nyuuzyou/wikis huggingface dataset:
Interval Divided into Subsets

== Theorem ==
Let $\mathbb I$ be a real interval.
Let $S$ and $T$ be non-empty subsets of $\mathbb I$ such that $\mathbb I \subseteq S \cup T$.
Then one of $S$ or $T$ contains an element at zero distance from the other.
== Proof ==
$\mathbb I \subseteq S \cup T \implies \forall x \in \mathbb I: x \in S \lor x \in T$ from the definition of union.
That is, every element of $\mathbb I$ belongs either to $S$ or to $T$.
The distance of an element $c \in \R$ from a subset $S$ of $\R$ is given as:
:$\ds \map d {c, S} = \map {\inf_{x \mathop \in S} } {\size {c - x} }$
Assume that $S$ and $T$ have no element in common, otherwise the result is trivial.
, suppose that $\exists s \in S, t \in T$ such that $s < t$.
(If not, then $\exists s \in S, t \in T$ such that $s > t$, and the following argument may be amended appropriately.)
Let $T_0 = \set {x: x \in T: x > s}$.
As $t \in T_0$ it follows that $T_0 \ne \O$.
Also, $T_0$ is bounded below by $s$.
Let $b = \map \inf {T_0}$.
If $b \notin T$ then $b \in S$.
But from Distance from Subset of Real Numbers‎, it follows that $\map d {b, T_0} = 0$.
Thus we have found an element in $S$ which is zero distance from $T$.
Otherwise, $b \in T$.
Then $b > s$, and the open interval $\openint s b$ is a non-empty subset of $S$.
Hence $b$ is an element in $T$ which is zero distance from $S$.
== Sources ==
* : $\S 2$: Continuum Property: Exercise $\S 2.13 \ (6)$
Category:Real Analysis
Interval between Local Maxima for Underdamped Free Vibration

== Theorem ==
Consider a physical system $S$ whose behaviour can be described with the second order ODE in the form:
:$(1): \quad \dfrac {\d^2 x} {\d t^2} + 2 b \dfrac {\d x} {\d t} + a^2 x = 0$
for $a, b \in \R_{>0}$.
Let $b < a$, so as to make $S$ underdamped.
:
Let $T$ be the period of oscillation of $S$.
Then the successive local maxima of $x$ occur for $t = 0, T, 2T, \ldots$
== Proof ==
Let the position of $S$ be described in the canonical form:
:$(1): \quad x = \dfrac {x_0 \, a} \alpha e^{-b t} \map \cos {\alpha t - \theta}$
where:
:$\alpha = \sqrt {a^2 - b^2}$.
:$\theta = \map \arctan {\dfrac b \alpha}$
From Period of Oscillation of Underdamped System is Regular, the period of oscillation $T$ is given by:
:$T = \dfrac {2 \pi} {a^2 - b^2}$
Differentiating { $t$:
From Derivative at Maximum or Minimum, the local maxima and local minima of $x$ occur at $x' = 0$:
It remains to be determined which of these points at which $x' = 0$ are local maxima.
This occurs when $x > 0$.
From Cosine of Angle plus Full Angle:
:$\cos x = \map \cos {2 \pi + x}$
We have that at $x = x_0$ at $t = 0$.
It is given that $x_0 > 0$.
So at $t = 0, 2 \pi, 4 \pi, \ldots$ we have that:
:$\cos \alpha t > 0$
Similarly, from Cosine of Angle plus Straight Angle:
:$\cos x = -\map \cos {\pi + x}$
So at $t = \pi, 3 \pi, 5 \pi, \ldots$ we have that:
:$\cos \alpha t < 0$
Thus we have that:
:$\alpha L = 2 \pi$
where $L$ is the value of $t$ between consecutive local maxima of $x$.
Thus:
:$L = \dfrac {2 \pi} {\alpha} = \dfrac {2 \pi} {a^2 - b^2} = T$
as required.
== Sources ==
* : $\S 3.20$: Problem $(2)$
Category:Mathematical Physics
Interval containing Prime Number of forms 4n - 1, 4n + 1, 6n - 1, 6n + 1

== Theorem ==
Let $n \in \Z$ be an integer such that $n \ge 118$.
Then between $n$ and $\dfrac {4 n} 3$ there exists at least one prime number of each of the forms:
:$4 m - 1, 4 m + 1, 6 m - 1, 6 m + 1$
== Proof ==
It is demonstrated that the result is true for $n = 118$:
:$\dfrac {4 \times 118} 3 = 157 \cdotp \dot 3$
The primes between $118$ and $157$ are:
So it can be seen that:
:the primes of the form $4 m - 1$ are:
::$127, 131, 139, 151$
:the primes of the form $4 m + 1$ are:
::$137, 139, 157$
:the primes of the form $6 m - 1$ are:
::$131, 137, 149$
:the primes of the form $6 m + 1$ are:
::$127, 139, 151, 157$
and so the conditions of the result are fulfilled for $118$.
But then note we have:
:$\dfrac {4 \times 117} 3 = 156$
and so the primes between $117$ and $157$ are the same ones as between $118$ and $157$, excluding $157$ itself.
Thus the conditions also hold for $117$.
It is puzzling as to why the condition $n \ge 118$ has been applied to the initial statement.
== Sources ==
* : $118$
* : $118$
Category:Prime Numbers
Interval of Definition of Power Tower Function

== Definition ==
The power tower function is defined on the closed interval $\closedint {e^{-e} } {e^{1/e} }$.
== Also see ==
* Euler's Number to Power of its Negative
* Euler's Number to Power of its Reciprocal
== Historical Note ==
== Sources ==
* : $0,06598803584 \ldots$
* : $1 \cdotp 444 \, 667 \, 861 \ldots$
* : $1 \cdotp 44466 \, 7861 \ldots$
Category:Powers
Category:Euler's Number
Interval of Ordered Set is Convex

== Theorem ==
Let $\struct {S, \preceq}$ be a totally ordered set.
Let $I$ be an interval: be it open, closed, or half-open in $S$.
Then $I$ is convex in $S$.
== Proof ==
Any interval can be represented as the intersection of two rays.
Thus by Ray is Convex and Intersection of Convex Sets is Convex Set (Order Theory), $I$ is convex.
Category:Order Theory
Interval of Totally Ordered Set is Convex

== Theorem ==
Let $\struct {S, \preccurlyeq}$ be a totally ordered set.
Let $I \subseteq S$ be an interval in $S$.
Then $I$ is convex.
== Proof ==
There are a number of cases to investigate.
=== Open Interval ===
Let $I = \openint a b$ be an open interval:
:$I = \set {x \in S: a \prec x \prec b}$
Let $s, t, x \in I$ such that $s \prec x \prec t$.
Then by definition:
:$a \prec s \prec x$
and:
:$x \prec t \prec b$
and so:
:$a \prec x \prec b$
and $x \in I$.
Thus we have:
:$\forall s, t, x, \in I: s \prec x \prec t \implies x \in I$
and $I$ is convex by definition.
=== Left Half-Open Interval ===
Let $I = \hointl a b$ be a left half-open interval:
:$I = \set {x \in S: a \prec x \preceq b}$
Let $s, t, x \in I$ such that $s \prec x \prec t$.
Then by definition:
:$a \prec s \prec x$
and:
:$x \prec t \preceq b$
and so:
:$a \prec x \preceq b$
and $x \in I$.
Thus we have:
:$\forall s, t, x, \in I: s \prec x \prec t \implies x \in I$
and $I$ is convex by definition.
=== Right Half-Open Interval ===
Let $I = \hointr a b$ be a right half-open interval:
:$I = \set {x \in S: a \preceq x \prec b}$
Let $s, t, x \in I$ such that $s \prec x \prec t$.
Then by definition:
:$a \preceq s \prec x$
and:
:$x \prec t \prec b$
and so:
:$a \preceq x \prec b$
and $x \in I$.
Thus we have:
:$\forall s, t, x, \in I: s \prec x \prec t \implies x \in I$
and $I$ is convex by definition.
=== Closed Interval ===
Let $I = \closedint a b$ be a closed interval:
:$I = \set {x \in S: a \preceq x \preceq b}$
Let $s, t, x \in I$ such that $s \prec x \prec t$.
Then by definition:
:$a \preceq s \prec x$
and:
:$x \prec t \preceq b$
and so:
:$a \preceq x \preceq b$
and $x \in I$.
Thus we have:
:$\forall s, t, x, \in I: s \prec x \prec t \implies x \in I$
and $I$ is convex by definition.
== Sources ==
* : Part $\text {II}$: Counterexamples: $39$. Order Topology: $1$
Category:Intervals
Category:Convex Sets (Order Theory)
Invalid Argument/Examples/I am Not Famous

== Example of Invalid Argument ==
The following is an invalid argument with true premises and a true conclusion:
:If I am President then I am famous.
:I am not President.
:Therefore I am not famous.
== Proof ==
This is an example of Denying the Antecedent.
It does not follow that just because I am not President then I am not famous.
The fact that I am indeed not famous is independent of the argument.
== Sources ==
* : $1$ Introduction: Logic and Language: $1.3$: Truth and Validity
Category:Examples of Invalid Arguments
Invalid Argument/Examples/Moon is Yellow

== Example of Invalid Argument ==
This argument is invalid:
:The moon is yellow.
:Therefore, the moon is made of cheese.
== Proof ==
Let $P$ denote the simple statement The moon is yellow..
Let $Q$ denote the simple statement The moon is made of cheese..
The argument can then be expressed as:
There is nothing connecting the two statements.
Hence this is not a valid argument form.
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.1$: Statements and connectives
Category:Examples of Invalid Arguments
Invalid Argument/Examples/Rockefeller was Not Famous

== Example of Invalid Argument ==
The following is an valid argument with true premises and a false conclusion:
:If Nelson Rockefeller was President then he was famous.
:Nelson Rockefeller was not President.
:Therefore Nelson Rockefeller was not famous.
== Proof ==
This is an example of Denying the Antecedent.
It is true that being President makes you famous, and it is also true that Nelson Rockefeller was not President.
However, it is not true that not being President makes you not famous.
== Sources ==
* : $1$ Introduction: Logic and Language: $1.3$: Truth and Validity
Category:Examples of Invalid Arguments
Invalid Argument/Examples/Socrates is Mortal

== Example of Invalid Argument ==
This argument is technically invalid:
: is a man.
:Therefore, is mortal.
== Proof ==
It may be thought that the conclusion may hence be deduced from the premise.
However, this does so purely because of the knowledge that:
:if $A$ is a man, then $A$ is mortal.
Hence while the conclusion follows from the premise, this does not happen purely by means of deduction from the argument itself.
Let $P$ denote the simple statement is a man..
Let $Q$ denote the simple statement is mortal..
The argument can then be expressed as:
But this is not a valid argument form.
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.1$: Statements and connectives: Example $1.1$
Category:Examples of Invalid Arguments
Invariance (Statistical Quantity)/Examples/t-Test

== Example of Invariance in context of Statistical Quantities ==
The $t$ statistic used in the $t$-test is an example of an invariance.
It is unchanged by a linear transformation of the sample values, such as replacing $x_i$ with $3 x_i + 7$.
== Sources ==
* 
* 
Category:Examples of Invariances (Statistical Quantities)
Invariance of Extremal Length under Conformal Mappings

== Theorem ==
Let $X, Y$ be Riemann surfaces (usually, subsets of the complex plane).
Let $\phi: X \to Y$ be a conformal isomorphism between $X$ and $Y$.
Let $\Gamma$ be a family of rectifiable curves (or, more generally, of unions of rectifiable curves) in $X$.
Let $\Gamma'$ be the family of their images under $\phi$.
Then $\Gamma$ and $\Gamma'$ have the same extremal length:
:$\map \lambda \Gamma = \map \lambda {\Gamma'}$
== Proof ==
Let $\rho'$ be a conformal metric on $Y$ in the sense of the definition of extremal length, given in local coordinates as:
:$\map {\rho'} z \size {\d z}$
Let $\rho$ be the metric on $X$ obtained as the pull-back of this metric under $\phi$.
That is, $\rho$ is given in local coordinates as:
:$\map {\rho'} {\map \phi w} \cdot \size {\map {\dfrac {\d \phi} {\d w} } w} \cdot \size {\d w}$
Then the area of $X$ with respect to $\rho$ and the area of $Y$ with respect to $\rho'$ are equal by definition:
:$\map A {\rho'} = \map A \rho$
Furthermore, if $\gamma \in \Gamma$ and $\gamma' := \map \phi \gamma$, then also:
:$\map L {\gamma, \rho} = \map L {\gamma', \rho'}$
and hence:
:$\map L {\Gamma, \rho} = \map L {\Gamma', \rho'}$
In summary, for any metric $\rho'$ on $Y$, there is a metric $\rho$ on $X$ such that:
:$\dfrac {\map L {\Gamma, \rho} } {\map A \rho} = \dfrac {\map L {\Gamma', \rho'} } {\map A {\rho'} }$
It thus follows from the definition of extremal length that:
:$\map \lambda \Gamma \ge \map \lambda {\Gamma'}$
The opposite inequality follows by exchanging the roles of $X$ and $Y$.
== Sources ==
* 
Category:Geometric Function Theory
Invariance of Pseudoinverse under Addition of Degenerate Transformation

== Theorem ==
Let $U, V$ be vector spaces over a field $K$.
Let $S: U \to V$ be a linear transformation.
Let $T: V \to U$ be a linear transformation.
Let $S$ and $T$ are pseudoinverse to each other.
Then $S + G_1$ and $T + G_2$ are pseudoinverse to each other, where:
:$G_1: U \to V$ is an arbitrary degenerate linear transformation
:$G_2: V \to U$ is an arbitrary degenerate linear transformation
== Proof ==
Let:
By , $G_3, G_4$ are degenerate.
Then:
which is degenerate in view of:
:Product with Degenerate Linear Transformation is Degenerate
:Right Product with Degenerate Linear Transformation is Degenerate
:Product with Degenerate Linear Transformation is Degenerate
Similarly:
which is degenerate.
== Sources ==
* : $2.2$: Index of a Linear Map
Category:Linear Algebra
Invariant Measure/Examples/Arcs on Unit Circle

== Example of Invariant Measure ==
Let $\CC$ denote the unit circle embedded in the complex plane:
:$\CC = \set {z \in \C: \cmod z = 1}$
The standard measure for arcs on $\CC$ is invariant under the mapping $\map T z = z^2$
== Sources ==
* 
* 
Category:Examples of Invariant Measures
Invariant Measure of Image under Bijection

== Theorem ==
Let $\struct {X, \Sigma, \mu}$ be a measure space.
Let $\theta: X \to X$ be an $\Sigma / \Sigma$-measurable bijection.
Let $\mu$ be an invariant measure.
Then:
:$\forall A \subseteq X: \map \mu {\theta \sqbrk A} = \map \mu A$
== Proof ==
== Sources ==
* 
* 
Category:Invariant Measures
Inverse Completion Less Zero of Integral Domain is Closed

== Theorem ==
Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$ and whose unity is $1_D$.
Let $\struct {K, \circ}$ be the inverse completion of $\struct {D, \circ}$.
Then $\struct {K^*, \circ}$ is closed, where $K^* = K \setminus \set {0_K}$.
== Proof ==
Let $\struct {K, \circ}$ be the inverse completion of $\struct {D, \circ}$.
We define $\struct {K, \circ}$ of $\struct {D, \circ}$ by means of the technique used in Inverse Completion of Integral Domain Exists.
The structure of $\struct {K, \circ}$ is such that each element of $\struct {K, \circ}$ is of the form $x \circ y^{-1}$, where $x \in D$ and $y \in D^*$.
From Zero of Inverse Completion of Integral Domain, $0_K$ is all elements of $D \times D^*$ of the form $\dfrac {0_D} x$.
Therefore the elements of $K^*$ are those of the form $x \circ y^{-1}$ where $x, y \in D^*$.
By Product of Division Products:
:$\dfrac a b \circ \dfrac c d = \dfrac {a \circ c} {b \circ d}$
As $\struct {D, +, \circ}$ is an integral domain, none of its non-zero elements are zero divisors.
Therefore:
:$\forall x, y \in D^*: x \circ y \ne 0_D$
So:
:$\forall \dfrac a b, \dfrac c d \in K^*: \dfrac {a \circ c} {b \circ d} \in K^*$
Hence the result.
Category:Integral Domains
Category:Inverse Completions
Inverse Completion Theorem

== Theorem ==
Every commutative semigroup containing cancellable elements admits an inverse completion.
== Proof ==
Let $\struct {S, \circ}$ be a commutative semigroup which has cancellable elements.
From Construction of Inverse Completion, we can construct an inverse completion $\struct {T', \oplus'}$ of $\struct {S', \oplus'}$, which is an isomorphic copy of $S$ under the mapping $\psi: S \to S'$.
By the Embedding Theorem, there exists a semigroup $\struct {T, \circ}$ which contains $\struct {S, \circ}$, and an isomorphism $\Psi$ from $\struct {T, \circ}$ to $\struct {T', \oplus'}$ which extends $\psi$.
Thus $T$ is an inverse completion of $S$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers: Theorem $20.3$
Category:Inverse Completions
Category:Semigroups
Category:Named Theorems
Inverse Completion is Commutative Monoid

== Theorem ==
Let $\left({S, \circ}\right)$ be a commutative semigroup.
Let $\left ({C, \circ}\right) \subseteq \left({S, \circ}\right)$ be the subsemigroup of cancellable elements of $\left({S, \circ}\right)$.
Let $\left({T, \circ'}\right)$ be an inverse completion of $\left({S, \circ}\right)$.
Then $\left({T, \circ'}\right)$ is a commutative monoid.
== Proof ==
From Inverse Completion is Commutative Semigroup:
:$\left({T, \circ'}\right)$ is a commutative semigroup.
By definition of inverse completion:
:$\forall x \in C: \exists x^{-1} \in C: x \circ x^{-1} \in T$
Thus by definition of inverse element:
:$e := x \circ x^{-1}$ is an identity element of $T$.
By Identity is Unique, $e$ is the only such element of $T$.
Thus $\left({T, \circ'}\right)$ is a semigroup with an identity element.
So, by definition, $\left({T, \circ'}\right)$ is a commutative monoid.
Category:Inverse Completions
Inverse Completion is Commutative Semigroup

== Theorem ==
Let $\struct {S, \circ}$ be a commutative semigroup.
Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the subsemigroup of cancellable elements of $\struct {S, \circ}$.
Let $\struct {T, \circ'}$ be an inverse completion of $\struct {S, \circ}$.
Then $T = S \circ' C^{-1}$, and is a commutative semigroup.
== Proof ==
From Subset Product defining Inverse Completion of Commutative Semigroup is Commutative Semigroup:
:$S \circ' C^{-1}$ is a commutative semigroup.
From Structure of Inverse Completion of Commutative Semigroup:
:$T = S \circ' C^{-1}$
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers: Theorem $20.1: \ 2^\circ$
Category:Inverse Completions
Inverse Completion is Unique

== Theorem ==
An inverse completion of a commutative semigroup is unique up to isomorphism.
== Proof ==
Let $T$ and $T'$ both be inverse completions of a commutative semigroup $S$ having cancellable elements.
Then from the Extension Theorem for Isomorphisms, there is a unique isomorphism $\phi: T \to T'$ satisfying $\forall x \in S: \map \phi x = x$.
Hence the result.
== Comment ==
Thus, when discussing inverse completions of a commutative semigroup with cancellable elements, we can talk about the inverse completion of such a semigroup.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers: Theorem $20.5$: Corollary
Category:Inverse Completions
Category:Semigroups
Inverse Completion of Commutative Semigroup is Abelian Group

== Theorem ==
Let $\struct {S, \circ}$ be a commutative semigroup
Let all the elements of $\struct {S, \circ}$ be cancellable.
Then an inverse completion of $\struct {S, \circ}$ is an abelian group.
== Proof ==
Let $\struct {S, \circ}$ be a commutative semigroup, all of whose elements are cancellable.
Let $\struct {T, \circ'}$ be an inverse completion of $\struct {S, \circ}$.
From Inverse Completion is Commutative Monoid:
:$\struct {T, \circ'} = \struct {S \circ' S^{-1}, \circ'}$
has been shown to be a commutative monoid.
Taking the group axioms in turn:
=== ===
As $\struct {T, \circ'}$ is a commutative semigroup, it is by definition closed.
=== ===
As $\struct {T, \circ'}$ is a commutative semigroup, it is by definition associative.
=== ===
Let $x \in S$.
Then $x^{-1} \in S^{-1}$ by definition.
As $\struct {T, \circ'} = \struct {S \circ' S^{-1}, \circ'}$ is closed:
:$x \circ' x^{-1} \in T$
This holds for all $x \in S$.
As $\struct {T, \circ'}$ is a commutative semigroup:
:$\exists e \in T: \forall x \in S: x \circ' x^{-1} = e^{-1} = x \circ' x$
Thus $\struct {T, \circ'}$ has an identity element.
=== ===
Every element of $S$ has an inverse in $S^{-1}$.
Therefore every element of $S$ and $S^{-1}$ is invertible.
From Inverse of Product in Associative Structure, every element of $S \circ' S^{-1}$ is therefore also invertible.
Thus every element of $T$ has an inverse.
All the group axioms are thus seen to be fulfilled, and so $\struct {T, \circ'}$ is a group.
Commutativity of $\circ$ has been demonstrated.
Hence the result, by definition of abelian group.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers: Theorem $20.2$
Category:Inverse Completions
Category:Abelian Groups
Inverse Completion of Commutative Semigroup is Inverse Completion of Itself

== Theorem ==
Let $\struct {S, \circ}$ be a commutative semigroup.
Let $\struct {C, \circ} \subseteq \struct {S, \circ}$ be the subsemigroup of cancellable elements of $\struct {S, \circ}$.
Let $\struct {T, \circ'}$ be an inverse completion of $\struct {S, \circ}$.
Then $\struct {T, \circ'}$ is its own inverse completion.
== Proof ==
Let $x \circ' y^{-1}$ be cancellable for $\circ'$, where $x \in S$ and $y \in C$.
We have that $y$ is invertible for $\circ'$.
So by Invertible Element of Associative Structure is Cancellable:
:$y$ is cancellable for $\circ'$.
Now by definition of inverse element:
:$x = \paren {x \circ' y^{-1} } \circ' y$
Thus $x$ is also cancellable for $\circ'$.
By Cancellable Elements of Semigroup form Subsemigroup, $x$ is cancellable for $\circ$.
So:
:$x \in S \implies x \in C$
Thus $x$ is invertible for $\circ'$.
Hence by Inverse of Product in Associative Structure:
:$x \circ' y$ is invertible for $\circ'$.
So every cancellable element of $\struct {T, \circ'}$ is invertible.
So, by definition, $\struct {T, \circ'}$ is its own inverse completion.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers: Theorem $20.1: \ 3^\circ$
Category:Inverse Completions
Inverse Completion of Integral Domain Exists

== Theorem ==
Let $\struct {D, +, \circ}$ be an integral domain whose zero is $0_D$ and whose unity is $1_D$.
Then an inverse completion of $\struct {D, \circ}$ can be constructed.
== Proof ==
From the definition of an integral domain:
:All elements of $D^* = D \setminus \set {0_D}$ are cancellable
:$\struct {D^*, \circ}$ is a commutative semigroup.
So by the Inverse Completion Theorem, there exists an inverse completion of $\struct {D, \circ}$.
From Construction of Inverse Completion, this is done as follows:
Let $\ominus$ be the congruence relation defined on $D \times D^*$ by:
:$\tuple {x_1, y_1} \ominus \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$
The fact that this is a congruence relation is proved in Equivalence Relation on Semigroup Product with Cancellable Elements.
Let $\struct {D \times D^*, \otimes}$ be the external direct product of $\struct {D, \circ}$ with $\struct {D^*, \circ}$, where $\otimes$ is the operation on $D \times D^*$ induced by $\circ$.
Let the quotient structure defined by $\ominus$ be $\struct {\dfrac {D \times D^*} \ominus, \otimes_\ominus}$.
where $\otimes_\ominus$ is the operation induced on $\dfrac {D \times D^*} \ominus$ by $\otimes$.
Let us use $D'$ to denote the quotient set $\dfrac {D \times D^*} {\ominus}$.
Let us use $\circ'$ to denote the operation $\otimes_\ominus$.
Thus $\struct {D', \circ'}$ is the inverse completion of $\struct {D, \circ}$.
An element of $D'$ is therefore an equivalence class of the congruence relation $\ominus$.
As the Inverse Completion is Unique up to isomorphism, it follows that we can define the structure $\struct {K, \circ}$ which is isomorphic to $\struct {D', \circ'}$.
An element of $D'$ is therefore an equivalence class of the congruence relation $\ominus$.
So an element of $K$ is the isomorphic image of an element $\eqclass {\paren {x, y} } \ominus$ of $\dfrac {D \times D^*} \ominus$.
Hence every element of $\struct {K, \circ}$ is of the form $x \circ y^{-1}$, where $x \in D$ and $y \in D^*$.
Alternatively, from the definition of division product, of the form $\dfrac x y$.
Hence we can therefore interpret any element of $\struct {K, \circ}$ as equivalence classes of elements of the form $\dfrac x y$.
Category:Integral Domains
Category:Inverse Completions
Inverse Completion of Natural Numbers

== Theorem ==
There exists an inverse completion of the natural numbers under addition.
== Proof ==
The set of natural numbers under addition can be denoted $\left ({\N, +}\right)$.
From Natural Numbers under Addition form Commutative Monoid, the algebraic structure $\left ({\N, +}\right)$ is a commutative monoid.
Therefore by definition of commutative monoid, $\left ({\N, +}\right)$ is a commutative semigroup.
From Natural Number Addition is Cancellable, all of the elements of $\left ({\N, +}\right)$ are cancellable.
The result follows from the Inverse Completion Theorem.
Category:Inverse Completions
Category:Natural Numbers
Inverse Cosecant of Imaginary Number

== Theorem ==
:$\map {\csc^{-1} } {i x} = i \csch^{-1} x$
== Proof ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: Relationship between Inverse Hyperbolic and Inverse Trigonometric Functions: $8.102$
* : $\S 14$: Hyperbolic Functions: Relationship between Inverse Hyperbolic and Inverse Trigonometric Functions: $14.102.$
Category:Inverse Cosecant
Category:Inverse Hyperbolic Cosecant