nyuuzyou/wikis huggingface dataset:
Natural Basis of Product Topology/Lemma 2

== Lemma for Natural Basis of Product Topology ==
Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $X$ be the cartesian product of $\family {X_i}_{i \mathop \in I}$:
:$\ds X := \prod_{i \mathop \in I} X_i$
Let $\BB$ be the set of cartesian products of the form $\ds\prod_{i \mathop \in I} U_i$ where:
:for all $i \in I : U_i \in \tau_i$
:for all but finitely many indices $i : U_i = X_i$
Then:
:$\forall B_1, B_2 \in \BB : B_1 \cap B_2 \in \BB$
== Proof ==
Let $B_1, B_2 \in \BB$.
Let $B_1 = \ds \prod_{i \mathop \in I} U_i$ where:
:for all $i \in I : U_i \in \tau_i$
:for all but finitely many indices $i : U_i = X_i$
Let $J_1$ be the finite set of indices such that:
:$J_1 = \set {j \in I : U_i \neq X_i}$
Let $B_2 = \ds \prod_{i \mathop \in I} V_i$ where:
:for all $i \in I : V_i \in \tau_i$
:for all but finitely many indices $i : V_i = X_i$
Let $J_2$ be the finite set of indices such that:
:$J_2 = \set{j \in I : V_i \neq X_i}$
Then:
Now:
and 
From Finite Union of Finite Sets is Finite, $J_1 \cup J_2$ is finite.
Then:
:for all but finitely many indices $i : U_i \cap V_i = X_i$
Thus $B_1 \cap B_2 \in \BB$
Category:Natural Basis of Product Topology
Natural Basis of Product Topology/Lemma 3

== Lemma for Natural Basis of Product Topology ==
Let $\family {\struct {X_i, \tau_i} }_{i \mathop \in I}$ be an indexed family of topological spaces where $I$ is an arbitrary index set.
Let $X$ be the cartesian product of $\family {X_i}_{i \mathop \in I}$:
:$\ds X := \prod_{i \mathop \in I} X_i$
Let $\BB$ be the set of cartesian products of the form $\ds \prod_{i \mathop \in I} U_i$ where:
:for all $i \in I : U_i \in \tau_i$
:for all but finitely many indices $i : U_i = X_i$
Then:
:$\ds \forall B \in \BB : B = \bigcap_{j \mathop \in J} \pr_j^{-1} \sqbrk {U_j}$
where:
:$\ds B = \prod_{i \mathop \in I} U_i$
:$J = \set{j \in I : U_i \ne X_i}$ is finite.
== Proof ==
Let $B \in \BB$.
Let $B = \ds \prod_{i \mathop \in I} U_i$
where
:for all $i \in I : U_i \in \tau_i$
:for all but finitely many indices $i : U_i = X_i$
Let $J = \set{j \in I : U_i \ne X_i}$.
Then $J$ is a finite set and:
:$\forall i \in I \setminus J : U_i = X_i$
For all $j \in J$, let:
:$\pr_j^{-1} \sqbrk {U_j} = \ds \prod_{i \mathop \in I} V^j_i$
where:
:$V^j_j = U_j$
:$\forall i \ne j : V^j_i = X_i$
Then:
To complete the proof it remains to show that:
:$\forall i \in I : U_i = \bigcap_{j \mathop \in J} V^j_i$
Let $i \in I$.
=== Case: $i \not \in J$ ===
Let $i \not \in J$.
Then:
=== Case: $i \in J$ ===
Let $i \in J$.
Then:
Thus:
:$\ds \forall i \in I : U_i = \bigcap_{j \mathop \in J} V^j_i$
The result follows.
Category:Natural Basis of Product Topology
Natural Frequency of Underdamped Cart attached to Wall by Spring

== Theorem ==
=== Problem Definition ===
Let $C$ be underdamped.
Let $C$ be pulled aside to $x = x_0$ and released from stationary at time $t = 0$.
Then the natural frequency of $C$ can be expressed as:
:$\nu = \dfrac 1 {2 \pi} \sqrt {\dfrac k m - \dfrac {c^2} {4 m^2} }$
== Proof ==
From Period of Oscillation of Underdamped Cart attached to Wall by Spring:
:$T = \dfrac {2 \pi} {\sqrt {\dfrac k m - \dfrac {c^2} {4 m^2} } }$
where $T$ is the period of oscillation of $C$.
Then by definition of the natural frequency of $C$:
== Sources ==
* : $\S 3.20$: Vibrations in Mechanical Systems: $(22)$
Category:Mechanics
Natural Frequency of Underdamped System

== Theorem ==
Consider a physical system $S$ whose behaviour can be described with the second order ODE in the form:
:$\dfrac {\d^2 x} {\d t^2} + 2 b \dfrac {\d x} {\d t} + a^2 x = 0$
for $a, b \in \R_{>0}$.
Let $b < a$, so as to make $S$ underdamped.
Then the natural frequency of $S$ is given by:
:$\nu = \dfrac {\sqrt {a^2 - b^2} } {2 \pi}$
== Proof ==
Let the position of $S$ be described in the canonical form:
:$(1): \quad x = \dfrac {x_0 \, a} \alpha e^{-b t} \map \cos {\alpha t - \theta}$
where $\alpha = \sqrt {a^2 - b^2}$.
From Period of Oscillation of Underdamped System is Regular:
:$T = \dfrac {2 \pi} {\sqrt {a^2 - b^2} }$
where $T$ is the period of the movement of $S$.
By definition of natural frequency:
:$\nu = \dfrac 1 T = \dfrac {\sqrt {a^2 - b^2} } {2 \pi}$
== Sources ==
* : $\S 3.20$: Vibrations in Mechanical Systems: $(22)$
Category:Mechanics
Natural Logarithm/Examples/10

== Example of Natural Logarithm ==
The natural logarithm of $10$ is approximately:
:$\ln 10 \approx 2 \cdotp 30258 \, 50929 \, 94045 \, 68401 \, 7991 \ldots$
== Sources ==
* : Table $1.1$. Mathematical Constants
* : Table $1.1$. Mathematical Constants
* : $\S 1$: Special Constants: $1.17$
* : $2 \cdotp 302 \, 585 \, 092 \, 994 \, 045 \, 684 \, 017 \, 991 \, 454 \, 684 \, 364 \, 207 \, 601 \ldots$
* : $2 \cdotp 30258 \, 50929 \, 94045 \, 68401 \, 79914 \, 54684 \, 36420 \, 7601 \ldots$
Category:Examples of Natural Logarithms
Natural Logarithm/Examples/3

== Example of Natural Logarithm ==
The natural logarithm of $3$ is:
:$\ln 3 = 1.09861 \, 22886 \, 68109 \, 69139 \, 5245 \ldots$
== Sources ==
* : Table $1.1$. Mathematical Constants
* : $\S 1$: Special Constants: $1.19$
Category:Examples of Natural Logarithms
Natural Logarithm Function is Differentiable/Proof 1

== Theorem ==
== Proof ==
This proof assumes the definition of $\ln$ as:
:$\ln x = \ds \int_1^x \frac 1 t \rd t$
As $\ln$ is defined as a definite integral, the result follows from the corollary to the first fundamental theorem of calculus.
Category:Natural Logarithm Function is Differentiable
Natural Logarithm as Derivative of Exponential at Zero

== Theorem ==
Let $\ln: \R_{>0}$ denote the real natural logarithm.
Then:
:$\ds \forall x \in \R_{>0}: \ln x = \lim_{h \mathop \to 0} \frac {x^h - 1} h$
== Proof ==
Fix $x \in \R_{>0}$.
Let $x > 1$.
From Power Function on Strictly Positive Base is Convex, $x^h$ is convex.
Thus for $0 < h < s$:
Further, $0 < \dfrac 1 x < 1$.
So, for $h < s < 0 \iff 0 < -s < -h$:
Hence $\dfrac {x^h - 1} h$ is increasing on $\R \setminus \set 0$.
Next:
So $\dfrac {x^h - 1} h$ is strictly positive on $\R_{>0}$.
In particular:
:$\dfrac {x^h - 1} h$ is bounded below (by $0$) and increasing on $\openint 0 \to$
:$\dfrac {x^h - 1} h$ is bounded above (by $\ds \inf_{h \mathop > 0} \frac {x^h - 1} h$) and increasing on $\openint \gets 0$
So from Limit of Increasing Function, $\ds \lim_{h \mathop \to 0^+} \frac {x^h - 1} h$ and $\ds \lim_{h \mathop \to 0^-} \frac {x^h - 1} h$ exist.
Further:
where $\sequence {n \paren {x^{1 / n} - 1 } }_{n \mathop \in \N}$ is now a real sequence.
Similarly:
Thus, for $x > 1$:
So from Limit iff Limits from Left and Right, for $x > 1$:
:$\ds \lim_{h \mathop \to 0} \frac {x^h - 1} h = \ln x$
Suppose instead that $0 < x < 1$.
From Ordering of Reciprocals:
:$\dfrac 1 x > 1$
Thus, from above:
Hence the result.
Category:Natural Logarithms
Natural Logarithm of 1 is 0/Proof 1

== Theorem ==
== Proof ==
We use the definition of the natural logarithm as an integral:
:$\ds \ln x = \int_1^x \frac {\d t} t$
From Integral on Zero Interval:
:$\ds \ln 1 = \int_1^1 \frac {\d t} t = 0$
Category:Natural Logarithm of 1 is 0
Natural Logarithm of 1 is 0/Proof 3

== Theorem ==
== Proof ==
We use the definition of the natural logarithm as the limit of a sequence:
:$\ds \ln x = \lim_{n \mathop \to \infty} n \paren {\sqrt [n] x - 1}$
Then:
Category:Natural Logarithm of 1 is 0
Natural Logarithm of 2 is Greater than One Half

== Lemma ==
:$\ln 2 \ge \dfrac 1 2$
where $\ln$ denotes the natural logarithm function.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Logarithm Tends to Infinity
== Sources ==
* : Appendix $A$: Properties of the Natural Logarithmic Function
Category:Natural Logarithms
Category:Inequalities
Category:Natural Logarithm of 2 is Greater than One Half
Natural Logarithm of 2 is Greater than One Half/Proof 1

== Lemma ==
== Proof ==
Let $f: \R_{>0} \to \R$ be the real function defined as:
:$\forall x \in \R_{>0}: \map f x = \dfrac 1 x$
From Real Rational Function is Continuous, $\map f x$ is a continuous real function, in particular on the closed interval $\closedint a b$.
Hence the Mean Value Theorem for Integrals can be applied:
There exists some $k \in \closedint 1 2$ such that:
Thus:
== Sources ==
* : Appendix $A$: Properties of the Natural Logarithmic Function
Category:Natural Logarithm of 2 is Greater than One Half
Natural Logarithm of e is 1

== Theorem ==
:$\ln e = 1$
where $\ln$ is the natural logarithm, $e$ is Euler's number, and $1$ is the identity element of multiplication.
== Proof ==
The definition of the Euler's number as the Base of Logarithm will be used.
Then the result follows directly.
== Also see ==
* Equivalence of Definitions of Euler's Number for other definitions of Euler's number
== Sources ==
* : $\S 1.2.2$: Numbers, Powers, and Logarithms: Exercise $17$
Category:Examples of Natural Logarithms
Natural Number Addition Commutativity with Successor

== Theorem ==
Let $\N$ be the natural numbers.
Then:
:$\forall m, n \in \N: m^+ + n = \paren {m + n}^+$
== Proof 1 ==
== Proof 2 ==
== Also defined as ==
Thus, in the context of 1-based natural numbers, this result can be written:
:$\forall m, n \in \N_{> 0}: \paren {m + 1} + n = \paren {m + n} + 1$
Category:Natural Number Addition
Category:Natural Number Addition Commutativity with Successor
Natural Number Addition Commutativity with Successor/Proof 1

== Theorem ==
== Proof ==
Proof by induction:
From definition of addition:
For all $n \in \N$, let $\map P n$ be the proposition:
:$\forall m \in \N: m^+ + n = \paren {m + n}^+$
=== Basis for the Induction ===
From definition of addition:
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 $\map P k$:
:$\forall m \in \N: m^+ + k = \paren {m + k}^+$
Then we need to show that $\map P {k^+}$ follows directly from $\map P k$:
:$\forall m \in \N: m^+ + k^+ = \paren {m + k^+}^+$
=== 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.
Therefore:
:$\forall m, n \in \N: m^+ + n = \paren {m + n}^+$
== Sources ==
* : $\S 13$: Arithmetic
Category:Natural Number Addition Commutativity with Successor
Natural Number Addition Commutativity with Successor/Proof 2

== Theorem ==
Let $\N$ be the natural numbers.
Then:
:$\forall m, n \in \N_{> 0}: \left({m + 1}\right) + n = \left({m + n}\right) + 1$
== Proof ==
Using the following axioms:
Proof by induction:
From Axiomatization of $1$-Based Natural Numbers, we have by definition that:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$\forall m \in \N_{> 0}: \paren {m + 1} + n = \paren {m + n} + 1$
=== Basis for the Induction ===
When $n = 1$, we have:
:$\paren {m + 1} + 1 = \paren {m + 1} + 1$
which holds trivially.
Thus $\map P 1$ 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 $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis $\map P k$:
:$\forall m \in \N: \paren {m + 1} + k = \paren {m + k} + 1$
Then we need to show that $\map P {k^+}$ follows directly from $\map P k$:
:$\forall m \in \N: \paren {m + 1} + \paren {k + 1} = \paren {m + \paren {k + 1} } + 1$
=== 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.
Therefore:
:$\forall m, n \in \N_{> 0}: \paren {m + 1} + n = \paren {m + n} + 1$
Category:Natural Number Addition Commutativity with Successor
Natural Number Addition Commutes with Zero

== Theorem ==
Let $\N$ be the natural numbers.
Then:
:$\forall n \in \N: 0 + n = n = n + 0$
== Proof ==
Proof by induction:
From definition of addition:
For all $n \in \N$, let $\map P n$ be the proposition:
:$0 + n = n = n + 0$
=== Basis for the Induction ===
By definition, we have:
:$0 + 0 = 0 = 0 + 0$
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 $\map P k$:
:$0 + k = k = k + 0$
Then we need to show that $\map P {k^+}$ follows directly from $\map P k$:
:$0 + k^+ = k^+ = k^+ + 0$
=== Induction Step ===
This is our induction step:
By definition:
:$k^+ + 0 = k^+$
So $\map P k \implies \map P {k^+}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \N: 0 + n = n = n + 0$
== Sources ==
* : $\S 13$: Arithmetic
Category:Natural Number Addition
Category:Examples of Commutative Operations
Natural Number Addition is Associative

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

== Theorem ==
== Proof ==
Consider the von Neumann construction of natural numbers $\N$, as elements of the minimally inductive set $\omega$.
We are to show that:
:$\paren {x + y} + n = x + \paren {y + n}$
for all $x, y, n \in \N$.
From the definition of addition, we have that:
Let $x, y \in \N$ be arbitrary.
For all $n \in \N$, let $\map P n$ be the proposition:
:$\paren {x + y} + n = x + \paren {y + n}$
=== 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, then it logically follows that $\map P {k^+}$ is true.
So this is our induction hypothesis:
:$\paren {x + y} + k = x + \paren {y + k}$
Then we need to show:
:$\paren {x + y} + \paren {k^+} = x + \paren {y + \paren {k^+} }$
=== 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.
== Sources ==
* : $\S 13$: Arithmetic
Category:Natural Number Addition is Associative
Natural Number Addition 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:
:$\paren {x + y} + n = x + \paren {y + n}$
=== Basis for the Induction ===
From Axiom $\text C$, we have by definition that:
:$\forall x, y \in \N_{> 0}: \paren {x + y} + 1 = x + \paren {y + 1}$
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 $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$\paren {x + y} + k = x + \paren {y + k}$
Then we need to show:
:$\paren {x + y} + \paren {k + 1} = x + \paren {y + \paren {k + 1} }$
=== 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 $2.2$
Category:Natural Number Addition is Associative
Natural Number Addition is Cancellable

== Theorem ==
Let $\N$ be the natural numbers.
Let $+$ be addition on $\N$.
Then:
:$\forall a, b, c \in \N: a + c = b + c \implies a = b$
:$\forall a, b, c \in \N: a + b = a + c \implies b = c$
That is, $+$ is cancellable on $\N$.
== Proof 1 ==
== Proof 2 ==
Category:Natural Number Addition
Category:Examples of Cancellable Operations
Category:Natural Number Addition is Cancellable
Natural Number Addition is Cancellable/Proof 2

== Theorem ==
== Proof ==
By Natural Number Addition is Commutative, we only need to prove the first statement.
Proof by induction.
Consider the natural numbers $\N$ defined in terms of Peano's Axioms.
From the definition of addition in Peano structure‎, we have that:
For all $n \in \N$, let $\map P n$ be the proposition:
:$\forall a, b \in \N: a + n = b + n \implies a = b$
=== Basis for the Induction ===
$\map P 0$ is the proposition:
:$\forall a, b \in \N: a + 0 = b + 0 \implies a = b$
which holds because of the definition of $+$.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \in \N$, then it logically follows that $\map P {\map s k}$ is true.
So this is our induction hypothesis $\map P k$:
:$\forall a, b \in \N: a + k = b + k \implies a = b$
Then we need to show that $\map P {\map s k}$ follows directly from $\map P k$:
:$\forall a, b \in \N: a + \map s k = b + \map s k \implies a = b$
=== Induction Step ===
This is our induction step:
So $\map P k \implies \map P {\map s k}$ and the result follows by the Principle of Mathematical Induction for Peano Structure.
Therefore:
:$\forall a, b \in \N: a + n = b + n \implies a = b$
Category:Natural Number Addition is Cancellable
Natural Number Addition is Cancellable for Ordering

== Theorem ==
Let $\N$ be the natural numbers.
Let $<$ be the strict ordering on $\N$.
Let $+$ be addition on $\N$.
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 $\N$ for $<$.
== Proof ==
Category:Natural Number Addition
Category:Examples of Cancellable Operations
Natural Number Addition is Closed

== Theorem ==
The operation of addition on the set of natural numbers $\N$ is closed:
:$\forall x, y \in \N: x + y \in \N$
== Proof ==
Follows directly from Natural Numbers under Addition form Commutative Monoid.
A monoid by definition is a semigroup.
Again by definition, the operation in a semigroup is closed.
== 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 $88$
* : $\S 8.9$
* : Chapter $1$: Numbers: Real Numbers: $1$
* : Chapter $1$: Complex Numbers: The Real Number System: $1$
Category:Natural Number Addition
Natural Number Addition is Commutative

== Theorem ==
The operation of addition on the set of natural numbers $\N$ is commutative:
:$\forall m, n \in \N: m + n = n + m$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
In the axiomatisation of $1$-based natural numbers, this is rendered:
:$\forall x, y \in \N_{> 0}: x + y = y + x$
== Sources ==
* : Chapter $\text I$: Introduction: $1$. The Continuum of Numbers (footnote ${}^*$)
* : Introduction $\S 4$: The natural numbers: $\text{A} 2$
* : $\S 4$: Number systems $\text{I}$: Peano's Axioms
* : $\S 1$: Numbers: $1.1$ Natural Numbers and Integers: Examples $1.1 \ \text {(a)}$
Category:Natural Number Addition is Commutative
Category:Natural Number Addition
Category:Commutative Law of Addition
Category:Examples of Commutative Operations
Natural Number Addition is Commutative/Proof 2

== Theorem ==
== Proof ==
Proof by induction.
Consider the natural numbers $\N$ defined as the elements of the minimally inductive st $\omega$.
From the definition of addition in $\omega$‎, we have that:
For all $n \in \N$, let $\map P n$ be the proposition:
:$\forall m \in \N: m + n = n + m$
=== Basis for the Induction ===
From Natural Number Addition Commutes with Zero, we have:
:$\forall m \in \N: m + 0 = m = 0 + m$
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 $\map P k$:
:$\forall m \in \N: m + k = k + m$
Then we need to show that $\map P {k^+}$ follows directly from $\map P k$:
:$\forall m \in \N: m + k^+ = k^+ + m$
=== 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:
:$\forall m, n \in \N: m + n = n + m$
== Sources ==
* : $\S 13$: Arithmetic
* : $\S 1$: Numbers: $1.1$ Natural Numbers and Integers: Theorem $1.3$
Category:Natural Number Addition is Commutative
Category:Proofs by Induction
Natural Number Addition is Commutative/Proof 3

== Theorem ==
The operation of addition on the set of natural numbers $\N_{> 0}$ is commutative:
:$\forall x, y \in \N_{> 0}: x + y = y + x$
== Proof ==
Using the following axioms:
Let $x \in \N_{> 0}$ be arbitrary.
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$x + n = n + x$
=== Basis for the Induction ===
From Natural Number Commutes with 1 under Addition, we have that:
:$\forall x \in \N_{> 0}: x + 1 = 1 + x$
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 $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$x + k = k + x$
Then we need to show:
:$x + \paren {k + 1} = \paren {k + 1} + x$
=== 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 $2.3$
Category:Natural Numbers/1-Based
Category:Natural Number Addition is Commutative
Natural Number Commutes with 1 under Addition

== Theorem ==
Let $n \in \N_{> 0}$ be a natural number.
Then $n$ commutes with $1$ under the operation of addition:
:$\forall n \in \N_{> 0}: n + 1 = 1 + n$
== Proof ==
Using the axiomatization:
For all $n \in \N_{> 0}$, let $\map P n$ be the proposition:
:$n + 1 = 1 + n$
=== Basis for the Induction ===
Setting $n = 1$ we have that:
:$1 + 1 = 1 + 1$
and so $\map P 1$ holds trivially.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$k + 1 = 1 + k$
Then we need to show:
:$\paren {k + 1} + 1 = 1 + \paren {k + 1}$
=== 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.
</onlyinclude>
== Sources ==
* : $\S 2.1$: Theorem $2.3$
Category:Natural Numbers/1-Based
Category:Natural Number Addition
Natural Number Functions are Uncountable

== Theorem ==
The set of all natural number one-variable functions $\set {f: \N \to \N}$ is uncountably infinite.
== Proof ==
Let $\Bbb F$ be the set of all functions from $\N$ to $\N$.
Clearly $\Bbb F$ is infinite because it contains for each $k \in \N$ the constant function $f_k: \N \to \N: \map {f_k} n = k$ and these are all different and (trivially) countably infinite in number.
Let $\Phi: \N \to \Bbb F$ be a function.
For each $n \in \N$ let $f_n: \N \to \N$ be the function $\map \Phi n$.
Let us define $g: \N \to \N$ by:
:$\map g n = \map {f_n} n + 1$
Then $g \in \Bbb F$, but $\forall n \in \N$, $\map g n \ne \map {f_n} n$ and so $g \ne f_n$.
Since $g$ is an element of $\Bbb F$ which is different from all the values taken by $\Phi$, it follows that $\Phi$ is not a surjection and hence not a bijection.
Thus no bijection exists between $\Bbb F$ and $\N$ and so $\Bbb F$ is not equivalent to $\N$.
Thus from Countably Infinite Iff Equivalent to Natural Numbers, $\Bbb F$ is uncountable.
Category:Mapping Theory
Category:Uncountable Sets
Natural Number Less than or Equal to Successor of Another

== Theorem ==
Let $\N$ be the natural numbers.
Let $m, n \in \N$ such that $m \le n^+$.
Then either:
:$(1): \quad m \le n$
or:
:$(2): \quad m = n^+$
== Proof ==
Let $m \le n^+$.
Suppose $m \le n$ is false.
Then:
:$n^+ \le m$
and because $m \le n^+$:
:$m = n^+$
== Sources ==
* : Chapter $3$: The Natural Numbers: $\S 5$ Applications to natural numbers: Theorem $5.8$
Category:Ordering on Natural Numbers