nyuuzyou/wikis huggingface dataset:
Natural Number is Ordinary Set

== Theorem ==
Let $n$ be a natural number.
Then $n$ is an ordinary set.
That is:
:$n \notin n$
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$n$ is an ordinary set.
=== Basis for the Induction ===
By the , $\O$ is a set.
From Empty Set is Ordinary:
:$\O \notin \O$
Thus $\map P 0$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$k$ is an ordinary set.
from which it is to be shown that:
:$k^+$ is an ordinary set.
=== Induction Step ===
This is the induction step:
By the Induction Hypothesis:
:$k \notin k$
From Successor Set of Ordinary Transitive Set is Ordinary:
:$k^+ \notin k^+$
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N: n$ is an ordinary set.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 3$ Derivation of the Peano postulates and other results: Theorem $3.2$
Category:Ordinary Sets
Category:Natural Numbers
Natural Number is Prime or has Prime Factor

== Theorem ==
Let $a$ be a natural number greater than $1$.
Then either:
:$a$ is a prime number
or:
:there exists a prime number $p \ne a$ such that $p \divides a$
where $\divides$ denotes is a divisor of.
== Proof ==
By definition of composite number $a$ is either prime or composite.
Let $a$ be prime.
Then the statement of the result is fulfilled.
Let $a$ be composite.
Then by :
:$\exists p: p \divides a$
where $p$ is a prime number.
The result follows by Proof by Cases.
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Prime Numbers
Natural Number is Superset of its Union

== Theorem ==
Let $n \in \N$ be a natural number as defined by the von Neumann construction.
Then:
:$\bigcup n \subseteq n$
== Proof ==
Let $n \in \N$.
From Natural Number is Transitive Set, $n$ is transitive.
From Class is Transitive iff Union is Subclass it follows directly that:
:$\bigcup n \subseteq n$
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 3$ Derivation of the Peano postulates and other results: Exercise $3.2 \ \text {(b)}$
Category:Natural Numbers
Natural Number is Transitive Set

== Theorem ==
Let $n$ be a natural number.
Then $n$ is a transitive set.
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$n$ is a transitive set.
=== Basis for the Induction ===
From Empty Class is Transitive, we have that $\O$ is a transitive class.
By the , $\O$ is a set.
Hence $\O$ is a transitive set.
Thus $\map P 0$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$k$ is a transitive set.
from which it is to be shown that:
:$k^+$ is a transitive set.
=== Induction Step ===
This is the induction step:
Let $k$ be a transitive set.
Then from Successor Set of Transitive Set is Transitive:
:$k^+$ is a transitive set.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N: n$ is a transitive set.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 3$ Derivation of the Peano postulates and other results: Theorem $3.1$
Category:Transitive Classes
Category:Natural Numbers
Natural Number is Union of its Successor

== Theorem ==
Let $n \in \N$ be a natural number as defined by the von Neumann construction.
Then:
:$\map \bigcup {n^+} = n$
== Proof ==
From Natural Number is Superset of its Union we have:
:$\bigcup n \subseteq n$
Then from Union with Superset is Superset‎:
:$\bigcup n \subseteq n \iff \paren {n \cup \bigcup n} = n$
and the result follows.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 3$ Derivation of the Peano postulates and other results: Exercise $3.2 \ \text {(d)}$
Category:Natural Numbers
Natural Number is not Subset of its Union

== Theorem ==
Let $n \in \N$ be a natural number as defined by the von Neumann construction.
Then, except in the degenerate case where $n = 0$, it is not the case that:
:$n \subseteq \bigcup n$
== Proof ==
First we note that from Union of Empty Set we have:
:$\bigcup \O = \O$
leading to:
:$\O \subseteq \bigcup \O$
thus disposing of the degenerate case.
Let $n \in \N$ such that $n \ne \O$.
By definition of the von Neumann construction:
:$n = \set {0, 1, 2, \ldots, n - 1}$
Thus, by definition, $m \in n$ for $m = 0, 1, 2, \ldots, n - 1$.
From Natural Numbers cannot be Elements of Each Other, it cannot therefore be the case that $n \in m$ for $m \in n$.
By definition of union of class:
:$\bigcup n = \set {x: \exists X \in n: x \in X}$
But it has been established that there is no $X \in n$ such that $n \in X$.
Hence the result.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 3$ Derivation of the Peano postulates and other results: Exercise $3.2 \ \text {(a)}$
Category:Natural Numbers
Natural Number m is Less than n iff m is an Element of n

== Theorem ==
Let $\omega$ be the set of natural numbers defined as the von Neumann construction.
Let $m, n \in \omega$.
Then:
:$m < n \iff m \in n$
That is, every natural number is the set of all smaller natural numbers.
== Proof ==
By definition of the ordering on von Neumann construction:
:$m \le n \iff m \subseteq n$
=== Necessary Condition ===
Let $m \in n$.
Then from Natural Number is Transitive Set:
:$m \subseteq n$
But if $m = n$, we would have:
:$m \in m$
which contradicts Natural Number is Ordinary Set.
So we have:
:$m \le n$ and $m \ne n$
that is:
:$m < n$
=== Sufficient Condition ===
Let $m < n$.
Then by Natural Number m is Less than n implies n is not Greater than Successor of n:
:$m^+ \le n$
But we have $m \in m^+$ by construction.
Hence:
:$m \in n$
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 5$ Applications to natural numbers: Theorem $5.6$
Category:Ordering on Natural Numbers
Natural Number m is Less than n implies n is not Greater than Successor of n

== Theorem ==
Let $\N$ be the natural numbers.
Let $m, n \in \N$.
Then:
:$m < n \implies m + 1 \le n$
== Proof using Naturally Ordered Semigroup ==
== Proof using Von Neumann Construction ==
Category:Ordering on Natural Numbers
Category:Natural Number m is Less than n implies n is not Greater than Successor of n
Natural Number m is Less than n implies n is not Greater than Successor of n/Proof using Naturally Ordered Semigroup

== Theorem ==
== Proof ==
Let $\N$ be considered as the naturally ordered semigroup:
:$\struct {\N, +, \le}$
The result follows from Sum with One is Immediate Successor in Naturally Ordered Semigroup.
Category:Natural Number m is Less than n implies n is not Greater than Successor of n
Natural Number m is Less than n implies n is not Greater than Successor of n/Proof using Von Neumann Construction

== Theorem ==
== Proof ==
Let $\N$ be defined as the von Neumann construction $\omega$.
By definition of the ordering on von Neumann construction:
:$m \le n \iff m \subseteq n$
From Von Neumann Construction of Natural Numbers is Minimally Inductive, $\omega$ is minimally inductive class under the successor mapping.
The result is then a direct application of Characteristics of Minimally Inductive Class under Progressing Mapping: Image of Proper Subset is Subset:
:$m \subset n \implies m^+ \subseteq n$
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 5$ Applications to natural numbers: Theorem $5.3 \ (2)$
Category:Natural Number m is Less than n implies n is not Greater than Successor of n
Natural Numbers Bounded Below under Addition form Commutative Semigroup

== Theorem ==
Let $m \in \N$ where $\N$ is the set of natural numbers.
Let $M \subseteq \N$ be defined as:
:$M := \set {x \in \N: x \ge m}$
That is, $M$ is the set of all natural numbers greater than or equal to $m$.
Then the algebraic structure $\struct {M, +}$ is a commutative semigroup.
== Proof ==
We have that:
:Natural Number Addition is Associative
:Natural Number Addition is Commutative
From Restriction of Associative Operation is Associative, $+$ is associative on $\struct {M, +}$.
From Restriction of Commutative Operation is Commutative, $+$ is commutative on $\struct {M, +}$.
It remains to be shown that $+$ is closed on $\struct {M, +}$.
Let $a, b \in M$.
Then $\exists r, s \in \N: a = m + r, b = m + s$.
Thus:
So $a + b \in M$ and so $+$ is closed on $\struct {M, +}$.
Category:Examples of Commutative Semigroups
Category:Natural Number Addition
Natural Numbers Set Equivalent to Ideals of Integers

== Theorem ==
Let $S$ be the set of all ideals of $\Z$.
Let the mapping $\psi: \N \to S$ be defined as:
:$\forall b \in \N: \map \psi b = \ideal b$
where $\ideal b$ is the principal ideal of $\Z$ generated by $b$.
Then $\psi$ is a bijection.
== Proof ==
First we show that $\psi$ is injective.
Suppose $0 < b < c$.
Then $b \in \ideal b$, but $b \notin \ideal c$, because from Principal Ideals of Integers‎, $c$ is the smallest positive integer in $\ideal c$.
Thus $\ideal b \ne \ideal c$.
It is also apparent that $b > 0 \implies \ideal b \ne \ideal 0$ as $\ideal 0 = \set 0$.
Thus $\psi$ is injective.
Surjectivity follows from Principal Ideals of Integers: every integer is the smallest strictly positive element of a principal ideal of $\Z$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm: Theorem $24.3$
Category:Ideal Theory
Category:Integers
Category:Natural Numbers
Natural Numbers are Comparable

== Theorem ==
Let $\N$ be the natural numbers.
Let $m, n \in \N$.
Then $m$ and $n$ are comparable by the ordering relation $\le$.
That is, either:
:$(1): \quad m \le n$
or:
:$(2): \quad n \le m$
or possibly both.
=== Strong Result ===
== Proof ==
Category:Ordering on Natural Numbers
Category:Natural Numbers are Comparable
Natural Numbers are Comparable/Proof

== Theorem ==
== Proof ==
Let $\N$ be defined as the minimally inductive set $\omega$.
By definition of the ordering on minimally inductive set:
:$m \le n \iff \begin {cases} m = n & \text {or} \\ m \in n & \end {cases}$
Thus it is sufficient to prove that exactly one of the following is the case:
:$(1): \quad m \in n$
:$(2): \quad m = n$
:$(3): \quad n \in m$
The proof operates by induction:
For each $n \in \omega$, let $\map S n$ be the set of all $m \in \omega$ which are comparable with $n$.
Let $S$ be the set of all $n \in \N$ for which $\map S n = \omega$.
Thus the proof is equivalent to demonstrating that $S = \omega$.
=== Basis for the Induction ===
First consider $\map S 0$.
Clearly $0 \in \map S 0$ as $0 = 0$.
Let $m \in \map S 0$.
Then $m \notin 0$ as $0 = \O$.
So either:
:$(1): \quad 0 = m$, in which case $0 \in m^+$ by definition of successor set
or:
:$(2): \quad 0 \in m$, in which case, because $m \in m^+$ by definition of successor set, again $0 \in m^+$.
In all cases, $m \in \map S 0 \implies m^+ \in \map S 0$.
So $\map S 0 = \omega$ by induction.
This is the basis for the induction.
=== Induction Hypothesis ===
The induction hypothesis is that:
:$\map S k = \omega$
for some $k \in \omega$.
It is now necessary to show that it follows that:
:$\map S {k^+} = \omega$
=== Induction Step ===
This is the induction step:
Consider the set $\map S {k^+}$, given that $\map S k = \omega$.
From the basis for the induction, we have that $k^+ \in \map S 0$.
That is, $k^+$ is comparable with $0$.
So $0$ is comparable with $k^+$ and so $0 \in \map S {k^+}$.
Suppose $m \in \map S {k^+}$.
Then either:
:$(1): \quad k^+ \in m$, in which case $k^+ \in m^+$
or:
:$(2): \quad k^+ = m$, in which case the same applies: $k^+ \in m^+$
or:
:$(3): \quad m \in k^+$.
In case $(3)$, either:
:$(3 \text a): \quad m = k$, in which case $m^+ = k^+$
or:
:$(3 \text b): \quad m \in k$.
Case $(3 \text b)$ is treated as follows.
We have that $m^+ \in \map S k$ by the induction hypothesis.
Therefore, either:
:$(4 \text a): \quad k \in m^+$
or:
:$(4 \text b): \quad k = m^+$
or:
:$(4 \text c): \quad m^+ \in k$.
$(4 \text a)$ is not compatible with $m \in k$, because either:
:$k \in m$
or
:$k = m$
and so $k \subseteq m$ which contradicts Finite Ordinal is not Subset of one of its Elements.
Both $(4 \text b)$ and $(4 \text c)$ imply that $m^+ \in n^+$.
Hence $\map S {k^+} = \omega$.
It follows by the Principle of Finite Induction that $S = \omega$.
It then follows from Finite Ordinal is not Subset of one of its Elements that it is not possible for more than one of:
:$(1): \quad m \in n$
:$(2): \quad m = n$
:$(3): \quad n \in m$
to be true at the same time.
== Sources ==
* : $\S 13$: Arithmetic
Category:Natural Numbers are Comparable
Natural Numbers are Comparable/Strong Result

== Theorem ==
Let $\N$ be the natural numbers.
Let $m, n \in \N$.
Then either:
:$(1): \quad m + 1 \le n$
or:
:$(2): \quad n \le m$
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 5$ Applications to natural numbers: Theorem $5.2$
Category:Natural Numbers are Comparable
Natural Numbers are Comparable/Strong Result/Proof 1

== Theorem ==
== Proof ==
Let $\N$ be defined as the von Neumann construction $\omega$.
By definition of the ordering on von Neumann construction:
:$m \le n \iff m \subseteq n$
From Von Neumann Construction of Natural Numbers is Minimally Inductive, $\omega$ is minimally inductive class under the successor mapping.
Then from Minimally Inductive Class under Progressing Mapping induces Nest, $\omega$ is a nest in which:
:$\forall m, n \in \omega: \map g x \subseteq y \lor y \subseteq x$
From the definition of $\map g x$ in this context:
:$\forall x \in \omega: \map g x = x^+$
That is:
:$\forall m, n \in \N: m + 1 \subseteq n \lor n \subseteq m$
whence the result.
Category:Natural Numbers are Comparable
Natural Numbers are Infinite

== Theorem ==
The set $\N$ of natural numbers is infinite.
== Proof ==
Let the mapping $s: \N \to \N$ be defined as:
:$\forall n \in \N: \map s n = n + 1$
$s$ is clearly an injection.
 $\N$ were finite.
By Equivalence of Mappings between Finite Sets of Same Cardinality it follows that $s$ is a surjection.
But:
:$\forall n \in \N: \map s n \ge 0 + 1 > 0$
So:
:$0 \notin \Img s$
and $s$ is not a surjection.
From this contradiction it is seen that $\N$ cannot be finite.
So, by definition, $\N$ is infinite.
== Sources ==
* : $\S 13$: Arithmetic
* : $\S 2.4$
* : Chapter $\text {III}$: The Natural Numbers: $\S 17$: Finite Sets: Theorem $17.8$
* 
* 
* : Chapter $1$: General Background: $\S 1$ What is infinity?
* : $1$: Set Theory and Logic: $\S 6$: Finite Sets: Corollary $6.4$
Category:Natural Numbers
Category:Infinite Sets
Natural Numbers are Non-Negative Integers

== Theorem ==
Let $m \in \Z$. Then:
:$(1): \quad m \in \N \iff 0 \le m$
:$(2): \quad m \in \N_{> 0} \iff 0 < m$
:$(3): \quad m \notin \N \iff -m \in \N_{> 0}$
That is, the natural numbers are precisely those integers which are greater than or equal to zero.
== Proof 1 ==
== Proof 2 ==
== Warning ==
== Sources ==
* : Chapter $2$: Integers and natural numbers: $\S 2.1$: The integers
Category:Integers
Category:Natural Numbers
Category:Natural Numbers are Non-Negative Integers
Natural Numbers are Non-Negative Integers/Proof 1

== Theorem ==
== Proof ==
Let $m \in \N$.
Then by definition of $0$:
:$0 \le m$
Conversely, let $m \in \Z: 0 \le m$.
Then:
:$\exists x, y \in \N: m = x - y$
Thus:
:$y \le m + y = x$
By Naturally Ordered Semigroup: NO 3:
:$\exists z \in \N: z + y = x = m + y$
From Naturally Ordered Semigroup: NO 2, $y$ is cancellable.
Hence:
$m = z \in \N_{> 0}$
Thus $(1)$ holds.
$(2)$ follows from $(1)$.
We infer from $(1)$ that:
:$m \notin \N \iff m < 0$
We infer from $(2)$ that:
:$-m > 0 \iff -m \in \N_{> 0}$
But by Ordering of Inverses in Ordered Monoid:
:$m < 0 \iff -m > 0$
Therefore $(3)$ also holds.
== Warning ==
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $20$. The Integers: Theorem $20.7$
Category:Natural Numbers are Non-Negative Integers
Natural Numbers are Non-Negative Integers/Proof 2

== Theorem ==
== Proof ==
Consider the formal definition of the integers: $x = \eqclass {a, b} {}$ is an equivalence class of ordered pairs of natural numbers.
Let $x \in \Z_{>0}$ be a (strictly) positive integer.
Thus by definition:
:$x > 0$
This is equivalent to the condition that $a > b$.
Hence:
:$x = \eqclass {b + u, b} {}$
for some $u \in \N_{>0}$.
It is immediate that:
:$\forall c \in \N: \eqclass {b + u, b} {} = \eqclass {c + u, c} {}$
Consider the mapping $\phi: \N_{>0} \to \Z_{>0}$ defined as:
:$\forall u \in \N_{>0}: \map \phi u = u'$
where $u' \in \Z$ be the (strictly) positive integer $\eqclass {b + u, b} {}$.
From the above we have that $\phi$ is well-defined.
Let $\eqclass {b + u, b} {} = \eqclass {c + v, c} {}$.
Then:
:$b + u + c = b + c + v$
and so:
:$u = v$
Hence $\phi$ is an injection.
Next note that all $u' \in \Z_{>0}$ can be expressed in the form:
:$u' = \eqclass {b + u, b} {}$
for arbitrary $b$.
Hence $\phi$ is a surjection.
Hence $\phi$ is an isomorphism.
By defining $\map \phi 0 = \eqclass {b, b} {}$ we see that $\N$ and $\Z_{\ge 0}$ are isomorphic.
== Warning ==
== Sources ==
* : Introduction $\S 5$: The system of integers
Category:Natural Numbers are Non-Negative Integers
Natural Numbers are Non-Negative Integers/Warning

== Natural Numbers are Non-Negative Integers: Warning ==
From a strictly purist point of view it is inaccurate to say that the natural numbers are the non-negative integers, as an integer is technically an element of an equivalence class composed of pairs of elements of $\N$, constructed as detailed in Construction of Inverse Completion.
However, because an Inverse Completion is Unique, it follows that the natural numbers can be considered to be a substructure of the integers from the Inverse Completion Theorem.
With that caveat in mind, the theorem holds.
Category:Natural Numbers are Non-Negative Integers
Natural Numbers cannot be Elements of Each Other

== Theorem ==
Let $m$ and $n$ be natural numbers.
Then it cannot be the case that both $m \in n$ and $n \in m$.
== Proof ==
 both $m \in n$ and $n \in m$.
We have $m \in n$
From Natural Number is Transitive Set:
:$m \subseteq n$
by definition of transitive.
Thus:
:$n \in m \subseteq n$
and so:
:$n \in n$
But from Natural Number is Ordinary Set:
:$n \notin n$
The result follows by Proof by Contradiction.
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 3$ Derivation of the Peano postulates and other results: Theorem $3.3$
Category:Ordinary Sets
Category:Natural Numbers
Natural Numbers form Commutative Semiring

== Theorem ==
The semiring of natural numbers $\struct {\N, +, \times}$ forms a commutative semiring.
== Proof ==
The algebraic structure $\struct {\N, +}$ is a commutative monoid from Natural Numbers under Addition form Commutative Monoid.
Then we have:
:$(1): \quad$ Natural Number Multiplication is Closed
:$(2): \quad$ Natural Number Multiplication is Associative
:$(3): \quad$ Natural Number Multiplication is Commutative
:$(4): \quad$ Natural Number Multiplication Distributes over Addition
Thus $\struct {\N, \times}$ forms an algebraic structure which is closed such that $\times$ is associative and commutative.
So by definition, $\struct {\N, \times}$ is a commutative semigroup.
The result follows from definition of commutative semiring and the distributivity of $\times$ over $+$.
Category:Natural Numbers
Category:Semirings
Natural Numbers form Inductive Set

== Theorem ==
Let $\N$ denote the natural numbers as subset of the real numbers $\R$.
Then $\N$ is an inductive set.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers
Category:Natural Numbers
Category:Inductive Sets
Category:Natural Numbers form Inductive Set
Natural Numbers form Inductive Set/Proof 1

== Theorem ==
== Proof ==
By definition of the natural numbers:
:$\N = \ds \bigcap \II$
where $\II$ is the collection of all inductive sets.
Suppose that $n \in \N$.
Then by definition of intersection:
:$\forall I \in \II: n \in I$
Because all these $I$ are inductive:
:$\forall I \in \II: n + 1 \in I$
Again by definition of intersection:
:$n + 1 \in \N$
Category:Natural Numbers form Inductive Set
Natural Numbers form Inductive Set/Proof 2

== Theorem ==
== Proof ==
By the given definition of the natural numbers:
:$\N = \bigcap \II$
where $\II$ is the collection of all inductive sets.
The result is a direct application of Intersection of Inductive Set as Subset of Real Numbers is Inductive Set.
== Sources ==
* : $1$: Set Theory and Logic: $\S 4$: The Integers and the Real Numbers: Exercise $3 \ \text{(b)}$
Category:Natural Numbers form Inductive Set
Natural Numbers form Subsemiring of Integers

== Theorem ==
The semiring of natural numbers $\struct {\N, +, \times}$ forms a subsemiring of the ring of integers $\struct {\Z, +, \times}$.
== Proof ==
We have that Natural Numbers form Commutative Semiring.
From Natural Numbers are Non-Negative Integers we have that $\N$ is a subset of $\Z$.
Hence the result, from the definition of subsemiring.
Category:Natural Numbers
Category:Integers
Natural Numbers have No Proper Zero Divisors

== Theorem ==
Let $\N$ be the natural numbers.
Then for all $m, n \in \N$:
:$m \times n = 0 \iff m = 0 \lor n = 0$
That is, $\N$ has no proper zero divisors.
== Proof ==
=== Necessary Condition ===
Suppose that $n = 0$ or $m = 0$.
Then from Zero is Zero Element for Natural Number Multiplication:
:$m \times n = 0$
=== Sufficient Condition ===
Let $m \times n = 0$.
, suppose $n \ne 0$.
But as:
:$m \times \paren {n - 1} \circ m = m \times n = 0$
it follows that:
:$0 \le m \le 0$
and so as $\le$ is antisymmetric, it follows that $m = 0$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.13: \ 1^\circ$
Category:Natural Numbers
Natural Numbers under Addition do not form Group

== Theorem ==
The algebraic structure $\struct {\N, +}$ consisting of the set of natural numbers $\N$ under addition $+$ is not a group.
=== Corollary ===
== Proof ==
From Natural Numbers under Addition form Commutative Monoid, $\struct {\N, +}$ has an identity element $0$.
However, for any $x \in \N$ such that $x \ne 0$ there exists no $y \in \N$ such that $x + y = 0$.
Thus the general element of $\struct {\N, +}$ has no inverse.
Hence the result by definition of group.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Example $7.1$
* : $\S 1.3$: Example $3$
* : $\S 1$: Some examples of groups
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups: Exercise $1$
Category:Natural Number Addition
Category:Examples of Groups
Natural Numbers under Addition do not form Group/Corollary

== Corollary to Natural Numbers under Addition do not form Group ==
The algebraic structure $\struct {\Z_{\ge 0}, +}$ consisting of the set of non-negative integers $\Z_{\ge 0}$ under addition $+$ does not form a subgroup of the additive group of integers.
== Proof ==
By Natural Numbers are Non-Negative Integers, $\struct {\Z_{\ge 0}, +}$ and $\struct {\N, +}$ are the same (or if not exactly the same, at least isomorphic).
The result follows from Natural Numbers under Addition do not form Group.
== Sources ==
* : Chapter $\text{II}$: Groups: Exercise $\text{E i}$
Category:Natural Number Addition
Natural Numbers under Addition form Commutative Monoid

== Theorem ==
The algebraic structure $\struct {\N, +}$ consisting of the set of natural numbers $\N$ under addition $+$ is a commutative monoid whose identity is zero.
== Proof ==
Consider the natural numbers $\N$ defined as the naturally ordered semigroup.
From the definition of the naturally ordered semigroup, it follows that $\struct {\N, +}$ is a commutative semigroup.
From the definition of zero, $\struct {\N, +}$ has $0 \in \N$ as the identity, hence is a monoid.
== Also see ==
* Natural Numbers under Addition do not form Group
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Example $7.1$
* : $\S 3.1$: Monoids
* : Chapter $1$: Basic Definitions and Results
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups: Exercise $1$
Category:Natural Number Addition
Category:Examples of Commutative Monoids