nyuuzyou/wikis huggingface dataset:
Relation between Two Ordinals/Corollary/Proof 2

== Corollary to Relation between Two Ordinals ==
== Proof ==
If either $S \subset T$ or $T \subset S$ then we invoke Ordinal Subset of Ordinal is Initial Segment, and the proof is complete.
 $S \not \subset T$ and $T \not \subset S$.
Now from Intersection is Subset, we have $S \cap T \subset T$ and $S \cap T \subset S$.
By Intersection of Two Ordinals is Ordinal‎, $S \cap T$ is an ordinal.
So by Ordinal Subset of Ordinal is Initial Segment, we have:
:$S \cap T = S_a$ for some $a \in S$
:$S \cap T = S_b$ for some $b \in T$
Then:
:$a = S_a = S \cap T = T_b = b$
But $a \in S, b \in T$.
Thus $a = b = S \cap T$.
But $S \cap T = S_a$, so:
:$x \in S \cap T \implies x \subset a$
In particular, this means $a \subset a$, which is a contradiction.
So either $S \subset T$ or $T \subset S$, and again we invoke Ordinal Subset of Ordinal is Initial Segment, and the proof is complete.
== Sources ==
* : $\S 1$: Naive Set Theory: $\S 1.7$: Well-Orderings and Ordinals: Theorem $1.7.9$
Category:Relation between Two Ordinals
Relation between Two Ordinals/Corollary/Proof 3

== Corollary to Relation between Two Ordinals ==
== Proof ==
We have that $S \ne T$
Therefore, from Relation between Two Ordinals either $S \subset T$ or $T \subset S$.
By Ordering on Ordinal is Subset Relation or Transitive Set is Proper Subset of Ordinal iff Element of Ordinal, either $S \in T$ or $T \in S$.
By definition, every element of an ordinal is an initial segment; hence the result.
Category:Relation between Two Ordinals
Relation between Volume Forms of Conformally Related Metrics on Oriented Riemannian Manifold

== Theorem ==
Let $M$ be an $n$-dimensional oriented Riemannian manifold.
Let $g_1$ and $g_2$ be conformally related Riemannian metrics.
That is, suppose:
:$g_2 = f g_1$
where $f \in \map {C^\infty} M$ is a positive smooth real function.
Let $\rd V_{g_1}$ and $\rd V_{g_2}$ be Riemannian volume forms associated with $g_1$ and $g_2$.
Then:
:$\ds \rd V_{g_2} = f^{\frac n 2} \rd V_{g_1}$
== Proof ==
== Sources ==
* : $\S 3$: Model Riemannian Manifolds. Euclidean Spaces
Category:Riemannian Geometry
Relation both Symmetric and Asymmetric is Null

== Theorem ==
Let $\RR$ be a relation in $S$ which is both symmetric and asymmetric.
Then:
:$\RR = \O$
== Proof ==
Let $\RR \ne \O$.
Then:
:$\exists \tuple {x, y} \in \RR$
Now, either $\tuple {y, x} \in \RR$ or $\tuple {y, x} \notin \RR$.
If $\tuple {y, x} \in \RR$, then $\RR$ is not asymmetric.
If $\tuple {y, x} \notin \RR$, then $\RR$ is not symmetric.
Therefore, if $\RR \ne \O$, $\RR$ can not be both symmetric and asymmetric.
Let $\RR = \O$.
Then:
:$\forall x, y \in S: \tuple {x, y} \notin \RR \land \tuple {y, x} \notin \RR$
From Biconditional as Disjunction of Conjunctions:
:$\tuple {p \land q} \lor \tuple {\neg p \land \neg q} \vdash p \iff q$
Thus:
:$\forall x, y \in S: \tuple {x, y} \in \RR \iff \tuple {y, x} \in \RR$
So the condition for symmetry is fulfilled, and $\RR = \O$ is symmetric.
As $\RR = \O$:
:$\forall x, y \in S: \tuple {x, y} \notin \RR$
From False Statement implies Every Statement $\neg p \vdash p \implies q$, we can deduce:
:$\forall x, y \in S: \tuple {x, y} \notin \RR$
therefore:
:$\forall x, y \in S: \tuple {x, y} \in \RR \implies \forall x, y \in S: \tuple {y, x} \notin \RR$
This is the definition of asymmetric.
Thus the only relation $\RR$ to be both symmetric and asymmetric is $\RR = \O$.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $5$ Properties of Relations
Category:Symmetric Relations
Category:Asymmetric Relations
Category:Null Relation
Relation is Antireflexive iff Disjoint from Diagonal Relation

== Theorem ==
Let $\RR \subseteq S \times S$ be a relation on a set $S$.
Then:
:$\RR$ is antireflexive
:$\Delta_S \cap \RR = \O$
where $\Delta_S$ is the diagonal relation.
== Proof ==
=== Necessary Condition ===
Let $\RR$ be an antireflexive relation.
Let $\tuple {x, y} \in \Delta_S \cap \RR$.
By definition:
:$\tuple {x, y} \in \Delta_S \implies x = y$
Likewise, by definition:
:$\tuple {x, y} \in \RR \implies x \ne y$.
Thus:
:$\Delta_S \cap \RR = \set {\tuple {x, y}: x = y \land x \ne y}$
and so:
:$\Delta_S \cap \RR = \O$
=== Sufficient Condition ===
Let $\Delta_S \cap \RR = \O$.
Then by definition:
:$\forall \tuple {x, y} \in \RR: \tuple {x, y} \notin \Delta_S$
Thus:
:$\not \exists \tuple {x, y} \in \RR: x = y$
Thus by definition, $\RR$ is antireflexive.
Category:Antireflexive Relations
Category:Diagonal Relation
Relation is Antisymmetric and Reflexive iff Intersection with Inverse equals Diagonal Relation

== Theorem ==
Let $\RR \subseteq S \times S$ be a relation on a set $S$.
Then $\RR$ is both antisymmetric and reflexive :
:$\RR \cap \RR^{-1} = \Delta_S$
where $\Delta_S$ denotes the diagonal relation.
== Proof ==
=== Necessary Condition ===
Let $\RR$ be both antisymmetric and reflexive.
Then:
=== Sufficient Condition ===
Let $\RR$ be such that:
:$\RR \cap \RR^{-1} = \Delta_S$
Then by definition of set equality:
:$\RR \cap \RR^{-1} \subseteq \Delta_S$
and so by Relation is Antisymmetric iff Intersection with Inverse is Coreflexive, $\RR$ is antisymmetric.
Also by definition of set equality:
:$\Delta_S \subseteq \RR \cap \RR^{-1}$
By Intersection is Subset:
:$\RR \cap \RR^{-1} \subseteq \RR$
By Subset Relation is Transitive:
:$\Delta_S \subseteq \RR$
So, by definition, $\RR$ is reflexive.
}
Category:Reflexive Relations
Category:Antisymmetric Relations
Category:Inverse Relations
Category:Diagonal Relation
Relation is Antisymmetric iff Intersection with Inverse is Coreflexive

== Theorem ==
Let $\RR$ be a relation on $S$.
Then:
:$\RR$ is antisymmetric
:
:$\RR \cap \RR^{-1}$ is coreflexive
where $\RR^{-1}$ is the inverse of $\RR$.
That is, :
:$\RR \cap \RR^{-1} \subseteq \Delta_S$
== Proof ==
=== Necessary Condition ===
Let $\RR$ be an antisymmetric relation.
Let $\tuple {a, b} \in \RR \cap \RR^{-1}$.
That means:
:$\tuple {a, b} \in \RR$
and
:$\tuple {a, b} \in \RR^{-1}$
which means, by definition of inverse relation:
:$\tuple {b, a} \in \RR$
But as $\RR$ is antisymmetric, that means $a = b$.
Thus:
:$\tuple {a, b} = \tuple {a, a}$
and so:
:$\tuple {a, b} \in \Delta_S$
where $\Delta_S$ is the diagonal relation.
Thus from the definition of subset:
:$\RR \cap \RR^{-1} \subseteq \Delta_S$
Hence, by definition, $\RR$ is coreflexive.
=== Sufficient Condition ===
Let $\RR \cap \RR^{-1}$ be coreflexive.
Hence by definition of coreflexive:
:$\RR \cap \RR^{-1} \subseteq \Delta_S$
Let $\tuple {a, b} \in \RR$ and $\tuple {b, a} \in \RR$.
That is, by definition of inverse relation:
:$\tuple {a, b} \in \RR$
and
:$\tuple {a, b} \in \RR^{-1}$
That is:
:$\tuple {a, b} \in \RR \cap \RR^{-1}$
But as $\RR \cap \RR^{-1} \subseteq \Delta_S$ it follows that $a = b$.
So by definition $\RR$ is antisymmetric.
== Sources ==
* : $\S 14$: Order
Category:Antisymmetric Relations
Category:Coreflexive Relations
Relation is Connected and Reflexive iff Total

== Theorem ==
Let $S$ be a set.
Let $\RR$ be a relation on $S$.
Then:
:$\RR$ is both a connected relation and a reflexive relation
:
:$\RR$ is a total relation.
== Proof ==
=== Necessary Condition ===
Let $\RR$ be a relation on $S$ which is both connected and reflexive.
Let $\tuple {a, b} \in S \times S$.
Suppose $a = b$.
Then as $\RR$ is reflexive, $\tuple {a, b} \in \RR$.
Suppose $a \ne b$.
Then as $\RR$ is connected, $\tuple {a, b} \in \RR \lor \tuple {b, a} \in \RR$.
That is:
:$\forall \tuple {a, b} \in S \times S: \tuple {a, b} \in \RR \lor \tuple {b, a} \in \RR$.
Thus $\RR$ is a total relation.
=== Sufficient Condition ===
Let $\RR$ be a total relation on $S$.
Then:
:$\forall \tuple {a, b} \in S \times S: \tuple {a, b} \in \RR \lor \tuple {b, a} \in \RR$
by definition.
Thus:
:$\forall \tuple {a, a} \in S \times S: \tuple {a, a} \in \RR$
and so $\RR$ is reflexive.
If $a \ne b$ the condition still holds, and so:
:$\forall \tuple {a, b} \in S \times S: a \ne b \implies \tuple {a, b} \in \RR \lor \tuple {b, a} \in \RR$
and so $\RR$ is connected.
Category:Total Relations
Category:Reflexive Relations
Category:Connected Relations
Relation is Connected iff Union with Inverse and Diagonal is Trivial Relation

== Theorem ==
Let $\RR$ be a relation on $S$.
Then $\RR$ is a connected relation :
:$\RR \cup \RR^{-1} \cup \Delta_S = S \times S$
where $\RR^{-1}$ is the inverse of $\RR$ and $\Delta_S$ is the diagonal relation.
== Proof ==
=== Necessary Condition ===
Let $\RR$ be a connected relation.
By definition of relation:
:$\RR \subseteq S \times S$
:$\RR^{-1} \subseteq S \times S$
:$\Delta_S \subseteq S \times S$
So from Union is Smallest Superset (and indeed, trivially):
:$\RR \cup \RR^{-1} \cup \Delta_S \subseteq S \times S$
Let $\tuple {a, b} \in S \times S$.
Suppose $a = b$.
Then $\tuple {a, b} \in \Delta_S$ by definition of the diagonal relation.
Then by Set is Subset of Union:
:$\tuple {a, b} \in \RR \cup \RR^{-1} \cup \Delta_S$
Suppose otherwise, that is, that $a \ne b$.
As $\RR$ is connected, either:
:$\tuple {a, b} \in \RR$
or:
:$\tuple {b, a} \in \RR$
From the definition of inverse relation, either:
:$\tuple {a, b} \in \RR$
or:
:$\tuple {a, b} \in \RR^{-1}$
That is:
:$\tuple {a, b} \in R \cup \RR^{-1}$
Again by Set is Subset of Union:
:$\tuple {a, b} \in \RR \cup \RR^{-1} \cup \Delta_S$
So, by definition of subset:
:$S \times S \subseteq \RR \cup \RR^{-1} \cup \Delta_S$
Hence, by definition of set equality:
:$\RR \cup \RR^{-1} \cup \Delta_S = S \times S$
=== Sufficient Condition ===
Let $\RR \cup \RR^{-1} \cup \Delta_S = S \times S$.
Let $\tuple {a, b} \in S \times S$.
Suppose $a \ne b$.
By definition of diagonal relation:
:$\tuple {a, b} \notin \Delta_S$
So, by definition of set union:
:$\tuple {a, b} \in \RR$
or:
:$\tuple {a, b} \in \RR^{-1}$
That is, by definition of inverse relation:
:$\tuple {a, b} \in \RR$
or:
:$\tuple {b, a} \in \RR$
So by definition $\RR$ is connected.
== Also see ==
* Relation is Total iff Union with Inverse is Trivial Relation
Category:Connected Relations
Category:Inverse Relations
Category:Trivial Relation
Relation is Equivalence iff Reflexive and Circular

== Theorem ==
Let $\RR \subseteq S \times S$ be a relation in $S$.
Then:
:$\RR$ is reflexive and circular
:
:$\RR$ is an equivalence relation.
== Proof ==
=== Sufficient Condition ===
Let $\RR$ be reflexive and circular.
Then from Reflexive Circular Relation is Equivalence:
:$\RR$ is an equivalence relation.
=== Necessary Condition ===
Let $\RR$ be an equivalence relation.
We have :
:$\RR$ is reflexive.
Then from Equivalence Relation is Circular:
:$\RR$ is circular.
Category:Equivalence Relations
Category:Circular Relations
Category:Reflexive Relations
Relation is Reflexive Symmetric and Antisymmetric iff Diagonal Relation

== Theorem ==
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a relation in $S$.
Then:
:$\RR$ is reflexive, symmetric and antisymmetric
:
:$\RR$ is the diagonal relation $\Delta_S$.
== Proof ==
=== Necessary Condition ===
Let $\RR$ is reflexive, symmetric and antisymmetric.
By definition of reflexive:
:$\Delta_S \subseteq \RR$
From Relation is Symmetric and Antisymmetric iff Coreflexive:
:$\RR \subseteq \Delta_S$
By definition of set equality:
:$\RR = \Delta_S$
=== Sufficient Condition ===
Let $\RR = \Delta_S$.
From Relation is Reflexive and Coreflexive iff Diagonal:
:$\RR$ is reflexive
and
:$\RR$ is coreflexive.
From Relation is Symmetric and Antisymmetric iff Coreflexive it follows that $\RR$ is both symmetric and antisymmetric.
Hence the result.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $5$ Properties of Relations: Exercise $5$
Category:Reflexive Relations
Category:Symmetric Relations
Category:Antisymmetric Relations
Category:Diagonal Relation
Relation is Reflexive and Coreflexive iff Diagonal

== Theorem ==
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a relation on $S$.
Then $\RR$ is reflexive and coreflexive :
: $\RR = \Delta_S$
where $\Delta_S$ is the diagonal relation.
== Proof ==
=== Necessary Condition ===
Let $\RR \subseteq S \times S$ be reflexive and coreflexive.
Then:
=== Necessary Condition ===
Let $\RR = \Delta_S$
By definition of set equality:
:$\Delta_S \subseteq \RR$
and
:$\RR \subseteq \Delta_S$
From $\Delta_S \subseteq \RR$ it follows by definition that $\RR$ is reflexive.
From $\RR \subseteq \Delta_S$ it follows by definition that $\RR$ is coreflexive.
Category:Reflexive Relations
Category:Coreflexive Relations
Relation is Set implies Domain and Image are Sets

== Theorem ==
Let $V$ be a basic universe.
Let $\RR \subseteq V \times V$ be a relation.
Let $\RR$ be a set.
Then $\Dom \RR$ and $\Img \RR$ are also sets.
== Proof ==
From Domain of Relation is Subclass of Union of Union of Relation:
:$\Dom \RR \subseteq \map \bigcup {\bigcup \RR}$
From Image of Relation is Subclass of Union of Union of Relation:
:$\Img \RR \subseteq \map \bigcup {\bigcup \RR}$
We are given that $\RR$ is a set.
From the :
:$\bigcup \RR$ is a set.
Applying the a second time:
:$\map \bigcup {\bigcup \RR}$ is a set.
We have that $\RR \subseteq V$, by definition of basic universe
From the , it follows that every subclass of $V$ is a set.
The result follows.
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 8$ Relations: Theorem $8.2$
Category:Relations
Relation is Symmetric and Antisymmetric iff Coreflexive

== Theorem ==
Let $S$ be a set.
Let $\RR \subseteq S \times S$ be a relation in $S$.
Then:
:$\RR$ is both symmetric and antisymmetric
:
:$\RR$ is coreflexive.
== Proof ==
=== Necessary Condition ===
Let $\RR$ be both symmetric and antisymmetric
Suppose $\RR \nsubseteq \Delta_S$.
Then:
:$\exists \tuple {x, y} \in \RR: x \ne y$
But then as $\RR$ is symmetric:
:$\tuple {y, x} \in \RR$
So we have:
:$\tuple {x, y} \in \RR$
and:
:$\tuple {y, x} \in \RR$
where $x \ne y$.
Thus $\RR$ is not antisymmetric, contrary to hypothesis.
So our assumption that $\RR \nsubseteq \Delta_S$ is false.
That is:
:$\RR \subseteq \Delta_S$
The result follows by definition of coreflexive.
=== Sufficient Condition ===
Let $\RR$ be coreflexive.
Then:
and so by definition $\RR$ is symmetric.
Let $\tuple {x, y} \in \RR$ and $\tuple {y, x} \in \RR$.
From the above, this can happen only when $x = y$.
That is, by definition, $\RR$ is antisymmetric.
So $\RR$ is both symmetric and antisymmetric.
Hence the result.
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $5$ Properties of Relations: Exercise $5$
* : $\S 1.19$: Some Important Properties of Relations: Exercise $10$
Category:Symmetric Relations
Category:Antisymmetric Relations
Category:Coreflexive Relations
Relation is Symmetric iff Inverse is Symmetric

== Theorem ==
A relation $\RR$ is symmetric its inverse $\RR^{-1}$ is also symmetric.
== Proof ==
Let $\RR$ be symmetric.
Then from Relation equals Inverse iff Symmetric:
:$\RR = \RR^{-1}$
The result follows.
== Sources ==
* : $\S 1.19$: Some Important Properties of Relations: Exercise $8$
Category:Symmetric Relations
Relation is Total iff Union with Inverse is Trivial Relation

== Theorem ==
Let $\RR$ be a relation on $S$.
Then $\RR$ is a total relation :
:$\RR \cup \RR^{-1} = S \times S$
where:
:$\RR^{-1}$ is the inverse of $\RR$.
:$S \times S$ is the trivial relation on $S$.
== Proof ==
=== Necessary Condition ===
Let $\RR$ be a total relation.
By definition of relation, both $\RR \subseteq S \times S$ and $\RR^{-1} \subseteq S \times S$.
So from Union is Smallest Superset (and indeed, trivially) $\RR \cup \RR^{-1} \subseteq S \times S$.
Let $\tuple {a, b} \in S \times S$.
As $\RR$ is total, either:
:$\tuple {a, b} \in \RR$
or:
:$\tuple {b, a} \in \RR$
From the definition of inverse relation, this means that either:
:$\tuple {a, b} \in \RR$
or:
:$\tuple {a, b} \in \RR^{-1}$
That is:
:$\tuple {a, b} \in \RR \cup \RR^{-1}$
and so by definition of subset:
:$S \times S \subseteq \RR \cup \RR^{-1}$
Hence, by definition of set equality:
:$\RR \cup \RR^{-1} = S \times S$
=== Sufficient Condition ===
Let $\RR \cup \RR^{-1} = S \times S$.
Let $\tuple {a, b} \in S \times S$.
Then by definition of set union:
:$\tuple {a, b} \in \RR$
or:
:$\tuple {a, b} \in \RR^{-1}$
That is, by definition of inverse relation:
:$\tuple {a, b} \in R$
or:
:$\tuple {b, a} \in R$
So by definition $\RR$ is total.
== Also see ==
* Relation is Connected iff Union with Inverse and Diagonal is Trivial Relation
== Sources ==
* : $\S 14$: Order
Category:Total Relations
Category:Inverse Relations
Category:Trivial Relation
Relation of Boubaker Polynomials to Chebyshev Polynomials

== Theorem ==
The Boubaker polynomials are related to Chebyshev polynomials through the equations:
:$\map {B_n} {2 x} = \dfrac {4 x} n \dfrac \d {\d x} \map {T_n} x - 2 \map {T_n} x$
:$\map {B_n} {2 x} = -2 \map {T_n} x + 4 x \map {U_{n - 1} } x$
where:
:$T_n$ denotes the Chebyshev polynomials of the first kind
:$U_n$ denotes the Chebyshev polynomials of the second kind.
== Proof ==
 
(Using Riordan Matrix)
The ordinary generating function of the Boubaker Polynomials can be expressed in terms of Riordan matrices:
:$\ds \sum_{n \geqslant 0} \map {B_n} t x^n = \frac {1 + 3x^2} {1 - x t + x^2} = \paren {1 + 3 x^2 \mid 1 + x^2} \paren {\dfrac 1 {1 - x t} }$ 
 
By considering the Riordan matrices of Chebyshev Polynomials of the first and second kind $T_n$ and $U_n$, respectively:
:$\map T {\dfrac 1 4 - \dfrac {x^2} 4 \mid \dfrac 1 2 + \dfrac 1 2 x^2}$
and:
:$\map T {\dfrac 1 2 \mid\dfrac 1 2 + \dfrac 1 2 x^2}$
and by setting:
:$\map T {1 + 3x^2 \mid 1 + x^2} = \map T {1 + 3x^2 \mid 1} \map T {\dfrac 1 2 \mid \dfrac 1 2 + \dfrac 1 2 x^2}$
we obtain finally:
:$\map {B_n} x = \map {U_n} {\dfrac x 2} + 3 \map {U_{n - 2} } {\dfrac x 2}$
which gives, due the relation between Chebyshev Polynomials of the first and second kind $T_n$ and $U_n$ and their derivatives:
 
:$\map {B_n} {2 x} = -2 \map {T_n} x + 4 x \map {U_{n - 1} } x$
:$\map {B_n} {2 x} = \dfrac {4 x} n \dfrac \d {\d x} \map {T_n} x - 2 \map {T_n} x$
== Also see ==
* Relation of Boubaker Polynomials to Dickson Polynomials
* Relation of Boubaker Polynomials to Fermat Polynomials
Category:Boubaker Polynomials
Category:Chebyshev Polynomials
Relation of Boubaker Polynomials to Dickson Polynomials

== Theorem ==
The Boubaker polynomials $B_n$ are linked to the Dickson polynomials by the relations:
:$\map {B_{n + 1} } x \map {B_{n + j} } x - \map {B_{n + j + 1} } x \map {B_n} x = \paren {3 x^2 + 4} \map {D_{n + 1} } {x, \dfrac 1 4}$
:$\map {B_n} x = \map {D_n} {2 x, \dfrac 1 4} + 4 \map {D_{n - 1} } {2 x, \dfrac 1 4}$
== Proof ==
== Also see ==
* Relation of Boubaker Polynomials to Fermat Polynomials
* Relation of Boubaker Polynomials to Chebyshev Polynomials
Category:Boubaker Polynomials
Relation of Boubaker Polynomials to Fermat Polynomials

== Theorem ==
The Boubaker polynomials are related to Fermat polynomials by:
:$\forall n \in \N: \map {B_n} x = \dfrac 1 {\paren {\sqrt 2}^n} \map {F_n} {\dfrac {2 \sqrt 2 x} 3} + \dfrac 3 {\paren {\sqrt 2}^{n - 2} } \map {F_{n - 2} } {\dfrac {2 \sqrt 2 x} 3}$
== Proof ==
(Using Riordan Matrix)
Since the ordinary generating function of the Boubaker Polynomials can be expressed in terms of Riordan matrices:
:$\ds \sum_{n \mathop \geqslant 0}^{} \map {B_n} t x^n = \frac {1 + 3x^2 }{1 - x t + x^2} = \paren {1 + 3 x^2 \biggl\lvert 1 + x^2} \paren {\frac 1 {1 - x t} }$ 
then, by writing differently:
:$\ds \map T {1 + 3x^2 \biggl\lvert 1 + x^2} = \map T {1 + 3x^2 \biggl \lvert 1} \map T {\frac 1 2 \biggl\lvert \frac 1 2 + \frac 1 2 x^2} \map T {2 \biggl\lvert 2}$
and:
:$\ds \map T {1 + 3x^2 \biggl\lvert 1 + x^2} = \map T {1 + 3x^2 \biggl\lvert 1} \map T {1 \biggl\lvert \sqrt 2} \map T {\frac 1 3 \biggl\lvert \frac 1 3 + \frac 2 3 x^2} \map T {3 \biggl\lvert \frac 3 {\sqrt 2} }$
and taking into account the Riordan matrix of Fermat polynomials:
:$\ds \map T {\frac 1 3 \biggl\lvert \frac 1 3 + \frac 1 3 x^2}$
we obtain finally:
:$\forall n \in \N: \map {B_n} x = \dfrac 1 {\paren {\sqrt 2}^n} \map {F_n} {\dfrac {2 \sqrt 2 x} 3} + \dfrac 3 {\paren {\sqrt 2}^{n - 2} } \map {F_{n - 2} } {\dfrac {2 \sqrt 2 x} 3}$
== Also see ==
* Relation of Boubaker Polynomials to Dickson Polynomials
* Relation of Boubaker Polynomials to Chebyshev Polynomials
== Sources ==
:A. Luzon et al.: Recurrence Relations for Polynomial Sequences via Riordan Matrices
Category:Boubaker Polynomials
Relation of Ratios to Products

== Theorem ==
That is:
:$a : b = c : d \iff ad = bc$
== Proof ==
Let $A, B, C, D$ be four (natural) numbers in proportion, so that $A : B = C : D$.
Let $A \times D = E$ and $B \times C = F$.
We need to show that $E = F$.
:
Let $A \times C = G$.
Then $A \times C = G$ and $A \times D = E$.
So from :
: $C : D = G : E$
But we have:
: $C : D = A : B$
: $A : B = G : E$
: $A \times C = G$
: $B \times C = F$
Then from :
: $A : B = G : F$
Further, we have that:
: $A : B = G : E$
Thus:
: $G : F = G : E$
So from :
: $E = F$
Now suppose that $E = F$.
We need to show that:
: $A : B = C : D$
Using the same construction, from :
: $G : E = G : F$
But from :
: $G : E = C : D$
Then from :
: $G : F = A : B$
So:
: $A : B = C : D$
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Ratios
Relation on Empty Set is Equivalence

== Theorem ==
Let $S = \O$, that is, the empty set.
Let $\RR \subseteq S \times S$ be a relation on $S$.
Then $\RR$ is the null relation and is an equivalence relation.
== Proof ==
As $S = \O$, we have from Cartesian Product is Empty iff Factor is Empty that $S \times S = \O$.
Then it follows that $\RR \subseteq S \times S = \O$.
=== Reflexivity ===
From the definition:
:$\RR = \O \implies \forall x \in S: \tuple {x, x} \notin \RR$
But as $\neg \, \exists x \in S$ it follows vacuously that $\RR$ is reflexive.
=== Symmetry ===
It follows vacuously that:
:$\tuple {x, y} \in \RR \implies \tuple {y, x} \in \RR$
and so $\RR$ is symmetric.
=== Transitivity ===
It follows vacuously that:
:$\tuple {x, y}, \tuple {y, z} \in \RR \implies \tuple {x, z} \in \RR$
and so $\RR$ is transitive.
It follows from the definition that $\RR$ is an equivalence relation.
Category:Examples of Equivalence Relations
Category:Empty Set
Relation on Set of Cardinality 1 is Symmetric and Transitive

== Theorem ==
Let $S$ be a set whose cardinality is equal to $1$:
:$\card S = 1$
Let $\odot \subseteq S \times S$ be a relation on $S$.
Then $\odot$ is both symmetric and transitive.
== Proof ==
, let $S = \set a$.
There are $2$ relations on $S$:
$(1): \quad \odot := \O$, which is the null relation on $S$.
From Null Relation is Antireflexive, Symmetric and Transitive, $\odot$ is antireflexive, symmetric and transitive.
Thus in this case $\odot$ is both symmetric and transitive.
$(2): \quad \odot := \set {\tuple {a, a} }$, which is the trivial relation on $S$.
From Trivial Relation is Equivalence, $\odot$ is reflexive, symmetric and transitive.
Thus in this case also, $\odot$ is both symmetric and transitive.
Category:Symmetric Relations
Category:Transitive Relations
Relation on Set of Cardinality 2 cannot be Non-Symmetric and Non-Transitive

== Theorem ==
Let $S$ be a set whose cardinality is equal to $2$:
:$\card S = 2$
Let $\odot \subseteq S \times S$ be a relation on $S$.
Then it is not possible for $\odot$ to be not symmetric and also not transitive.
== Proof ==
, let $S = \set {a, b}$.
Let $\odot$ not be symmetric.
 $\odot$ is not transitive.
As $\odot$ is not symmetric:
:$\exists \tuple {x, y} \in \odot: \tuple {y, x} \notin \odot$
Thus there are two possibilities:
:$\exists \tuple {a, b} \in \odot: \tuple {b, a} \notin \odot$
:$\exists \tuple {b, a} \in \odot: \tuple {a, b} \notin \odot$
, let $\tuple {a, b} \in \odot$.
As $\odot$ is not transitive, either:
:$\exists \tuple {x, a} \in \odot: \tuple {x, b} \notin \odot$
or:
:$\exists \tuple {b, y} \in \odot: \tuple {a, y} \notin \odot$
In the first case, $x \ne b$ as that would make $\odot$ symmetric.
In the second case, $y \ne a$ as that would also make $\odot$ symmetric.
Hence $x = a$ and $y = b$, and we have either:
:$\exists \tuple {a, a} \in \odot: \tuple {a, b} \notin \odot$
or:
:$\exists \tuple {b, b} \in \odot: \tuple {a, b} \notin \odot$
But it has been established that $\tuple {a, b} \in \odot$.
From this contradiction it follows that $\odot$ is transitive.
Hence it is not possible for $\odot$ to be neither symmetric nor transitive.
Category:Non-Symmetric Relations
Category:Non-Transitive Relations