nyuuzyou/wikis huggingface dataset:
Electron Charge to Mass Ratio

== Physical Law ==
The ratio of the charge to the mass of an electron is given by:
== Demonstration ==
Let it be assumed that the cathode rays consist of particles $p$ which have charge $\E$ and mass $m_\E$.
A beam of cathode rays is passed through a region in which an electric field and a magnetic field are arranged to be perpendicular to each other.
In this arrangement, the forces on the cathode rays caused by the electric field and magnetic field can be made to be equal and opposite.
Let the electric field strength of the electric field be $\mathbf E$.
Let the magnetic field strength of the magnetic field be $\mathbf B$.
Let $\mathbf F_e$ be the force on $p$ caused by the electric field.
Let $\mathbf F_m$ be the force on $p$ caused by the magnetic field.
Let the speed of the cathode rays be $v$.
From Electric Force on Charged Particle:
:$\mathbf F_e = \mathbf E \E$
From Magnetic Force on Moving Charged Particle:
:$\mathbf F_m = \mathbf B v \E$
$\mathbf E$ and $\mathbf B$ are adjusted so as to make sure the beam passes through undeflected.
Thus:
Thus we have determined the speed of the cathode rays.
Next we subject the beam to only one of the fields, and observe the deflection of the beam.
Let the beam be subject only to the electric field.
We arrange that $\mathbf E$ is generated by applying a potential difference between two plates whose dimensions we can determine.
Let the length of the region of space through which $\mathbf E$ acts be $l$.
As before, let $\mathbf F_e$ be the force on $p$ caused by the electric field.
Let $t$ be the time during which $p$ is between the plates.
$\mathbf F_e$ will act on $p$ over $t$.
Let $\mathbf a$ be the acceleration of $p$ caused by $\mathbf E$ perpendicular to the direction of the beam.
From Newton's Laws of Motion:
:$\mathbf F_e = m_\E \mathbf a$
This acceleration acts over time $t = \dfrac l v$.
Hence the component of velocity $v_t$ imparted to $p$ perpendicular to the direction of the beam is:
:$v_t = \dfrac {\E \mathbf E l} {m_\E v}$
Hence the change of direction $\theta$ of $p$ is given by:
All the values on the are known.
From these we can calculate the value of $\dfrac \E {m_\E}$.
== Historical Note ==
== Sources ==
* : $2$. Physical Constants and Conversion Factors: Table $2.3$ Adjusted Values of Constants
::which gives the mantissas of these figures as:
:::$1 \cdotp 758 \, 802 \, 8$ with an uncertainty of $\pm 54$ corresponding to the $2$ least significant figures
:::$5 \cdotp 272 \, 759$ with an uncertainty of $\pm 16$ corresponding to the $2$ least significant figures
* 
* : $3.$ Physical Constants
* : $1$: Simple models of the atom: $\text {1-3}$: The Electrical Structure of Matter
Category:Electrons
Electron Charge to Mass Ratio/Historical Note

== Historical Note on Electron Charge to Mass Ratio ==
When an electrode is charged to a high electric potential in a high vacuum, it is observed to emit rays.
This had been established over the course of several decades of experiments by a number of workers.
These rays were called cathode rays.
It was the work of in $1897$ to attempt to establish some basic facts about these cathode rays.
== Sources ==
* : $1$: Simple models of the atom: $\text {1-3}$: The Electrical Structure of Matter
Category:Historical Notes
Electron has Smallest Negative Electric Charge Possible

== Physical Law ==
The electric charge on the electron is the negative charge with the smallest possible magnitude that can be directly observed.
== Justification ==
Electric charge has been observed as being quantized.
The quantum of positive charge is defined as being the elementary charge.
The electron bears a negative charge exactly equal in magnitude with the positive charge on the proton.
== Sources ==
* : Chapter $\text {I}$: Preliminary Survey: $\S 1$. Electrostatics
Category:Electric Charge
Category:Electrons
Element Commutes with Product of Commuting Elements

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Let $x, y, z \in S$.
Let $x$ commute with both $y$ and $z$.
Then $x$ commutes with $y \circ z$.
=== General Theorem ===
== Proof ==
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 2$: Compositions: Exercise $2.16$
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses: Theorem $4.5: \ 1^\circ$
Category:Commutativity
Category:Semigroups
Category:Element Commutes with Product of Commuting Elements
Element Commutes with Product of Commuting Elements/General Theorem

== Theorem ==
Let $\struct {S, \circ}$ be a semigroup.
Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$ be a sequence of terms of $S$.
Let $b \in S$ such that $b$ commutes with $a_k$ for each $k \in \closedint 1 n$.
Then $b$ commutes with $a_1 \circ \cdots \circ a_n$.
== Proof ==
The proof proceeds by induction on $n$, the length of $\sequence {a_k}_{1 \mathop \le k \mathop \le n}$.
=== Basis for the Induction ===
The case $n = 1$ states that:
:$b \circ a_1 = a_1 \circ b$
That is, $b$ commutes with $a_1$, which holds by hypothesis.
This constitutes the basis for the induction.
=== Induction Hypothesis ===
Fix $n \in \N$ with $n \ge 1$.
Assume that the theorem holds for sequences of length $n$.
This is our induction hypothesis.
=== Induction Step ===
This is our induction step:
Let $\sequence {a_k}_{1 \mathop \le k \mathop \le n + 1}$ be a sequence of length $n + 1$ in $S$.
Suppose that $b$ commutes with $a_k$ for all $k \in \closedint 0 {n + 1}$.
Then (suppressing brackets as $\circ$ is associative):
We conclude that the theorem holds for sequences of length $n + 1$.
The result follows by the Principle of Mathematical Induction.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 18$: Induced $N$-ary Operations: Theorem $18.6$
Category:Element Commutes with Product of Commuting Elements
Element Depends on Independent Set iff Union with Singleton is Dependent

== Theorem ==
Let $M = \struct {S, \mathscr I}$ be a matroid.
Let $X \in \mathscr I$.
Let $x \in S : x \notin X$.
Then:
:$x \in \map \sigma X$ $X \cup \set x$ is dependent.
== Proof ==
=== Necessary Condition ===
Let $x \in \map \sigma X$.
By definition of the closure:
:$x$ depends on $X$.
From Union with Disjoint Singleton is Dependent if Element Depends on Subset:
:$X \cup \set x$ is dependent.
=== Sufficient Condition ===
Let $X \cup \set x$ be dependent.
==== Lemma ====
From Max Operation Yields Supremum of Operands:
:$\map \rho {X \cup \set x} = \max \set {\size A : A \subseteq X \cup \set x \land A \in \mathscr I} \le \size X$
By assumption:
:$X \in \mathscr I$
From Max Operation Yields Supremum of Operands:
:$\size X \le \map \rho {X \cup \set x}$
Thus:
:$\size X = \map \rho {X \cup \set x}$
From Rank of Independent Subset Equals Cardinality:
:$\map \rho X = \size X$
Thus:
:$\map \rho X = \map \rho {X \cup \set x}$
It follows that $x$ depends on $X$ by definition.
So:
:$x \in \map \sigma X$
Category:Matroid Dependence
Category:Matroid Dependent Subsets
Category:Element Depends on Independent Set iff Union with Singleton is Dependent
Element Depends on Independent Set iff Union with Singleton is Dependent/Lemma

== Theorem ==
Let $M = \struct{S, \mathscr I}$ be a matroid.
Let $X \in \mathscr I$.
Let $x \in S : x \notin X$.
Let $X \cup \set x$ be dependent.
Let $A \in \mathscr I$ such that $A \subseteq X \cup \set x$.
Then:
:$\size A \le \size X$
== Proof ==
==== Case 1: $x \in A$ ====
Let $x \in A$.
We have:
:
:$A \setminus \set x = X$
Then:
So:
:$X \cup \set x$ is independent.
This contradicts:
:$X \cup \set x$ is dependent.
So:
:$A \setminus \set x \subsetneq X$
Then:
So:
:$\size A \le \size X$
==== Case 2: $x \notin A$ ====
Let $x \notin A$.
Then:
From Intersection with Subset is Subset:
:$A \subseteq X$
From Cardinality of Subset of Finite Set:
:$\size A \le \size X$
In either case:
:$\size A \le \size X$
Category:Element Depends on Independent Set iff Union with Singleton is Dependent
Element Well Inside Itself Iff Has Complement

== Theorem ==
Let $L = \struct{S, \vee, \wedge, \preceq}$ be a distributive lattice with greatest element $\top$ and smallest element $\bot$.
Let $\eqslantless$ denote the well inside relation on $L$.
Then:
:$\forall a \in S : a \eqslantless a \iff a$ has a complement
== Proof ==
Follows immediately from:
:* Definition:Well Inside Relation
:* Definition:Complement (Lattice Theory)
== Sources ==
* : Chapter $\text {III}$: Compact Hausdorff Spaces, $\S1.1$
Category:Well Inside Relation
Element equals to Supremum of Infima of Open Sets that Element Belongs implies Topological Lattice is Continuous

== Theorem ==
Let $L = \left({S, \preceq, \tau}\right)$ be a complete Scott topological lattice.
Let
:$\forall x \in S: x = \sup \left\{ {\inf X: x \in X \in \sigma\left({L}\right)}\right\}$
Then $L$ is continuous.
== Proof ==
Thus by Way Below Closure is Directed in Bounded Below Join Semilattice:
:$\forall x \in S:x^\ll$ is directed.
Thus by definition pf complete lattice:
:$L$ is up-complete.
Let $x \in S$.
Define $W := \left\{ {\inf X: x \in X \in \sigma\left({L}\right)}\right\}$
By definition of way below closure:
:$x$ is upper bound for $x^\ll$
By definition of supremum:
:$\sup \left({x^\ll}\right) \preceq x$
We will prove that
:$W \subseteq x^\ll$
Let $d \in W$.
By definition of $W$:
:$\exists V: d = \inf V \land x \in V \land V \in \sigma\left({L}\right)$
By Scott Topology equals to Scott Sigma:
:$\tau = \sigma\left({L}\right)$
By definition:
:$V$ is open.
By Infimum of Open Set is Way Below Element in Complete Scott Topological Lattice:
:$d \ll x$
Thus by definition of way below closure:
:$d \in x^\ll$
By Supremum of Subset:
:$\sup W \preceq \sup\left({x^\ll}\right)$
By assumption:
:$x \preceq \sup\left({x^\ll}\right)$
Hence by definition of antisymmetry:
:$x = \sup \left({x^\ll}\right)$
== Sources ==
* 
* 
Category:Continuous Lattices
Category:Topological Order Theory
Element has Idempotent Power in Finite Semigroup

== Theorem ==
Let $\struct {S, \circ}$ be a finite semigroup.
For every element in $\struct {S, \circ}$, there is a power of that element which is idempotent.
That is:
:$\forall x \in S: \exists i \in \N: x^i = x^i \circ x^i$
== Proof ==
From Finite Semigroup Equal Elements for Different Powers, we have:
:$\forall x \in S: \exists m, n \in \N: m \ne n: x^m = x^n$
Let $m > n$.
Let $n = k, m = k + l$.
Then:
: $\forall x \in S: \exists k, l \in \N: x^k = x^{k + l}$
Now we show that:
:$x^k = x^{k + l} \implies x^{k l} = x^{k l} \circ x^{k l}$
That is, that $x^{k l}$ is idempotent.
First:
From here we can easily prove by induction that:
: $\forall n \in \N: x^k = x^{k + n l}$
In particular:
: $x^k = x^{k + k l} = x^{k \paren {l + 1} }$
There are two cases to consider:
:$(1): \quad$ If $l = 1$, then $x^k = x^{k \paren {l + 1} } = x^{2 k} = x^k \circ x^k$, and $x^{k l} = x^k$ is idempotent.
:$(2): \quad$ If $l > 1$, then:
and again, $x^{k l}$ is idempotent.
== Sources ==
* : Chapter $5$: Semigroups: Exercise $4$
Category:Semigroups
Category:Idempotence
Element in Bounded Metric Space has Bound

== Theorem ==
Let $M = \struct {X, d}$ be a metric space.
Let $M' = \struct {Y, d_Y}$ be a subspace of $M$.
Let $M'$ be bounded in $M$.
Then:
:$\forall a' \in X: \exists K' \in \R: \forall x \in Y: \map d {x, a'} \le K'$
That is, if there is one element of $X$ which satisfies the condition for $Y$ to be bounded in $M$, they all do.
== Proof ==
Let $a \in X$ such that $\exists K \in \R: \forall x \in Y: \map d {x, a} \le K$.
Let $a' \in X$.
== Sources ==
* : $2$: Continuity generalized: metric spaces: $2.2$: Examples: Definitions $2.2.12$
Category:Bounded Metric Spaces
Element in Coset iff Product with Inverse in Subgroup

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Let $H$ be a subgroup of $G$.
Let $x, y \in G$.
==== Element in Left Coset iff Product with Inverse in Subgroup ====
==== Element in Right Coset iff Product with Inverse in Subgroup ====
Category:Cosets
Element in Image of Preimage under Mapping

== Theorem ==
Let $f: S \to T$ be a mapping.
Then:
:$\forall y \in T: \in f \sqbrk {f^{-1} \sqbrk y} = \set y$
== Proof ==
A mapping is by definition a relation.
Therefore applies:
:$B \subseteq \Img S \implies \paren {f \circ f^{-1} } \sqbrk B = B$
Thus:
:$\set y \subseteq T \implies f^{-1} \sqbrk {f \sqbrk {\set y} } = \set y$
Hence the result.
== Sources ==
* : $\S 1.12$: Set Inclusions for Image and Inverse Image Sets: Theorem $12.7 \ \text{(c)}$
Category:Mapping Theory
Element in Integral Domain is Divisor iff Principal Ideal is Superset

== Theorem ==
Let $\struct {D, +, \circ}$ be an integral domain.
Let $\ideal x$ denote the principal ideal of $D$ generated by $x$.
Let $x, y \in \struct {D, +, \circ}$.
Then:
:$x \divides y \iff \ideal y \subseteq \ideal x$
where $x \divides y$ denotes that $x$ is a divisor of $y$.
== Proof ==
Let that $x \divides y$.
Then by definition of divisor: 
Conversely:
So:
:$x \divides y \iff \ideal y \subseteq \ideal x$
== Sources ==
* : $\S 62.3$ Factorization in an integral domain: $\text{(i)}$
Category:Integral Domains
Category:Principal Ideals of Rings
Category:Factorization
Element in Integral Domain is Unit iff Principal Ideal is Whole Domain

== Theorem ==
Let $\struct {D, +, \circ}$ be an integral domain.
Let $U_D$ be the group of units of $D$.
Let $\ideal x$ be the principal ideal of $D$ generated by $x$.
Let $x, y \in \struct {D, +, \circ}$.
Then:
:$x \in U_D \iff \ideal x = D$
== Proof ==
Conversely:
== Sources ==
* : $\S 62.3$ Factorization in an integral domain: $\text{(ii)}$
Category:Integral Domains
Category:Principal Ideals of Rings
Element in Left Coset iff Product with Inverse in Subgroup

== Theorem ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Let $x, y \in G$.
Let $y H$ denote the left coset of $H$ by $y$.
Then:
:$x \in y H \iff x^{-1} y \in H$
== Proof ==
</onlyinclude>
== Also see ==
* Element in Right Coset iff Product with Inverse in Subgroup
== Sources ==
* : $\S 42.6 \ \text {(1L)}$ Another approach to cosets
* : Chapter $7$: Cosets and Lagrange's Theorem: Exercise $7 \ \text{(i)}$
Category:Cosets
Element in Preimage of Image under Mapping

== Theorem ==
Let $f: S \to T$ be a mapping.
Let $f^{-1} \sqbrk {\map f x}$ denote the preimage of $\map f x$ under $f$.
Then:
:$\forall x \in S: x \in f^{-1} \sqbrk {\map f x}$
== Proof ==
A mapping is by definition a left-total relation.
Therefore Preimage of Image under Left-Total Relation is Superset applies:
:$A \subseteq S \implies A \subseteq f^{-1} \sqbrk {f \sqbrk A}$
Thus:
:$\set x \subseteq S \implies \set x \subseteq f^{-1} \sqbrk {f \sqbrk A}$
Hence the result.
== Sources ==
* : $\S 1.12$: Set Inclusions for Image and Inverse Image Sets: Theorem $12.7 \ \text{(a)}$
Category:Preimages under Mappings
Element in Right Coset iff Product with Inverse in Subgroup

== Theorem ==
Let $G$ be a group.
Let $H$ be a subgroup of $G$.
Let $x, y \in G$.
Let $H \circ y$ denote the right coset of $H$ by $y$.
Then:
:$x \in H y \iff x y^{-1} \in H$
== Proof ==
Let $\struct {G, *}$ be the opposite group of $G$.
Then:
:$x \in H y \iff x \in y * H$
:$x y^{-1} \in H \iff y^{-1} * x \in H$ 
Since $H$ is closed under inverses:
:$x y^{-1} \in H \iff x^{-1} * y \in H$ 
By Element in Left Coset iff Product with Inverse in Subgroup:
:$x \in y * H \iff x^{-1} * y \in H$
Hence the result.
== Also see ==
* Element in Left Coset iff Product with Inverse in Subgroup
== Sources ==
* : $\S 6.1$. The quotient sets of a subgroup: Lemma $\text{(i)}$
* : $\S 42.6 \ \text {(1R)}$ Another approach to cosets
Category:Cosets
Element in its own Equivalence Class

== Theorem ==
Let $\RR$ be an equivalence relation on a set $S$.
Then every element of $S$ is in its own $\RR$-class:
:$\forall x \in S: x \in \eqclass x \RR$
== Proof ==
== Sources ==
* : Appendix $\text{A}.3$: Equivalence Relations: Theorem $\text{A}.2$
* : $\S 17.1$: Equivalence classes
Category:Equivalence Classes
Category:Fundamental Theorem on Equivalence Relations
Element is Finite iff Element is Compact in Lattice of Power Set

== Theorem ==
Let $X$ be a set.
Let $L = \struct {\powerset X, \cup, \cap, \subseteq}$ be a lattice of power set.
Let $x \in \powerset X$.
Then $x$ is a finite set $x$ is a compact element.
== Proof ==
=== Sufficient Condition when Empty ===
The case when $x = \O$
By Empty Set is Bottom of Lattice of Power Set:
:$x = \bot$
where $\bot$ denotes the bottom of $L$.
By Bottom is Way Below Any Element:
:$x \ll x$
where $\ll$ denotes the way below relation.
Thus by definition
:$x$ is a finite set $x$ is a compact element.
=== Sufficient Condition when Non-Empty ===
The case when $x \ne \O$:
Let $x$ be a finite set.
We will prove that
:for every a set $Y$ of subsets of $X$ such that $x \subseteq \bigcup Y$
::then there exists a finite subset $Z$ of $Y$: $x \subseteq \bigcup Z$
Let $Y$ be a set of subsets of $X$ such that
:$x \subseteq \bigcup Y$
By definitions of union and subset:
:$\forall e \in x: \exists u \in Y: e \in u$
By Axiom of Choice:
:$\exists f:x \to Y: \forall e \in x: e \in \map f e$
Define:
:$Z = f \sqbrk x$
By Image of Mapping from Finite Set is Finite:
:$Z$ is a finite set.
By definition of image of mapping:
:$Z \subseteq Y$
It remains to prove that
:$x \subseteq \bigcup Z$
Let $e \in x$.
By definitions of $f$ and image of mapping:
:$e \in \map f e \in Z$
Thus by definition of union:
:$e \in \bigcup Z$
By Way Below in Lattice of Power Set:
:$x \ll x$
Thus by definition of compact:
:$x$ is compact.
=== Necessary Condition when Non-Empty ===
Let $x$ be a compact element.
By definition of subset:
:$Y := \set {\set y: y \in x}$ is set of subsets of $X$.
We will prove that
:$x \subseteq \bigcup Y$
Let $y \in x$.
By definition of $Y$:
:$\set y \in Y$
By definition of singleton:
:$y \in \set y$
Thus by definition of union:
:$y \in \bigcup Y$
By definition of compact:
:$x \ll x$
By Way Below in Lattice of Power Set:
:there exists finite subset $Z$ of $Y$: $x \subseteq \bigcup Z$
By Singleton is Finite:
:$\forall A \in Z: A$ is a finite set.
By Finite Union of Finite Sets is Finite:
:$\bigcup Z$ is a finite set.
Thus by Subset of Finite Set is Finite:
:$x$ is a finite set.
== Sources ==
* 
* 
Category:Way Below Relation
Category:Power Set
Element is Loop iff Member of Closure of Empty Set

== Theorem ==
Let $M = \struct{S, \mathscr I}$ be a matroid.
Let $x \in S$.
Then:
:$x$ is a loop $x \in \map \sigma \O$
where $\map \sigma \O$ denotes the closure of the empty set.
== Proof ==
From Element is Loop iff Rank is Zero:
:$x$ is a loop $\map \rho {\set x} = 0$
where $\rho$ is the rank function of $M$.
Now:
The result follows.
== Sources ==
* Chapter $1.$ $\S 4.$ Loops and parallel elements
Category:Matroid Loops
Category:Matroid Closure
Element is Loop iff Singleton is Circuit

== Theorem ==
Let $M = \struct{S, \mathscr I}$ be a matroid.
Let $x \in S$.
Then:
:$x$ is a loop $\set x$ is a circuit
== Proof ==
=== Necessary Condition ===
Let $x$ be a loop.
By definition of a loop:
:$\set x$ is a dependent subset of $S$
Let $A \subseteq \set x$ be a dependent subset.
From Power Set of Singleton:
:$\powerset {\set x} = \set{\O, \set x}$
By matroid axiom $(\text I 1)$:
:$\O$ is an independent subset
Then:
:$A = \set x$
It follows that:
:$\set x$ is a minimal dependent subset of $S$.
Then $\set x$ be a circuit by definition. 
=== Sufficient Condition ===
Let $\set x$ be a circuit.
By definition of a circuit:
:$\set x$ is a minimal dependent subset of $S$.
In particular, $\set x$ is a dependent subset of $S$.
Then $\set x$ is a loop by definition.
== Sources ==
* Chapter $1.$ $\S 4.$ Loops and parallel elements
Category:Matroid Loops
Category:Matroid Circuits
Element is Meet Irreducible iff Complement of Element is Irreducible

== Theorem ==
Let $T = \struct {S, \tau}$ be a topological space.
Let $P = \struct {\tau, \preceq}$ be an ordered set
where $\mathord \preceq = \mathord \subseteq \cap \paren {\tau \times \tau}$
Let $A \in \tau$ such that
:$A \ne \top_P$
where $\top_P$ denotes the greatest element in $P$.
Then $A$ is meet irreducible in $P$ $\relcomp S A$ is irreducible
where $\relcomp S A$ denotes the relative complement of $A$ relative to $S$.
== Proof ==
=== Sufficient Condition ===
Let $A$ be meet irreducible in $P$.
By Top in Ordered Set of Topology:
:$A \ne S$
By Relative Complement of Empty Set and Relative Complement of Relative Complement:
:$\relcomp S A \ne \O$
Thus by definition:
:$\relcomp S A$ is non-empty.
Thus by definition:
:$\relcomp S A$ is closed.
Let $B, C$ be closed subsets of $S$ such that
:$\relcomp S A = B \cup C$
By De Morgan's Laws (Set Theory)/Relative Complement/Complement of Union and Relative Complement of Relative Complement:
:$A = \relcomp S B \cap \relcomp S C$
By definition of closed set:
:$\relcomp S B \in \tau$ and $\relcomp S C \in \tau$
By definition of topological space:
:$\relcomp S B \cap \relcomp S C \in \tau$
By Meet in Inclusion Ordered Set:
:$\relcomp S B \wedge \relcomp S C = A$
By definition of meet irreducible:
:$A = \relcomp S B$ or $A = \relcomp S C$
Thus by Relative Complement of Relative Complement:
:$\relcomp S A = B$ or $\relcomp S A = C$
=== Necessary Condition ===
This follows from Complement of Element is Irreducible implies Element is Meet Irreducible.
== Sources ==
* 
* 
Category:Topology
Category:Meet Irreducible Elements
Element is Member of Base iff Not Loop

== Theorem ==
Let $M = \struct{S, \mathscr I}$ be a matroid.
Let $\mathscr B$ denote the set of all bases of $M$.
Let $x \in S$.
Then:
:$\exists B \in \mathscr B: x \in B$ $x$ is not a loop
== Proof ==
=== Necessary Condition ===
Let $B \in \mathscr B$ such that $x \in B$.
From Singleton of Element is Subset:
:$\set x \subseteq B$
By definition of a base:
:$B \in \mathscr I$
From matroid axiom $(\text I 2)$:
:$\set x \in \mathscr I$
Then $\set x$ is not a dependent subset by definition.
It follows that $x$ is not a loop by definition.
=== Sufficient Condition ===
Let $x$ not be a loop.
By definition of a loop:
:$x$ is not a dependent subset
By definition of a dependent subset:
:$x \in \mathscr I$
From Independent Subset is Contained in Base:
:$\exists B \in \mathscr B: \set x \subseteq B$
By definition of a subset:
:$x \in B$
== Sources ==
* Chapter $1.$ $\S 4.$ Loops and parallel elements
Category:Matroid Bases
Category:Matroid Loops
Element is Unit iff its Euclidean Valuation equals that of 1

== Theorem ==
Let $\struct {D, +, \times}$ be a Euclidean domain whose zero is $0$, and unity is $1$.
Let the valuation function of $D$ be $\nu$.
Let $a \in D$.
Then:
:$a$ is a unit of $D$ $\map \nu a = \map \nu 1$
== Proof ==
For $a \in D$ we have that:
:$\map \nu 1 \le \map \nu {1 a} = \map \nu a$
by definition of Euclidean valuation.
Let $a$ be a unit of $D$.
Then:
:$\exists b \in D: a b = 1$
Then:
:$\map \nu a \le \map \nu {a b} = \map \nu 1$
and so:
:$\map \nu a = \map \nu 1$
Let $\map \nu a = \map \nu 1$.
We can write this as:
:$\map \nu {1 a} = \map \nu 1$
and it follows from Euclidean Valuation of Non-Unit is less than that of Product that $a$ is a unit of $D$.
== Sources ==
* : Chapter $6$: Polynomials and Euclidean Rings: $\S 30$. Unique Factorization: Theorem $58 \ \text{(ii)}$
Category:Euclidean Domains
Element of *-Algebra Uniquely Decomposes into Hermitian Elements

== Theorem ==
Let $\struct {A, \ast}$ be a $\ast$-algebra over $\C$. 
Let $a \in A$.
Then there exists unique Hermitian elements $b, c \in A$ such that:
:$a = b + i c$
In particular, $b = \map \Re a$ and $c = \map \Im a$ where $b$ and $c$ are the real and imaginary parts of $a$ respectively.
== Proof ==
=== Proof of Existence === 
Let:
:$b = \map \Re a = \dfrac 1 2 \paren {a + a^\ast}$
and:
:$c = \map \Im a = \dfrac 1 {2 i} \paren {a - a^\ast}$
Then we have using $(\text C^\ast 2)$ and $(\text C^\ast 1)$: 
:$b^\ast = \dfrac 1 2 \paren {a^\ast + a^{\ast \ast} } = \dfrac 1 2 \paren {a + a^\ast} = b$
and:
We therefore have: 
=== Proof of Uniqueness ===
Suppose that $b_1, c_1, b_2, c_2 \in A$ are Hermitian elements such that:
:$a = b_1 + i c_1 = b_2 + i c_2$
Then, we have:
:$b_1 - b_2 = i \paren {c_2 - c_1}$
Using $(\text C^\ast 4)$ and $(\text C^\ast 2)$, we have:
:$b_1 - b_2 = -i \paren {c_2 - c_1} = -\paren {b_1 - b_2}$
So we have $b_1 = b_2$. 
Hence we have $i \paren {c_2 - c_1} = 0$.
Hence $c_1 = c_2$.
Hence we have the desired uniqueness.
== Sources ==
* : $2.1$: $C^\ast$-Algebras
Category:Hermitian Elements of *-Algebras