nyuuzyou/wikis huggingface dataset:
Equiangular Right Triangles are Similar/Proof 2

== Theorem ==
== Proof ==
Let $ABC$ be an arbitrary right triangle with $\angle ABC$ a right angle.
Construct a straight line from $A$ parallel to $BC$.
Construct a second straight line from $C$ parallel to $AB$, meeting the first straight line at $D$.
By Quadrilateral is Parallelogram iff Both Pairs of Opposite Sides are Equal or Parallel:
:$\Box ABCD$ is a parallelogram.
By definition of rectangle, and from Rectangle is Parallelogram:
:$\Box ABCD$ is a rectangle.
By construction, $AC$ is a diameter of $\Box ABCD$.
:
Let $H$ be an arbitrary point on $AC$.
Constuct $EHF$ parallel to $BC$ intersecting $AB$ at $F$ and $CD$ at $H$.
Constuct $DGI$ parallel to $AB$ intersecting $AD$ at $G$ and $BC$ at $I$.
Hence by construction $EHF$ and $DGI$ are both parallel to the sides of $\Box ABCD$.
By Triangle Side-Side-Side Congruence:
:$\triangle ABC \cong \triangle ADC$
so:
:$\angle BAC = \angle DCA$
:$\angle DAC = \angle ACB$
By Parallelism implies Equal Corresponding Angles:
:$\angle AHF = \angle ACB$
Because the diameter of each is shared, by Triangle Angle-Side-Angle Congruence:
:$\triangle AFH \cong \triangle AGH$
:$\triangle HIC \cong \triangle HEC$
For the same reasons and also by construction: 
:$\triangle ABC \cong \triangle ADC$
By definition of congruence, $\Box ABCD$ is composed of two parts of equal area:
:$\map \AA {\triangle AFH} + \map \AA {\Box FBIH} + \map \AA {\triangle HIC}$
:$\map \AA {\triangle AGH} + \map \AA {\Box GHED} + \map \AA {\triangle HEC}$
By subtraction, we obtain:
:$\map \AA {\Box FBIH} = \map \AA {\Box GHED}$
From Area of Rectangle:
:$FH \cdot HI = GH \cdot HE$
Rearranging:
:$\dfrac {FH} {GH} = \dfrac {HE} {HI}$
Thus, corresponding sides within each right triangle have equal ratios.
and also corresponding sides compared between two right triangle have equal ratios:
:$\dfrac {FH} {HE} = \dfrac {GH} {HI}$
We can extend this result to the hypotenuse:
In the above diagram, the altitude of $\triangle ABC$ is $CD$.
By Perpendicular in Right-Angled Triangle makes two Similar Triangles:
:$\triangle ADC \sim \triangle CDB$
By what we have shown above:
:$\dfrac {AC} {BC} = \dfrac {AD} {CD} = \dfrac {CD} {BD}$
Rearranging:
:$\dfrac {BC} {CD} = \dfrac {AC} {AD}$
which yields the result.
Category:Equiangular Right Triangles are Similar
Equiangular Triangle is Equilateral

== Theorem ==
Let $\triangle ABC$ be equiangular.
Then $\triangle ABC$ is an equilateral triangle.
== Proof ==
Let $\triangle ABC$ be equiangular.
By definition of equiangular polygon, any two of the internal angles of $\triangle ABC$ are equal.
, let $\angle ABC = \angle ACB$.
Then by Triangle with Two Equal Angles is Isosceles, $AB = AC$.
As the choice of equal angles was arbitrary, it follows that any two sides of $\triangle ABC$ are equal.
Hence all $3$ sides of $\triangle ABC$ are equal.
Hence the result by definition of equilateral triangle.
== Also see ==
* Equilateral Triangle is Equiangular, of which this is the converse.
== Sources ==
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.13$: Corollary
* : Appendix $\text{A}.5$: Theorems and Proofs: Problem Set $\text A.5$: $32$
* 
* 
Category:Equilateral Triangles
Equiangular Triangles are Similar

== Theorem ==
Let two triangles have the same corresponding angles.
Then their corresponding sides are proportional.
Thus, by definition, such triangles are similar.
== Proof ==
Let $\triangle ABC, \triangle DCE$ be equiangular triangles such that:
:$\angle ABC = \angle DCE$
:$\angle BAC = \angle CDE$
:$\angle ACB = \angle CED$
:
Let $BC$ be placed in a straight line with $CE$.
From Two Angles of Triangle are Less than Two Right Angles $\angle ABC + \angle ACB$ is less than two right angles.
As $\angle ACB = \angle DEC$, it follows that $\angle ABC + \angle DEC$ is also less than two right angles.
So from the Parallel Postulate, $BA$ and $ED$, when produced, will meet.
Let this happen at $F$.
We have that $\angle ABC = \angle DCE$.
So from Equal Corresponding Angles implies Parallel Lines:
:$BF \parallel CD$
Again, we have that $\angle ACB = \angle CED$.
Again from Equal Corresponding Angles implies Parallel Lines:
:$AC \parallel FE$
Therefore by definition $\Box FACD$ is a parallelogram.
Therefore from Opposite Sides and Angles of Parallelogram are Equal $FA = DC$ and $AC = FD$.
Since $AC \parallel FE$, it follows from Parallel Transversal Theorem that:
:$BA : AF = BC : CE$
But $AF = CD$ so:
:$BA : AF = BC : CE$
From Proportional Magnitudes are Proportional Alternately:
:$AB : BC = DC : CE$
Since $CD \parallel BF$, from Parallel Transversal Theorem:
:$BC : CE = FD : DE$
But $FD = AC$ so $BC : CE = AC : DE$.
So from Proportional Magnitudes are Proportional Alternately, $BC : CA = CE : ED$.
It then follows from Equality of Ratios Ex Aequali that $BA : AC = CD : DE$.
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Triangles
Category:Similar Triangles
Equiangular Triangles are Similar/Proof 2

== Theorem ==
Equiangular Triangles are Similar
== Proof ==
Let $\triangle ABC$ be any triangle.
Let $\triangle ADE$ be equiangular with it.
Draw the triangles so they coincide at one vertex and the bases are parallel.
By Equal Alternate Angles implies Parallel Lines:
:$BC \parallel DE$
Draw the altitude $AFG$ to both $\triangle ABC$ and $\triangle ADE$. 
(For an obtuse triangle, the altitude need not lie inside $\triangle ADE$). 
$\triangle ABF$, $\triangle ADG$, $\triangle ACF$ and $\triangle AEF$ are all right triangles.
By what we had above:
$\triangle ABF$ and $\triangle ADG$ are equiangular.
The same is true of $\triangle ACF$ and $\triangle AEG$.
By Equiangular Right Triangles are Similar:
:$\dfrac {AB} {AD} = \dfrac {AF} {AG}$
:$\dfrac {AC} {AE} = \dfrac {AF} {AG}$
Rearranging:
:$\dfrac {AB} {AD} = \dfrac {AC} {AE}$
We have equal ratios of sides.
But we can do this for any two sides of the two triangles.
Category:Triangles
Equidecomposability Unaffected by Union

== Theorem ==
Let $\set {S_1, \ldots, S_m}, \set {T_1, \ldots, T_m}$ be sets of sets in $\R^n$ such that:
:for each $k \in \set {1, \dots, m}, S_k$ and $T_k$ are equidecomposable. 
Then the set $\ds S = \bigcup_{i \mathop = 1}^m S_i$ is equidecomposable with $\ds T = \bigcup_{i \mathop = 1}^m T_i$.
== Proof ==
We have for each $k \in \set{1, \dots, m}$ a decomposition $\set {A_{k, 1}, \cdots, A_{k, l_k} }$ and set of isometries $\phi_{i, j}: \R^n \to \R^n$ such that:
:$\ds S_k = \bigcup_{a \mathop = 1}^{l_k} \map {\phi_{k, a} } {A_{k, a} }$
and similarly for $T_k$ and some isometries $\theta_{i, j}: \R^n \to \R^n$:
:$\ds T_k = \bigcup_{a \mathop = 1}^{l_k} \map {\theta_{k, a} } {A_{k, a} }$
Thus:
:$\ds S = \bigcup_{k \mathop = 1}^m \bigcup_{i \mathop = 1}^{l_k} \map {\phi_{k, i} } {A_{k, i} }$
and:
:$\ds T = \bigcup_{k \mathop = 1}^m \bigcup_{i \mathop = 1}^{l_k} \map {\theta_{k, i} } {A_{k, i} }$
This yields equivalent decompositions of $S$ and $T$.
Category:Equidecomposable Sets
Category:Topology
Equidecomposability is Equivalence Relation

== Theorem ==
The property of being equidecomposable is an equivalence relation on the power set $\powerset {\R^n}$.
== Proof ==
=== Relexivity ===
A set is necessarily equidecomposable with itself; the same decomposition and set of isometries suffice for $A$ as for $A$.
=== Symmetry ===
There is no order to the relation of being equidecomposable; symmetry follows.
=== Transitivity ===
Suppose $A, B, C \subset \R^n$ are sets such that $A, B$ are equidecomposable and $B, C$ are equidecomposable.
Let $X_1, \dots, X_m$ be a decomposition of $A, B$ together with isometries $\mu_1, \dots, \mu_m, \nu_1, \dots, \nu_m: \R^n \to \R^n$ such that:
:$\ds A = \bigcup_{i \mathop = 1}^m \map {\mu_i} {X_i}$ 
and 
:$\ds B = \bigcup_{i \mathop = 1}^m \map {\nu_i} {X_i}$
Further let $Y_1, \dots, Y_p$ together with $\xi_1, \dots, \xi_p, \tau_1, \dots, \tau_p$ be sets and isometries such that:
:$\ds B = \bigcup_{i \mathop = 1}^p \map {\xi_i} {Y_i}$ 
and:
:$\ds C = \bigcup_{i \mathop = 1}^p \map {\tau_i} {Y_i}$
Consider the sets
:$Z_{i, j} = \map {\nu_i} {X_i} \cap \map {\xi_j} {Y_j}$
where $1 \le i \le m$ and $1 \le j \le p$.
We have:
:$\ds \bigcup_{i \mathop = 1}^m \bigcup_{j \mathop = 1}^p \map {\paren {\mu_i \circ \nu_i^{-1} } } {Z_{i, j} } = \bigcup_{i \mathop = 1}^m \bigcup_{j \mathop = 1}^p \map {\paren {\mu_i \circ \nu_i^{-1} } } {\map {\nu_i} {X_i} \cap \map {\xi_j} {Y_j} } = \bigcup_{i \mathop = 1}^m \map {\paren {\mu_i \circ \nu_i^{-1} \circ \nu_i} } {X_i} = \bigcup_{i \mathop = 1}^m \map {\mu_i} {X_i} = A$
:$\ds \bigcup_{j \mathop = 1}^p \bigcup_{i \mathop = 1}^m \map {\paren {\tau_j \circ \xi_j^{-1} } } {Z_{i, j} } = \bigcup_{j \mathop = 1}^p \bigcup_{i \mathop = 1}^m \map {\paren {\tau_j \circ \xi_j^{-1} } } {\map {\nu_i} {X_i} \cap \map {\xi_j} {Y_j} } = \bigcup_{j \mathop = 1}^p \map {\paren {\tau_j \circ \xi_j^{-1} \circ \xi_j} } {Y_j} = \bigcup_{j \mathop = 1}^p \map {\tau_j} {Y_j} = C$
so $Z_{i, j}$ together with the isometries $\mu_i \circ \nu_i^{-1}, \tau_j \circ \xi_j^{-1}$ is a decomposition of $A$ and $C$.
Hence these two are equidecomposable.
Category:Equidecomposable Sets
Category:Topology
Category:Examples of Equivalence Relations
Equidecomposable Nested Sets

== Theorem ==
Let $A, B, C$ be sets such that $A$ and $C$ are equidecomposable and $A \subseteq B \subseteq C$.
Then $B$ and $C$ are equidecomposable.
== Proof ==
Let $\set {X_k}_{k \mathop = 1}^n$ be a decomposition of $A$ and $C$, so that there are isometries $\set {\phi_k}_{k \mathop = 1}^n$ and $\set {\psi_k}_{k \mathop = 1}^n$ such that:
$\ds A = \bigcup_{k \mathop = 1}^n \map {\phi_k} {X_k}$
$\ds C = \bigcup_{k \mathop = 1}^n \map {\psi_k} {X_k}$
Let $Y_k = \map {\psi_k^{-1} } {B \cap \map {\psi_k} {X_k} }$.
Then:
:$\ds \bigcup_{k \mathop = 1}^n \map {\psi_k} {Y_k} = B \cap \bigcup_{k \mathop = 1}^n \map {\psi_k} {X_k} = B \cap C = B$
Let $Z_k = \map {\phi^{-1} } {A \cap \map {\phi_k} {X_k} }$.
Then:
:$\ds \bigcup_{k \mathop = 1}^n \map {\phi_k} {Z_k} = A \cap \bigcup_{k \mathop = 1}^n \map {\phi_k} {X_k} = A \cap C = A$
Category:Equidecomposable Sets
Category:Topology
Equidistance is Independent of Betweenness

== Theorem ==
Let $\GG$ be a formal systematic treatment of geometry containing only:
:The language and axioms of first-order logic, and the disciplines preceding it
:The undefined terms of Tarski's Geometry (excluding equidistance) 
:Some or all of Tarski's Axioms of Geometry.
In $\GG$, equidistance $\equiv$ is necessarily an undefined term with respect to betweenness $\mathsf B$.
 
== Proof ==
Our assertion is that $\equiv$ cannot be defined in terms of $\mathsf B$.
 that it can.
Call this assumption $\paren A$.
If $\paren A$ holds, it must hold in all systems.
Let one such system be $\tuple {\R^2, \mathsf B_1, \equiv_1}$ where:
:$\R^2$ is the cartesian product of the set of real numbers with itself
:$\mathsf B_1$ is a ternary relation of betweenness
:$\equiv_1$ is a quaternary relation of equidistance
Let $\GG$ be the discipline preceding the given discipline, where $\GG$ is as defined above (excluding both $\equiv$ and $\mathsf B$).
Define $\mathsf B_1$ as follows:
Define $\equiv_1$ as follows:
Now, define the isomorphism $\phi$ on $\struct {\R^2, \mathsf B_2, \equiv_2}$ as:
:$\phi: \R^2 \to \R^2$ on $\struct {\R^2, \mathsf B_1, \equiv_1}, \tuple {x_1, x_2} \mapsto \tuple {x_1, 2 x_2}$
Now consider the system:
:$\struct {\R^2, \mathsf B_2, \equiv_2}$
where $\mathsf B_2$ and $\equiv_2$ are the relations defined as above, but on the elements in the images of $\mathsf B_1$ and $\equiv_1$, respectively.
Observe that $\mathsf B_1$ and $\mathsf B_2$ coincide, because in:
:$\paren {x_1 - y_1} \cdot \paren {2 y_2 - 2 z_2} = \paren {2 x_2 - 2 y_2} \cdot \paren {y_1 - z_1} \land$
:$\paren {0 \le \paren {x_1 - y_1} \cdot \paren {y_1 - z_1} } \land \paren {0 \le \paren {2 x_2 - 2 y_2} \cdot \paren {2 y_2 - 2 z_2} }$
we can simply factor out the $2$ and divide both sides of the equality of inequality by $2$.
But consider the elements:
:$p_1 = \tuple {0, 0}$
:$p_2 = \tuple {0, 1}$
:$p_3 = \tuple {1, 0}$
Observe that $p_1 p_2 \equiv_1 p_1 p_3$:
:$\paren {0 - 0}^2 + \paren {0 - 1}^2 = \paren {0 - 1}^2 + \paren {0 - 0}^2$
But $\map \neg {p_1 p_2 \equiv_2 p_1 p_3}$:
:$\paren {0 - 0}^2 + \paren {0 - 2}^2 \ne \paren {0 - 1}^2 + \paren {0 - 0}^2$
But both $\struct {\R^2, \mathsf B_1, \equiv_1}$ and $\struct {\R^2, \mathsf B_2, \equiv_2}$ are both models of $\GG$.
Recall that if $\paren A$ holds, it must hold in all systems.
But it does not.
Hence $\paren A$ is false, from Proof by Contradiction.
== Also see ==
* Betweenness Not Independent of Equidistance, which states that there are models where one can define $\mathsf B$ in terms of $\equiv$.
== Sources ==
* 
Category:Geometry
Equidistance of Ellipse equals Major Axis

== Theorem ==
Let $K$ be an ellipse whose foci are $F_1$ and $F_2$.
Let $P$ be an arbitrary point on $K$.
Let $d$ be the constant distance such that:
:$d_1 + d_2 = d$
where:
:$d_1 = P F_1$
:$d_2 = P F_2$
Then $d$ is equal to the major axis of $K$.
== Proof ==
:
By the equidistance property of $K$:
:$d_1 + d_2 = d$
applies to all points $P$ on $K$.
Thus it also applies to the two vertices $V_1$ and $V_2$:
:$V_1 F_1 + V_1 F_2 = d$
:$V_2 F_1 + V_2 F_2 = d$
Adding:
:$V_1 F_1 + V_2 F_1 + V_1 F_2 + V_2 F_2 = 2 d$
But:
:$V_1 F_1 + V_2 F_1 = V_1 V_2$
:$V_1 F_2 + V_2 F_2 = V_1 V_2$
and so:
:$2 V_1 V_2 = 2 d$
By definition, the major axis is $V_1 V_2$.
Hence the result.
== Sources ==
* 
Category:Ellipses
Equidistance of Hyperbola equals Transverse Axis

== Theorem ==
Let $K$ be an hyperbola whose foci are $F_1$ and $F_2$.
Let $P$ be an arbitrary point on $K$.
Let $d$ be the constant distance such that:
:$\size {d_1 - d_2} = d$
where:
:$d_1 = P F_1$
:$d_2 = P F_2$
Then $d$ is equal to the transverse axis of $K$.
== Proof ==
 
:
By the equidistance property of $K$:
:$\size {d_1 - d_2} = d$
applies to all points $P$ on $K$.
Thus it also applies to the two vertices $V_1$ and $V_2$.
Observing the signs of $\size {d_1 - d_2}$ as appropriate:
:$V_1 F_2 - V_1 F_1 = d$
:$V_2 F_1 - V_2 F_2 = d$
Adding:
:$\paren {V_1 F_2 - V_2 F_2} + \paren {V_2 F_1 - V_1 F_1} = 2 d$
But:
:$V_1 F_2 - V_2 F_2 = V_1 V_2$
:$V_2 F_1 - V_1 F_1 = V_1 V_2$
and so:
:$2 V_1 V_2 = 2 d$
By definition, the transverse axis of $K$ is $V_1 V_2$.
Hence the result.
Category:Hyperbolas
Equilateral Pentagon is Equiangular if Three Angles are Equal

== Theorem ==
== Proof ==
:
Let $ABCDE$ be an equilateral pentagon.
Let $A, B, C$ be three vertices of $ABCDE$ taken in order such that $\angle A = \angle B = \angle C$.
Let $AC, BE, FD$ be joined.
We have that $CB$ and $BA$ are equal to $BA$ and $AE$ respectively.
We also have that $\angle CBA = \angle BAE$.
Therefore from :
:$AC = BE$
and:
:$\triangle ABC = \triangle ABE$
Thus:
:$\angle BCA = \angle BEA$
:$\angle ABE = \angle CAB$
So from :
:$AF = CF$
But we have that:
:$AC = BE$
Therefore:
:$FC = FE$
But:
:$CD = DE$
Therefore:
:$FC = FE$
and
:$CD = ED$
while $FD$ is common.
So from :
:$\angle FCD = \angle FED$
But we have already proved that:
:$\angle BCA = \angle AEB$
Therefore:
:$\angle BCD = \angle AED$
But :
:$\angle BCD = \angle A = \angle B$
Therefore:
:$\angle AED = \angle A = \angle B$
Similarly it can be proved that:
:$\angle CDE = \angle A = \angle B = \angle C$
Therefore $ABCDE$ is equiangular.
Next, suppose that $A, C, D$ are the three vertices of $ABCDE$ such that $\angle A = \angle C = \angle D$.
Let $BD$ be joined.
We have that $BA = BC$ and $AE = CD$.
Also, they contain equal angles.
Therefore from :
:$BE = BD$
and:
:$\triangle ABE = \triangle BCD$
Therefore:
:$\angle AEB = \angle CDB$
But $BE = BD$.
Therefore from :
:$\angle BED = \angle BDE$
Therefore:
:$\angle AED = \angle DDE$
But :
:$\angle CDE = \angle A = \angle C$
Therefore:
:$\angle AED = \angle A = \angle C$
For the same reason:
:$\angle ABC = \angle A = \angle C = \angle D$
Therefore $ABCDE$ is equiangular.
== Sources ==
* : Book $\text{XIII}$. Propositions
Category:Pentagons
Equilateral Polygon is Equiangular/Mistake

== Source Work ==
:
:equiangular
== Mistake ==
:Any equilateral plane figure is also equiangular, and so is regular ...
== Correction ==
As is demonstrated in Equilateral Polygon is not necessarily Equiangular, the rhombus is the simplest counterexample:
== Sources ==
* 
Category:Mistakes/Dictionary of Mathematics
Equilateral Triangle is Equiangular

== Theorem ==
Let $\triangle ABC$ be an equilateral triangle.
Then $\triangle ABC$ is also equiangular, with each internal angle equal to $60 \degrees$.
== Proof ==
Let $\triangle ABC$ be an equilateral triangle.
By definition of equilateral triangle, any two of the sides of $\triangle ABC$ are equal.
, let $AB = AC$.
Then by Isosceles Triangle has Two Equal Angles:
:$\angle ABC = \angle ACB$
As the choice of equal sides was arbitrary, it follows that every two of the internal angles of $\triangle ABC$ are equal.
Hence all $3$ internal angles of $\triangle ABC$ are equal.
From Sum of Angles of Triangle equals Two Right Angles, the internal angles of $\triangle ABC$ all add up to $180 \degrees$.
Hence it follows that one such internal angle is $\dfrac {180 \degrees} 3$, or $60 \degrees$.
== Also see ==
* Equiangular Triangle is Equilateral, of which this is the converse.
== Sources ==
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.12$: Corollary $2$
* : Appendix $\text{A}.5$: Theorems and Proofs: Problem Set $\text A.5$: $32$
* 
* 
* 
* 
* 
* 
* 
Category:Equilateral Triangles
Equilibrant/Examples/100kg at 150, 75kg at 60, 50kg at -45

== Example of Equilibrant ==
Three forces $\mathbf F_1, \mathbf F_2, \mathbf F_3$ act on a particle $B$ at a point $P$ embedded in the complex plane:
:
The equilibrant $\mathbf E$ of $\mathbf F_1, \mathbf F_2, \mathbf F_3$ is:
:$\mathbf E = \polar {80.8 \, \mathrm {kg}, -80.2 \degrees}$
== Proof 1 ==
== Proof 2 ==
Category:Examples of Equilibrants
Equilibrant/Examples/100kg at 150, 75kg at 60, 50kg at -45/Proof 2

== Example of Equilibrant ==
== Proof ==
:
Using the Parallelogram Law to add $\mathbf F_1$ and $\mathbf F_2$ we arrive at $\mathbf F_1 + \mathbf F_2$.
Again using the Parallelogram Law to add $\mathbf F_1 + \mathbf F_2$ and $\mathbf F_3$ we arrive at $\mathbf F_1 + \mathbf F_2 + \mathbf F_3$.
The equilibrant $\mathbf E$ can then be found by taking the negative of $\mathbf F_1 + \mathbf F_2 + \mathbf F_3$.
== Sources ==
* : $1$: Complex Numbers: Supplementary Problems: Polar Form of Complex Numbers: $86 \ \text {(a)}$
Category:Examples of Equilibrants
Equivalence Class/Examples

== Examples of Equivalence Classes ==
=== Same Age Relation ===
=== People Born in Same Year ===
=== People with Same Height ===
=== Months that Start on the Same Day of the Week ===
=== $z^4 = w^4$ on Complex Numbers ===
=== $\sin \dfrac {\pi x} 6 = \sin \dfrac {\pi y} 6$ over $\Z$ ===
=== Congruence Modulo $\N_{< m}$ ===
Category:Examples of Equivalence Classes
Equivalence Class/Examples/Congruence Modulo Initial Segment of Natural Numbers

== Example of Equivalence Class ==
Let $m \in \Z_{>0}$ be a (strictly) positive integer.
Let $\N_{<m}$ denote the initial segment of the natural numbers $\N$:
:$\N_{<m} = \set {0, 1, \ldots, m - 1}$
Let $\RR_m$ denote the equivalence relation:
:$\forall x, y \in \Z: x \mathrel {\RR_m} y \iff \exists k \in \Z: x = y + k m$
For each $a \in \N_{<m}$, let $\eqclass a m$ be the equivalence class of $a \in \N_{<m}$ under $\RR_m$ is the set:
:$\eqclass a m := \set {a + z m: z \in \Z}$
== Proof ==
From Congruence Modulo Integer is Equivalence Relation, $\RR_m$ is an equivalence relation.
From the Division Theorem:
:$\forall n \in \Z: \exists! z, a \in \Z: n = a + z m, a \in \N_{<m}$
Hence the result.
== Examples ==
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets
Category:Examples of Equivalence Classes
Equivalence Class/Examples/Congruence Modulo Initial Segment of Natural Numbers/Examples/4

== Example of Equivalence Class ==
Let $m \in \Z_{>0}$ be a (strictly) positive integer.
Let $\N_{<m}$ denote the initial segment of the natural numbers $\N$:
:$\N_{<m} = \set {0, 1, \ldots, m - 1}$
Let $\RR_m$ denote the equivalence relation:
:$\forall x, y \in \Z: x \mathrel {\RR_m} y \iff \exists k \in \Z: x = y + k m$
For each $a \in \N_{<m}$, let $\eqclass a m$ be the equivalence class of $a \in \N_{<m}$ under $\RR_m$:
:$\eqclass a m := \set {a + z m: z \in \Z}$
Then:
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 9$: Compositions Induced on the Set of All Subsets
Category:Examples of Equivalence Classes
Equivalence Class/Examples/Equal Fourth Powers over Complex Numbers

== Example of Equivalence Class ==
Let $\C$ denote the set of complex numbers.
Let $\RR$ denote the equivalence relation on $\C$ defined as:
:$\forall w, z \in \C: z \mathrel \RR w \iff z^4 = w^4$
Then the equivalence class of $1 + i \sqrt 3$ under $\RR$ is:
:$\eqclass {1 + i \sqrt 3} \RR = \set {1 + i \sqrt 3, -1 - i \sqrt 3, -\sqrt 3 + i, \sqrt 3 - i}$
== Proof ==
From Equivalence Relation Examples: Equal Fourth Powers over Complex Numbers, $\RR$ is an equivalence relation.
We have that:
:$1 + i \sqrt 3 = 2 \cis \dfrac \pi 3$
Hence $\paren {1 + i \sqrt 3}^4 = 16 \cis \dfrac {4 \pi} 3$
So:
:$\eqclass {1 + i \sqrt 3} \RR = \set {2 \map \cis {\dfrac \pi 3 + \dfrac {n \pi} 2}: n \in \set {0, 1, 2, 3} }$
Hence the result.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.1 \ \text{(a)}$
Category:Examples of Equivalence Classes
Equivalence Class/Examples/Equal Sine of pi x over 6 on Integers

== Example of Equivalence Class ==
Let $\Z$ denote the set of integers.
Let $\RR$ denote the relation on $\Z$ defined as:
:$\forall x, y \in \Z: x \mathrel \RR y \iff \sin \dfrac {\pi x} 6 = \sin \dfrac {\pi y} 6$
Then the equivalence class of $1$ under $\RR$ is:
:$\eqclass 1 \RR = \set {1 + 12 k: k \in \Z} \cup \set {5 + 12 k: k \in \Z}$
== Proof ==
From Equivalence Relation Examples: Equal $\sin \dfrac {\pi x} 6$ on Integers, $\RR$ is an equivalence relation.
We have that:
:$\eqclass 1 \RR = \set {x \in \Z: \sin \dfrac {\pi x} 6 = \sin \dfrac \pi 6}$
That is:
:$\eqclass 1 \RR = \set {x \in \Z: \sin \dfrac {\pi x} 6 = \dfrac 1 2}$
We have that:
:$\sin \dfrac \pi 6 = \dfrac 1 2$
and:
:$\sin \dfrac {5 \pi} 6 = \dfrac 1 2$
The result follows from Sine and Cosine are Periodic on Reals.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 11$: Quotient Structures: Exercise $11.1 \ \text{(b)}$
Category:Examples of Equivalence Classes