nyuuzyou/wikis huggingface dataset:
Inverse Image under Order Embedding of Strict Upper Closure of Image of Point

== Theorem ==
Let $\struct {S, \preceq}$ and $\struct {T, \preceq'}$ be ordered sets.
Let $\phi: S \to T$ be an order embedding of $\struct {S, \preceq}$ into $\struct {T, \preceq'}$
Let $p \in S$.
Then:
:$\map {\phi^{-1} } {\map \phi p^{\succ'} } = p^\succ$
where $\cdot^\succ$ and $\cdot^{\succ'}$ represent strict upper closure with respect to $\preceq$ and $\preceq'$, respectively.
== Proof ==
Let $x \in \map {\phi^{-1} } {\map \phi p^{\succ'} }$.
By the definition of inverse image:
:$\map \phi x \in \map \phi p^{\succ'}$
By the definition of strict upper closure:
:$\map \phi p \prec' \map \phi x$
Since $\phi$ is an order embedding:
:$p \prec x$
Thus by the definition of strict upper closure:
:$x \in p^\succ$
and so:
:$\map {\phi^{-1} } {\map \phi p^{\succ'} } \subseteq p^\succ$
Let $x \in p^\succ$.
By the definition of strict upper closure:
:$p \prec x$
Since $\phi$ is an order embedding:
:$\map \phi p \prec' \map \phi x$
Thus by the definition of strict upper closure:
:$\map \phi x \in \map \phi p^{\succ'}$
Thus by the definition of inverse image:
:$x \in \map {\phi^{-1} } {\map \phi p^{\succ'} }$
and so:
:$p^\succ \subseteq \map {\phi^{-1} } {\map \phi p^{\succ'} }$
Thus by definition of set equality:
:$\map {\phi^{-1} } {\map \phi p^{\succ'} } = p^\succ$
Category:Order Embeddings
Category:Upper Closures
Inverse Integral Operator is Linear if Unique

== Theorem ==
Let $T$ be an integral operator.
Let $f$ be an integrable real function on a domain appropriate to $T$.
Let $F = \map T f$ and $G = \map T g$.
Let $T$ have a unique inverse $T^{-1}$.
Then $T^{-1}$ is a linear operator:
:$\forall p, q \in \R: \map {T^{-1} } {p F + q G} = p \map {T^{-1} } F + q \map {T^{-1} } G$
== Proof ==
Let:
:$x_1 = \map {T^{-1} } F$
:$x_2 = \map {T^{-1} } G$
Thus:
:$F = \map T {x_1}$
:$G = \map T {x_2}$
Then for all $p, q \in \R$:
and so $x = p F + q G$ is a solution to the equation:
:$\map T x = p F + q G$
But this equation has only one solution:
:$x = \map {T^{-1} } {p F + q G}$
Thus $p F + q G$ must coincide with the above:
:$p \map {T^{-1} } F + q \map {T^{-1} } G = \map {T^{-1} } {p F + q G}$
which proves that $T^{-1}$ is a linear operator.
== Sources ==
* : Chapter $2$: $\S 1.4$: The inverse of an operator
* : $\S 1.1$. The Idea of an Integral Transform
Category:Integral Transforms
Inverse Mapping/Examples/Arbitrary Finite Set with Itself

== Example of Compositions of Mappings ==
Let $X = Y = \set {a, b}$.
Consider the mappings from $X$ to $Y$:
We have that:
:$f_1$ is the inverse mapping of itself
:$f_4$ is the inverse mapping of itself
:the inverse of neither $f_2$ nor $f_3$ are mappings.
== Sources ==
* : Chapter $2$: Maps and relations on sets: Example $2.13$
Category:Examples of Inverse Mappings
Inverse Mapping/Examples/Real Cube Function

== Examples of Inverse Mappings ==
Let $f: \R \to \R$ be the mapping defined on the set of real numbers as:
:$\forall x \in \R: \map f x = x^3$
The inverse of $f$ is:
:$\forall y \in \R: \inv f y = \sqrt [3] y$
== Proof ==
From Bijection Example: Real Cube Function $f$ is a bijection.
By definition of the cube root:
:$\sqrt [3] y := \set {x \in \R: x^3 = y}$
From Inverse Mapping is Bijection, it follows that $f^{-1}$ is likewise a bijection.
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Composition of Functions
Category:Examples of Inverse Mappings
Inverse Mapping/Examples/x^2-4x+5

== Examples of Inverse Mappings ==
Let $f: \R \to \R$ be the real function defined as:
:$\forall x \in \R: f \paren x = x^2 - 4 x + 5$
Consider the following bijective restrictions of $f$:
The inverse of $f_1$ is:
:$\forall y \in \hointr 1 \to: \inv {f_1} y = 2 - \sqrt {y - 1}$
The inverse of $f_2$ is:
:$\forall y \in \hointr 1 \to: \inv {f_2} y = 2 + \sqrt {y - 1}$
== Proof ==
Let $y = \map f x$.
Then:
$f_1$ is the bijective restriction of $f$ where $x \le 2$.
Hence the negative square root is taken of $\sqrt {y - 1}$, and so:
:$\inv {f_1} y = 2 - \sqrt {y - 1}$
$f_2$ is the bijective restriction of $f$ where $x \le 2$.
Hence the positive square root is taken of $\sqrt {y - 1}$, and so:
:$\inv {f_2} y = 2 + \sqrt {y - 1}$
== Sources ==
* : Chapter $4$: Mappings: Exercise $12 \ \text{(iv)}$
Category:Examples of Inverse Mappings
Inverse Mapping in Induced Structure of Homomorphism to Abelian Group

== Theorem ==
Let $\struct {S, \circ}$ be an algebraic structure.
Let $\struct {T, \oplus}$ be an abelian group.
Let $f$ be a homomorphism from $S$ into $T$.
Let $f^*$ be the pointwise inverse of $f$.
Then $f^*$ is a homomorphism from $\struct {S, \circ}$ into $\struct {T, \oplus}$.
== Proof ==
Let $\struct {T, \oplus}$ be an abelian group.
Let $x, y \in S$.
Then:
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 13$: Compositions Induced on Cartesian Products and Function Spaces: Theorem $13.7: \ 2^\circ$
Category:Homomorphisms (Abstract Algebra)
Category:Abelian Groups
Category:Pointwise Operations
Inverse Mapping is Bijection

== Theorem ==
Let $S$ and $T$ be sets.
Let $f: S \to T$ and $g: T \to S$ be inverse mappings of each other.
Then $f$ and $g$ are bijections.
== Proof ==
From Inverse is Mapping implies Mapping is Injection and Surjection:
:$f$ is both an injection and a surjection.
Again from Inverse is Mapping implies Mapping is Injection and Surjection:
:$g$ is both an injection and a surjection.
The result follows by definition of bijection.
== Also see ==
* Mapping is Injection and Surjection iff Inverse is Mapping
== Sources ==
* : $1$. Preliminary ideas; sets, vector spaces: $1.1$. Sets
* : Chapter $1$: Theory of Sets: $\S 9$: Inverse Functions, Extensions, and Restrictions
Category:Bijections
Category:Inverse Mappings
Inverse Mapping is Unique

== Theorem ==
Let $f: S \to T$ be a mapping.
If $f$ has an inverse mapping, then that inverse mapping is unique.
That is, if:
:$f$ and $g$ are inverse mappings of each other
and
:$f$ and $h$ are inverse mappings of each other
then $g = h$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Exercise $1.3: \ 12 \ \text{(a)}$
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{G}$
* : $\S 2$: Sets and functions: Inverse images and inverse functions
* : $\S 1$: Naive Set Theory: $\S 1.6$: Functions
* : Chapter $3$: Elementary consequences of the definitions: Proposition $3.2$: Remark
Category:Inverse Mappings
Category:Inverse Mapping is Unique
Inverse Mapping on Group of Units in Unital Banach Algebra is Continuous

== Theorem ==
Let $\struct {A, \norm {\, \cdot \,} }$ be a unital Banach algebra.
Let $\map G A$ be the group of units of $A$.
Define $\phi : \map G A \to \map G A$ by: 
:$\map \phi x = x^{-1}$
for each $x \in \map G A$. 
Then $\phi$ is continuous. 
== Proof ==
Let $x \in \map G A$ and $y \in A$ be such that: 
:$\ds \norm {x - y} < \frac 1 {\norm {x^{-1} } }$
As shown in Group of Units in Unital Banach Algebra is Open, we have $y \in \map G A$ and: 
:$\norm {1 - x^{-1} y} < 1$
Then, from Element of Unital Banach Algebra Close to Identity is Invertible, we have: 
:$\ds \norm {\paren {x^{-1} y}^{-1} } = \norm {y^{-1} x} \le \frac 1 {1 - \norm {1 - x^{-1} y} }$
Using the algebra norm property, we have: 
:$\ds \frac 1 {1 - \norm {1 - x^{-1} y} } = \frac 1 {1 - \norm {x^{-1} \paren {x - y} } } \le \frac 1 {1 - \norm {x^{-1} } \norm {x - y} }$
We now have: 
Then we have: 
Taking $y \to x$, we have $\norm {x - y} \to 0$ and so: 
:$\ds \frac {\norm {x^{-1} }^2 \norm {x - y} } {1 - \norm {x^{-1} } \norm {x - y} } \to 0$
so that: 
:$\norm {y^{-1} - x^{-1} } \to 0$ as $y \to x$. 
So $\phi$ is continuous.
Category:Unital Banach Algebras
Inverse Morphism is Unique

== Theorem ==
Let $\mathbf C$ be a metacategory.
Let $f: C \to D$ be an isomorphism of $\mathbf C$.
Then $f$ admits a unique inverse morphism $g: D \to C$.
== Proof ==
Since $f$ is an isomorphism, it admits at least one inverse morphism.
Now let $g, g': D \to C$ be two inverse morphisms for $f$.
Then:
In conclusion, $g = g'$.
Hence the result.
== Sources ==
* : $\S 2.8$: Exercise $2.3$
Category:Category Theory
Inverse Relation/Examples/Divisor

== Example of Inverse Relation ==
Let $\Z_{>0}$ denote the set of (strictly) positive integers.
Let $\RR$ denote the relation on $\Z_{>0}$ defined as:
:$\forall a, b \in \Z_{>0}: \tuple {a, b} \in \RR \iff a \divides b$
where $a \divides b$ denotes that $a$ is a divisor of $b$.
The inverse relation of $\RR$ is defined as:
:$\forall a, b \in \Z_{>0}: \tuple {a, b} \in \RR^{-1} \iff b \divides a$
That is, $a$ is a multiple of $b$.
Thus, $\tuple {1, 1}$, $\tuple {6, 2}$, $\tuple {12, 3}$ are all elements of the inverse relation of $\RR$.
== Sources ==
* 
* 
Category:Examples of Inverse Relations
Inverse Relation Equal iff Subset

== Theorem ==
Let $S$ and $T$ be sets. Let $\RR \subseteq S \times T$ be a relation.
If $\RR$ is a subset or superset of its inverse, then it equals its inverse.
That is, the following are equivalent:
== Proof ==
=== $(1)$ implies $(2)$ ===
Suppose $\RR \subseteq \RR^{-1}$.
Then:
Hence $\RR^{-1} \subseteq \RR$.
=== $(2)$ implies $(1)$ ===
Since $(1)$ implies $(2)$ was already established above, interpreting it on $\RR^{-1}$ yields:
:$\RR^{-1} \subseteq \paren {\RR^{-1} }^{-1}$
By Inverse of Inverse Relation, $\paren {\RR^{-1} }^{-1} = \RR$.
Hence $(2)$ implies $(1)$.
=== $(3)$ $(1)$ and $(2)$ ===
This is precisely the definition of set equality.
Category:Inverse Relations
Inverse Relation Functor is Contravariant Functor

== Theorem ==
Let $\mathbf{Rel}$ be the category of relations.
Let $C: \mathbf{Rel} \to \mathbf{Rel}$ be the inverse relation functor.
Then $C$ is a contravariant functor.
== Proof ==
For any set $X$. we have:
For any two relations $\RR \subseteq X \times Y$ and $\SS \subseteq Y \times Z$, we have:
Hence $C$ is shown to be a contravariant functor.
== Sources ==
* : $\S 1.9$: Exercise $1 \,\text{(c)}$
Category:Category of Relations
Category:Functors
Inverse Relation Properties

== Theorem ==
Let $\RR$ be a relation on a set $S$.
If $\RR$ has any of the properties:
:Reflexive
:Antireflexive
:Non-reflexive
:Symmetric
:Asymmetric
:Antisymmetric
:Non-symmetric
:Transitive
:Antitransitive
:Non-transitive
then its inverse $\RR^{-1}$ has the same properties.
== Proof ==
=== Reflexivity ===
=== Antireflexivity ===
=== Non-Reflexivity ===
=== Symmetry ===
=== Asymmetry ===
=== Antisymmetry ===
=== Non-Symmetry ===
=== Transitivity ===
=== Antitransitivity ===
=== Non-Transitivity ===
Category:Inverse Relations
Inverse Relation is Left and Right Inverse iff Bijection

== Theorem ==
Let $\RR \subseteq S \times T$ be a relation on a cartesian product $S \times T$.
Let:
:$I_S$ be the identity mapping on $S$
:$I_T$ be the identity mapping on $T$.
Let $\RR^{-1}$ be the inverse relation of $\RR$.
Then $\RR$ is a bijection :
:$\RR^{-1} \circ \RR = I_S$
and
:$\RR \circ \RR^{-1} = I_T$
where $\circ$ denotes composition of relations.
== Proof ==
=== Necessary Condition ===
Let $\RR \subseteq S \times T$ be such that:
:$\RR^{-1} \circ \RR = I_S$
and
:$\RR \circ \RR^{-1} = I_T$.
From Left and Right Inverse Relations Implies Bijection, it follows that $\RR$ is a bijection.
=== Sufficient Condition ===
Suppose $\RR$ is a bijection.
From Bijective Relation has Left and Right Inverse we have that:
:$\RR^{-1} \circ \RR = I_S$ and
:$\RR \circ \RR^{-1} = I_T$.
Hence the result.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 5$: Composites and Inverses of Functions: Exercise $5.8 \ \text{(f)}$
Category:Inverse Relations
Category:Bijections
Inverse Relational Structures of Isomorphic Structures are Isomorphic

== Theorem ==
Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be relational structures.
Let ${\RR_1}^{-1}$ and ${\RR_2}^{-1}$ be the inverses of $\RR_1$ and $\RR_2$ respectively.
Let $f: \struct {S, \RR_1} \to \struct {T, \RR_2}$ be a relation isomorphism.
Then $f: \struct {S, {\RR_1}^{-1} } \to \struct {T, {\RR_2}^{-1} } $ is also a relation isomorphism.
== Proof ==
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.26$
Category:Relation Isomorphisms
Category:Inverse Relations
Inverse Row form of Cayley Table for Group/Examples/S3

== Example of Inverse Row form of Cayley Table for Group ==
The Cayley table of the symmetric group on $3$ letters can be written in inverse row form as:
:$\begin{array}{c|cccccc}
\circ & e & p & q & r & s & t \\
\hline
e & e & p & q & r & s & t \\
p^{-1} & q & e & p & t & r & s \\
q^{-1} & p & q & e & s & t & r \\
r^{-1} & r & t & s & e & q & p \\
s^{-1} & s & r & t & p & e & q \\
t^{-1} & t & s & r & q & p & e \\
\end{array}$
== Sources ==
* : Chapter $\text {I}$: The Group Concept: $\S 5$: The Multiplication Table
Category:Examples of Cayley Tables
Category:Symmetric Group on 3 Letters
Inverse Secant of Imaginary Number

== Theorem ==
:$\sec^{-1} x = \pm i \sech^{-1} x$
== Proof ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.100$: Relationship between Inverse Hyperbolic and Inverse Trigonometric Functions
Category:Inverse Secant
Category:Inverse Hyperbolic Secant
Inverse Sine of Imaginary Number

== Theorem ==
:$\map {\sin^{-1} } {i x} = i \sinh^{-1} x$
== Proof ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.92$: Relationship between Inverse Hyperbolic and Inverse Trigonometric Functions
Category:Inverse Sine
Category:Inverse Hyperbolic Sine
Inverse Tangent of Imaginary Number

== Theorem ==
:$\map {\tan^{-1} } {i x} = i \tanh^{-1} x$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 8$: Hyperbolic Functions: $8.96$: Relationship between Inverse Hyperbolic and Inverse Trigonometric Functions
Category:Inverse Tangent
Category:Inverse Hyperbolic Tangent
Category:Inverse Tangent of Imaginary Number
Inverse Tangent of i

== Theorem ==
The inverse tangent of $i$ is not defined.
== Proof ==
 $\tan z_0 = i$.
This contradicts the theorem Sum of Squares of Sine and Cosine:
:$\sin^2 z_0 + \cos^2 z_0 = 1$
Hence the result by Proof by Contradiction.
== Sources ==
* : $\S 4.6$. The Logarithm: Examples: $\text {(iv)}$
Category:Inverse Tangent
Inverse Trigonometric Function of Reciprocal

== Theorem ==
For all $x \in \R$ for which the expressions above are defined:
=== Arcsine of Reciprocal equals Arccosecant ===
=== Arccosecant of Reciprocal equals Arcsine ===
=== Arccosine of Reciprocal equals Arcsecant ===
=== Arcsecant of Reciprocal equals Arccosine ===
=== Arctangent of Reciprocal equals Arccotangent ===
=== Arccotangent of Reciprocal equals Arctangent ===
Category:Inverse Trigonometric Functions
Category:Reciprocals
Inverse for Complex Addition

== Theorem ==
Let $z = x + i y \in \C$ be a complex number.
Let $-z = -x - i y \in \C$ be the negative of $z$.
Then $-z$ is the inverse element of $z$ under the operation of complex addition:
:$\forall z \in \C: \exists -z \in \C: z + \paren {-z} = 0 = \paren {-z} + z$
== Proof ==
From Complex Addition Identity is Zero, the identity element for $\struct {\C, +}$ is $0 + 0 i$.
Then:
Similarly for $\paren {-x - i y} + \paren {x + i y}$.
== Sources ==
* : $1$: Complex Numbers: Axiomatic Foundations of the Complex Number System: $8$
* : $1$ The complex plane: Complex numbers $\S 1.2$ The algebraic structure of the complex numbers
Category:Complex Addition
Category:Examples of Inverse Elements
Inverse for Complex Multiplication

== Theorem ==
Each element $z = x + i y$ of the set of non-zero complex numbers $\C_{\ne 0}$ has an inverse element $z^{-1}$ under the operation of complex multiplication:
:$\forall z \in \C_{\ne 0}: \exists z^{-1} \in \C_{\ne 0}: z \times z^{-1} = 1 + 0 i = z^{-1} \times z$
This inverse can be expressed as:
:$\dfrac 1 z := \dfrac {x - i y} {x^2 + y^2} = \dfrac {\overline z} {z \overline z}$
where $\overline z$ is the complex conjugate of $z$.
== Proof ==
Similarly for $\dfrac {x - i y} {x^2 + y^2} \paren {x + i y}$.
So the inverse of $x + i y \in \struct {\C_{\ne 0}, \times}$ is $\dfrac {x - i y} {x^2 + y^2}$.
As $x^2 + y^2 > 0 \iff x, y \ne 0$ the inverse is defined for all $z \in \C: z \ne 0 + 0 i$.
From the definition, the complex conjugate $\overline z$ of $z = x + i y$ is $x - i y$.
From the definition of the modulus of a complex number, we have:
:$\cmod z = \sqrt {a^2 + b^2}$
From Modulus in Terms of Conjugate, we have that:
:$\cmod z^2 = z \overline z$
Hence the result:
:$\dfrac 1 z = \dfrac {\overline z} {z \overline z}$
== Examples ==
== Sources ==
* : $\S 1.2$. The Algebraic Theory: $(1.15)$
* : $3$: Elementary Analytic Methods: $3.7$ Complex Numbers and Functions: Powers: $3.7.25$
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses
* : Chapter $\text{II}$: Groups: Subgroups
* : $1$: Complex Numbers: Axiomatic Foundations of the Complex Number System: $9$
* : $1$: Complex Numbers: Supplementary Problems: Axiomatic Foundations of Complex Numbers: $78$
* : $1$ The complex plane: Complex numbers $\S 1.2$ The algebraic structure of the complex numbers
* : Chapter $2$: Mathematical Background: $2.1$ The Complex Field $C$
* 
Category:Complex Multiplication
Category:Examples of Inverse Elements
Inverse for Complex Multiplication/Examples/3+2i

== Example of Inverse for Complex Multiplication ==
:$\dfrac 1 {3 + 2 i} = \dfrac 3 {13} + \dfrac {2 i} {13}$
== Proof ==
== Sources ==
* : $\S 1$. Algebraic Theory of Complex Numbers: Exercise $1 \ \text{(ii)}$
Category:Examples of Inverse Elements
Inverse for Integer Addition

== Theorem ==
Each element $x$ of the set of integers $\Z$ has an inverse element $-x$ under the operation of integer addition:
:$\forall x \in \Z: \exists -x \in \Z: x + \paren {-x} = 0 = \paren {-x} + x$
== Proof ==
Let us define $\eqclass {\tuple {a, b} } \boxtimes$ as in the formal definition of integers.
That is, $\eqclass {\tuple {a, b} } \boxtimes$ is an equivalence class of ordered pairs of natural numbers under the congruence relation $\boxtimes$.
$\boxtimes$ is the congruence relation defined on $\N \times \N$ by:
:$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1$
In order to streamline the notation, we will use $\eqclass {a, b} {}$ to mean $\eqclass {\tuple {a, b} } \boxtimes$, as suggested.
From the method of construction, the element $\eqclass {a, a + x} {}$ has an inverse $\eqclass {a + x, a} {}$ where $a$ and $x$ are elements of the natural numbers $\N$.
Thus:
So $\eqclass {a, a + x} {}$ has the inverse $\eqclass {a + x, a} {}$.
== Sources ==
* : Introduction $\S 5$: The system of integers
* : $\S 2.5$: Theorem $2.25$
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers: $\mathbf Z. \, 4$
* : $\S 1$
Category:Integer Addition
Category:Examples of Inverse Elements
Inverse for Rational Addition

== Theorem ==
Each element $x$ of the set of rational numbers $\Q$ has an inverse element $-x$ under the operation of rational number addition:
:$\forall x \in \Q: \exists -x \in \Q: x + \paren {-x} = 0 = \paren {-x} + x$
== Proof ==
Let $x = \dfrac a b$ where $b \ne 0$.
We take the definition of rational numbers as the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
From Existence of Field of Quotients, we have that the inverse of $\dfrac a b$ for $+$ is $\dfrac {-a} b$:
From Negative of Division Product, we have that:
:$-\dfrac a b = \dfrac {-a} b = \dfrac a {-b}$
So $\dfrac a b$ has a unique and unambiguous inverse for $+$.
Category:Rational Addition
Category:Examples of Inverse Elements
Inverse for Rational Multiplication

== Theorem ==
Each element $x$ of the set of non-zero rational numbers $\Q_{\ne 0}$ has an inverse element $\dfrac 1 x$ under the operation of rational number multiplication:
:$\forall x \in \Q_{\ne 0}: \exists \dfrac 1 x \in \Q_{\ne 0}: x \times \dfrac 1 x = 1 = \dfrac 1 x \times x$
== Proof ==
From the definition, the field $\struct {\Q, +, \times}$ of rational numbers is the field of quotients of the integral domain $\struct {\Z, +, \times}$ of integers.
From Rational Multiplication Identity is One, the identity for $\struct {\Q, \times}$ is $1 = \dfrac 1 1 = \dfrac p p$ where $p \in \Z$ and $p \ne 0$.
From the definition of product inverses in field of quotients, we have that:
:$\paren {\dfrac p q}^{-1} = \dfrac q p$
which can be demonstrated directly:
Note that this is defined only when $p \ne 0$ and $q \ne 0$.
Now let $x = \dfrac p q \in \Q$.
We define $\dfrac 1 x$ as $1 \times \paren {\dfrac p q}^{-1}$.
Hence:
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses
Category:Rational Multiplication
Category:Examples of Inverse Elements
Inverse for Real Addition

== Theorem ==
Each element $x$ of the set of real numbers $\R$ has an inverse element $-x$ under the operation of real number addition:
:$\forall x \in \R: \exists -x \in \R: x + \paren {-x} = 0 = \paren {-x} + x$
== Proof ==
We have:
Similarly for $\paren {-\eqclass {\sequence {x_n} } {} } + \eqclass {\sequence {x_n} } {}$.
Thus the inverse of $x \in \struct {\R, +}$ is $-x$.
Category:Real Addition
Category:Examples of Inverse Elements
Inverse for Real Multiplication

== Theorem ==
Each element $x$ of the set of non-zero real numbers $\R_{\ne 0}$ has an inverse element $\dfrac 1 x$ under the operation of real number multiplication:
:$\forall x \in \R_{\ne 0}: \exists \dfrac 1 x \in \R_{\ne 0}: x \times \dfrac 1 x = 1 = \dfrac 1 x \times x$
== Proof ==
By the definition of real number:
:$\forall \epsilon \in \R_{>0}: \exists t \in \N: \forall i > t: \size {x_i - x} < \epsilon$
Let $\epsilon = \size x$.
This is possible because $x \ne 0$, since it is required that $\epsilon > 0$.
Construct a sequence $\sequence {y_n}$ as follows:
:$y_n = \begin {cases} \dfrac 1 {x_n} & n > t \\ 0 & n \le t \end {cases}$
Let $u_n$ denote the Heaviside step function with parameter $t$.
We have:
Similarly for $\eqclass {\sequence {y_n} } {} \times \eqclass {\sequence {x_n} } {}$.
So the inverse of $x \in \struct {\R_{\ne 0}, \times}$ is $x^{-1} = \dfrac 1 x = y$.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 4$: Neutral Elements and Inverses
Category:Real Multiplication
Category:Examples of Inverse Elements
Inverse in Affine Group of One Dimension

== Theorem ==
Let $\map {\operatorname {Af}_1} \R$ denote the $1$-dimensional affine group on $\R$.
Let $f_{a b} \in \map {\operatorname {Af}_1} \R$.
Let $c = \dfrac 1 a$ and $d = \dfrac {-b} a$.
Then $f_{c d} \in \map {\operatorname {Af}_1} \R$ is the inverse of $f_{a b}$.
== Proof ==
As $a \in \R_{\ne 0}$ by definition of $\map {\operatorname {Af}_1} \R$ it follows that $\dfrac 1 a \in \R_{\ne 0}$ and $\dfrac {-b} a \in \R$.
So let $c = \dfrac 1 a$ and $d = \dfrac {-b} a$.
Then:
Similarly:
Hence the result.
== Sources ==
* : $\S 3.2$: Groups; the axioms: Exercise $(2)$
Category:Affine Groups
Inverse in Group is Unique

== Theorem ==
Let $\struct {G, \circ}$ be a group.
Then every element $x \in G$ has exactly one inverse:
:$\forall x \in G: \exists_1 x^{-1} \in G: x \circ x^{-1} = e^{-1} = x^{-1} \circ x$
where $e$ is the identity element of $\struct {G, \circ}$.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Also see ==
* Inverse in Monoid is Unique
== Sources ==
* : Chapter $1$: Rings - Definitions and Examples: $1$: The definition of a ring: Definitions $1.1 \ \text{(b)}$
* : $\S 1$: Exercise $5$
Category:Group Theory
Category:Inverse Elements
Category:Inverse in Group is Unique
Inverse in Group is Unique/Proof 2

== Theorem ==
== Proof ==
Let $\struct {G, \circ}$ be a group whose identity element is $e$.
By , every element of $G$ has at least one inverse.
Suppose that:
:$\exists b, c \in G: a \circ b = e, a \circ c = e$
that is, that $b$ and $c$ are both inverse elements of $a$.
Then:
So $b = c$ and hence the result.
== Sources ==
* : $\S 1.4$: Lemma $5$
* : Chapter $\text{II}$: Groups: The Group Property
* : Chapter $3$: Elementary consequences of the definitions: Proposition $3.2$
Category:Inverse in Group is Unique