nyuuzyou/wikis huggingface dataset:
Relative Complement of Relative Complement

== Theorem ==
The relative complement of the relative complement of a set is itself:
:$\relcomp S {\relcomp S T} = T$
Thus, considered as a mapping on the power set of $S$:
:$\complement_S: \powerset S \to \powerset S: \relcomp S T = S \setminus T$
$\complement_S$ is an involution.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 5$: Complements and Powers
* : A Note on Set-Theoretic Concepts: $(4)$
* : Chapter $1$: Set Theory: $1.3$: Set operations
* : $\S 1.6$. Difference and complement
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Theorem $3.2$
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems: $\text{(i)}$
* : $\S 1$: Naive Set Theory: $\S 1.2$: Operations on Sets: Exercise $1.2.2 \ \text{(i)}$
Category:Relative Complement
Category:Relative Complement of Relative Complement
Relative Complement of Relative Complement/Proof 1

== Theorem ==
== Proof ==
By the definition of relative complement:
:$\relcomp S {\relcomp S T} = S \setminus \paren {S \setminus T}$
Let $t \in T$.
Then by the definition of set difference:
:$t \notin S \setminus T$
Since $t \in T$ and $T \subseteq S$, by the definition of subset:
:$t \in S$
Thus:
:$t \in \paren {S \setminus \paren {S \setminus T} }$
Suppose instead that:
:$t \in \paren {S \setminus \paren {S \setminus T} }$
Then:
:$t \in S$
and:
:$\neg \paren {t \in \paren {S \setminus T} }$
Thus:
:$\neg \paren {\paren {t \in S} \land \neg \paren {t \in T} }$
By Conditional is Equivalent to Negation of Conjunction with Negative:
:$t \in S \implies t \in T$
By Modus Ponendo Ponens:
:$t \in T$
Category:Relative Complement of Relative Complement
Relative Complement of Relative Complement/Proof 2

== Theorem ==
== Proof ==
The definition of the relative complement requires that:
:$T \subseteq S$
But from Intersection with Subset is Subset‎:
:$T \subseteq S \iff T \cap S = T$
Thus:
:$\relcomp S {\relcomp S T} = T$
follows directly.
Category:Relative Complement of Relative Complement
Relative Complement with Self is Empty Set

== Theorem ==
The relative complement of a set in itself is the empty set:
:$\relcomp S S = \O$
== Proof ==
== Also see ==
* Relative Complement of Empty Set
== Sources ==
* : $\S 5$: Complements and Powers
* : A Note on Set-Theoretic Concepts
* : Chapter $1$: Set Theory: $1.3$: Set operations
* : Chapter $\text I$: Algebraic Structures: $\S 3$: Unions and Intersections of Sets: Theorem $3.2$
* : $\S 2$: Sets and functions: Sets
* : $\S 1$. Sets; inclusion; intersection; union; complementation; number systems
Category:Relative Complement
Category:Empty Set
Relative Cumulative Frequency/Examples/Arbitrary Example 1

== Example of Relative Cumulative Frequency ==
Consider the sample:
:$2, 5, 3, 3, 3, 5, 3, 6, 2, 3, 9, 5$
There are $12$ observations in this sample.
The relative cumulative frequency of the observation $5$ is $\dfrac {10} {12} = \dfrac 5 6$.
== Sources ==
* 
* 
Category:Examples of Relative Cumulative Frequency
Relative Difference between Infinite Set and Finite Set is Infinite

== Theorem ==
Let $S$ be an infinite set.
Let $T$ be a finite set.
Then $S \setminus T$ is an infinite set.
== Proof ==
 $S \setminus T$ is a finite set.
Then:
By Union of Finite Sets is Finite, $S \cup T$ is a finite set.
By Subset of Finite Set is Finite, $S$ must also be a finite set.
But $S$ is an infinite set.
This is a contradiction.
Therefore $S \setminus T$ is an infinite set.
Category:Set Difference
Category:Infinite Sets
Relative Efficiency/Examples/Arbitrary Example 1

== Example of Relative Efficiency ==
Let $T_1$ and $T_2$ be estimators of a parameter $\theta$.
Let $n_1$ and $n_2$ respectively be the sample sizes needed to give $T_1$ and $T_2$ the same variance.
Let $n_1 = 10$ and $n_2 = 20$.
Then the relative efficiency of $T_1$ $T_2$ is $\dfrac {20} {10} = 2$.
Thus $T_1$ is twice as efficient as $T_2$.
== Sources ==
* 
* 
Category:Examples of Relative Efficiency
Relative Frequency/Examples/Arbitrary Example 1

== Example of Relative Frequency ==
Consider the sample:
:$2, 5, 3, 3, 3, 5, 3, 6, 2, 3, 9, 5$
There are $12$ observations in this sample.
The relative frequency of the observation $3$ is $\dfrac 5 {12}$.
The relative frequency of the observation $9$ is $\dfrac 1 {12}$.
== Sources ==
* 
* 
Category:Examples of Relative Frequency
Relative Frequency is Probability Measure

== Theorem ==
The relative frequency model is a probability measure.
== Proof ==
We check all the Kolmogorov axioms in turn:
=== First Axiom ===
Let $n$ be the number of times a certain event $\omega$ is observed to happen.
Let $n'$ be the number of times $\omega$ is observed not to happen.
By Law of Excluded Middle and Principle of Non-Contradiction, $\omega$ either happened or did not, and not both.
Therefore $n + n'$ is the total number of observations.
It is supposed that at least one observation is actually made, so that $n + n' \ne 0$.
The relative frequency model says that the probability of $\omega$ occurring can be defined as:
:$\map \Pr \omega = \dfrac n {n + n'}$
the numerator and denominator of which are positive integers such that $n + n' \ge n$.
If $\omega$ is observed never to happen, then $n = 0$ and $\map \Pr \omega = \dfrac 0 {0 + n'} = 0$. 
If $\omega$ is observed to always happen, then $n' = 0$ and $\map \Pr \omega = \dfrac n {n + 0} = 1$.
Otherwise, if $n, n' \ne 0$, from Mediant is Between:
:$0 = \dfrac 0 {0 + n'} < \dfrac n {n + n'} < \dfrac n {n + 0} = 1$
Thus $\Pr$ is bounded:
:$0 \le \map \Pr {\cdot} \le 1$
=== Second Axiom ===
By hypothesis:
:$\map \Pr \Omega = \dfrac {n + n'} {n + n'} = 1$
=== Third Axiom ===
This is a proof by induction.
==== Basis for the Induction ====
The case $j = 2$ is verified as follows:
Let $A$ and $B$ be two pairwise disjoint events.
Let $p$ and $q$ be the number of times $A$ and $B$ have been observed, respectively.
Let $n$ be the total number of trials observed. 
By the definition of pairwise disjoint, $A$ and $B$ never happened at the same time.
By the same reasoning as the proof for the first axiom, in all $n$ observations:
:$A$ happened $p$ times
:$B$ happened $q$ times
:$A \lor B$ happened $p + q$ times.
By hypothesis:
This is the basis for the induction.
==== Induction Hypothesis ====
Let $A_1, A_2, \ldots, A_j$ be $j$ pairwise disjoint events.
By the definition of the relative frequency model, $j$ is finite.
Assume:
:$\ds \map \Pr {\bigcup_{i \mathop = 1}^j A_i} = \map \Pr {A_1} + \map \Pr {A_2} + \cdots + \map \Pr {A_j}$
This is our induction hypothesis.
==== Induction Step ====
This is our induction step:
Let $A_1, A_2, \ldots, A_j, A_{j + 1}$ be $j+1$ pairwise disjoint events.
Define $C = A_1 \lor A_2 \lor A_3 \lor \cdots \lor A_j$.
Then $C$ and $A_{j + 1}$ are also pairwise disjoint.
By the base case: 
:$\map \Pr {C \cup A_{j + 1} } = \map \Pr C + \map \Pr {A_{j + 1} }$
By the definition of $C$, this equation is logically equivalent to:
:$\ds \map \Pr {\bigcup_{i \mathop = 1}^{j + 1} A_i} = \sum_{i \mathop = 1}^{j + 1} \map \Pr {A_i}$
By the definition of the relative frequency model, $j + 1$ is finite.
The result follows by the Principle of Mathematical Induction.
== Sources ==
* : $\S 1.3, \ \S 1.4$
Category:Probability Theory
Category:Proofs by Induction
Category:Statistics
Relative Homotopy is Equivalence Relation

== Theorem ==
Let $X$ and $Y$ be topological spaces.
Let $K \subseteq X$ be a (possibly empty) subset of $X$.
Let $\map \CC {X, Y}$ be the set of all continuous mappings from $X$ to $Y$.
Define a relation $\sim$ on $\map \CC {X, Y}$ as:
:$f \sim g$ $f$ and $g$ are homotopic relative to $K$.
Then $\sim$ is an equivalence relation.
== Proof ==
We examine each condition for equivalence.
=== Reflexivity ===
For any function $f: X \to Y$, define $H: X \times \closedint 0 1 \to Y$ by $\map H {x, t} := \map f x$.
This yields a homotopy between $f$ and itself.
Also, trivially, if $x \in K$ and $t \in \closedint 0 1$, then:
:$\map f x = H \left({x, t}\right)$
so that $H$ is a homotopy relative to $K$.
Thus $\sim$ is a reflexive relation.
=== Symmetry ===
Given a $K$-relative homotopy:
:$H: X \times \closedint 0 1 \to Y$ 
from $\map f x = \map H {x, 0}$ to $\map g x = \map H {x, 1}$, the function:
:$\map G {x, t} = \map H {x, 1 - t}$
is a $K$-relative homotopy from $g$ to $f$.
Thus $\sim$ is a symmetric relation.
=== Transitivity ===
Suppose that $f \sim g$ and $g \sim h$.
Let $F, G: X \times \closedint 0 1 \to Y$ be $K$-relative homotopies between $f$ and $g$, $g$ and $h$, respectively.
Define $H: X \times \closedint 0 1 \to Y$ by:
:$\map H {x, t} := \begin {cases} \map F {x, 2 t} & : 0 \le t \le \dfrac 1 2 \\ \map G {x, 2 t - 1} & : \dfrac 1 2 \le t \le 1 \end{cases}$
By Continuous Mapping on Finite Union of Closed Sets, $H$ is a $K$-relative homotopy between $f$ and $h$.
Thus $\sim$ is a transitive relation.
Having verified all three conditions, it follows that $\sim$ is an equivalence relation.
== Sources ==
* 
* 
Category:Homotopy Theory
Relative Lengths of Chords of Circles

== Theorem ==
Of chords in a circle, the diameter is the greatest, and of the rest the nearer to the center is always greater than the more remote.
== Proof ==
:
Let $ABCD$ be a clrcle, let $AD$ be its diameter and $E$ the center.
Let $BC$ and $FG$ be chords of $ABCD$, where $BC$ is nearer to the center than $FG$.
Let $EH$ and $EK$ be drawn perpendicular to $BC$ and $FG$ respectively.
Because $BC$ is nearer to the center than $FG$, it follows from Book III: Definition 5 that $EK$ is longer than $EH$.
Position $L$ on $EK$ so that $EL = EH$, and draw $MN$ through $L$ perpendicular to $FG$.
Join $EM$, $EN$, $EF$ and $EG$.
Since $EH = EL$ it follows from Equal Chords in Circle that $BC = MN$.
Also, since $AE = EM$ and $ED = EN$, $AD = ME + EN$.
But $ME + EN > MN$ from Sum of Two Sides of Triangle Greater than Third Side, and $MN = BC$.
So $AD > BC$.
Thus we have that the diameter is greater than any other chord of a circle.
Since $ME = FE$ and $EN = EG$ and $\angle MEN > \angle FEG$, from the Hinge Theorem it follows that $MN > FG$.
But $MN = BC$.
So a chord nearer the center is greater than one more remote.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Circles
Relative Lengths of Lines Inside Circle

== Theorem ==
== Proof ==
:
Let $ABCE$ be a circle, and let $AE$ be a diameter of it.
Let $D$ be the center of the circle, and let $F$ be a point on $AE$ different from $D$.
Let $FB, FC, FG$ be drawn from $F$ to the edge of the circle.
Then $FA$ is the greatest, $FE$ is the least, and (according to the diagram above) of the others, $FB$ is greater than $FC$, while $FC$ is greater than $FG$.
The proof is as follows.
Join $BD, CD, GD$.
From Sum of Two Sides of Triangle Greater than Third Side, we have that $DB + DF > BF$.
But $AD = BD$, so $AF > BF$.
Since $BD = CD$ and $FD$ is common, $BD + DF = CD + CF$.
But $\angle BDF > \angle CDF$, so from the Hinge Theorem $BF > CF$.
For the same reason, $CF > GF$.
Again, since $GF + FD > DG$, and $DG = DE$, $GF + FD > DE$.
Subtract $FD$ from each, and we see that $GF > FE$.
Therefore $FA$ is the greatest, followed by, in order of size, $FB, FC, FG$ and finally $FE$ is the smallest.
----
Also, from the point $F$ only two equal straight lines fall on the circle $ABCE$, one on either side of the least line $FE$.
Construct the angle $\angle FDH$ equal to $\angle GDF$ on the opposite side of $DF$, and join $FH$.
Since $GD = DH$ and $DF$ is common, and $\angle FDH = \angle GDF$ from Triangle Side-Angle-Side Congruence, $\triangle FDH = \triangle GDF$, and so $FG = FH$.
Another straight line equal to $FG$ will not fall on the circle from $F$.
For if this is possible, let $FK$ be that straight line.
Since $FK = FG$, and $FH = FG$, it follows that $FK = FH$.
But this contradicts what we proved earlier.
Therefore only one such straight line can be constructed.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Circles
Relative Lengths of Lines Outside Circle

== Theorem ==
== Construction ==
:
Let $\bigcirc ABC$ be a circle.
Let $D$ be a point outside $\bigcirc ABC$.
Using , let $M$ be the center of $\bigcirc ABC$.
Let $E$, $F$, and $C$ be points on the concave circumference of $\bigcirc ABC$
Using , draw straight lines $DM$, $DC$, $DE$, and $DF$.
Using , extend $DM$ to $A$ on $\bigcirc ABC$.
Place $H$, $L$, $K$, and $G$ on the intersections of $\bigcirc ABC$ and $DC$, $DF$, $DE$, and $DA$ respectively.
Then:
:of the straight lines falling on the concave circumference $AEFC$, the line $DA$ through the center is the greatest, with $DE > DF > DC$
:of the straight lines falling on the convex circumference $HLKG$, the line $DG$ is the least, with $DK < DL < DH$.
== Proof ==
Using , join $ME$, $MF$, $MC$, $MH$, $ML$, and $MK$.
Since $AM = EM$, add $DM$ to each, so $AD = EM + MD$.
But from , we know that $EM + MD > ED$.
Hence it follows that:
:$AD > ED$
Since this is true for any $E$, $AD$ is the greatest line from $D$ to the concave circumference.
We have that:
:$EM = FM$
:$DM$ is common
:$\angle EMD > \angle FMD$.
Hence, from the , it follows that:
:$ED > FD$
Similarly we can show that:
:$FD > CD$
Thus:
:$DA > DE > DF > DC$
From , we know that:
:$MK + KD > MD$
Because $MG = MK$, we have:
:$KD > GD$.
Because $K$ is arbitrary, it follows that $GD$ is the least line from $D$ to the convex circumference.
We have that $K$ is inside $\triangle MLD$.
Hence from :
:$ML + LD > MK + KD$.
Then because $MK = ML$:
:$DK < DL$
Similarly we can show that $DL < DH$.
Thus:
:$DG < DK < DL < DH$
----
From the point $D$ only two equal straight lines fall on $\bigcirc ABC$: one on each side of the line $DA$.
From , construct the angle $\angle DMB = \angle KMD$ on the straight line $MD$ at the point $M$.
Using , join $DB$.
We have that:
:$MK = MB$
:$MD$ is common.
Hence from :
:$DK = DB$
Another straight line equal to $DK$ will not fall on $\bigcirc ABC$ from $D$.
For if this is possible, let $DN$ be this straight line.
Then $DN = DB$.
But from what we proved above, either:
:$DN > DB$
or:
:$DN < DB$
depending on where $N$ falls on $\bigcirc ABC$.
This is a contradiction.
Therefore the point $N$ cannot exist as described.
It follows that no more than two equal straight lines fall on $\bigcirc ABC$ from $D$: one on each side of $DA$.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Circles
Relative Likelihood of n Sixes on 6n Dice

== Problem ==
Which is more likely:
:to throw at least $1$ six with $6$ dice
:to throw at least $2$ sixes with $12$ dice
:to throw at least $3$ six with $18$ dice?
== Solution ==
The chance of getting $1$ six and $5$ other outcomes in a particular order is $\paren {\dfrac 1 6} \paren {\dfrac 5 6}^5$.
We need to multiply by the number of orders for $1$ six and $5$ non-sixes.
Therefore the probability of exactly $1$ six is:
:$\dbinom 6 1 \paren {\dfrac 1 6} \paren {\dfrac 5 6}^5$
Similarly, the probability of exactly $x$ sixes when $6$ dice are thrown is:
:$\dbinom 6 x \paren {\dfrac 1 6}^x \paren {\dfrac 5 6}^{6 - x}$
for $x = 0, 1, 2, 3, 4, 5, 6$.
The probability of $x$ sixes for $n$ dice is:
:$\dbinom n x \paren {\dfrac 1 6}^x \paren {\dfrac 5 6}^{n - x}$
for $x = 0, 1, 2, \ldots, n$.
according to the binomial distribution.
The probability of $1$ of more sixes with $6$ dice is the complement of the probability of $0$ sixes:
:$1 - \dbinom 6 0 \paren {\dfrac 1 6}^0 \paren {\dfrac 5 6}^6 \approx 0.665$
When $6 n$ dice are rolled, the probability of $n$ or more sixes is:
:$\ds \sum_{x \mathop = n}^{6 n} \dbinom {6 n} x \paren {\dfrac 1 6}^x \paren {\dfrac 5 6}^{6 n - x}$
which equals:
:$1 - \ds \sum_{n \mathop = 0}^{n - 1} \dbinom {6 n} x \paren {\dfrac 1 6}^x \paren {\dfrac 5 6}^{6 n - x}$
Calculating the probabilities, we get:
:
$6 n$	$n$ 	$\map P {\text {$n$ or more sixes} }$
$6$ 	$1$ 	$0.665$
$12$ 	$2$ 	$0.619$
$18$ 	$3$ 	$0.597$
$24$ 	$2$ 	$0.584$
$30$ 	$5$ 	$0.576$
$96$ 	$16$ 	$0.542$
$600$	$100$	$0.517$
$900$	$150$	$0.514$
It is clear that the gambler does better by betting on $1$ six on $6$ dice than $2$ on $12$ or $3$ on $18$.
== Historical Note ==
== Sources ==
* 
* : Prince Rupert's Cube: $129$
Category:Classic Problems
Category:Probability Theory
Relative Likelihood of n Sixes on 6n Dice/Mistake

== Source Work ==
: 
:The Puzzles:
::Prince Rupert's Cube
:::$129$ (Solution)
== Mistake ==
:Our short table shows the probabilities, rounded to three decimals, of obtaining the mean number or more sixes when $6 n$ dice are tossed.
:
$6 n$	$n$ 	$\map P {\text {$n$ or more sixes} }$
$6$ 	$1$ 	$0.665$
$12$ 	$2$ 	$0.619$
$18$ 	$3$ 	$0.597$
$24$ 	$2$ 	$0.584$
$30$ 	$5$ 	$0.576$
$96$ 	$6$ 	$0.542$
$600$	$100$	$0.517$
$900$	$150$	$0.514$
== Correction ==
The $6$th line is incorrect.
For $96$ dice, you are looking to score at least $16$ sixes, not $6$, as $\dfrac {96} 6 = 16$ not $6$.
== Sources ==
* : Prince Rupert's Cube: $129$
Category:Mistakes/Curious and Interesting Puzzles
Relative Matrix of Composition of Linear Transformations

== Theorem ==
Let $R$ be a ring with unity.
Let $M, N, P$ be free $R$-modules of finite dimension $m, n, p > 0$ respectively.
Let $\AA, \BB, \CC$ be ordered bases of $M, N, P$.
Let $f: M \to N$ and $g : N \to P$ be linear transformations, and $g \circ f$ be their composition.
Let $\mathbf M_{f, \BB, \AA}$ and $\mathbf M_{g, \CC, \BB}$ be their matrices relative to $\AA, \BB$ and $\BB, \CC$ respectively.
Then the matrix of $g \circ f$ relative to $\AA$ and $\CC$ is:
:$\mathbf M_{g \mathop \circ f, \CC, \AA} = \mathbf M_{g, \CC, \BB} \cdot \mathbf M_{f, \BB, \AA}$
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Product of Change of Basis Matrices, an analogous result for change of basis
* Composition of Linear Transformations is Linear Transformation
Category:Relative Matrices of Linear Transformations
Category:Relative Matrix of Composition of Linear Transformations
Relative Matrix of Composition of Linear Transformations/Proof 1

== Theorem ==
== Proof ==
Let $m \in M$, and $\sqbrk m_\AA$ be its coordinate vector with respect to $\AA$.
On the one hand:
On the other hand:
Thus:
:$\forall m \in M: \paren {\mathbf M_{g \mathop \circ f, \CC, \AA} - \mathbf M_{g, \CC, \BB} \cdot \mathbf M_{f, \BB, \AA} } \cdot \sqbrk m_\AA = 0$
The result follows.
Category:Relative Matrix of Composition of Linear Transformations
Relative Matrix of Composition of Linear Transformations/Proof 2

== Theorem ==
== Proof ==
Let:
:$\AA = \sequence {a_m}$
:$\BB = \sequence {b_n}$
:$\CC = \sequence {c_p}$
Let:
:$\sqbrk \alpha_{m n} = \sqbrk {f; \sequence {b_n}, \sequence {a_m} }$
and:
:$\sqbrk \beta_{n p} = \sqbrk {g; \sequence {c_p}, \sequence {b_n} }$
Then:
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices: Theorem $29.1$
Category:Relative Matrix of Composition of Linear Transformations
Relative Matrix of Linear Transformation/Examples/Differentiation

== Example of Relative Matrix of Linear Transformation ==
Let $\map {P_3} \R$ and $\map {P_4} \R$ denote the sets of real polynomials of degree $3$ and $4$ respectively.
Let $\sequence {I^J}_{0 \mathop \le j < n}$ denote the ordered bases of $\map {P_3} \R$ and $\map {P_4} \R$ where $n = 3$ and $n = 4$ respectively.
Let $D: p \to D p$ denote the operation of differentiation on a real polynomial $p$.
Let $\mathbf D := \sqbrk {D; \sequence {I^J}_{0 \mathop \le j < 3}, \sequence {I^J}_{0 \mathop \le j < 4} }$ denote the relative matrix of the linear transformation that is $D$.
Then:
:$\mathbf D = \begin {bmatrix} 0 & 1 & 0 & 0 \\ 0 & 0 & 2 & 0 \\ 0 & 0 & 0 & 3 \end {bmatrix}$
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices
Category:Examples of Relative Matrices of Linear Transformations
Relative Prime Modulo Tensor is Zero

== Theorem ==
Let $p \in \Z_{>0}$ and $q \in \Z_{>0}$ be positive coprime integers.
Let $\Z / p \Z$ and $\Z / q \Z$ be $\Z$-modules.
Then:
:$\Z / p \Z \otimes_\Z \Z / q\Z = 0$
where $\otimes_\Z$ denotes tensor product over integers.
== Proof ==
By Bézout's Identity there exists $a, b \in \Z$ such that $a p + b q = 1$.
Then for $s \otimes_\Z t \in \Z / p \Z \otimes \Z / q \Z$:
== Also see ==
* Tensor Product of Quotient Rings, a generalization
Category:Tensor Algebra
Relative Pseudocomplement Preserves Order

== Theorem ==
Let $\struct {S, \vee, \wedge, \preceq}$ be a Brouwerian lattice.
Let $a, b, c \in S$.
Let $b \preceq c$.
Then
:$a \to b \preceq a \to c$
where $x \to y$ denotes the relative pseudocomplement of $x$ with respect to $y$.
== Proof ==
We have:
Category:Brouwerian Lattices
Relative Pseudocomplement and Shift Mapping form Galois Connection in Brouwerian Lattice

== Theorem ==
Let $\struct {S, \vee, \wedge, \preceq}$ be a Brouwerian lattice.
Let $a$ be an element of $S$.
Let $g, d: S \to S$ be mappings such that:
:$\forall s \in S: \map g s = a \to s$
and
:$\forall s \in S: \map d s = a \wedge s$
Then $\struct {g, d}$ is a Galois connection.
== Proof ==
By Brouwerian Lattice iff Shift Mapping is Lower Adjoint:
:$d$ is lower adjoint
By definition of lower adjoint:
:$\exists g': S \to S: \struct {g', d}$ is Galois connection
By Galois Connection is Expressed by Maximum:
:$\forall s \in S: \map {g'} s = \map \max {d^{-1} \sqbrk {s^\preceq} }$
By definition of image of set:
:$\forall s \in S: \map {g'} s = \map \max {\set {x \in S: \map d x \in s^\preceq} }$
By definition of lower closure of element:
:$\forall s \in S: \map {g'} s = \map \max {\set {x \in S: \map d x \preceq s} }$
By definition of $d$:
:$\forall s \in S: \map {g'} s = \map \max {\set {x \in S: a \wedge x \preceq s} }$
By definition of relative pseudocomplement:
:$\forall s \in S: \map {g'} s = a \to s = \map g s$
By Equality of Mappings:
:$g = g'$
Hence
:$\struct {g, d}$ is a Galois connection.
== Sources ==
* 
* 
Category:Galois Connections
Category:Brouwerian Lattices
Relative Sizes of Angles in Segments

== Theorem ==
In a circle:
:the angle in a semicircle is right
:the angle in a segment greater than a semicircle is acute
:the angle in a segment less than a semicircle is obtuse.
Further:
:the angle of a segment greater than a semicircle is obtuse
:the angle of a segment less than a semicircle is acute.
== Proof ==
:
Let $ABCD$ be a circle whose diameter is $BC$ and whose center is $E$.
Join $AB$, $AC$, $AD$, $DC$ and $AE$.
Let $BA$ be produced to $F$.
Since $BE = EA$, from Isosceles Triangle has Two Equal Angles it follows that $\angle ABE = \angle BAE$.
Since $CE = EA$, from Isosceles Triangle has Two Equal Angles it follows that $\angle ACE = \angle CAE$.
So from $\angle BAC = \angle ABE + \angle ACE = \angle ABC + \angle ACB$.
But from Sum of Angles of Triangle Equals Two Right Angles $\angle FAC = \angle ABC + \angle ACB$.
So $\angle BAC = \angle FAC$, and so from Book I Definition 10 each one is a right angle.
So the angle in the semicircle $BAC$ is a right angle.
From Two Angles of Triangle are Less than Two Right Angles, in $\triangle ABC$, $\angle ABC + \angle BAC$ is less than two right angles.
As $\angle BAC$ is a right angle, it follows that $\angle ABC$ is less than a right angle.
It is also the angle in a segment $ABC$ greater than a semicircle.
Therefore the angle in a segment greater than a semicircle is acute.
We have that $ABCD$ is a cyclic quadrilateral.
From Opposite Angles of Cyclic Quadrilateral sum to Two Right Angles, $\angle ABC + \angle ADC$ equals two right angles.
As $\angle ABC$ is less than a right angle, it follows that $\angle ADC$ is greater than a right angle.
But $\angle ADC$ is the angle in a segment $ADC$ less than a semicircle.
Therefore the angle in a segment less than a semicircle is obtuse.
Since the angle contained by the straight lines $BA$ and $AC$ is a right angle, the angle contained by the arc $ABC$ and the straight line $AC$ is greater than a right angle.
So the angle of a segment greater than a semicircle is obtuse.
Since the angle contained by the straight lines $AC$ and $AF$ is a right angle, the angle contained by the arc $ADC$ and the straight line $AC$ is less than a right angle.
So the angle of a segment less than a semicircle is acute.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Circles