nyuuzyou/wikis huggingface dataset:
Addition on 1-Based Natural Numbers is Cancellable

== Theorem ==
Let $\N_{> 0}$ be the $1$-based natural numbers.
Let $+$ be addition on $\N_{>0}$.
Then:
:$\forall a, b, c \in \N_{>0}: a + c = b + c \implies a = b$
:$\forall a, b, c \in \N_{>0}: a + b = a + c \implies b = c$
That is, $+$ is cancellable on $\N_{>0}$.
== Proof ==
By Ordering on $1$-Based Natural Numbers is Trichotomy, one and only one of the following holds:
:$a = b$
:$a < b$
:$b < a$
Suppose $a < b$.
Then by Ordering on $1$-Based Natural Numbers is Compatible with Addition:
:$a + c < b + c$
By Ordering on $1$-Based Natural Numbers is Trichotomy, this contradicts the fact that $a + c = b + c$.
Similarly, suppose $b > a$.
Then by Ordering on $1$-Based Natural Numbers is Compatible with Addition:
:$b + c < a + c$
By Ordering on $1$-Based Natural Numbers is Trichotomy, this also contradicts the fact that $a + c = b + c$.
The only other possibility is that $a = b$.
So 
:$\forall a, b, c \in \N_{>0}: a + c = b + c \implies a = b$
and so $+$ is right cancellable on $\N_{>0}$.
From Natural Number Addition is Commutative and Right Cancellable Commutative Operation is Left Cancellable:
:$\forall a, b, c \in \N_{>0}: a + b = a + c \implies b = c$
So $+$ is both right cancellable and left cancellable on $\N_{>0}$.
Hence the result.
== Sources ==
* : Introduction $\S 4$: The natural numbers: $\text{A} 3$
* : $2.2$: Theorem $2.12$
Category:Natural Numbers/1-Based
Addition on 1-Based Natural Numbers is Cancellable for Ordering

== Theorem ==
Let $\N_{> 0}$ be the $1$-based natural numbers.
Let $<$ be the strict ordering on $\N_{>0}$.
Let $+$ be addition on $\N_{>0}$.
Then:
:$\forall a, b, c \in \N_{>0}: a + c < b + c \implies a < b$
:$\forall a, b, c \in \N_{>0}: a + b < a + c \implies b < c$
That is, $+$ is cancellable on $\N_{>0}$ for $<$.
== Proof ==
By Ordering on $1$-Based Natural Numbers is Trichotomy, one and only one of the following holds:
:$a = b$
:$a < b$
:$b < a$
Let $a + c < b + c$.
Suppose $a = b$.
Then by Ordering on $1$-Based Natural Numbers is Compatible with Addition:
:$a + c = b + c$
By Ordering on $1$-Based Natural Numbers is Trichotomy, this contradicts the fact that $a + c < b + c$.
Similarly, suppose $b < a$.
Then by Ordering on $1$-Based Natural Numbers is Compatible with Addition:
:$b + c < a + c$
By Ordering on $1$-Based Natural Numbers is Trichotomy, this also contradicts the fact that $a + c < b + c$.
The only other possibility is that $a < b$.
So 
:$\forall a, b, c \in \N_{>0}: a + c = b + c \implies a < b$
and so $+$ is right cancellable on $\N_{>0}$ for $<$.
Let $a + b < a + c$.
Suppose $b = c$.
Then by Ordering on $1$-Based Natural Numbers is Compatible with Addition:
:$a + b = a + c$
By Ordering on $1$-Based Natural Numbers is Trichotomy, this contradicts the fact that $a + b < a + c$.
Similarly, suppose $c < b$.
Then by Ordering on $1$-Based Natural Numbers is Compatible with Addition:
:$a + c < a + b$
By Ordering on $1$-Based Natural Numbers is Trichotomy, this also contradicts the fact that $a + b < a + c$.
The only other possibility is that $b < c$.
So 
:$\forall a, b, c \in \N_{>0}: a + b < a + c \implies b < c$
and so $+$ is left cancellable on $\N_{>0}$ for $<$.
From Natural Number Addition is Commutative and Right Cancellable Commutative Operation is Left Cancellable:
:$\forall a, b, c \in \N_{>0}: a + b = a + c \implies b = c$
So $+$ is both right cancellable and left cancellable on $\N_{>0}$.
Hence the result.
== Sources ==
* : $2.2$: Theorem $2.14 \ \text{(i)}$
Category:Natural Numbers/1-Based
Addition on Numbers has no Zero Element

== Theorem ==
On all the number systems:
* natural numbers $\N$
* integers $\Z$
* rational numbers $\Q$
* real numbers $\R$
* complex numbers $\C$
there exists no zero element for addition.
== Proof ==
Suppose $z$ is a zero element for addition in a standard number system $\F$.
Then:
As $n$ is arbitrary, and therefore not always $0$, it follows there can be no such $z$.
== Also see ==
* Identity Element of Addition on Numbers
* Zero Element of Multiplication on Numbers
== Sources ==
* : $\S 4.3$. Units and zeros: Example $74$
Category:Numbers
Category:Addition
Category:Zero Elements
Additive Function (Algebra)/Examples

== Examples of Additive Functions ==
In the field of abstract algebra, this operation can be seen to be a endomorphism on $\struct {S, +}$.
In the field of linear algebra, it can be seen that a linear transformation is additive.
When the domain is the set of real numbers, this is the Cauchy Functional Equation.
=== Example: $\map f x = 3 x$ ===
=== Square Root is not Additive ===
Category:Examples of Additive Functions
Additive Function is Linear for Rational Factors

== Theorem ==
Let $f: \R \to \R$ be an additive function.
Then:
:$\forall r \in \Q, x \in \R: \map f {x r} = r \map f x$
== Proof ==
Trivially, we have:
:$\forall x \in \R: \map f {1 \cdot x} = 1 \map f x$
Next, suppose that:
:$\map f {n x} = n \map f x$
By additivity of $f$, we have:
Hence by the Principle of Mathematical Induction:
:$\forall n \in \N, x \in \R: \map f {n x} = n \map f x$
As Additive Function is Odd Function and Odd Function of Zero is Zero, we conclude:
:$\forall p \in \Z, x \in \R: \map f {p x} = p \map f x$
Let $p \ne 0$.
By substituting $y = p x$, the above gives:
:$\forall p \in \Z \setminus \set 0, y \in \R: \map f y = p \map f {\dfrac y p}$
In other words:
:$\forall p \in \Z \setminus \set 0, y \in \R: \map f {\dfrac y p} = \dfrac 1 p \map f y$
Given $p, q \in \Z, q \ne 0$, we have:
Therefore we conclude:
:$\forall r \in \Q, x \in \R: \map f {r x} = r \map f x$
Category:Additive Functions
Additive Function is Odd Function

== Theorem ==
Let $f: \R \to \R$ be an additive function.
Then $f$ is an odd function.
== Proof ==
From Additive Function of Zero is Zero:
:$\map f 0 = 0$
Thus, for all $x \in \R$, we have:
It follows that the function $f$ is odd:
:$\forall x \in \R: \map f {-x} = -\map f x$.
Category:Additive Functions
Category:Odd Functions
Additive Function is Strongly Additive

== Theorem ==
Let $\SS$ be an algebra of sets.
Let $f: \SS \to \overline \R$ be an additive function on $\SS$.
Then $f$ is also strongly additive.
That is:
:$\forall A, B \in \SS: \map f {A \cup B} + \map f {A \cap B} = \map f A + \map f B$
== Proof 1 ==
== Proof 2 ==
Category:Additive Function is Strongly Additive
Category:Set Systems
Category:Additive Functions
Additive Function is Strongly Additive/Proof 1

== Theorem ==
== Proof ==
From Set Difference and Intersection form Partition:
:$A$ is the union of the two disjoint sets $A \setminus B$ and $A \cap B$
:$B$ is the union of the two disjoint sets $B \setminus A$ and $A \cap B$.
So, by the definition of additive function:
:$\map f A = \map f {A \setminus B} + \map f {A \cap B}$
:$\map f B = \map f {B \setminus A} + \map f {A \cap B}$
We also have from Set Difference is Disjoint with Reverse that:
:$\paren {A \setminus B} \cap \paren {B \setminus A} = \O$
From Sum of Additive Function Values is Well-Defined, it follows that $\map f A + \map f B$ is well-defined.
Hence:
Hence the result.
Category:Additive Function is Strongly Additive
Additive Function is Strongly Additive/Proof 2

== Theorem ==
== Proof ==
Recall that $\sqcup$ denotes the disjoint union.
First, since:
we have:
:$\paren 1 : \quad \map f {A \cup B} = \map f A + \map f {B \setminus A}$
Secondly, since
we have:
:$\paren 2 : \quad \map f B = \map f {B \setminus A} + \map f {A \cap B}$
On the other hand:
so that:
:$\map f {A \cup B} = \map f {A \cap B} + \map f {\paren {A \cup B} \setminus \paren {A \cap B} }$
In particular, we have neither:
:$\map f {A \cup B} = + \infty \land \map f {A \cap B} = - \infty$
nor:
:$\map f {A \cup B} = - \infty \land \map f {A \cap B} = + \infty$
Thus the sum in $\overline {\R}$:
:$\map f {A \cup B} + \map f {A \cap B}$
is well-defined.
Finally:
Category:Additive Function is Strongly Additive
Additive Function on Empty Set is Zero

== Theorem ==
Let $\AA$ be an algebra of sets.
Let $f: \AA \to \overline \R$ be an additive function on $\AA$.
Then $\map f \O = 0$.
== Proof ==
From Properties of Algebras of Sets:
:$\O \in \AA$
Let $X \in \AA$.
Then:
Category:Set Systems
Additive Group and Multiplicative Group of Field are not Isomorphic

== Theorem ==
Let $\struct {F, +, \times}$ be a field whose zero is $0_F$ and whose unity is $1_F$.
Let $\struct {F, +}$ denote the additive group of $F$.
Let $\struct {F_{\ne 0_F}, \times}$ denote the multiplicative group of $F$.
Then $\struct {F, +}$ and $\struct {F_{\ne 0_F}, \times}$ are not isomorphic to each other.
== Proof ==
 $\phi: \struct {F_{\ne 0_F}, \times} \to \struct {F, +}$ is an isomorphism.
By definition:
:$0_F$ is the identity of $\struct {F, +}$
and
:$1_F$ is the identity of $\struct {F_{\ne 0_F}, \times}$.
We have that:
and so by definition $F$ has characteristic $2$.
Let $x \in \struct {F_{\ne 0_F}, \times}$.
Then:
As $\phi$ is an isomorphism, it is also a monomorphism.
From Kernel is Trivial iff Monomorphism:
:$\map \ker \phi = \set {1_F}$
and so $x^2 = 1$.
Thus $x = 1$ and so $\order F = 2$.
Thus $\order {F_{\ne 0_F} } = 1$.
So $\phi$ is a bijection from a set of cardinality $1$ to a set of cardinality $2$.
So $\phi$ cannot be a bijection and so cannot be an isomorphism.
Hence the result by Proof by Contradiction.
== Sources ==
* : Chapter $9$: Rings: Exercise $16$
Category:Field Theory
Category:Group Isomorphisms
Additive Group of Complex Numbers is Direct Product of Reals with Reals

== Theorem ==
Let $\struct {\C, +}$ be the additive group of complex numbers.
Let $\struct {\R, +}$ be the additive group of real numbers.
Then the direct product $\struct {\R, +} \times \struct {\R, +}$ is isomorphic with $\struct {\C, +}$.
== Proof ==
Let us define the mapping $\phi: \R^2 \to \C$ as:
:$\forall \tuple {x, y} \in \R^2: \phi: \tuple {x, y} = x + y i$
We will show that $\phi$ is a group isomorphism.
=== $\phi$ is a group homomorphism ===
=== $\phi$ is bijective ===
We show that $\phi^{-1}: z \mapsto \tuple {\map \Re z, \map \Im z}$ is the inverse of $\phi$.
We have:
Similarly:
Therefore:
Hence $\phi^{-1}$ is the inverse of $\phi$.
Hence $\phi$ is a group isomorphism, and thus the direct product $\struct {\R, +} \times \struct {\R, +}$ is isomorphic with $\struct {\C, +}$.
== Sources ==
* : Chapter $\text{II}$: Groups: Direct Products
Category:Additive Group of Complex Numbers
Category:Additive Group of Real Numbers
Category:Group Direct Products
Additive Group of Integers is Countably Infinite Abelian Group

== Theorem ==
The set of integers under addition $\struct {\Z, +}$ forms a countably infinite abelian group.
== Proof ==
From Integers under Addition form Abelian Group, $\struct {\Z, +}$ is an abelian group.
From Integers are Countably Infinite, the set of integers can be placed in one-to-one correspondence with the set of natural numbers.
Hence by definition, the underlying set of $\struct {\Z, +}$ is countably infinite.
== Sources ==
* : Chapter $2$: Subgroups and Cosets: $\S 38$
* : Chapter $1$: Definitions and Examples: Example $1.4$
* : $\S 1.1$: Example $1$
Category:Abelian Groups
Category:Infinite Groups
Category:Additive Group of Integers
Additive Group of Integers is Normal Subgroup of Complex

== Theorem ==
Let $\struct {\Z, +}$ be the additive group of integers.
Let $\struct {\C, +}$ be the additive group of complex numbers.
Then $\struct {\Z, +}$ is a normal subgroup of $\struct {\C, +}$.
== Proof ==
From Additive Group of Integers is Normal Subgroup of Reals, $\struct {\Z, +} \lhd \struct {\R, +}$.
From Additive Group of Reals is Subgroup of Complex, $\struct {\R, +} \lhd \struct {\C, +}$.
Thus $\struct {\Z, +} \le \struct {\C, +}$.
From Complex Numbers under Addition form Infinite Abelian Group, $\struct {\C, +}$ is abelian.
From Subgroup of Abelian Group is Normal, it follows that $\struct {\Z, +} \lhd \struct {\C, +}$.
Category:Additive Group of Integers
Category:Additive Group of Complex Numbers
Category:Examples of Normal Subgroups
Additive Group of Integers is Normal Subgroup of Rationals

== Theorem ==
Let $\struct {\Z, +}$ be the additive group of integers.
Let $\struct {\Q, +}$ be the additive group of rational numbers.
Then $\struct {\Z, +}$ is a normal subgroup of $\struct {\Q, +}$.
== Proof ==
From Additive Group of Integers is Subgroup of Rationals, $\struct {\Z, +}$ is a subgroup of $\struct {\Q, +}$.
From Rational Numbers under Addition form Infinite Abelian Group, $\struct {\Q, +}$ is an abelian group.
From Subgroup of Abelian Group is Normal it follows that $\struct {\Z, +}$ is a normal subgroup of $\struct {\Q, +}$.
Category:Additive Group of Integers
Category:Additive Group of Rational Numbers
Category:Examples of Normal Subgroups
Additive Group of Integers is Normal Subgroup of Reals

== Theorem ==
Let $\struct {\Z, +}$ be the additive group of integers.
Let $\struct {\R, +}$ be the additive group of real numbers.
Then $\struct {\Z, +}$ is a normal subgroup of $\struct {\R, +}$.
== Proof ==
From Additive Group of Integers is Subgroup of Reals, $\struct {\Z, +}$ is a normal subgroup of $\struct {\R, +}$.
As the additive group of real numbers is abelian, from Subgroup of Abelian Group is Normal it follows that $\struct {\Z, +} \lhd \struct {\R, +}$.
== Sources ==
* : Chapter $6$: Cosets: Exercise $14$
Category:Additive Group of Integers
Category:Additive Group of Real Numbers
Category:Examples of Normal Subgroups
Additive Group of Integers is Subgroup of Rationals

== Theorem ==
Let $\struct {\Z, +}$ be the additive group of integers.
Let $\struct {\Q, +}$ be the additive group of rational numbers.
Then $\struct {\Z, +}$ is a subgroup of $\struct {\Q, +}$.
== Proof ==
Recall that Integers form Integral Domain.
The set $\Q$ of rational numbers is defined as the field of quotients of the integers.
The fact that the integers are a subgroup of the rationals follows from the work done in proving the Existence of Field of Quotients from an integral domain.
== Sources ==
* : $\S 3.2$: Groups; the axioms: Examples of groups $\text{(i)}$
* : Chapter $4$: Subgroups: Example $4.3$
Category:Additive Group of Integers
Category:Additive Group of Rational Numbers
Category:Examples of Subgroups
Additive Group of Integers is Subgroup of Reals

== Theorem ==
Let $\struct {\Z, +}$ be the additive group of integers.
Let $\struct {\R, +}$ be the additive group of real numbers.
Then $\struct {\Z, +}$ is a subgroup of $\struct {\R, +}$.
== Proof ==
From Additive Group of Integers is Subgroup of Rationals, $\struct {\Z, +}$ is a subgroup of $\struct {\Q, +}$.
From Additive Group of Rationals is Normal Subgroup of Reals, $\struct {\Q, +}$ is a subgroup of $\struct {\R, +}$.
Thus $\struct {\Z, +}$ is a subgroup of $\struct {\R, +}$.
== Sources ==
* : $\S 5.2$. Subgroups: Example $91$
Category:Additive Group of Integers
Category:Additive Group of Real Numbers
Additive Group of Rational Numbers is not Isomorphic to Multiplicative Group of Rational Numbers

== Theorem ==
Let $\struct {\Q, +}$ be the additive group of rational numbers.
Let $\struct {\Q_{\ne 0}, \times}$ be the multiplicative group of rational numbers.
Then $\struct {\Q_{\ne 0}, \times}$ is not isomorphic to $\struct {\Q, +}$.
== Proof ==
A direct application of Additive Group and Multiplicative Group of Field are not Isomorphic.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures: Exercise $6.10$
Category:Examples of Group Isomorphisms
Category:Multiplicative Group of Rational Numbers
Category:Additive Group of Rational Numbers
Additive Group of Rationals is Normal Subgroup of Complex

== Theorem ==
Let $\struct {\Q, +}$ be the additive group of rational numbers.
Let $\struct {\C, +}$ be the additive group of complex numbers.
Then $\struct {\Q, +}$ is a normal subgroup of $\struct {\C, +}$.
== Proof ==
From Additive Group of Rationals is Normal Subgroup of Reals, $\struct {\Q, +} \lhd \struct {\R, +}$.
From Additive Group of Reals is Normal Subgroup of Complex, $\struct {\R, +} \lhd \struct {\C, +}$.
Thus $\struct {\Q, +} \le \struct {\C, +}$.
From Complex Numbers under Addition form Infinite Abelian Group, $\struct {\C, +}$ is abelian.
From Subgroup of Abelian Group is Normal, it follows that $\struct {\Q, +} \lhd \struct {\C, +}$.
Category:Additive Group of Rational Numbers
Category:Additive Group of Complex Numbers
Category:Examples of Normal Subgroups
Additive Group of Rationals is Normal Subgroup of Reals

== Theorem ==
Let $\struct {\Q, +}$ be the additive group of rational numbers.
Let $\struct {\R, +}$ be the additive group of real numbers.
Then $\struct {\Q, +}$ is a normal subgroup of $\struct {\R, +}$.
== Proof ==
From the definition of real numbers, $\Q$ is a subset of $\R$.
As $\struct {\R, +}$ is a group, and $\struct {\Q, +}$ is a group, it follows from the definition of subgroup that $\struct {\Q, +}$ is a subgroup of $\struct {\R, +}$.
As $\struct {\R, +}$ is abelian, it follows from Subgroup of Abelian Group is Normal that $\struct {\Q, +}$ is normal in $\struct {\R, +}$.
== Sources ==
* : $\S 3.2$: Groups; the axioms
* : Chapter $4$: Subgroups: Example $4.3$
Category:Rational Addition
Category:Additive Group of Real Numbers
Category:Additive Group of Rational Numbers
Category:Examples of Normal Subgroups
Additive Group of Real Numbers is Not Isomorphic to Multiplicative Group of Real Numbers

== Theorem ==
Let $\struct {\R, +}$ denote the additive group of real numbers.
Let $\struct {\R_{\ne 0}, \times}$ denote the multiplicative group of real numbers.
Then $\struct {\R, +}$ is not isomorphic to $\struct {\R_{\ne 0}, \times}$.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : Chapter $7$: Homomorphisms: Exercise $5$
Category:Additive Group of Real Numbers is Not Isomorphic to Multiplicative Group of Real Numbers
Category:Multiplicative Group of Real Numbers
Category:Additive Group of Real Numbers
Category:Examples of Isomorphisms (Abstract Algebra)
Additive Group of Real Numbers is Not Isomorphic to Multiplicative Group of Real Numbers/Proof 1

== Theorem ==
== Proof ==
Consider the element $-1 \in \struct {\R_{\ne 0}, \times}$.
We have that:
:$-1 \times -1 = 1$
From Real Multiplication Identity is One it follows that $-1$ is of order $2$ in $\struct {\R_{\ne 0}, \times}$.
From Group Isomorphism Preserves Order of Group Element, it is sufficient to demonstrate that there exists no element of order $2$ in $\struct {\R, +}$.
From Real Addition Identity is Zero, that means finding $x$ such that $x + x = 0$.
But:
:$x + x = 0 \implies x = 0$
and as $0$ is the identity of $\struct {\R, +}$, there is indeed no element of order $2$ in $\struct {\R, +}$.
The result follows from Group Isomorphism Preserves Order of Group Element.
Category:Additive Group of Real Numbers is Not Isomorphic to Multiplicative Group of Real Numbers
Additive Group of Real Numbers is Not Isomorphic to Multiplicative Group of Real Numbers/Proof 2

== Theorem ==
== Proof ==
From Real Numbers form Field, $\struct {\R, +, \times}$ forms a field.
The result then follows as an example of Additive Group and Multiplicative Group of Field are not Isomorphic.
Category:Additive Group of Real Numbers is Not Isomorphic to Multiplicative Group of Real Numbers
Additive Group of Real Numbers is Not Isomorphic to Multiplicative Group of Real Numbers/Proof 3

== Theorem ==
== Proof ==
There are two element of $\struct {\R_{\ne 0}, \times}$ which are self-inverse, that is, $-1$ and $1$.
However, there is only one element of $\struct {\R, +}$ which is self-inverse, that is, $0$.
 there exists an isomorphism $f: \struct {\R_{\ne 0}, \times} \to \struct {\R, +}$.
Then:
So we have demonstrated that $f$ is not an injection.
Hence $f$ is not a bijection and so not an isomorphism.
It follows from Proof by Contradiction that there can be no such isomorphism.
Hence the result.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 6$: Isomorphisms of Algebraic Structures
Category:Additive Group of Real Numbers is Not Isomorphic to Multiplicative Group of Real Numbers
Additive Group of Reals is Normal Subgroup of Complex

== Theorem ==
Let $\struct {\R, +}$ be the additive group of real numbers.
Let $\struct {\C, +}$ be the additive group of complex numbers.
Then $\struct {\R, +}$ is a normal subgroup of $\struct {\C, +}$.
== Proof ==
From Additive Group of Reals is Subgroup of Complex, $\struct {\R, +}$ is a subgroup of $\struct {\C, +}$.
Then from Complex Numbers under Addition form Infinite Abelian Group, $\struct {\C, +}$ is abelian.
The result follows from Subgroup of Abelian Group is Normal.
Category:Additive Group of Real Numbers
Category:Additive Group of Complex Numbers
Category:Examples of Normal Subgroups
Additive Group of Reals is Subgroup of Complex

== Theorem ==
Let $\struct {\R, +}$ be the additive group of real numbers.
Let $\struct {\C, +}$ be the additive group of complex numbers.
Then $\struct {\R, +}$ is a subgroup of $\struct {\C, +}$.
== Proof ==
Let $x, y \in \C$ such that $x = x_1 + 0 i, y = y_1 + 0 i$.
As $x$ and $y$ are wholly real, we have that $x, y \in \R$.
Then $x + y = \paren {x_1 + y_1} + \paren {0 + 0} i$ which is also wholly real.
Also, the inverse of $x$ is $-x = -x_1 + 0 i$ which is also wholly real.
We have that $\R$ is non-empty.
Thus by the Two-Step Subgroup Test, $\struct {\R, +}$ is a subgroup of $\struct {\C, +}$.
== Sources ==
* : $\S 3.2$: Groups; the axioms: Examples of groups $\text{(i)}$
* : Chapter $4$: Subgroups: Example $4.3$
Category:Additive Group of Real Numbers
Category:Additive Group of Complex Numbers
Additive Groups of Integers and Integer Multiples are Isomorphic

== Theorem ==
Let $n \in \Z_{> 0}$ be a strictly positive integer.
Let $\struct {n \Z, +}$ denote the additive group of integer multiples.
Let $\struct {\Z, +}$ denote the additive group of integers.
Then $\struct {n \Z, +}$ is isomorphic to $\struct {\Z, +}$.
== Proof ==
We have that:
:Infinite Cyclic Group is Isomorphic to Integers.
:Integer Multiples under Addition form Infinite Cyclic Group.
:Infinite Cyclic Group is Unique up to Isomorphism
Hence the result.
== Sources ==
* : Chapter $7$: Homomorphisms: Exercise $5$
Category:Additive Groups of Integer Multiples
Category:Additive Group of Integers
Category:Infinite Cyclic Group
Additive Inverse in Ring of Bounded Continuous Real-Valued Functions

== Theorem ==
Let $\struct {S, \tau}$ be a topological space.
Let $\R$ denote the real number line.
Let $\struct {\map {C^*} {S, \R}, +, *}$ be the ring of bounded continuous real-valued functions from $S$.
Let $f \in \map {C^*} {S, \R}$.
Then:
:the additive inverse of $f$ is the pointwise negation $-f$ defined by:
::$\forall s \in S : \map {\paren{-f}} s = - \map f s$
== Proof ==
Let $\struct{\map C {S, \R}, +, *}$ be the ring of continuous real-valued functions from $S$.
From Ring of Bounded Continuous Functions is Subring of Continuous Real-Valued Functions:
:$\struct{\map {C^*} {S, \R}, +, *}$ is a subring of $\struct{\map C {S, \R}, +, *}$
From Additive Inverse in Ring of Continuous Real-Valued Functions:
:the additive inverse of $f$ in $\struct{\map C {S, \R}, +, *}$ is the pointwise negation $-f$ defined by:
::$\forall s \in S : \map {\paren{-f}} s = - \map f s$
From Inverses in Subgroup:
:$-f$ is the additive inverse of $f$
== Also see ==
* Ring of Bounded Continuous Functions is Subring of Continuous Real-Valued Functions
* Zero of Ring of Bounded Continuous Real-Valued Functions
* Unity of Ring of Bounded Continuous Real-Valued Functions
* Ring of Bounded Continuous Real-Valued Functions is Commutative
== Sources ==
: Chapter $1$: Functions on a Topological Space, $\S 1.4$
Category:Rings of Bounded Continuous Real-Valued Functions