nyuuzyou/wikis huggingface dataset:
Singleton of Greatest Element in Two is Completely Prime Filter

== Theorem ==
Let $\struct{\mathbf 2, \vee, \wedge, \preceq}$ denote the (Boolean Lattice) $\mathbf 2$.
Then:
:$\set \top$ is a completely prime filter
== Proof ==
By definition of (Boolean Lattice) $\mathbf 2$:
:$\mathbf 2 := \set {\bot, \top}$
endowed with the logical operations $\lor$, $\land$ defined by the following Cayley tables:
:$\begin{array}{c|cc}
 \lor & \bot & \top \\ \hline
 \bot & \bot & \top \\
 \top & \top & \top
\end{array} \qquad \begin{array}{c|cc}
 \land & \bot & \top \\ \hline
 \bot & \bot & \bot \\
 \top & \bot & \top
\end{array}$
and ordering $\preceq$ defined by:
:$\preceq = \set{ \tuple{\bot, \bot}, \tuple{\bot, \top}, \tuple{\top, \top}}$.
From Two is a Locale:
:$\struct{\mathbf 2, \vee, \wedge, \preceq}$ is a locale
By definition of locale:
:$\struct{\mathbf 2, \vee, \wedge, \preceq}$ is a complete lattice
From Characterization of Completely Prime Filter in Complete Lattice, it is sufficient to show:
:$(1)\quad\forall A \subseteq \mathbf 2 : \bigvee A \in \set \top \iff \paren{\exists a \in A : a \in \set \top}$
:$(2)\quad\forall $ finite $A \subseteq \mathbf 2 : \bigwedge A \in \set \top \iff \paren{\forall a \in A : a \in \set \top}$
where:
:* $\bigvee A$ denotes the supremum of $A$ in $\mathbf 2$
:* $\bigwedge A$ denotes the infimum of $A$ in $\mathbf 2$
From Power Set of Doubleton:
:$\powerset {\mathbf 2} = \set {\O, \set \bot, \set \top, \set {\bot, \top} }$
=== Statement $(1)$ ===
We have:
It follows that:
:$(1)\quad\forall A \subseteq \mathbf 2 : \bigvee A \in \set \top \iff \paren{\exists a \in A : a \in \set \top}$
=== Statement $(2)$ ===
We have:
It follows that:
:$(2)\quad\forall $ finite $A \subseteq \mathbf 2 : \bigwedge A \in \set \top \iff \paren{\forall a \in A : a \in \set \top}$
From Characterization of Completely Prime Filter in Complete Lattice:
:$\set \top$ is a completely prime filter
Category:Completely Prime Filters
Singleton of Power Set less Empty Set is Minimal Subset

== Theorem ==
Let $S$ be a set which is non-empty.
Let $\CC = \powerset S \setminus \O$, that is, the power set of $S$ without the empty set.
Let $x \in S$.
Then $\set x$ is a minimal element of the ordered structure $\struct {\CC, \subseteq}$.
== Proof ==
Let $y \in \CC$ such that $y \subseteq \set x$.
We have that $\O \notin \CC$.
Therefore:
:$\exists z \in S: z \in y$
But as $y \subseteq \set x$ it follows that:
:$z \in \set x$
and so by definition of singleton:
:$z = x$
and so:
:$y = \set x$
and so:
:$y = x$
Thus, by definition, $\set x$ is a minimal element of $\struct {\CC, \subseteq}$.
== Sources ==
* : $\S 14$: Order
Category:Power Set
Category:Singletons
Singleton of Set is Filter in Lattice of Power Set

== Theorem ==
Let $X$ be a set.
Let $L = \struct {\powerset X, \cup, \cap, \subseteq}$ be an inclusion lattice of power set of $X$.
Then $\set X$ is a filter on $L$.
== Proof ==
By Singleton is Directed and Filtered Subset:
:$\set X$ is filtered.
We will prove that
:$\set X$ is an upper section.
Let $x \in \set X$, $y \in \powerset X$ such that:
:$x \subseteq y$
By definition of singleton:
:$x = X$
By definition of power set:
:$y \subseteq X$
By definition of set equality:
:$y = X$
Thus:
:$y \in \set X$
Thus by definition of filter in ordered set:
:$\set X$ is a filter.
== Sources ==
* 
Category:Power Set
Singleton of Subset is Element of Powerset of Powerset

== Theorem ==
Let $S \subseteq T$ where $S$ and $T$ are both sets.
Then:
:$\set S \in \powerset {\powerset T}$
where $\powerset T$ denotes the power set of $T$.
== Proof ==
== Sources ==
* : $\S 2$. Sets of sets
Category:Power Set
Category:Subsets
Singular Absolutely Continuous Measure is Zero

== Theorem ==
Let $\struct {X, \Sigma}$ be a measurable space. 
Let $\mu$ be a measure on $\struct {X, \Sigma}$.
Let $\nu$ be a signed measure or complex measure on $\struct {X, \Sigma}$ such that:
:$\nu$ is mutually singular with respect to $\mu$
:$\nu$ is absolutely continuous with respect to $\mu$.
Then $\nu = 0$. 
== Proof ==
If $\nu$ is a measure, we take $\cmod \nu = \nu$ for the sake of convenience. 
Since $\nu$ is mutually singular with respect to $\mu$, there exists a $\mu$-null set $N \subseteq X$ such that:
:$\map {\cmod \nu} {N^c} = 0$
Let $A \in \Sigma$.
We have:
:$\map {\cmod \nu} A = \map {\cmod \nu} {A \cap N} + \map {\cmod \nu} {A \cap N^c}$
From Null Sets Closed under Subset, we have:
:$\map \mu {A \cap N} = 0$ 
:$\map {\cmod \nu} {A \cap N^c} = 0$ 
Since $\nu$ is absolutely continuous with respect to $\mu$, we have:
:$\map {\cmod \nu} {A \cap N} = 0$ 
Hence we have $\map {\cmod \nu} A = 0$. 
From Characterization of Null Sets of Variation of Signed Measure and Characterization of Null Sets of Variation of Complex Measure, we have $\map \nu A = 0$. 
Since $A \in \Sigma$ was arbitrary we have $\nu = 0$. 
Category:Mutually Singular Measures
Category:Absolutely Continuous Measures
Category:Signed Measures
Category:Complex Measures
Singular Random Variable is not Absolutely Continuous

== Theorem ==
Let $\struct {\Omega, \Sigma, \Pr}$ be a probability space. 
Let $X$ be a singular random variable on $\struct {\Omega, \Sigma, \Pr}$.
Then $X$ is not absolutely continuous. 
 
== Proof ==
Let $P_X$ be the probability distribution of $X$. 
Let $\map \BB \R$ be the Borel $\sigma$-algebra on $\R$.
Let $\lambda$ be the Lebesgue measure for $\struct {\R, \map \BB \R}$. 
From the definition of an absolutely continuous random variable, we have that $X$ is absolutely continuous :
:$P_X$ is absolutely continuous with respect to $\lambda$. 
That is: 
:for all $A \in \map \BB \R$ with $\map \lambda A = 0$ we have $\map {P_X} A = 0$.
where $\map \BB \R$ is the Borel $\sigma$-algebra on $\R$.
Since $X$ is singular:
:there exists $B \in \map \BB \R$ with $\map \lambda B = 0$ and $\map \Pr {X \in B} = 1$.
That is, from the definition of probability distribution: 
:there exists $B \in \map \BB \R$ with $\map \lambda B = 0$ and $\map {P_X} B = 1$.
So:
:for all $A \in \map \BB \R$ with $\map \lambda A = 0$ we have $\map {P_X} A = 0$.
does not hold and so $X$ is not absolutely continuous.
Category:Singular Random Variables
Singular Value Decomposition/Examples/Arbitrary Example 1

== Example of Singular Value Decomposition ==
Let $\mathbf A$ be the matrix specified as:
:$\mathbf A := \begin {pmatrix} 2/3 & 0 \\ 5/6 & 1/2 \\ 1/3 & 1 \end {pmatrix}$
Then $\mathbf A$ has the following singular value decomposition:
:$\begin {pmatrix} 2/3 & 0 \\ 5/6 & 1/2 \\ 1/3 & 1 \end {pmatrix} = \begin {pmatrix} 1/3 & 2/3 & 2/3 \\ 2/3 & 1/3 & -2/3 \\ 2/3 & -2/3 & 1/3 \end {pmatrix} \begin {pmatrix} \sqrt 2 & 0 \\ 0 & 1 / \sqrt 2 \\ 0 & 0 \end {pmatrix} \begin {pmatrix} 1 / \sqrt 2 & 1 / \sqrt 2 \\ 1 / \sqrt 2 & -1 / \sqrt 2 \end {pmatrix}$
== Sources ==
* 
Category:Examples of Singular Value Decompositions
Six Regular 4-Dimensional Polytopes

== Theorem ==
There exist exactly six $4$-dimensional regular polytopes:
:the pentatope (also known as the $4$-simplex)
:the tesseract
:the $16$-cell
:the $24$-cell
:the $120$-cell
:the $600$-cell
== Proof ==
== Sources ==
* : $6$
* : $6$
* 
* 
Category:Polytopes
Category:6
Category:4
Sixth Apotome/Example

== Example of Sixth Apotome==
Let $a, b \in \set {x \in \R_{>0} : x^2 \in \Q}$ be two rationally expressible numbers such that $a - b$ is an apotome.
By definition, $a - b$ is a sixth apotome :
:$(1): \quad a \notin \Q$
:$(2): \quad b \notin \Q$
:$(3): \quad \dfrac {\sqrt {a^2 - b^2}} a \notin \Q$
where $\Q$ denotes the set of rational numbers.
Let $a = \sqrt 7$ and $b = \sqrt 5$.
Then:
Therefore $\sqrt 7 - \sqrt 5$ is a sixth apotome.
Category:Examples of Apotome
Sixth Power as Sum of 7 Sixth Powers

== Theorem ==
The smallest known integer whose $6$th power can be expressed as the sum of $7$ smaller $6$th powers is $1141$:
:$1141^6 = 74^6 + 234^6 + 402^6 + 474^6 + 702^6 + 894^6 + 1077^6$
== Proof ==
== Sources ==
* : $1141$
* : $1141$
Category:Sixth Powers
Category:1141
Size of Complete Graph

== Theorem ==
Let $K_n$ denote the complete graph of order $n$ where $n \ge 0$.
The size of $K_n$ is given by:
:$\size {K_n} = \dfrac {n \paren {n - 1} } 2$
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$\size {K_n} = \dfrac {n \paren {n - 1} } 2$
First we explore the degenerate case $\map P 0$:
Thus $\map P 0$ is seen to hold.
=== 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:
:$\size {K_k} = \dfrac {k \paren {k - 1} } 2$
from which it is to be shown that:
:$\size {K_{k + 1} } = \dfrac {\paren {k + 1} k} 2$
=== Induction Step ===
This is the induction step:
Let $K_{k + 1}$ be constructed by adding a new vertex $v_{k + 1}$ to the complete graph $K_k$.
To do so, it is necessary to add an edge to join $v_{k + 1}$ to every vertex of $K_k$.
Thus there are a total of $k$ edges more in $K_{k + 1}$ than there are in $K_k$.
So:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: \size {K_n} = \dfrac {n \paren {n - 1} } 2$
Category:Complete Graphs
Category:Proofs by Induction
Size of Conjugacy Class is Index of Normalizer

== Theorem ==
Let $G$ be a group.
Let $x \in G$.
Let $\conjclass x$ be the conjugacy class of $x$ in $G$.
Let $\map {N_G} x$ be the normalizer of $x$ in $G$.
Let $\index G {\map {N_G} x}$ be the index of $\map {N_G} x$ in $G$.
The number of elements in $\conjclass x$ is $\index G {\map {N_G} x}$.
== Proof ==
The number of elements in $\conjclass x$ is the number of conjugates of the set $\set x$.
From Number of Distinct Conjugate Subsets is Index of Normalizer, the number of distinct subsets of a $G$ which are conjugates of $S \subseteq G$ is $\index G {\map {N_G} S}$.
The result follows.
== Sources ==
* : $1$: Subgroups: $1. \text{T}.3$
* : Chapter $2$: Conjugacy, Normal Subgroups, and Quotient Groups: $\S 51$
Category:Conjugacy Classes
Category:Normalizers
Size of Graph/Examples/Arbitrary Order 4

== Example of Size of Graph ==
Let $G = \struct {V, E}$ be the graph defined as:
:$V = \set {v_1, v_2, v_3, v_4}$.
:$E = \set {\set {\tuple {v_1, v_2}, \tuple {v_2, v_1} }, \set {\tuple {v_1, v_3}, \tuple {v_3, v_1} }, \set {\tuple {v_2, v_3}, \tuple {v_3, v_2} }, \set {\tuple {v_3, v_4}, \tuple {v_4, v_3} } }$
The size of $G$ is the cardinality of $E$:
:$\card E = 4$
== Sources ==
* : Chapter $1$: Mathematical Models: $\S 1.3$: Graphs
Category:Examples of Size of Graph
Size of Graph/Examples/Order 3 Graphs

== Theorem ==
Let $G$ be a simple graph of order $3$.
Then it is possible for $G$ to have a size of $0$, $1$, $2$ or $3$.
Examples of each are presented below:
:
== Sources ==
* : Chapter $1$: Mathematical Models: $\S 1.3$: Graphs: Problem $17$
Category:Examples of Size of Graph
Size of Linearly Independent Subset is at Most Size of Finite Generator

== Theorem ==
Let $R$ be a division ring.
Let $V$ be an $R$-vector space.
Let $F \subseteq V$ be a finite generator of $V$ over $R$.
Let $L \subseteq V$ be linearly independent over $R$.
Then:
:$\size L \le \size F$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Vector Spaces
Category:Size of Linearly Independent Subset is at Most Size of Finite Generator
Size of Linearly Independent Subset is at Most Size of Finite Generator/Proof 1

== Theorem ==
== Proof ==
We first consider the case where $L$ is finite.
Let $S \subseteq \N$ be the set of all $n \in \N$ such that:
:For every finite generator $F$ of $V$, if $\card {L \setminus F} \le n$, then $\card L \le \card F$
where:
:$L \setminus F$ denotes the set difference between $L$ and $F$
:$\card L$ and $\card F$ denote the cardinality of $L$ and $F$ respectively.
We use the Principle of Finite Induction to prove that $S = \N$.
=== Basis of the Induction ===
Let $\card {L \setminus F} \le 0$.
Then from Cardinality of Empty Set:
:$L \setminus F = \O$
By Set Difference with Superset is Empty Set:
:$L \subseteq F$
By Cardinality of Subset of Finite Set:
:$\card L \le \card F$
Hence:
:$0 \in S$
This is the basis for the induction.
=== Induction Hypothesis ===
It is to be shown that if $k \in S$ where $k \ge 1$, then it follows that $k + 1 \in S$.
This is the induction hypothesis:
:For every finite generator $F$ of $V$, if $\card {L \setminus F} \le k$, then $\card L \le \card F$
It is to be demonstrated that it follows that:
:For every finite generator $F$ of $V$, if $\card {L \setminus F} \le k + 1$, then $\card L \le \card F$
=== Induction Step ===
This is the induction step:
Assume the induction hypothesis that $n \in S$.
Let $F$ be a finite generator of $V$ such that:
:$\card {L \setminus F} = n + 1$
Let $v \in L \setminus F$.
Let $L' = L \cap \paren {F \cup \set v}$.
By Intersection is Subset:
:$L' \subseteq L$
By Subset of Linearly Independent Set is Linearly Independent, it follows that $L'$ is linearly independent over $R$.
Also by Intersection is Subset:
:$L' \subseteq F \cup \set v$
Therefore, by Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set:
:there exists a basis $B$ of $V$ such that:
::$L' \subseteq B \subseteq F \cup \set v$
Since $v \notin F$ is a linear combination of $F$, it follows that $F \cup \set v$ is linearly dependent over $R$.
Therefore:
:$B \subsetneq F \cup \set v$
By Cardinality of Subset of Finite Set:
:$\card B < \card {F \cup \set v} = \card F + 1$
Hence:
:$\card B \le \card F$
We have that:
Since $n \in S$:
:$\card L \le \card B \le \card F$
Hence:
:$n + 1 \in S$
and so the induction step has been completed.
By Set Difference is Subset:
:$L \setminus F \subseteq L$
From Subset of Finite Set is Finite:
:$L \setminus F$ is finite.
Therefore, we can apply the fact that $S = \N$ to conclude that:
:$\card L \le \card F$
Let $L$ be infinite.
Then by Set is Infinite iff exist Subsets of all Finite Cardinalities, there exists a finite subset $L' \subseteq L$ such that:
:$\card {L'} = \card F + 1$
By Subset of Linearly Independent Set is Linearly Independent, it follows that $L'$ is linearly independent over $R$.
It is proven above that this is impossible.
Hence the result.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 4$. Vector Spaces: Theorem $4.1$
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.9$
Category:Size of Linearly Independent Subset is at Most Size of Finite Generator
Category:Proofs by Induction
Size of Linearly Independent Subset is at Most Size of Finite Generator/Proof 2

== Theorem ==
== Proof ==
Let $S \subseteq \N$ be the set of all natural numbers $n \in \N$ such that:
:For any finite generator $F$ of $V$ over $R$, if $\card {F \cap L} \ge n$, then $\card L \le \card F$.
It is to be demonstrated that $S = \N$.
That is, that $\card {F \cap L} \ge n \implies \card L \le \card F$ for all $n \in \N$.
By Intersection is Subset and Cardinality of Subset of Finite Set, we have $\card {F \cap L} \le \card F$.
Hence, it is vacuously true that $\card F + 1 \in S$.
Therefore, $S$ is non-empty.
From the well-ordering principle, $S$ has a smallest element $N$.
If $N = 0$, the theorem immediately follows.
 $N \ge 1$.
Let $\card {F \cap L} \ge N - 1$.
If $L \subseteq F$, the theorem follows from Cardinality of Subset of Finite Set.
Otherwise, there exists a $v \in L$ such that $v \notin F$.
Let $F' = F \cup \set v$.
By Intersection is Subset, we have $F' \cap L \subseteq L$, so it follows by Subset of Linearly Independent Set is Linearly Independent that $F' \cap L$ is linearly independent over $R$.
Also, by Intersection is Subset:
:$F' \cap L \subseteq F'$
We have that $F'$ is a generator of $V$ over $R$.
By Vector Space has Basis Between Linearly Independent Set and Finite Spanning Set, there exists a basis $B$ for $V$ such that:
:$F' \cap L \subseteq B \subseteq F'$
Since $v \notin F$ is a linear combination of $F$, it follows that $F'$ is linearly dependent over $R$.
By the definition of a basis:
:$B \subsetneq F'$
By Cardinality of Subset of Finite Set and Cardinality is Additive Function:
:$\card B < \card {F'} = \card F + 1$
Hence:
:$\card B \le \card F$
We have that:
Since $N \in S$, it follows by the definition of a basis that:
:$\card L \le \card B \le \card F$
Hence:
:$N - 1 \in S$
But this contradicts the assumption that $N$ is the smallest element of $S$.
Therefore:
:$N = 0$
Hence the result.
Category:Size of Linearly Independent Subset is at Most Size of Finite Generator
Size of Linearly Independent Subset is at Most Size of Finite Generator/Proof 3

== Theorem ==
== Proof ==
Let $\alpha_1, \alpha_2, \ldots, \alpha_n$ be a generator of $V$.
Let $\xi_1, \xi_2, \ldots, \xi_r$ be a linearly independent set of elements of $V$.
Hence the sequence $\sequence {\xi_1, \alpha_1, \alpha_2, \ldots, \alpha_n}$ is a linearly dependent sequence of elements of $V$.
One of these elements, which cannot be $\xi_1$, is a linear combination of the preceding elements.
Let this elements be $\alpha_i$.
So we can omit $\alpha_i$ from that sequence, and the remaining set is still a generator of $V$.
Therefore $\xi_2$ is a linear combination of these.
Thus $\sequence {\xi_2, \xi_1, \alpha_1, \alpha_2, \ldots, \alpha_{i - 1}, \alpha_{i + 1}, \ldots, \alpha_n}$ is a linearly dependent sequence of elements of $V$.
Again, one of them is a linear combination of the preceding elements.
This cannot be $\xi_2$, as $\xi_2$ has no preceding elements.
Neither can it be $\xi_1$, as $\xi_1$ and $\xi_2$ are linearly independent.
Thus we can omit whichever $\alpha_j$ it is, and we have a new set which is a generator of $V$.
This consists of $\xi_1$, $\xi_2$ and whichever $n - 2$ of the remaining elements of $\set {\alpha_1, \alpha_2, \ldots, \alpha_n}$.
After $p$ such steps, we have a set which is a generator of $V$ which consists of:
:$\xi_1, \xi_2, \ldots, \xi_p$ and $n - p$ of the elements of $\set {\alpha_1, \alpha_2, \ldots, \alpha_n}$.
 suppose $n < r$.
Then when $p = n$, the remaining set which is a generator of $V$ consists of:
:$\xi_1, \xi_2, \ldots, \xi_n$
and there is at least one more element $\xi_{n + 1}$.
This is a linear combination of $\set {\xi_1, \xi_2, \ldots, \xi_n}$.
But this contradicts the supposition that $\set {\xi_1, \xi_2, \ldots, \xi_n, \xi_{n + 1} }$ is a linearly independent set.
Hence, by Proof by Contradiction, $n \ge r$.
The result follows.
== Sources ==
* : Chapter $7$: Vector Spaces: $\S 33$. Definition of a Basis: Theorem $66$
Category:Size of Linearly Independent Subset is at Most Size of Finite Generator
Size of Regular Graph in terms of Degree and Order

== Theorem ==
Let $G = \struct {V, E}$ be an $r$-regular graph of order $p$.
Let $q$ denote the size of $G$.
Then:
:$q = \dfrac {p r} 2$
when such an $r$-regular graph exists.
If an $r$-regular graph of order $p$ does exist, then $p r$ is an even integer.
== Proof ==
Let $G$ be of order $p$, size $q$ and $r$-regular.
Then by definition $G$ has:
:$p$ vertices each of which is of degree $r$.
:$q$ edges.
Thus:
If $p r$ is odd, that means $p$ is odd and $r$ is odd.
That means there is an odd number of odd vertices.
From the corollary to the Handshake Lemma, this is not possible
Thus for $G$ to exist, it is necessary for $p r$ to be even.
The result follows.
== Sources ==
* : Chapter $2$: Elementary Concepts of Graph Theory: $\S 2.1$: The Degree of a Vertex: Problem $9$
Category:Regular Graphs
Size of Star Graph

== Theorem ==
Let $S_n$ denote the star graph of order $n$ where $n > 0$.
The size of $S_n$ is given by:
:$\size {S_n} = n - 1$
== Proof ==
By definition of star graph, one vertex is adjacent to all the $n - 1$ other vertices.
So the size of $S_n$ is at least $n - 1$.
All the other vertices are of degree $1$, and the edges they are incident with are the ones joining them to the distinguished vertex.
Those edges have already been counted.
So the size of $S_n$ is exactly $n - 1$.
Category:Star Graphs
Category:Star Graphs
Size of Surface of Regular Dodecahedron

== Theorem ==
== Proof ==
:
Let $ABCDE$ be the regular pentagon which is the face of a regular dodecahedron.
Let the circle $ABCDE$ be circumscribed around the pentagon $ABCDE$.
Let $F$ be the center of the circle $ABCDE$.
Let $FG$ be the perpendicular dropped from $F$ to $CD$.
Let $CF$ and $FD$ be joined.
Then:
where $\map \AA {ABCDE}$ denotes the area of the regular pentagon $ABCDE$.
The result follows from the definition of a regular dodecahedron as having $12$ such faces.
== Sources ==
* : The So-Called Book $\text{XIV}$, by Hypsicles
Category:Regular Dodecahedra
Size of Surface of Regular Icosahedron

== Theorem ==
== Proof ==
:
Let $ABC$ be the equilateral triangle which is the face of a regular icosahedron.
Let the circle $ABC$ be circumscribed around the equilateral triangle $ABC$.
Let $D$ be the center of the circle $ABC$.
Let $DE$ be the perpendicular dropped from $D$ to $BC$.
Let $BD$ and $CD$ be joined.
Then:
The result follows from the definition of a regular icosahedron as having $20$ such faces.
== Sources ==
* : The So-Called Book $\text{XIV}$, by Hypsicles
Category:Regular Icosahedra
Size of y-1 lt n and Size of y+1 gt 1 over n

== Theorem ==
Let $T_n \subseteq \R$ be the subset of the set of real numbers $\R$ defined as:
:$T_n = \set {y: \size {y - 1} < n \land \size {y + 1} > \dfrac 1 n}$
Then:
:$T_n = \openint {1 - n} {-1 - \dfrac 1 n} \cup \openint {-1 + \dfrac 1 n} {1 + n}$
== Proof ==
:
First note that:
We have:
Category:Unions of Families
Category:Intersections of Families
Sizes of Pyramids of Same Height with Polygonal Bases are as Bases

== Theorem ==
== Proof ==
:
Let there be two pyramids of the same height whose bases are the polygons $ABCDE$ and $FGHKL$ and whose apices are $M$ and $N$.
It is to be demonstrated that the ratio of $ABCDE$ to $FGHKL$ equals the ratio of pyramid $ABCDEM$ to pyramid $FGHKLN$.
Let $AC, AD, FH, FK$ be joined.
We have that $ABCM$ and $ACDM$ are tetrahedra of the same height.
So by :
:the ratio of $ABCM$ to $ACDM$ equals the ratio of $\triangle ABC$ to $\triangle ACD$.
From :
:$ABCD : \triangle ACD = ABCDM : ACDM$
But by :
:the ratio of $ACDM$ to $ADEM$ equals the ratio of $\triangle ACD$ to $\triangle ADE$.
Therefore from :
:$ABCD : \triangle ADE = ABCDM : ADEM$
Again from :
:$ABCDE : \triangle ADE = ABCDEM : ADEM$
Similarly it can be shown that:
:$FGHKL : \triangle FGH = FGHKLN : FGHN$
We have that $ADEM$ and $FGHN$ are two tetrahedra of the same height.
Therefore from :
:the ratio of $ADEM$ to $FGHN$ equals the ratio of $\triangle ADE$ to $\triangle FGH$.
But we have:
:$\triangle ADE : ABCDE = ADEM : BCDEM$
Therefore by :
:$ABCDE : \triangle FGH = ABCDEM : FGHN$
We also have:
:$\triangle FGH : FGHKL = FGHN : FGHKLN$
Further by :
:$ABCDE : FGHKL = ABCDEM : FGHKLN$
== Sources ==
* : Book $\text{XII}$. Propositions
Category:Pyramids