nyuuzyou/wikis huggingface dataset:
Relation Isomorphism Preserves Transitivity

== Theorem ==
Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be relational structures.
Let $\struct {S, \RR_1}$ and $\struct {T, \RR_2}$ be (relationally) isomorphic.
Then $\RR_1$ is a transitive relation $\RR_2$ is a transitive relation.
== Proof ==
Let $\phi: S \to T$ be a relation isomorphism.
By Inverse of Relation Isomorphism is Relation Isomorphism it follows that $\phi^{-1}: T \to S$ is also a relation isomorphism.
, it is therefore sufficient to prove only that if $\RR_1$ is transitive, then $\RR_2$ is also transitive.
So, suppose $\RR_1$ is a transitive relation.
Let $y_1, y_2, y_3 \in T$ such that $y_1 \mathrel {\RR_2} y_2$ and $y_2 \mathrel {\RR_2} y_3$.
Let $x_1 = \map {\phi^{-1} } {y_1}$, $x_2 = \map {\phi^{-1} } {y_2}$ and $x_3 = \map {\phi^{-1} } {y_3}$.
As $\phi$ is a bijection it follows from Inverse Element of Bijection that:
:$y_1 = \map \phi {x_1}$
:$y_2 = \map \phi {x_2}$
:$y_3 = \map \phi {x_3}$
As $\phi^{-1}$ is a relation isomorphism it follows that:
:$x_1 = \map {\phi^{-1} } {y_1} \mathrel {\RR_1} \map {\phi^{-1} } {y_2} = x_2$
:$x_2 = \map {\phi^{-1} } {y_2} \mathrel {\RR_1} \map {\phi^{-1} } {y_3} = x_3$
As $\RR_1$ is a transitive relation it follows that:
:$x_1 \mathrel {\RR_1} x_3$
As $\phi$ is a relation isomorphism it follows that:
:$y_1 = \map \phi {x_1} \mathrel {\RR_2} \map \phi {x_3} = y_3$
Hence if $y_1 \mathrel {\RR_2} y_2$ and $y_2 \mathrel {\RR_2} y_3$, then also:
:$y_1 \mathrel {\RR_2} y_3$
Hence, $\RR_2$ is a transitive relation, by definition.
Hence the result.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.9 \ \text{(c)}$
Category:Relation Isomorphisms
Category:Transitive Relations
Relation Isomorphism is Equivalence Relation

== Theorem ==
Relation isomorphism is an equivalence relation.
== Proof ==
Let $\struct {S_1, \RR_1}$, $\struct {S_2, \RR_2}$ and $\struct {S_3, \RR_3}$ be relational structures.
Let $S \cong T$ denote the relation that $S$ is (relation) isomorphic to $T$.
Checking in turn each of the criteria for equivalence:
=== Reflexive ===
The fact that relation isomorphism is reflexive follows immediately from Identity Mapping is Relation Isomorphism.
=== Symmetric ===
Suppose $S_1 \cong S_2$.
Let $\phi: S_1 \to S_2$ be a relation isomorphism from $S_1$ to $S_2$.
Then by Inverse of Relation Isomorphism is Relation Isomorphism, $\phi^{-1}: S_2 \to S_1$ is also a relation isomorphism.
So $S_2 \cong S_1$ and so relation isomorphism is symmetric.
=== Transitive ===
Suppose $S_1 \cong S_2$ and $S_2 \cong S_3$.
Let:
:$\alpha: S_1 \to S_2$ be a relation isomorphism from $S_1$ to $S_2$
:$\beta: S_2 \to S_3$ be a relation isomorphism from $S_2$ to $S_3$.
From Composite of Relation Isomorphisms is Relation Isomorphism, the composite mapping $\beta \circ \alpha$ is also a relation isomorphism.
That is, $S_1 \cong S_3$.
So relation isomorphism is transitive.
Category:Graph Theory
Category:Equivalence Relations
Category:Relation Isomorphisms
Category:Examples of Equivalence Relations
Relation Isomorphism preserves Lattice Structure

== Theorem ==
Let $\struct {A, \RR}$ and $\struct {B, \SS}$ be relational structures which are relationally isomorphic.
Let $\struct {A, \RR}$ be a lattice.
Then $\struct {B, \SS}$ is also a lattice.
== Proof ==
Let $\struct {A, \RR}$ be a lattice.
Recall the definition:
From Relation Isomorphism preserves Ordering:
:$\SS$ is an ordering on $B$.
Let $\phi: \struct {A, \RR} \to \struct {B, \SS}$ be a relation isomorphism.
We need to show that for all $x, y \in A$, the ordered set $\struct {\set {\map \phi x, \map \phi y}, \SS}$ admits both a supremum and an infimum. 
Let $x, y \in A$.
Then $\struct {\set {x, y}, \RR}$ admits both a supremum and an infimum.
Let $c = \map \sup {\set {x, y}, \RR}$.
Then by definition of supremum:
:$\forall a \in \set {x, y}: a \mathrel \RR c$
:$\forall d \in A: c \mathrel \RR d$
where $d$ is an upper bound of $\struct {\set {x, y}, \RR} \subseteq A$.
Now consider the image of $\set {x, y}$ under $\phi$.
By definition of relation isomorphism:
:$\forall \map \phi s \in \set {\map \phi x, \map \phi y}: \map \phi c \mathrel \SS \map \phi s$
:$\forall \map \phi d \in B: \map \phi c \mathrel \SS \map \phi d$
where $\map \phi d$ is an upper bound of $\struct {\set {\map \phi x, \map \phi y}, \SS} \subseteq B$.
So by definition of supremum:
:$\map \phi c = \map \sup {\set {\map \phi x, \map \phi y}, \SS}$
That is, $\struct {\set {\map \phi x, \map \phi y}, \SS}$ admits a supremum.
Using a similar technique it can be shown that:
:If $c = \map \inf {\set {x, y}, \SS}$, then:
:$\map \phi c = \map \inf {\set {\map \phi x, \map \phi y}, \SS}$
Hence $\struct {\set {\map \phi x, \map \phi y}, \SS}$ admits both a supremum and an infimum.
That is, $\SS$ is a lattice ordering and so $\struct {B, \SS}$ is a lattice.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.10$
Category:Relation Isomorphisms
Category:Lattice Theory
Relation Isomorphism preserves Ordering

== Theorem ==
Let $\struct {A, \RR}$ and $\struct {B, \SS}$ be relational structures which are relationally isomorphic.
Let $\struct {A, \RR}$ be an ordered set.
Then $\struct {B, \SS}$ is also an ordered set.
== Proof ==
Let $\struct {A, \RR}$ be an ordered set.
Recall the definition:
From Relation Isomorphism Preserves Reflexivity:
:$\SS$ is reflexive.
From Relation Isomorphism Preserves Antisymmetry:
:$\SS$ is antisymmetric.
From Relation Isomorphism Preserves Transitivity:
:$\SS$ is transitive.
So by definition $\struct {B, \SS}$ is an ordered set.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.10$
Category:Relation Isomorphisms
Category:Orderings
Relation Isomorphism preserves Total Ordering

== Theorem ==
Let $\struct {A, \RR}$ and $\struct {B, \SS}$ be relational structures which are relationally isomorphic.
Let $\struct {A, \RR}$ be a totally ordered set.
Then $\struct {B, \SS}$ is also a totally ordered set.
== Proof ==
Let $\struct {A, \RR}$ be a totally ordered set.
Recall the definition:
From Relation Isomorphism preserves Ordering:
:$\SS$ is an ordering.
Let $\phi: \struct {A, \RR} \to \struct {B, \SS}$ be a relation isomorphism.
Let $x, y \in A$.
Then either $x \mathrel \RR y$ or $y \mathrel \RR x$.
From the definition of relation isomorphism, either:
:$\map \phi x \mathrel \SS \map \phi y$
or:
:$\map \phi y \mathrel \SS \map \phi x$
and so by definition $\SS$ is also a total ordering.
So by definition $\struct {B, \SS}$ is also a totally ordered set.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.10$
Category:Relation Isomorphisms
Category:Total Orderings
Relation Isomorphism preserves Well-Ordering

== Theorem ==
Let $\struct {A, \RR}$ and $\struct {B, \SS}$ be relational structures which are relationally isomorphic.
Let $\struct {A, \RR}$ be a well-ordered set.
Then $\struct {B, \SS}$ is also a well-ordered set.
== Proof ==
Let $\struct {A, \RR}$ be a well-ordered set.
Recall the definition:
By Well-Ordering is Total Ordering, $\RR$ is a total ordering.
From Relation Isomorphism preserves Total Ordering:
:$\SS$ is a total ordering on $B$.
Let $\phi: \struct {A, \RR} \to \struct {B, \SS}$ be a relation isomorphism.
Let $A' \subseteq A$.
As $\struct {A, \RR}$ is a well-ordered set:
:$\exists m_1 \in A': \forall a \in A': a = m_1 \lor \neg \paren {a \mathrel \RR m_1}$
Now consider the image $m_2$ of $m_1$ in $B$:
:$m_2 = \map \phi {m_1}$
 $\struct {B, \SS}$ is not a well-ordered set.
Let there exist $B' \subseteq B$ such that:
:$\exists x \in B': x \ne m_2, x \mathrel \SS m_2$
By Inverse of Relation Isomorphism is Relation Isomorphism, $\phi^{-1}$ is a relation isomorphism.
Hence:
:$\map {\phi^{-1} } x \ne \map {\phi^{-1} } {m_2}, \map {\phi^{-1} } x \mathrel \SS \map {\phi^{-1} } {m_2}$
But this contradicts the fact that $\struct {A, \RR}$ is well-ordered.
Hence by Proof by Contradiction $\struct {B, \SS}$ must be a well-ordered set.
== Sources ==
* : Chapter $\text {III}$: The Natural Numbers: $\S 14$: Orderings: Exercise $14.10$
Category:Relation Isomorphisms
Category:Well-Orderings
Relation Partitions Set iff Equivalence

== Theorem ==
Let $\RR$ be a relation on a set $S$.
Then $S$ can be partitioned into subsets by $\RR$ $\RR$ is an equivalence relation on $S$.
The partition of $S$ defined by $\RR$ is the quotient set $S / \RR$.
== Proof ==
== Sources ==
* : Introduction $\S 3$: Equivalence relations
* : $\S 7$: Relations
* : Chapter $\text{I}$: Sets and Functions: Relations: Theorem $5$
* : $\S 1.2$: Theorem $1.11$
* : $0$: Some Conventions and some Basic Facts
* : Chapter $1$: Power Set
Category:Equivalence Relations
Category:Quotient Sets
Category:Set Partitions
Category:Relation Partitions Set iff Equivalence
Relation Partitions Set iff Equivalence/Proof

== Theorem ==
== Proof ==
Let $\RR$ be an equivalence relation on $S$.
From the Fundamental Theorem on Equivalence Relations, we have that the equivalence classes of $\RR$ form a partition.
Let $S$ be partitioned into subsets by a relation $\RR$.
From Relation Induced by Partition is Equivalence, $\RR$ is an equivalence relation.
== Sources ==
* : Chapter $0$: Relations: Theorem $6$
* : $\S 7$: Relations
* : $\S 0.3$: Theorem $5$
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{E}$
* : $\S 1.4$: Decomposition of a set into classes. Equivalence relations: Theorem $4$
* : $\S 6$. Indexed families; partitions; equivalence relations: Theorem $6.4$
Category:Equivalence Relations
Category:Quotient Sets
Category:Relation Partitions Set iff Equivalence
Relation Reflexivity

== Theorem ==
Every relation has exactly one of these properties: it is either:
:reflexive,
:antireflexive or
:non-reflexive.
== Proof ==
A reflexive relation can not be antireflexive, and vice versa:
:$\tuple {x, x} \in \RR \iff \neg \paren {\tuple {x, x} \notin \RR}$
By the definition of non-reflexive, a reflexive relation can not also be non-reflexive.
So a reflexive relation is neither antireflexive nor non-reflexive.
An antireflexive relation can be neither reflexive (see above) nor non-reflexive (by the definition of non-reflexive).
By its own definition, if a relation is neither reflexive nor antireflexive, then it is non-reflexive.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $5$ Properties of Relations
Category:Reflexive Relations
Relation Segment is Increasing

== Theorem ==
Let $S$ be a set.
Let $\RR, \QQ$ be relations on $S$ such that
:$\RR \subseteq \QQ$
Let $x \in S$.
Then
:$x^\RR \subseteq x^\QQ$
where $x^\RR$ denotes the $\RR$-segment of $x$.
== Proof ==
Let $y \in x^\RR$.
By definition of $\RR$-segment:
:$\tuple {y, x} \in \RR$
By definition of subset:
:$\tuple {y, x} \in \QQ$
Thus by definition of $\QQ$-segment:
:$y \in x^\QQ$
== Sources ==
* 
Category:Relation Theory
Relation Segment of Auxiliary Relation is Ideal

== Theorem ==
Let $L = \struct {S, \vee, \preceq}$ be a bounded below join semilattice.
Let $R$ be auxiliary relation on $S$.
Let $x \in S$.
Then
:$x^R$ is ideal in $L$
where $x^R$ denotes the $R$-segment of $x$.
== Proof ==
=== Non-empty ===
By definition of auxiliary relation:
:$\tuple {\bot, x} \in R$
where $\bot$ denotes the smallest element in $L$.
By definition of relation segment:
:$\bot \in x^R$
Thus by definition:
:$x^R$ is non-empty.
=== Directed ===
Let $y, z \in x^R$.
By definition of relation segment:
:$\tuple {y, x} \in R$ and $\tuple {z, x} \in R$
By definition of auxiliary relation:
:$\tuple {y \vee z, x} \in R$
By definition of relation segment:
:$y \vee z \in x^R$
Thus by Join Succeeds Operands:
:$y \preceq y \vee z$ and $z \preceq y \vee z$
Thus by definition:
:$x^R$ is directed.
=== Lower ===
Let $y \in x^R, z \in S$ such that
:$z \preceq y$
By definition of relation segment:
:$\tuple {y, x} \in R$
By definition of reflexivity:
:$x \preceq x$
By definition of auxiliary relation:
:$\tuple {z, x} \in R$
Thus by definition of relation segment:
:$z \in x^R$
Thus by definition:
:$x^R$ is lower section.
Thus by definition:
:$x^R$ is ideal in $L$.
== Sources ==
* 
* 
* 
* 
Category:Auxiliary Relations
Relation Symmetry

== Theorem ==
Every non-null relation has exactly one of these properties: it is either:
:symmetric,
:asymmetric or
:non-symmetric.
== Proof ==
From Relation both Symmetric and Asymmetric is Null, the empty set is both symmetric and asymmetric.
This is why the assumption that $\RR \ne \O$.
Let $\RR$ be symmetric.
Then $\RR$ is not asymmetric by definition.
Also, $\RR$ is not non-symmetric by definition.
Let $\RR$ be asymmetric.
Then $\RR$ is not symmetric by definition.
Also, $\RR$ is not non-symmetric by definition.
If $\RR$ is not symmetric, and $\RR$ is not asymmetric, then $\RR$ is non-symmetric by definition.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $5$ Properties of Relations
Category:Symmetric Relations
Relation between Adjacent Best Rational Approximations to Root 2

== Theorem ==
Consider the Sequence of Best Rational Approximations to Square Root of 2:
:$\sequence S := \dfrac 1 1, \dfrac 3 2, \dfrac 7 5, \dfrac {17} {12}, \dfrac {41} {29}, \dfrac {99} {70}, \dfrac {239} {169}, \dfrac {577} {408}, \ldots$
Let $\dfrac {p_n} {q_n}$ and $\dfrac {p_{n + 1} } {q_{n + 1} }$ be adjacent terms of $\sequence S$.
Then:
:$\dfrac {p_{n + 1} } {q_{n + 1} } = \dfrac {p_n + 2 q_n} {p_n + q_n}$
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$\dfrac {p_{n + 1} } {q_{n + 1} } = \dfrac {p_n + 2 q_n} {p_n + q_n}$
=== Basis for the Induction ===
$\map P 1$ is the case:
Thus $\map P 1$ 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 $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\dfrac {p_{k + 1} } {q_{k + 1} } = \dfrac {p_k + 2 q_k} {p_k + q_k}$
from which it is to be shown that:
:$\dfrac {p_{k + 2} } {q_{k + 2} } = \dfrac {p_{k + 1} + 2 q_{k + 1} } {p_{k + 1} + q_{k + 1} }$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{>0}: \dfrac {p_{n + 1} } {q_{n + 1} } = \dfrac {p_n + 2 q_n} {p_n + q_n}$
== Sources ==
* : $1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$
* : $1 \cdotp 41421 \, 35623 \, 73095 \, 04880 \, 16887 \, 24209 \, 69807 \, 85697 \ldots$
Category:Square Root of 2
Relation between Direction Cosines

== Theorem ==
Let $\LL$ be a line embedded in a Cartesian $3$-space.
Let $l$, $m$ and $n$ be the direction cosines of $\LL$ the $x$-axis, $y$-axis and $z$-axis respectively.
Then:
:$l^2 + m^2 + n^2 = 1$
== Proof ==
== Sources ==
* 
* 
Category:Direction Cosines
Relation between Equations for Hypocycloid and Epicycloid

== Theorem ==
Consider the hypocycloid defined by the equations:
:$x = \paren {a - b} \cos \theta + b \map \cos {\paren {\dfrac {a - b} b} \theta}$
:$y = \paren {a - b} \sin \theta - b \map \sin {\paren {\dfrac {a - b} b} \theta}$
By replacing $b$ with $-b$, this converts to the equations which define an epicycloid:
:$x = \paren {a + b} \cos \theta - b \map \cos {\paren {\dfrac {a + b} b} \theta}$
:$y = \paren {a + b} \sin \theta - b \map \sin {\paren {\dfrac {a + b} b} \theta}$
== Proof ==
== Sources ==
* : Chapter $\text {B}.21$: The Cycloid: Problem $13$
Category:Hypocycloids
Category:Epicycloids
Relation between Floor and Ceiling

== Theorem ==
Let $x \in \R$ be a real number.
Let $\left \lfloor {x}\right \rfloor$ be the floor of $x$, and $\left \lceil {x}\right \rceil$ be the ceiling of $x$.
Then the following results apply:
=== Floor of Ceiling is Ceiling ===
=== Ceiling of Floor is Floor ===
=== Floor equals Ceiling iff Integer ===
=== Floor of Negative equals Negative of Ceiling ===
=== Ceiling of Negative equals Negative of Floor ===
Category:Floor Function
Category:Ceiling Function
Relation between Geodesic and Exponential Map

== Theorem ==
Let $\struct {M, g, \nabla}$ be a Riemannian or pseudo-Riemannian manifold without boundary endowed with the Levi-Civita connection.
Let $TM$ be the tangent bundle of $M$.
Let $\exp$ be the exponential map.
Let $I \subseteq \R$ be an open real interval.
Let $\gamma_v : I \to M$ be the unique maximal geodesic such that:
:$\map {\gamma '} 0 = v$
where $v \in TM$.
Then:
:$\forall t \in I : \forall v \in TM : \map {\gamma_v} t = \map \exp {t v}$
== Proof ==
== Sources ==
* : $\S 5$: The Levi-Civita Connection. The Exponential Map
Category:Exponential Maps
Relation between Normal Neighborhood Charts on Riemannian Manifold

== Theorem ==
Let $\struct {M, g}$ be an $n$-dimensional Riemannian manifold.
Let $\tuple {x^i}$ and $\tuple {\tilde x^j}$ be normal neighborhood charts on $M$.
Let $\map O {n, \R}$ be the orthogonal group.
Let $A^j_i \in \map O {n, \R}$ be a constant matrix.
Then:
:$\forall \tuple {x^i} : \forall \tuple {\tilde x^i} : \exists A^j_i \in \map O {n, \R} : \tilde x^j = A^j_i x^i$
== Proof ==
== Sources ==
* : $\S 5$: The Levi-Civita Connection. Normal Neighborhoods and Normal Coordinates
Category:Normal Neighborhoods
Relation between P-Product Metric and Chebyshev Distance on Real Vector Space

== Theorem ==
For $n \in \N$, let $\R^n$ be a Euclidean space.
Let $p \in \R_{\ge 1}$.
Let $d_p$ be the $p$-product metric on $\R^n$. 
Let $d_\infty$ be the Chebyshev distance on $\R^n$.
Then
:$\forall x, y \in \R^n: \map {d_\infty} {x, y} \le \map {d_p} {x, y} \le n^{1/p} \map {d_\infty} {x, y}$
== Proof ==
By definition of the Chebyshev distance on $\R^n$, we have:
:$\ds \map {d_\infty} {x, y} = \max_{i \mathop = 1}^n \size {x_i - y_i}$
where $x = \tuple {x_1, x_2, \ldots, x_n}$ and $y = \tuple {y_1, y_2, \ldots, y_n}$.
Let $j$ be chosen so that:
:$\ds \size {x_j - y_j} = \max_{i \mathop = 1}^n \size {x_i - y_i}$
Then:
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 2$: Metric Spaces: Exercise $3$
Category:Chebyshev Distance
Category:P-Product Metrics on Real Vector Space are Topologically Equivalent
Relation between Product and Box Topology

== Product and Box Topology ==
=== Product Topology ===
=== Box Topology ===
== Relation between Product and Box Topology ==
=== Product Topology ===
* Product Space is Product in Category of Topological Spaces where it is shown that the product topology is the topology on the Cartesian product of topological spaces that defines the categorical product in the category of topological spaces. The product topology is therefore important in a categorical sense.
* Product Topology is Coarsest Topology such that Projections are Continuous where it is shown that the Product topology is the coarsest topology on the cartesian product of topological spaces for which the projections are continuous.
==== Natural Basis of Product Topology ====
* Natural Basis of Product Topology where it is shown that the natural basis on $X$ is the set $\BB$ 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 $i : U_i = X_i$
* Natural Basis of Product Topology of Finite Product where it is shown that the natural basis on $\ds X = \prod_{k \mathop = 1}^n X_k$ is:
:::$\BB = \set{\ds\prod_{k \mathop = 1}^n U_k : \forall k : U_k \in \tau_k}$
=== Box Topology ===
* Basis for Box Topology where it is shown that the extension of the characterisation of the natural basis for the Product topology on a finite cartesian product to an arbitrary product is a basis and therefore the box topology is a topology.
* Box Topology may not form Categorical Product in the Category of Topological Spaces, the box topology on the cartesian product of topological spaces does not always give us the categorical product in the category of topological spaces.
* Box Topology may not be Coarsest Topology such that Projections are Continuous where it is shown that the box topology is not in general the coarsest topology on the cartesian product of topological spaces for which the projections are continuous. 
* Box Topology contains Product Topology where it is shown that the box topology is a finer topology than the product topology.
* Box Topology on Finite Product Space is Product Topology where it is shown that the box topology and the product topology are the same topology on a finite cartesian product of topological spaces. So the box topology is only of interest for an arbitrary or infinite product of topological spaces.
Category:Product Topology
Category:Box Topology
Relation between Signed and Unsigned Stirling Numbers of the First Kind

== Theorem ==
Let $m, n \in \Z_{\ge 0}$ be positive integers.
Then:
:$\ds {n \brack m} = \paren {-1}^{n + m} \map s {n, m}$
where:
:$\ds {n \brack m}$ denotes an unsigned Stirling number of the first kind
:$\map s {n, m}$ denotes a signed Stirling number of the first kind.
== Proof ==
Category:Stirling Numbers
Relation between Square of Fibonacci Number and Square of Lucas Number

== Theorem ==
Let $F_n$ denote the $n$th Fibonacci number.
Let $L_n$ denote the $n$th Lucas number.
Then:
:$5 {F_n}^2 + 4 \paren {-1}^n = {L_n}^2$
== Proof ==
Let:
:$\phi = \dfrac {1 + \sqrt 5} 2$
:$\hat \phi = \dfrac {1 - \sqrt 5} 2$
Note that we have:
Then:
Hence the result.
== Sources ==
* : $11$
* : $11$
Category:Fibonacci Numbers
Category:Lucas Numbers
Relation between Square of Fibonacci Number and Square of Lucas Number/Mistake

== Source Work ==
:
:The Dictionary
::$11$
== Mistake ==
:Squaring the Fibonacci numbers, then alternately subtracting and adding $4$, produces the squares of the Lucas numbers:
::$5 \times 1^2 - 4 = 1^2 \qquad 5 \times 1^2 + 4 = 3^2$
::$5 \times 2^2 - 4 = 2^2 \qquad 5 \times 3^2 + 4 = 7^2 \qquad$ and so on.
== Correction ==
The expression:
:$5 \times 2^2 - 4 = 2^2$
should read:
:$5 \times 2^2 - 4 = 4^2$
and it is noted that $4$ is indeed the third Lucas number.
Interestingly and oddly, this is correct in the first edition of of $1986$.
== Sources ==
* : $11$
Category:Mistakes/Curious and Interesting Numbers
Relation between Two Ordinals

== Theorem ==
Let $S$ and $T$ be ordinals.
Then either $S \subseteq T$ or $T \subseteq S$.
=== Corollary ===
== Proof ==
 the claim is false.
That is, by De Morgan's laws: Conjunction of Negations:
:$\paren {\neg \paren {S \subseteq T} } \land \paren {\neg \paren {T \subseteq S} }$
where $\neg$ denotes logical not and $\land$ denotes logical and.
Now from Intersection is Subset, we have $S \cap T \subseteq S$ and $S \cap T \subseteq T$.
Also, by Intersection with Subset is Subset, we have $S \cap T \ne S$ and $S \cap T \ne T$.
That is, $S \cap T \subsetneq S$ and $S \cap T \subsetneq T$.
Note that by Intersection of Two Ordinals is Ordinal‎, $S \cap T$ is an ordinal.
So by Transitive Set is Proper Subset of Ordinal iff Element of Ordinal:
:$S \cap T \in S$
:$S \cap T \in T$
But then $S \cap T \in S \cap T$, which contradicts Ordinal is not Element of Itself.
Category:Relation between Two Ordinals
Category:Ordinals
Relation between Two Ordinals/Corollary

== Corollary to Relation between Two Ordinals ==
Let $S$ and $T$ be ordinals.
If $S \ne T$, then either $S$ is an initial segment of $T$, or vice versa.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Relation between Two Ordinals