum Games
Non-Zero-Sum Game/Examples/Trade

== Example of Non-Zero-Sum Game==
Consider two parties trading: both may simultaneously gain by a transaction.
It can be assumed that the buyer attaches a value to goods purchased which is higher than the price paid.
At the same time, the price paid to the vendor can also be assumed to be higher than the value, to the vendor, of those goods.
Thus, at the end of this transaction, both parties have a positive payoff.
== Sources ==
* : $\S 2$: The Two-Person, Zero-Sum Game with Equilibrium Points
Category:Examples of Non-Zero-Sum Games
Non-Zero-Sum Game as Zero-Sum Game

== Theorem ==
Let $G$ be a non-zero-sum game for $n$ players.
Then $G$ can be modelled as a zero-sum game for $n + 1$ players.
== Proof ==
At each outcome, the total payoff of $G$ will be an amount which will (for at least one outcome) not be zero
Let an $n + 1$th player be introduced to $G$ who has one move:
:$(1): \quad$ Select any player $m$.
:$(2): \quad$ If the total payoff of $G$ is $+k$, receive in payment $k$ from player $m$.
:$(3): \quad$ If the total payoff of $G$ is $-k$, pay $k$ to player $m$.
This new game is the same as $G$ but with an extra (dummy) player, and is now zero-sum.
== Examples ==
== Sources ==
* : Chapter $\text{I}$: An Outline of the Theory of Games: $2$
* 
Category:Non-Zero-Sum Game as Zero-Sum Game
Category:Zero-Sum Games
Non-Zero-Sum Game as Zero-Sum Game/Examples/Casino

== Example of Non-Zero-Sum Game as Zero-Sum Game ==
Consider a casino with $n$ players.
This is a non-zero-sum game.
However, if you add the banker as the $n + 1$th player, this then becomes a zero-sum game.
== Sources ==
* 
Category:Non-Zero-Sum Game as Zero-Sum Game
Non-Zero C*-Algebra contains Non-Zero Hermitian Element

== Theorem ==
Let $\tuple {A, \ast, \norm {\, \cdot \,} }$ be a $\text C^\ast$ algebra such that:
:$A \ne \set { {\mathbf 0}_A}$
where ${\mathbf 0}_A$ is the zero vector of $A$. 
Then there exists a Hermitian element $x \in A \setminus \set { {\mathbf 0}_A}$. 
== Proof 1 ==
 
== Proof 2 ==
Category:C*-Algebras
Category:Hermitian Elements of *-Algebras
Category:Non-Zero C*-Algebra contains Non-Zero Hermitian Element
Non-Zero C*-Algebra contains Non-Zero Hermitian Element/Proof 1

== Theorem ==
== Proof ==
Let $a \in A \setminus \set { {\mathbf 0}_A}$.
From Product of Element in *-Star Algebra with its Star is Hermitian, $a a^\ast$ is Hermitian.
We just need to show that $a a^\ast \ne {\mathbf 0}_A$.
By the $\text C^\ast$ identity, we have:
:$\norm {a a^\ast} = \norm a^2 \ne 0$ 
from , since $a \ne {\mathbf 0}_A$.
Hence applying again, we have $a a^\ast \ne {\mathbf 0}_A$.
Category:Non-Zero C*-Algebra contains Non-Zero Hermitian Element
Non-Zero C*-Algebra contains Non-Zero Hermitian Element/Proof 2

== Theorem ==
== Proof ==
 that ${\mathbf 0}_A$ is the only Hermitian element of $A$. 
Let $a \in A$. 
From Element of *-Algebra Uniquely Decomposes into Hermitian Elements, there exists Hermitian elements $b, c$ such that:
:$a = b + ic$ 
Since ${\mathbf 0}_A$ is the only Hermitian element of $A$, we have $b = {\mathbf 0}_A$ and $c = {\mathbf 0}_A$.
We would then have $a = {\mathbf 0}_A$, contradicting that $A \ne \set { {\mathbf 0}_A}$. 
Hence there are Hermitian elements of $A$ other than ${\mathbf 0}_A$.
Category:Non-Zero C*-Algebra contains Non-Zero Hermitian Element
Non-Zero Complex Numbers are Closed under Multiplication

== Theorem ==
The set of non-zero complex numbers is closed under multiplication.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Example $8.2$
Category:Complex Multiplication
Category:Algebraic Closure
Category:Non-Zero Complex Numbers are Closed under Multiplication
Non-Zero Complex Numbers are Closed under Multiplication/Proof 1

== Theorem ==
== Proof ==
Recall that Complex Numbers form Field under the operations of addition and multiplication.
By definition of a field, the algebraic structure $\struct {\C_{\ne 0}, \times}$ is a group.
Thus, by definition, $\times$ is closed in $\struct {\C_{\ne 0}, \times}$.
Category:Non-Zero Complex Numbers are Closed under Multiplication
Non-Zero Complex Numbers are Closed under Multiplication/Proof 2

== Theorem ==
== Proof ==
Let $z_1, z_2 \in \C_{\ne 0}$.
Then by definition of complex number:
:$z_1 = x_1 + i y_1, z_2 = x_2 + i y_2$
for some $x_1, y_1, x_2, y_2 \in \R$ such that:
:$x_1 \ne 0$ or $y_1 \ne 0$
:$x_2 \ne 0$ or $y_2 \ne 0$
Expressing $z_1$ and $z_2$ in exponential form (although polar form is equally adequate):
:$z_1 = r_1 e^{i \theta_1}$ and $z_2 = r_2 e^{i \theta_2}$
for some $r_1, r_2, \theta_1, \theta_2 \in \R$.
Then by Product of Complex Numbers in Polar Form:
:$z_1 \times z_2 = \paren {r_1 \times r_2} e^{i \paren {\theta_1 + \theta_2} }$
By definition of exponential form:
:$r_1 = \sqrt {x_1^2 + y_1^2}$
:$r_2 = \sqrt {x_2^2 + y_2^2}$
Thus $r_1 > 0$ and $r_2 > 0$.
Hence $r_1 \times r_2 > 0$ and so $z_1 \times z_2 \ne 0$.
== Sources ==
* : Chapter $\text{II}$: Groups: Subgroups
* : $1$: Complex Numbers: Supplementary Problems: Fundamental Operations with Complex Numbers: $59$
Category:Non-Zero Complex Numbers are Closed under Multiplication
Non-Zero Complex Numbers are Closed under Multiplication/Proof 3

== Theorem ==
== Proof ==
Equivalently this is to say:
:$z_1 z_2 = 0 \implies z_1 = 0 \lor z_2 = 0$
Let $z_1 z_2 = 0$.
, let $\tuple {x_2, y_2} \ne \tuple {0, 0}$.
 also that $\tuple {x_1, y_1} \ne \tuple {0, 0}$.
Then:
From this contradiction it follows that:
:$\tuple {x_1, y_1} = \tuple {0, 0}$
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Axiomatic Foundations of Complex Numbers: $75$
Category:Non-Zero Complex Numbers are Closed under Multiplication
Non-Zero Complex Numbers under Multiplication form Group

== Theorem ==
Let $\C_{\ne 0}$ be the set of complex numbers without zero, that is:
:$\C_{\ne 0} = \C \setminus \set 0$
The structure $\struct {\C_{\ne 0}, \times}$ is a group.
== Proof ==
Taking the group axioms in turn:
=== ===
Non-Zero Complex Numbers are Closed under Multiplication.
=== ===
Complex Multiplication is Associative.
=== ===
From Complex Multiplication Identity is One, the identity element of $\struct {\C_{\ne 0}, \times}$ is the complex number $1 + 0 i$.
=== ===
From Inverse for Complex Multiplication‎, the inverse of $x + i y \in \struct {\C_{\ne 0}, \times}$ is:
:$\dfrac 1 z = \dfrac {x - i y} {x^2 + y^2} = \dfrac {\overline z} {z \overline z}$
where $\overline z$ is the complex conjugate of $z$.
== Sources ==
* : Chapter $\text{II}$: Groups: Subgroups
* : $\S 1$: Some examples of groups: Example $1.07$
* : Chapter $1$: Introductory Concepts: $1$. Basic Building Blocks: $2$. GROUP: Example $5$
Category:Complex Multiplication
Category:Examples of Groups
Non-Zero Complex Numbers under Multiplication form Infinite Abelian Group

== Theorem ==
Let $\C_{\ne 0}$ be the set of complex numbers without zero, that is:
:$\C_{\ne 0} = \C \setminus \set 0$
The structure $\struct {\C_{\ne 0}, \times}$ is an infinite abelian group.
== Proof ==
From Non-Zero Complex Numbers under Multiplication form Group, $\struct {\C_{\ne 0}, \times}$ is a group.
Then we have:
:Complex Multiplication is Commutative
and:
:Complex Numbers are Uncountable.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Example $7.2$
* : $\S 34$. Examples of groups: $(1)$
* : $\S 3.2$: Groups; the axioms: Examples of groups $\text{(ii)}$
* : Chapter $1$: Definitions and Examples: Example $1.5$
Category:Complex Multiplication
Category:Examples of Abelian Groups
Category:Examples of Infinite Groups
Non-Zero Dilation of Absorbing Set is Absorbing

== Theorem ==
Let $\GF \in \set {\R, \C}$.
Let $X$ be a vector space over $\GF$. 
Let $A \subseteq X$ be an absorbing set.
Let $\lambda \in \GF \setminus \set 0$. 
Then $\lambda A$ is absorbing.
== Proof ==
Let $x \in X$.
Then there exists $t \in \R_{> 0}$ such that: 
:$x \in \alpha A$ for $\cmod \alpha \ge t$.
That is: 
:$\ds x \in \frac \alpha \lambda \paren {\lambda A}$ for $\cmod \alpha \ge t$.
Since the map $\alpha \mapsto \alpha/\lambda$ is a bijection from $\cmod \alpha \ge t$ to $\cmod \alpha \ge t/\cmod \lambda$, we have:
:$\ds x \in \beta \paren {\lambda A}$ for $\cmod \beta \ge \dfrac t {\cmod \lambda}$.
So $\lambda A$ is absorbing.
Category:Absorbing Sets
Category:Dilation of Subsets of Vector Spaces
Non-Zero Elements of Division Ring form Group

== Theorem ==
Let $\struct {R, +, \circ}$ be a division ring.
Then $\struct {R^*, \circ}$ is a group.
== Proof ==
A division ring by definition is a ring with unity, and therefore not null.
A division ring by definition has no zero divisors, so $\struct {R^*, \circ}$ is a semigroup.
$1_R \in \struct {R^*, \circ}$ and so the identity of $\circ$ is in $\struct {R^*, \circ}$.
By the definition of a division ring, each element of $\struct {R^*, \circ}$ is a unit, and therefore has a unique inverse in $\struct {R^*, \circ}$.
Thus $\struct {R^*, \circ}$ is a semigroup with an identity and inverses and so is a group.
Category:Division Rings
Category:Group Theory