nyuuzyou/wikis huggingface dataset:
Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods/Lemma

== Theorem ==
Let $\struct {S, \tau}$ be a Hausdorff space.
Let $C$ be a compact subspace of $S$.
Let $x \in S \setminus C$.
Then there exist open sets $U$ and $V$ such that $x \in U$, $C \subseteq V$, and $U \cap V = \O$.
== Proof ==
Let $\FF$ be the set of all ordered pairs $\tuple {A, B}$ such that $A$ and $B$ are open, $x \in A$, and $A \cap B = \O$.
As a set of ordered pairs, $\FF$ constitutes a relation on $\tau$:
:$\FF \subseteq \tau \times \tau$
By the definition of Hausdorff space, for each $y \in C$ there exists an element $\tuple {A, B} \in \FF$ such that $y \in B$.
Thus the image of $\FF$ covers $C$.
By the definition of compactness, there exists a finite subset $\GG \subseteq \FF$ such that:
:$\ds C \subseteq V = \bigcup \Img \GG$
Then $\Preimg \GG$ is also finite, so by the definition of a topology:
:$\ds U = \bigcap \Preimg \GG$ is open.
Then $x \in U$, $C \subseteq V$, and $U \cap V = \O$.
Category:Compact Spaces
Category:Hausdorff Spaces
Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods/Proof 1

== Theorem ==
== Proof ==
Let $\FF$ be the set of all ordered pairs $\tuple {Z, W}$ such that:
:$Z, W \in \tau$
:$V_1 \subseteq Z$
:$Z \cap W = \O$
By the lemma, $\Img \FF$ covers $V_2$.
By the definition of compact space, there exists a finite subset $K$ of $\Img \FF$ which also covers $V_2$.
By the definition of topology, $\ds \bigcup K$ is open.
By the Principle of Finite Choice, there exists a bijection $\GG \subseteq \FF$ such that $\Img \GG = K$.
Then $\GG$, and hence its preimage, will be finite.
Let $\ds J = \bigcap \Preimg \GG$.
By Intersection is Largest Subset, $V_1 \subseteq J$.
By the definition of a topology, $J$ is open.
Then $\ds \bigcup K$ and $J$ are disjoint open sets such that $\ds V_2 \subseteq \bigcup K$ and $V_1 \subseteq J$.
Category:Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods
Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods/Proof 2

== Theorem ==
== Proof ==
We first suppose that $V_1 = \set x$ and $x \not \in V_2$. 
Since $T$ is Hausdorff:
:for each $y \in V_2$ there exists an open neighborhood $O^1_y$ of $x$ and an open neighborhood $O^2_y$ of $y$ such that $O^1_y \cap O^2_y = \O$.
Hence we have:
:$\ds V_2 \subseteq \bigcup_{y \mathop \in V_2} O_y^2$ 
Since $V_2$ is compact, there exists $y_1, \ldots, y_n \in V_2$ such that:
:$\ds V_2 \subseteq \bigcup_{i \mathop = 1}^n O_{y_i}^2$
Now let:
:$\ds U_2 = \bigcup_{i \mathop = 1}^n O_{y_i}^2$
and:
:$\ds U_1 = \bigcap_{i \mathop = 1}^n O_{y_i}^1$
We have that $U_1$ and $U_2$ are open, $x \in U_1$ and $V_2 \subseteq U_2$. 
Further, if $x \in U_2$ then $x \in O_{y_i}^2$ for some $i$. 
Since $O_{y_i}^2 \cap O_{y_i}^1 = \O$, it follows that $x \not \in O_{y_i}^1$. 
Hence $x \not \in U_1$. 
So $U_1 \cap U_2 = \O$.
Hence $U_1$ and $U_2$ are disjoint open sets with $V_1 \subseteq U_1$ and $V_2 \subseteq U_2$. 
Hence we have proven the claim in the case $V_1 = \set x$. 
Now take $V_1$ to be general.
For each $x \in V_1$, there exists open sets $U_{1, x}$ and $U_{2, x}$ such that $U_{1, x} \cap U_{2, x} = \O$, $x \in U_{1, x}$ and $V_2 \subseteq U_{2, x}$.
In particular:
:$\ds V_1 \subseteq \bigcup_{x \in V_1} U_{1, x}$
Since $V_1$ is compact, there exists $x_1, \ldots, x_n \in V_1$ such that:
:$\ds V_1 \subseteq \bigcup_{i \mathop = 1}^n U_{1, x_i}$
Now set:
:$\ds U_1 = \bigcup_{i \mathop = 1}^n U_{1, x_i}$
and:
:$\ds U_2 = \bigcap_{i \mathop = 1}^n U_{2, x_i}$
We have $U_1 \cap U_2 = \O$ by an identical argument to above. 
Further, $U_1$ and $U_2$ are both open.
Since $V_2 \subseteq U_{2, x_i}$ for each $i$ it follows that $V_2 \subseteq U_2$.
Hence $U_1$ and $U_2$ are the desired open sets.
Category:Disjoint Compact Sets in Hausdorff Space have Disjoint Neighborhoods
Disjoint Family of Sets/Examples/3 Arbitrary Sets

== Examples of Disjoint Families of Sets ==
Let $I = \set {1, 2, 3}$ be an indexing set.
Let:
Then the family of sets $\family {S_i}_{i \mathop \in I}$ is disjoint, but not pairwise disjoint.
== Proof ==
Let $S = {a, b, c}$, and so:
:$S_1, S_2, S_3 \subseteq S$
We have that:
Thus there is no element of $S$ which is also an element of all of $S_1$, $S_2$ and $S_3$.
That is:
:$\ds \bigcap_{i \mathop \in I} S_i = \set {x: \forall i \in I: x \in S_i} = \O$
That is:
:$\family {S_i}_{i \mathop \in I}$ is disjoint.
However, note that:
Thus it is noted that while $\family {S_i}_{i \mathop \in I}$ is disjoint, it is not the case that $\family {S_i}_{i \mathop \in I}$ is pairwise disjoint.
Category:Disjoint Families of Sets
Disjoint Independent Events means One is Void

== Theorem ==
Let $A$ and $B$ be events in a probability space.
Suppose $A$ and $B$ are:
:disjoint
:independent.
Then either $\map \Pr A = 0$ or $\map \Pr B = 0$.
That is, if two events are disjoint and independent, at least one of them can't happen.
== Proof ==
For $A$ and $B$ to be independent:
:$\map \Pr {A \cap B} = \map \Pr A \, \map \Pr B$
For $A$ and $B$ to be disjoint:
:$\map \Pr {A \cap B} = 0$
So:
:$\map \Pr A \, \map \Pr B = 0$
Hence the result.
== Comment ==
If this makes you scratch your head in bewilderment, consider what it means.
For $A$ and $B$ to be independent, then whether one event occurs has no bearing on whether the other one occurs or not.
But suppose we know that they are disjoint.
That is, when one of them happens, the other can't happen.
Then they can't be independent, because the occurrence of one has a direct relation to the happening (or in this case, non-happening) of the other.
== Sources ==
* : $\S 1.7$: Independent Events: Exercise $21$
Category:Probability Theory
Disjoint Permutations Commute

== Theorem ==
Let $S_n$ denote the symmetric group on $n$ letters.
Let $\rho, \sigma \in S_n$ such that $\rho$ and $\sigma$ are disjoint.
Then $\rho \sigma = \sigma \rho$.
== Proof ==
Let $\rho$ and $\sigma$ be disjoint permutations.
Let $i \in \Fix \rho$.
Then:
:$\map {\sigma \rho} i = \map \sigma i$
whereas:
:$\map {\rho \sigma} i = \map \rho {\map \sigma i}$
 $\map \sigma i \notin \Fix \rho$.
Then because $\sigma$ and $\rho$ are disjoint it follows that:
But it was previously established that $i \in \Fix \rho$.
This is a contradiction.
Therefore:
:$\map \sigma i \in \Fix \rho$
and so:
:$\map {\rho \sigma} i = \map \sigma i = \map {\sigma \rho} i$
Let $i \notin \Fix \rho$.
Then:
: $i \in \Fix \sigma$
and the same proof can be performed with $\rho$ and $\sigma$ exchanged.
== Sources ==
* : Appendix: Elementary set and number theory
* : Chapter $2$: The Symmetric Groups: $\S 79$
* : Chapter $2$: The Symmetric Groups: $\S 79 \gamma$
* : Chapter $9$: Permutations: Proposition $9.8$
Category:Disjoint Permutations
Category:Commutativity
Category:Symmetric Groups
Disjoint Union Preserves Domination

== Theorem ==
Let $I$ be an indexing set.
For all $i \in I$, let $A_i$ and $B_i$ be sets such that $A_i \preccurlyeq B_i$.
Here, $\preccurlyeq$ denotes domination.
Then:
:$\ds \bigsqcup_{i \mathop \in I} A_i \preccurlyeq \bigsqcup_{i \mathop \in I} B_i$
where $\bigsqcup$ denotes disjoint union.
== Proof ==
By definition of domination, for all $i \in I$, there exists an injection $\iota_i: A_i \to B_i$.
Thus the mapping $\ds \iota : \bigsqcup_{i \mathop \in I} A_i \to \bigsqcup_{i \mathop \in I} B_i$ defined by:
:$\map \iota {x, i} = \tuple {\map {\iota_i} x, i}$
is an injection.
Category:Set Theory
Disjoint Union is Coproduct in Category of Sets

== Theorem ==
Let $\mathbf{Set}$ be the category of sets.
Let $S$ and $T$ be sets.
Then their disjoint union $S \sqcup T$ is a coproduct in $\mathbf{Set}$.
== Proof ==
We have the implicit mappings $i_1: S \to S \sqcup T$ and $i_2: T \to S \sqcup T$ defined by:
:$i_1 \left({s}\right) = \left({s, 1}\right)$
:$i_2 \left({t}\right) = \left({t, 2}\right)$
Now given a set $V$ and mappings $f: S \to V$ and $g: T \to V$, there is to be a unique $\left[{f, g}\right]: S \sqcup T \to V$ such that:
:$f = \left[{f, g}\right] \circ i_1$
:$g = \left[{f, g}\right] \circ i_2$
Define $\left[{f, g}\right]$ by:
:$\left[{f, g}\right] \left({x, \delta}\right) := \begin{cases}
 f \left({x}\right) & \text{if $\delta = 1$} \\
 g \left({x}\right) & \text{if $\delta = 2$}\end{cases}$
Now it is immediate that $\left[{f, g}\right]$ so defined satisfies the two conditions above.
Furthermore, these conditions fix $\left[{f, g}\right]$ uniquely, since every $\left({x, \delta}\right) \in S \sqcup T$ has $\delta = 1$ or $\delta = 2$.
Hence the result, by definition of coproduct.
== Sources ==
* : $\S 3.2$: Example $3.4$
Category:Category of Sets
Disjunction/Examples/Arbitrary Example 1

== Example of Disjunction ==
The statement:
:Either $2 + 2 = 4$ or $6$ is an odd number
is a disjunction whose truth value is true, because one of the disjuncts is true .
The statement:
:Either $2 + 2 = 4$ or $6$ is not an odd number
is a disjunction whose truth value is true, because both of the disjuncts is true .
The statement:
:Either $2 + 2 = 5$ or $6$ is an odd number
is a disjunction whose truth value is false, because neither of the disjuncts is true .
== Sources ==
* 
* 
Category:Examples of Disjunctions
Disjunction/Examples/Waiver of Premiums

== Example of Disjunction ==
:Premiums will be waived in the event of sickness or unemployment
is a disjunction whose disjuncts are:
:Premiums will be waived in the event of sickness
:Premiums will be waived in the event of unemployment
the unspoken understanding being that in the event of both sickness and unemployment, premiums will also be waived.
== Sources ==
* : $2$ Arguments Containing Compound Statements: $2.1$: Simple and Compound Statements
Category:Examples of Disjunctions
Disjunction and Conditional

== Theorems ==
=== Modus Tollendo Ponens ===
=== Rule of Material Implication ===
Both of the above come in negative forms:
Disjunction is definable through implication:
:$p \lor q \dashv \vdash \paren {p \implies q} \implies q$
=== Alternative rendition ===
They can alternatively be rendered as:
They can be seen to be logically equivalent to the forms above.
== Proof ==
=== Comment ===
Note that this:
* $\neg \paren {\neg p \implies q} \dashv \vdash \neg \paren {p \lor q}$
can be proved in both directions without resorting to the Law of Excluded Middle.
All the others:
* $p \lor q \vdash \neg p \implies q$
* $\neg p \lor q \vdash p \implies q$
* $\neg \paren {p \implies q} \vdash \neg \paren {\neg p \lor q}$
are not reversible in intuitionistic logic.
== Proof by Truth Table ==
We apply the Method of Truth Tables to the propositions in turn.
As can be seen by inspection, in all cases the truth values under the main connectives match for all boolean interpretations.
$\begin {array} {|cccc||ccccc|} \hline
\neg & (p & \lor & q) & \neg & (\neg & p & \implies & q) \\
\hline
\T & \F & \F & \F & \T & \T & \F & \F & \F \\
\F & \F & \T & \T & \F & \T & \F & \T & \T \\
\F & \T & \T & \F & \F & \F & \T & \T & \F \\
\F & \T & \T & \T & \F & \F & \T & \T & \T \\
\hline
\end{array}$
$\begin {array} {|cccc||ccccc|} \hline
\neg & (p & \implies & q) & \neg & (\neg & p & \lor & q) \\
\hline
\F & \F & \T & \F & \F & \T & \F & \T & \F \\
\F & \F & \T & \T & \F & \T & \F & \T & \T \\
\T & \T & \F & \F & \T & \F & \T & \F & \F \\
\F & \T & \T & \T & \F & \F & \T & \T & \T \\
\hline
\end{array}$
Category:Disjunction
Category:Conditional
Disjunction has no Inverse

== Theorem ==
Let $\lor$ denote the disjunction operation of propositional logic.
Then there exists no binary logical connective $\circ$ such that:
:$(1): \quad \forall p, q \in \left\{{T, F}\right\}: \left({p \lor q}\right) \circ q = p$
== Proof ==
Let $q$ be true.
Then $p \lor q = T$, whatever truth value $p$ holds.
Either $T \circ T = T$ or $T \circ T = F$, but not both.
So if $q = T$ either:
:$\left({p \land q}\right) \circ q = T$
or:
:$\left({p \land q}\right) \circ q = F$
If the first, then $(1)$ does not hold when $p = F$.
If the second, then $(1)$ does not hold when $p = T$.
Hence there can be no such $\circ$.
Category:Disjunction
Disjunction in terms of NAND

== Theorem ==
:$p \lor q \dashv \vdash \paren {p \uparrow p} \uparrow \paren {q \uparrow q}$
where $\lor$ denotes logical disjunction and $\uparrow$ denotes logical NAND.
== Proof ==
== Sources ==
* : $\S 2.5$: Further Logical Constants
* : Chapter $1$: Sets and mappings: $\S 1.1$: The need for logic: Exercise $(5)$
Category:Logical NAND
Category:Disjunction
Disjunction of Conditional and Converse

== Theorem ==
Given any two statements, one of them implies the other.
:$\vdash \paren {p \implies q} \lor \paren {q \implies p}$
That is, given any conditional, either it is true or its converse is.
== Proof 1 ==
== Proof by Truth Table ==
== Also see ==
* Smullyan's Drinking Principle
* Paradoxes of Material Implication, in which category this result is sometimes grouped
== Sources ==
* : $\S \text{II}.13$: Symbolism of sentential calculus
* : $\S 1.2.1$: Rules for natural deduction: Exercises $1.5: \ 2 \ \text{(e)}$
Category:Disjunction of Conditional and Converse
Category:Conditional
Category:Disjunction
Disjunction of Conditional and Converse/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth values under the main connective is true for all boolean interpretations, proving a tautology.
$\begin{array}{|ccccccc|} \hline
(p & \implies & q) & \lor & (q & \implies & p) \\
\hline
\F & \T & \F & \T & \F & \T & \F \\
\F & \T & \T & \T & \T & \F & \F \\
\T & \F & \F & \T & \F & \T & \T \\
\T & \T & \T & \T & \T & \T & \T \\
\hline
\end{array}$
Category:Disjunction of Conditional and Converse
Disjunction of Conditionals

== Theorem ==
:$\vdash \paren {p \implies q} \lor \paren {q \implies r}$
== Proof ==
== Sources ==
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Theorem $\text{T58}$
* : $\S 1.2.1$: Rules for natural deduction: Exercises $1.5: \ 2 \ \text{(f)}$
Category:Disjunction
Category:Conditional
Disjunction of Finite State Machines

== Theorem ==
Let $\Sigma$ be a finite set.
Let $F_1 = \tuple {S_1, A_1, I_1, \Sigma, T_1}$ and $F_2 = \tuple {S_2, A_2, I_2, \Sigma, T_2}$ be finite state machines.
Then there exists a finite state machine $F$ such that $\map L F = \map L {F_1} \cup \map L {F_2}$.
== Proof ==
Construct $F = \tuple {S_1 \times S_2, A_1 \times S_2 \cup S_1 \times A_2, \tuple {I_1, I_2}, \Sigma, T}$ where:
:$\map T {s_1, s_2, \Sigma} = \tuple {\map {T_1} {s_1}, \map {T_2} {s_2} }$
The machines are simultaneously emulated in the first and second components of the state tuple.
When the input word is exhausted, $F$ accepts at least one of $F_1$ and $F_2$ is in an accepting state, as required.
Category:Abstract Machines
Disjunction with Contradiction/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations.
$\begin{array}{|c|ccc||c|ccc|} \hline
\bot & p & p & \lor & \bot \\
\hline
\F & \F & \F & \F & \F \\
\F & \T & \T & \T & \F \\
\hline
\end{array}$
Category:Disjunction with Contradiction
Disjunction with Tautology

== Theorem ==
A disjunction with a tautology:
: $p \lor \top \dashv \vdash \top$
== Proof by Natural Deduction ==
== Proof by Truth Table ==
== Sources ==
* : $\S 2.3.3$
Category:Disjunction
Category:Tautology
Category:Disjunction with Tautology
Disjunction with Tautology/Proof 2

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, in each case, the truth values in the appropriate columns match for all boolean interpretations.
$\begin{array}{|c|c||ccc|} \hline
p & \top & p & \lor & \top \\
\hline
F & T & F & T & T \\
T & T & T & T & T \\
\hline
\end{array}$
Category:Disjunction with Tautology
Disjunctive Normal Form/Examples/Arbitrary Example 2

== Examples of Disjunctive Normal Form ==
:$\paren {\neg p \land q \land r} \lor \paren {\paren {p \lor \neg q} \land r} \lor \paren {\neg r}$
is not in disjunctive normal form because there is a disjunction buried in the second conjunction.
Category:Examples of Disjunctive Normal Form
Disjunctive Normal Form/Examples/Arbitrary Example 4

== Examples of Disjunctive Normal Form ==
:$\paren {p \land q \land r \land \neg r} \lor \paren {q \land \neg q} \lor \paren {q \land p \land \neg p}$
is in disjunctive normal form.
It is immediate that the above forms a contradiction.
== Sources ==
* 
Category:Examples of Disjunctive Normal Form
Dispersion Point in Particular Point Space

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a particular point space.
Then $p$ is dispersion point of $T$.
== Proof ==
Let $H = S \setminus \set p$.
Let $T_H = \struct {H, \tau_H}$ be the topological subspace induced on $H$ by $\tau_p$.
From Particular Point Space less Particular Point is Discrete, the space $T_H$ is discrete.
We have Discrete Space is Locally Connected.
Thus from Totally Disconnected and Locally Connected Space is Discrete we have that $S \setminus \set p$ is totally disconnected.
Hence the result, from definition of dispersion point.
== Sources ==
* : Part $\text {II}$: Counterexamples: $8 \text { - } 10$. Particular Point Topology: $11$
Category:Dispersion Points
Category:Particular Point Topology
Dispersion Point of Excluded Point Space

== Theorem ==
Let $T = \struct {S, \tau_{\bar p} }$ be an excluded point space.
Then $p$ is a dispersion point of $T$.
== Proof ==
We have that the Excluded Point Topology is Open Extension Topology of Discrete Topology.
So $S \setminus \set p$ is a discrete space.
Then a discrete space is totally disconnected.
The result follows from definition of dispersion point.
== Sources ==
* : Part $\text {II}$: Counterexamples: $13 \text { - } 15$. Excluded Point Topology: $5$
Category:Dispersion Points
Category:Excluded Point Topology
Displacement of Particle under Force

== Theorem ==
Let $P$ be a particle of constant mass $m$.
Let the position of $P$ at time $t$ be specified by the position vector $\mathbf r$.
Let a force applied to $P$ be represented by the vector $\mathbf F$.
Then the motion of $P$ can be given by the differential equation:
:$\mathbf F = m \dfrac {\d^2 \mathbf r} {\d t^2}$
or using Newtonian notation:
:$\mathbf F = m \ddot {\mathbf r}$
== Proof ==
== Sources ==
* : Chapter $1$: Equilibrium: $1.1$ Introduction
Category:Classical Mechanics
Dissection of 2 Equal Squares and One Smaller Square into One Square

== Problem ==
Dissect $2$ equal squares and one smaller square, and use the pieces to make $1$ large square.
== Solution ==
:
Bisect the $2$ larger squares along their diagonals.
Arrange the right triangles so generated around the outside of the smaller square as shown.
Join the $4$ right angles of the $4$ right triangles.
The overlapping bits fit into the gaps.
== Sources ==
* : Abul Wafa (): $40$
Category:Dissections
Dissection of 2 Regular Hexagons into One

== Problem ==
Dissect $2$ regular hexagons, and use the pieces to make $1$ larger regular hexagon.
== Solution ==
:
Dissect one of the regular hexagons into $6$ equal triangles as shown on the left.
Arrange these triangles around the outside of the other regular hexagon as shown on the right.
Join the $6$ obtuse angles of those $6$ triangles to make a larger regular hexagon.
The overlapping bits are then cut off and are fitted into the gaps.
== Sources ==
* 
* : Abul Wafa (): $41$
Category:Dissections
Dissection of 2 Regular Hexagons into One/Mistake

== Source Work ==
: 
:The Puzzles:
::Abul Wafa (): $41$
:::Solution
== Mistake ==
:How can two regular hexagons, of different sizes, be dissected into seven pieces which fit together to make one, larger, regular hexagon?
== Correction ==
The number of pieces needed in the construction given is in fact $13$.
The smaller regular hexagon is $1$ piece.
The larger regular hexagon is cut into $6$ triangles.
Each of these triangles is then each cut into $2$ pieces which are then assembled around the outside of the smaller regular hexagon
Thus a total of $13$ pieces.
== Sources ==
* : Abul Wafa (): $41$
Category:Mistakes/Curious and Interesting Puzzles
Dissection of 3 Equal Squares into One Square

== Problem ==
Dissect $3$ equal squares and use the pieces to make $1$ large square.
== Solution ==
:
Bisect $2$ of the squares along their diagonals.
Arrange the right triangles so generated around the outside of the $3$rd square as shown.
Join the $4$ right angles of the $4$ right triangles.
The overlapping bits fit into the gaps.
== Tessellation ==
== Historical Note ==
== Sources ==
* : Abul Wafa (): $39$
Category:Dissections
Category:Dissection of 3 Equal Squares into One Square
Dissection of 3 Equal Triangles and 1 Smaller Similar Triangle into One

== Problem ==
Dissect $3$ equal triangles and $1$ smaller triangle, similar to those $3$, and assemble the pieces into one large triangle.
== Solution ==
:
Note that the larger triangle is not similar to its components.
== Sources ==
* 
* : Abul Wafa (): $42$
Category:Similar Triangles
Category:Dissections
Dissection of Polygon into Triangles with Chords

== Theorem ==
The number of different ways $k$ a convex $n$-sided polygon can be divided into triangles using chords, not counting reflections and rotations as different, is given for the first few $n$ as follows:
:
$n$ 	$k$
$3$ 	$1$
$4$ 	$1$
$5$ 	$1$
$6$ 	$3$
$7$ 	$4$
$8$ 	$12$
$9$ 	$27$
$10$	$82$
$11$	$228$
$12$	$733$
$13$	$2282$
== Proof ==
== Sources ==
* : $27$
Category:Polygons
Category:Dissections
Dissection of Polygon into Triangles with Chords counting Isometries

== Theorem ==
The number of different ways $k$ a convex $n$-sided polygon can be divided into triangles using chords, counting reflections and rotations as different, is given for the first few $n$ as follows:
:
$n$ 	$k$
$3$ 	$1$
$4$ 	$2$
$5$ 	$5$
$6$ 	$14$
$7$ 	$42$
$8$ 	$132$
$9$ 	$429$
$10$	$1430$
$11$	$4862$
These are the Catalan numbers.
== Proof ==
== Sources ==
* : $42$
* : $27$
* : $42$
* 
Category:Polygons
Category:Dissections
Category:Catalan Numbers
Dissection of Rectangle into 9 Distinct Integral Squares

== Theorem ==
Let $R$ be a rectangle.
Let $R$ be divided into $n$ squares which all have different lengths of sides.
Then $n \ge 9$.
The smallest rectangle with integer sides that can be so divided into squares with integer sides is $32 \times 33$.
:
== Proof ==
== Sources ==
* : $9$
* : $9$
Category:Recreational Mathematics
Category:Rectangles
Category:Squares
Category:9
Category:Dissections
Dissection of Square into 8 Acute Triangles

== Theorem ==
A square can be dissected into $8$ acute triangles.
== Construction ==
Let $\Box ABCD$ be a square.
:
Let semicircles be constructed on $AD$ and $BC$ so as to meet in the center of $\Box ABCD$.
Let $E$ and $F$ be the midpoints of $CD$ and $AB$ respectively.
Let semicircles be constructed on $DE$ and $EC$
Let $G$ and $H$ be points on opposite sides of the straight line $EF$ such that both $G$ and $H$ are within the region bordered by the $4$ semicircles.
Then $\triangle AGF$, $\triangle FGH$, $\triangle HFB$, $\triangle BHC$, $\triangle CHE$, $\triangle EHG$, $\triangle DEG$, $\triangle DGA$ are $8$ acute triangles which compose $\Box ABCD$.
== Proof ==
Category:Squares
Category:Dissections
Category:Acute Triangles
Distance-Preserving Image Isometric to Domain for Metric Spaces

== Theorem ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $\phi: M_1 \to M_2$ be a distance-preserving mapping.
Then:
:$\phi: M_1 \to \Img \phi$
is an isometry.
== Proof ==
Let $M_1 = \struct {A_1, d_1}$ and $M_2 = \struct {A_2, d_2}$ be metric spaces.
Let $\phi$ be a distance-preserving mapping from $M_1$ to $M_2$.
Let $A = \Img \phi$ be the image of $\phi$.
By Subspace of Metric Space is Metric Space, $\struct {A, d_2}$ is a metric space.
As $\phi$ is a distance-preserving mapping, by Distance-Preserving Mapping is Injection of Metric Spaces it is injective.
From Restriction of Injection is Injection, $\phi: M_1 \to \Img \phi$ is an injection.
From Restriction of Mapping to Image is Surjection, $\phi: M_1 \to \Img \phi$ is a surjection.
Thus $\phi \to \Img \phi$ is by definition a bijection.
Thus $\phi: M_1 \to \Img \phi$ is a bijective distance-preserving mapping.
Hence, by definition, $\phi: M_1 \to \Img \phi$ is an isometry.
Category:Isometries (Metric Spaces)
Category:Metric Spaces