nyuuzyou/wikis huggingface dataset:
Natural Number Multiplication Distributes over Addition

== Theorem ==
The operation of multiplication is distributive over addition on the set of natural numbers $\N$:
:$\forall x, y, z \in \N:$
::$\paren {x + y} \times z = \paren {x \times z} + \paren {y \times z}$
::$z \times \paren {x + y} = \paren {z \times x} + \paren {z \times y}$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
In the Axiomatization of 1-Based Natural Numbers, this is rendered:
:$\forall x, y, z \in \N_{> 0}:$
::$\paren {x + y} \times z = \paren {x \times z} + \paren {y \times z}$
::$z \times \paren {x + y} = \paren {z \times x} + \paren {z \times y}$
== Sources ==
* : Chapter $\text I$: Introduction: $1$. The Continuum of Numbers (footnote ${}^*$)
* : Introduction $\S 4$: The natural numbers: $\text{D}$
* : $\S 4$: Number systems $\text{I}$: Peano's Axioms
* : $\S 1$: Numbers: $1.1$ Natural Numbers and Integers: Examples $1.1 \ \text {(f)}$
Category:Natural Number Addition
Category:Natural Number Multiplication
Category:Natural Number Multiplication Distributes over Addition
Category:Examples of Distributive Operations
Natural Number Multiplication Distributes over Addition/Proof 2

== Theorem ==
== Proof ==
We are to show that:
:$\forall x, y, z \in \N: \paren {x + y} \times z = \paren {x \times z} + \paren {y \times z}$
From the definition of natural number multiplication, we have by definition that:
Let $x, y \in \N$ be arbitrary.
For all $z \in \N$, let $\map P z$ be the proposition:
:$\forall x, y \in \N: \paren {x + y} \times z = \paren {x \times z} + \paren {y \times z}$
=== Basis for the Induction ===
$\map P 0$ is the case:
and so $\map P 0$ holds.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k^+}$ is true.
So this is our induction hypothesis:
:$\forall x, y \in \N: \paren {x + y} \times k = \paren {x \times k} + \paren {y \times k}$
Then we need to show:
:$\forall x, y \in \N: \paren {x + y} \times k^+ = \paren {x \times k^+} + \paren {y \times k^+}$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k^+}$ and the result follows by the Principle of Mathematical Induction:
:$\forall x, y, z \in \N: \paren {x + y} \times n = \paren {x \times z} + \paren {y \times z}$
Next we need to show that:
:$\forall x, y, z \in \N: z \times \paren {x + y} = \paren {z \times x} + \paren {z \times y}$
So:
Thus we have proved:
:$\forall x, y, z \in \N: z \times \paren {x + y} = \paren {z \times x} + \paren {z \times y}$
== Sources ==
* : $\S 13$: Arithmetic
Category:Natural Number Multiplication Distributes over Addition
Category:Proofs by Induction
Natural Number Multiplication Distributes over Addition/Proof 3

== Theorem ==
== Proof ==
Using the following axioms:
=== Left Distributive Law for Natural Numbers ===
First we show that:
:$n \times \paren {x + y} = \paren {n \times x} + \paren {n \times y}$
=== Right Distributive Law for Natural Numbers ===
Then we show that:
:$\paren {x + y} \times n = \paren {x \times n} + \paren {y \times n}$
The result follows.
== Sources ==
* : $\S 2.1$
Category:Natural Numbers/1-Based
Category:Natural Number Multiplication Distributes over Addition
Natural Number Multiplication is Associative

== Theorem ==
The operation of multiplication on the set of natural numbers $\N$ is associative:
:$\forall x, y, z \in \N: \paren {x \times y} \times z = x \times \paren {y \times z}$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
In the Axiomatization of 1-Based Natural Numbers, this is rendered:
:$\forall x, y, z \in \N_{> 0}: \paren {x \times y} \times z = x \times \paren {y \times z}$
== Sources ==
* : Chapter $\text I$: Introduction: $1$. The Continuum of Numbers (footnote ${}^*$)
* : Introduction $\S 4$: The natural numbers: $\text{M} 1$
* : $\S 4$: Number systems $\text{I}$: Peano's Axioms
* : $\S 1$: Numbers: $1.1$ Natural Numbers and Integers: Examples $1.1 \ \text {(f)}$
Category:Natural Number Multiplication
Category:Examples of Associative Operations
Category:Natural Number Multiplication is Associative
Category:Associative Law of Multiplication
Natural Number Multiplication is Associative/Proof 1

== Theorem ==
== Proof ==
From Index Laws for Semigroup: Product of Indices we have:
:$+^{z \times y} x = \map {+^z} {+^y x}$
By definition of multiplication, this amounts to:
:$x \times \paren {z \times y} = \paren {x \times y} \times z$
From Natural Number Multiplication is Commutative, we have:
:$x \times \paren {z \times y} = x \times \paren {y \times z}$
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.12$
Category:Natural Number Multiplication is Associative
Natural Number Multiplication is Associative/Proof 2

== Theorem ==
== Proof ==
We are to show that:
:$\paren {x \times y} \times n = x \times \paren {y \times n}$
for all $x, y, n \in \N$.
From the definition of natural number multiplication, we have that:
Let $x, y \in \N$ be arbitrary.
For all $n \in \N$, let $\map P n$ be the proposition:
:$\paren {x \times y} \times n = x \times \paren {y \times n}$
=== Basis for the Induction ===
$\map P 0$ is the case:
and so $\map P 0$ holds.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show 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 our induction hypothesis:
:$\paren {x \times y} \times k = x \times \paren {y \times k}$
Then we need to show:
:$\paren {x \times y} \times \paren {k + 1} = x \times \paren {y \times \paren {k + 1} }$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
== Sources ==
* : $\S 13$: Arithmetic
Category:Natural Number Multiplication is Associative
Category:Proofs by Induction
Natural Number Multiplication is Associative/Proof 3

== Theorem ==
== Proof ==
Using the following axioms:
Let $x, y \in \N_{> 0}$ be arbitrary.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\paren {x \times y} \times n = x \times \paren {y \times n}$
=== Basis for the Induction ===
$\map P 1$ is the case:
and so $\map P 1$ holds.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k^+}$ is true.
So this is our induction hypothesis:
:$\paren {x \times y} \times k = x \times \paren {y \times k}$
Then we need to show:
:$\paren {x \times y} \times \paren {k + 1} = x \times \paren {y \times \paren {k + 1} }$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
== Sources ==
* : $\S 2.1$: Theorem $2.7$
Category:Natural Numbers/1-Based
Category:Natural Number Multiplication is Associative
Natural Number Multiplication is Cancellable

== Theorem ==
Let $\N$ be the natural numbers.
Let $\times$ be multiplication on $\N$.
Then:
:$\forall x, y, z \in \N_{>0}: x \times z = y \times z \implies x = y$
:$\forall x, y, z \in \N_{>0}: x \times y = x \times z \implies y = z$
That is, $\times$ is cancellable on $\N_{>0}$.
== Proof ==
By Natural Number Multiplication is Commutative, proving one of the assertions suffices.
From Ordering on Natural Numbers is Compatible with Multiplication it follows that:
:$\forall x, y, z \in \N: x < y \iff x \times z < y \times z$
Interchanging $x$ and $y$, we obtain:
:$\forall x, y, z \in \N: y < x \iff y \times z < x \times z$
From Ordering on Natural Numbers is Trichotomy, the only remaining possibility is:
:$\forall x, y, z \in \N: x = y \iff x \times z = y \times z$
Hence, $\times$ is cancellable on $\N_{>0}$.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.13$: Corollary
* : $\S 1$: Numbers: $1.1$ Natural Numbers and Integers: Examples $1.1 \ \text {(h)}$
Category:Examples of Cancellable Operations
Category:Natural Number Multiplication
Natural Number Multiplication is Cancellable for Ordering

== Theorem ==
Let $\N$ be the natural numbers.
Let $\times$ be multiplication on $\N$.
Let $<$ be the strict ordering on $\N$.
Then:
:$\forall a, b, c \in \N: a \times c < b \times c \implies a < b$
:$\forall a, b, c \in \N: a \times b < a \times c \implies b < c$
That is, $\times$ is cancellable on $\N$ for $<$.
== Proof ==
Category:Examples of Cancellable Operations
Category:Natural Number Multiplication
Natural Number Multiplication is Closed

== Theorem ==
Let $m$ and $n$ be natural numbers.
Then:
:$m \times n \in \N$
where $\times$ denotes natural number multiplication.
== Proof ==
Let $g: \N \to \N$ be defined by:
:$\map g n = m + n$
Applying the Principle of Recursive Definition to $0$ and $g$, we get the following function $f: \N \to \N$:
:$\map f n = \begin{cases} 0 & : n = 0 \\ m + \map f k & : n = k + 1 \end{cases}$
which is seen to be equivalent to $m \times n$ for all $m,n \in \N$.
Hence $m \times n \in \N$.
== Sources ==
* : Chapter $\text I$: Introduction: $1$. The Continuum of Numbers: $1$. The System of Rational Numbers and the Need for its Extension
* : $\S 5.1$. Subsets closed to an operation: Example $88$
* : $\S 8.17$
* : Chapter $1$: Numbers: Real Numbers: $1$
* : Chapter $1$: Complex Numbers: The Real Number System: $1$
Category:Natural Number Multiplication
Natural Number Multiplication is Commutative

== Theorem ==
The operation of multiplication on the set of natural numbers $\N$ is commutative:
:$\forall x, y \in \N: x \times y = y \times x$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
In the Axiomatization of 1-Based Natural Numbers, this is rendered:
:$\forall x, y \in \N_{> 0}: x \times y = y \times x$
== Euclid's Proof ==
== Sources ==
* : Chapter $\text I$: Introduction: $1$. The Continuum of Numbers (footnote ${}^*$)
* : Introduction $\S 4$: The natural numbers: $\text{M} 2$
* : $\S 4$: Number systems $\text{I}$: Peano's Axioms
* : $\S 1$: Numbers: $1.1$ Natural Numbers and Integers: Examples $1.1 \ \text {(f)}$
Category:Commutative Law of Multiplication
Category:Natural Number Multiplication
Category:Examples of Commutative Operations
Category:Natural Number Multiplication is Commutative
Natural Number Multiplication is Commutative/Euclid's Proof

== Theorem ==
== Proof ==
Let $A, B$ be two (natural) numbers, and let $A$ by multiplying $B$ make $C$, and $B$ by multiplying $A$ make $D$.
We need to show that $C = D$.
:
We have that $A \times B = C$.
So $B$ measures $C$ according to the units of $A$.
But the unit $E$ also measures $A$ according to the units in it.
So $E$ measures $A$ the same number of times that $B$ measures $C$.
Therefore from $E$ measures $B$ the same number of times that $A$ measures $C$.
We also have that $A$ measures $D$ according to the units of $B$ since $B \times A = D$.
But the unit $E$ also measures $B$ according to the units in it.
Therefore $E$ measures $B$ the same number of times that $A$ measures $D$.
But we also have that $E$ measures $B$ the same number of times that $A$ measures $C$.
So $A$ measures $C$ and $D$ the same number of times.
Therefore $C = D$.
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Natural Number Multiplication is Commutative
Natural Number Multiplication is Commutative/Proof 1

== Theorem ==
== Proof ==
Natural number multiplication is recursively defined as:
:$\forall m, n \in \N: \begin{cases} m \times 0 & = 0 \\ m \times \paren {n + 1} & = m \times n + m \end{cases}$
From the Principle of Recursive Definition, there is only one mapping $f$ satisfying this definition; that is, such that:
:$\forall n \in \N: \begin{cases} \map f 0 = 0 \\ \map f {n + 1} = \map f n + m \end{cases}$
Consider now $f'$ defined as $\map {f'} n = n \times m$.
Then by Zero is Zero Element for Natural Number Multiplication:
:$\map {f'} 0 = 0 \times m = 0$
Furthermore:
showing that $f'$ also satisfies the definition of $m \times n$.
By the Principle of Recursive Definition it follows that:
:$m \times n = n \times m$
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 16$: The Natural Numbers: Theorem $16.10$
Category:Natural Number Multiplication is Commutative
Natural Number Multiplication is Commutative/Proof 2

== Theorem ==
== Proof ==
Proof by induction:
From the definition of natural number multiplication, we have that:
For all $n \in \N$, let $\map P n$ be the proposition:
:$\forall m \in \N: m \times n = n \times m$
=== Basis for the Induction ===
From Zero is Zero Element for Natural Number Multiplication:
:$\forall m \in \N: m \times 0 = 0 = 0 \times m$
Thus $\map P 0$ is seen to be true.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 0$, then it logically follows that $\map P {k^+}$ is true.
So this is our induction hypothesis $\map P k$:
:$\forall m \in \N: m \times k = k \times m$
Then we need to show that $\map P {k^+}$ follows from $\map P k$:
:$\forall m \in \N: m \times k^+ = k^+ \times m$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {k^+}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall m, n \in \N: m \times n = n \times m$
== Sources ==
* : $\S 13$: Arithmetic
Category:Natural Number Multiplication is Commutative
Natural Number Multiplication is Commutative/Proof 3

== Theorem ==
== Proof ==
Using the following axioms:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\forall a \in \N_{> 0}: a \times n = n \times a$
=== Basis for the Induction ===
$\map P 1$ is the case:
and so $\map P 1$ holds.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show 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 our induction hypothesis:
:$\forall a \in \N: a \times k = k \times a$
Then we need to show:
:$\forall a \in \N: a \times \paren {k + 1} = \paren {k + 1} \times a$
=== Induction Step ===
This is our induction step:
The result follows by the Principle of Mathematical Induction.
== Sources ==
* : $\S 2.1$: Theorem $2.6$
Category:Natural Numbers/1-Based
Category:Natural Number Multiplication is Commutative
Natural Number Multiplication is Unique

== Theorem ==
Let $\N$ denote the natural numbers.
Let $M: \N \times \N \to \N$ be the mapping defined as the multiplication operation:
:$\forall \tuple {x, y} \in \N \times \N: \map M {x, y} = \begin {cases} 0 & : y = 0 \\ \paren {\map M {x, r} } + r & : y = r + 1 \end {cases}$
where $+$ denotes natural number addition.
Then $M$ exists and is unique.
== Proof ==
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 8$ Definition by finite recursion: Exercise $8.4 \ \text {(b)}$
Category:Natural Number Multiplication
Natural Number Ordering is Preserved by Successor Mapping

== Theorem ==
Let $\N$ be the natural numbers.
Let $m, n \in \N$.
Then:
:$n \le m \implies n^+ \le m^+$
== 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.
From Successor Mapping on Natural Numbers is Progressing, this successor mapping is a progressing mapping.
From Characteristics of Minimally Inductive Class under Progressing Mapping: Mapping Preserves Subsets:
:$\forall m, n \in \omega: m \subseteq n \implies m^+ \subseteq n^+$
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 5$ Applications to natural numbers: Theorem $5.3 \ (3)$
Category:Ordering on Natural Numbers
Natural Number Ordering is Transitive

== Theorem ==
Let $m, n, k \in \N$ where $\N$ is the set of natural numbers.
Let $<$ be the relation defined on $\N$ such that:
:$m < n \iff m \in n$
where $\N$ is defined as the minimally inductive set $\omega$.
Then:
:$k < m, m < n \implies k < n$
That is: $<$ is a transitive relation.
== Proof ==
Let $k < m, m < n$.
By definition it follows that $k \in m, m \in n$.
We have from Element of Finite Ordinal iff Subset that:
:$k \in m \iff k \subseteq m$
:$m \in n \iff m \subseteq n$
It follows from Subset Relation is Transitive that $k \subseteq n$.
Hence the result.
Category:Natural Numbers
Natural Number Power is of Exponential Order Epsilon

== Theorem == 
Let $n \in \N$ be a natural number.
Then: 
:$t \mapsto t^n$
is of exponential order $\epsilon$ for any $\epsilon > 0$ arbitrarily small in magnitude.
== Proof ==
The proof proceeds by induction on $n$, where $n$ is the degree of the polynomial.
=== Basis for the Induction ===
When $n = 0$, the mapping is a constant mapping.
By Constant Function is of Exponential Order Zero and Raising Exponential Order, the mapping is of exponential order $\epsilon$.
This is the basis for the induction.
=== Induction Hypothesis ===
Fix $k \in \N$ with $k \ge 0$.
Assume:
:$t^k \in \EE_\epsilon$
That is, 
:$\forall t \ge M : \size {t^k} < K e^{\epsilon t}$
For some $M > 0$, $K > 0$, and for any $\epsilon > 0$ arbitrarily small.
This is our induction hypothesis.
=== Induction Step ===
We have:
:$x^k$ is of exponential order epsilon by the induction hypothesis
:Identity is of Exponential Order Epsilon
:Product of Functions of Exponential Order
Therefore, $x^{k + 1}$ is also of exponential order epsilon.
The result follows by the Principle of Mathematical Induction.
Category:Exponential Order
Natural Number Subtraction is not Closed

== Theorem ==
The operation of subtraction on the natural numbers is not closed.
== Proof ==
By definition of natural number subtraction:
:$n - m = p$
where $p \in \N$ such that $n = m + p$.
However, when $m > n$ there exists no $p \in \N$ such that $n = m + p$.
== Sources ==
* : Chapter $\text I$: Introduction: $1$. The Continuum of Numbers: $1$. The System of Rational Numbers and the Need for its Extension
* : $\S 5.1$. Subsets closed to an operation: Example $88$
* : Chapter $1$: Integral Domains: $\S 2$. Operations: Example $1$
* : $\S 27$. Binary operations
Category:Subtraction
Category:Natural Numbers
Natural Number has Same Prime Factors as Integer Power

== Theorem ==
Let $x$ be a natural number such that $x > 1$.
Let $n \ge 1$ be a (strictly) positive integer.
The $n$th power of $x$ has the same prime factors as $x$.
== Proof ==
Let $p$ a prime number such that $p$ divides $x^n$.
This is possible because $x > 1$, so $x^n > 1$, hence $x^n$ has prime divisors due to Fundamental Theorem of Arithmetic.
To prove the statement, we need to show $p$ divides $x$.
We will prove this statement by the Principle of Mathematical Induction on $n$.
=== Basis of the Induction ===
 
We have $n = 1$
Clearly, since $p$ divides $x^1 = x$, then $p$ divides $x$.
=== Inductive Step ===
Suppose that for a given $n$, if $p$ divides $x^n$ then $p$ divides $x$. 
Then, if $p$ divides $x^{n+1}$, by definition of prime number, either $p$ divides $x^n$ or $p$ divides $x$.
If $p$ divides $x^n$, we get from induction hypothesis that $p$ divides $x$.
The other case trivially leads to our conclusion.
Hence the result, by Principle of Mathematical Induction.
Conversely, let $p$ a prime number such that it divides $x$. Then, $p$ divides $x * x^{n-1} = x^n$, as required.
Category:Natural Numbers
Natural Number is Divisor or Multiple of Divisor of Another

== Theorem ==
== Proof ==
Let $A, BC$ be two (natural) numbers and let $BC < A$.
We need to show that $BC$ is either an aliquot part or aliquant part of $A$.
That is, either $BC$ is a divisor of $A$, or it is a multiple of some divisor of $A$.
:
$A$ and $BC$ are either coprime or they are not.
First, suppose $A$ and $BC$ are coprime.
Then if $BC$ be divided into the units in it, each unit of $BC$ will be some aliquot part of $A$, so that $BC$ is an aliquant part of $A$.
Next, let $A$ and $BC$ not be coprime.
Then $BC$ either divides $A$ or it does not.
If $BC$ divides $A$ then $BC$ is an aliquot part of $A$.
But if not, then let $D$ be the GCD of $A$ and $BC$ by Euclid's algorithm.
Let $BC$ be divided into the numbers equal to $D$, namely $BE, EF, FC$.
Then each of the numbers $BE, EF, FC$ is also an aliquot part of $A$.
That is, $BC$ is an aliquant part of $A$.
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Number Theory
Category:Divisors
Natural Number is Not Equal to Successor

== Theorem ==
Let $\N_{> 0}$ be the $1$-based natural numbers:
:$\N_{> 0} = \set {1, 2, 3, \ldots}$
Then:
:$\forall n \in \N_{> 0}: n \ne n + 1$
== Proof 1 ==
== Proof 2 ==
Category:Natural Numbers/1-Based
Natural Number is Not Equal to Successor/Proof 1

== Theorem ==
== Proof ==
Using the following axioms:
Proof by induction:
For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition:
:$n \ne n + 1$
=== Basis for the Induction ===
$P \left({1}\right)$ is the case:
:$1 \ne 1 + 1$
From axiom $E$:
:$E: \quad \forall a, b \in \N_{> 0}: \text{ exactly 1 of these three holds: } a = b \lor \left({\exists x \in \N_{> 0}: a + x = b}\right) \lor \left({\exists y \in \N_{> 0}: a = b + y}\right)$
Putting $a = b = 1$ this means:
:$1 = 1$
or
:$1 = 1 + 1$
As $1 = 1$ it follows that $1 \ne 1 + 1$ and so $P \left({1}\right)$ holds.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 2$, then it logically follows that $P \left({k+1}\right)$ is true.
So this is our induction hypothesis:
:$k \ne k + 1$
Then we need to show:
:$k + 1 \ne \left({k + 1}\right) + 1$
=== Induction Step ===
This is our induction step:
As $k \ne k + 1$ from the induction hypothesis, it follows that:
:$k + 1 \ne \left({k + 1}\right) + 1$
So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N_{> 0}: n \ne n + 1$
== Sources ==
* : Exercise $2.1: \ 7$
Category:Natural Numbers/1-Based
Natural Number is Not Equal to Successor/Proof 2

== Theorem ==
== Proof ==
Using the following axioms:
From axiom $E$:
:$E: \quad \forall a, b \in \N_{> 0}: \text{ exactly 1 of these three holds: } a = b \lor \left({\exists x \in \N_{> 0}: a + x = b}\right) \lor \left({\exists y \in \N_{> 0}: a = b + y}\right)$
Hence, taking $a = n$ and $b = n + 1$, we see that since:
:$a + 1 = b$
it follows that the middle condition of the three holds:
:$\exists x \in \N_{>0}: a + x = b$
Therefore, since exactly one the three holds it must be that:
:$n \ne n + 1$
as desired.
Category:Natural Numbers/1-Based
Natural Number is Ordinal

== Theorem ==
Let $n \in \N$ be a natural number.
Then $n$ is an ordinal.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers: Theorem $1.5$
Category:Natural Number is Ordinal
Category:Ordinals
Category:Natural Numbers
Natural Number is Ordinal/Proof 1

== Theorem ==
== Proof ==
Consider the class of all ordinals $\On$.
From Class of All Ordinals is Minimally Superinductive over Successor Mapping, $\On$ is superinductive.
Hence $\On$ is inductive.
From the von Neumann construction of the natural numbers, $\N$ is identified with the minimally inductive set $\omega$.
By definition of minimally inductive set:
:$\omega \subseteq \On$
and the result follows.
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers: Theorem $1.5$
Category:Natural Number is Ordinal
Natural Number is Ordinal/Proof 2

== Theorem ==
== Proof ==
From the von Neumann construction of the natural numbers, $\N$ is identified with the minimally inductive set $\omega$.
From Superinductive Class under Successor Mapping contains All Ordinals, it follows by the Principle of Mathematical Induction that every natural number is an ordinal.
== Sources ==
* : Chapter $5$: Ordinal Numbers: $\S 1$ Ordinal numbers: Theorem $1.5$
Category:Natural Number is Ordinal