nyuuzyou/wikis huggingface dataset:
Conditions for Transformation to be Canonical

== Theorem ==
Let:
:$\ds J_1 \sqbrk {\sequence {y_i}_{1 \mathop \le i \mathop \le n}, \sequence {p_i}_{1 \mathop \le i \mathop \le n} } = \int_a^b \paren {\sum_{i \mathop = 1}^n p_i y_i'-H} \rd x$
:$\ds J_2 \sqbrk {\sequence {Y_i}_{1 \mathop \le i \mathop \le n}, \sequence {P_i}_{1 \mathop \le i \mathop \le n} } = \int_a^b \paren {\sum_{i \mathop = 1}^n P_i Y_i'-H^*} \rd x$
be functionals.
Then $\paren {\sequence {y_i}_{1 \mathop \le i \mathop \le n}, \sequence{p_i}_{1 \mathop \le i \mathop \le n}, H} \to \paren {\sequence{Y_i}_{1 \mathop \le i \mathop \le n}, \sequence {P_i}_{1 \mathop \le i \mathop \le n}, H^*}$ is a canonical transformation if:
:$\ds \sum_{i \mathop = 1}^n p_i y_i' - H = \sum_{i \mathop = 1}^n P_i Y_i' - H^* \pm \dfrac {\d \Phi} {\d x}$
 
and:
:$\ds p_i = \mp \frac {\partial \Phi} {\d y_i}, \quad P_i = \pm \frac {\partial \Phi} {\d Y_i}, \quad H = H^* \mp \frac {\partial \Phi} {\partial x}$
== Proof ==
By Conditions for Integral Functionals to have same Euler's Equations, functionals:
:$\ds \int_a^b F_1 \rd x$ 
and:
:$\ds \int_a^b F_2 \rd x = \int_a^b \paren {F_1 \pm \frac {\d \Phi} {\d x} } \rd x$ 
have same Euler's equations.
Express the first one in canonical variables $\paren {x, \sequence{y_i}_{1 \mathop \le i \mathop \le n}, \sequence{p_i}_{1 \mathop \le i \mathop \le n}, H}$ and the second one in $\paren {x, \sequence{Y_i}_{1 \mathop \le i \mathop \le n}, \sequence {P_i}_{1 \mathop \le i \mathop \le n}, H^*}$:
:$\ds \int_a^b F_1 \rd x = \int_a^b \paren {\sum_{i \mathop = 1}^n p_i y_i' - H} \rd x$
:$\ds \int_a^b F_2 \rd x = \int_a^b \paren {\sum_{i \mathop = 1}^n P_i Y_i' - H^*} \rd x$
However:
:$\ds \int_a^b \paren {F_2 - F_1} \rd x = \int_a^b \pm \frac {\d \Phi} {\d x} \rd x$
Inserting new expressions for $F_1$ and $F_2$ yields:
:$\ds \int_a^b \paren {\sum_{i \mathop = 1}^n P_i Y_i'-H^* - \sum_{i \mathop = 1}^n p_i y_i' + H} \rd x = \int_a^b \pm \frac {\d \Phi} {\d x}\rd x$
This is satisfied, if integrands are equal.
Transform the coordinates to $\tuple {x, \sequence {y_i}_{1 \mathop \le i \mathop \le n}, \sequence {Y_i}_{1 \mathop \le i \mathop \le n} }$ and write out the full derivative of $\Phi$:
:$\ds \sum_{i \mathop = 1}^n P_i Y_i' - H^* - \sum_{i \mathop = 1}^n p_i y_i' + H = \pm \frac {\partial \Phi} {\partial x} \pm \sum_{i \mathop = 1}^n \frac {\partial \Phi} {\partial y_i} y_i' \pm \sum_{i \mathop = 1}^n \frac {\partial \Phi} {\partial Y_i} Y_i'$
Collect terms multiplied by the same the coordinates together:
:$\ds \sum_{i \mathop = 1}^n Y_i' \paren {\pm \frac {\partial \Phi} {\partial Y_i} - P_i} + \sum_{i \mathop = 1}^n y_i' \paren {\pm \frac {\partial \Phi} {\partial y_i} + p_i} + \paren {\pm \frac {\partial \Phi} {\partial x} - H + H^*} = 0$
This has to hold for arbitrary values of independent coordinates $\tuple {x, \sequence {y_i}_{1 \mathop \le i \mathop \le n}, \sequence {Y_i}_{1 \mathop \le i \mathop \le n} }$.
Hence:
:$p_i = \mp \dfrac {\partial \Phi} {\d y_i}, \quad P_i = \pm \dfrac {\partial \Phi} {\d Y_i}, \quad H^* = H \mp \dfrac {\partial \Phi} {\partial x}$
== Sources ==
* : $\S 4.19$: Canonical Transformations
Category:Calculus of Variations
Conditions for Uniqueness of Left Inverse Mapping

== Theorem ==
Let $S$ and $T$ be sets such that $S \ne \O$.
Let $f: S \to T$ be an injection.
Then a left inverse mapping of $f$ is in general not unique.
Uniqueness occurs under either of two circumstances:
:$(1): \quad S$ is a singleton
:$(2): \quad f$ is a bijection.
== Proof ==
If $f$ is a bijection, then by definition $f$ is also a surjection.
Then:
:$T \setminus \Img f = \O$
:and we have that $g = f^{-1}$.
As $f^{-1}$ is uniquely defined $g$ is itself unique.
If $S$ is a singleton then there can only be one mapping $g: T \to S$:
:$\forall t \in T: \map g t = s$
If $f$ is not a bijection, then as it is an injection it can not be a surjection.
Then:
:$T \setminus \Img f \ne \O$
Let $t \in T \setminus \Img f$.
We can now choose any $x_0 \in S$ such that $\map g t = x_0$.
If $S$ is not a singleton, such an $x_0$ is not unique.
Hence the result.
== Warning ==
== Examples ==
== Also see ==
* Surjection iff Right Inverse/Non-Uniqueness
== Sources ==
* : $\S 0.4$: Theorem $6$
* : Chapter $\text{I}$: Sets and Functions: Composition of Functions
* : Chapter $1$: Sets and mappings: $\S 1.3$: Mappings: Exercise $6$
* : Appendix $\text{A}.7$: Inverses: Proposition $\text{A}.7.1$
Category:Left Inverse Mappings
Category:Conditions for Uniqueness of Left Inverse Mapping
Conditions for Uniqueness of Left Inverse Mapping/Examples/Arbitrary Example

== Example of Conditions for Uniqueness of Left Inverse Mapping ==
Let $S = \set {0, 1}$.
Let $T = \set {a, b, c}$.
Let $f: S \to T$ be defined as:
:$\forall x \in S: \map f x = \begin {cases} a & : x = 0 \\ b & : x = 1 \end {cases}$
Then $f$ has $2$ distinct left inverses.
== Proof ==
Let $g_0: T \to S$ be the mapping defined as:
Let $g_1: T \to S$ be the mapping defined as:
We have that:
:$\Cdm {g_0} = \Dom f = \Cdm {g_1}$
and that $f$ is an injection.
Hence we can construct:
and:
Hence both $g_0$ and $g_1$ are distinct left inverses of $f$.
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Exercise $\text{G}$
Category:Conditions for Uniqueness of Left Inverse Mapping
Conditions on Rational Solution to Polynomial Equation

== Theorem ==
Let $P$ be the polynomial equation:
:$a_n z^n + a_{n - 1} z^{n - 1} + \cdots + a_1 z + a_0 = 0$
where $a_0, \ldots, a_n$ are integers.
Let $\dfrac p q$ be a root of $P$ expressed in canonical form.
Then $p$ is a divisor of $a_0$ and $q$ is a divisor of $a_n$.
== Proof ==
By definition of the canonical form of a rational number, $p$ and $q$ are coprime.
Substitute $\dfrac p q$ for $z$ in $P$ and multiply by $q^n$:
:$(1): \quad a_n p^n + a_{n - 1} p^{n - 1} q + \cdots + a_1 p q^{n - 1} + a_0 q^n = 0$
Dividing $(1)$ by $p$ gives:
:$(2): \quad a_n p^{n - 1} + a_{n - 1} p^{n - 2} q + \cdots + a_1 q^{n - 1} = -\dfrac {a_0 q^n} p$
The of $(2)$ is an integer and therefore so is the .
We have that $p$ and $q$ are coprime.
By Euclid's Lemma it follows that $p$ divides $a_0$.
Similarly, dividing $(1)$ by $q$ gives:
:$(3): \quad -\dfrac {a_n p^n} q = a_{n - 1} p^{n - 1} + \cdots + a_1 p q^{n - 2} + a_0 q^{n - 1}$
By Euclid's Lemma it follows that $q$ divides $a_n$.
== Sources ==
* : $1$: Complex Numbers: Solved Problems: Polynomial Equations: $33$
Category:Polynomial Equations
Conditions under which Commutative Semigroup is Group

== Theorem ==
Suppose the following:
Then $\struct {S, \circ}$ is a group.
== Proof ==
Some lemmata:
=== Lemma 1 ===
=== Lemma 2 ===
=== Lemma 3 ===
First we note that if $S = \O$, then $(1)$ and $(2)$ hold vacuously.
But from Group is not Empty, $\struct {S, \circ}$ cannot then be a group.
Hence the stipulation that $S$ is non-empty.
Next we note that if $S = \set e$ is a singleton, $\struct {S, e}$ is the trivial group.
So in the following, it is assumed that there are at least two elements of $S$.
Taking the group axioms in turn:
=== and ===
We have that $\struct {S, \circ}$ is a semigroup.
Hence the semigroup axioms are fulfilled:
:
:
Hence and are fulfilled by $\struct {S, \circ}$.
=== ===
Let $x \in S$.
Let $e \in S$ be the element of $S$ asserted by $(1)$ such that:
:$e \circ x = x$
From Lemma $1$, this is the only such element of $S$ with this property.
Let $z \in S$ such that $z \ne x$ be arbitrary.
By $(1)$ there exists $e' \in S$ such that:
:$e' \circ z = z$
By Lemma $3$ we have that:
:$e' = e$
As $z$ is arbitrary, it follows that:
:$\exists e \in S: \forall z \in S: e \circ z = z$
As $\struct {S, \circ}$ is a commutative semigroup, it follows also that:
:$\forall z \in S: z \circ e = z$
Thus $e$ has been shown to be an identity of $\struct {S, \circ}$.
From Identity is Unique, $e$ is the only such element of $S$ which is an identity.
Hence $e$ is the identity of $\struct {S, \circ}$.
=== ===
We have that $e$ is the identity element of $\struct {S, \circ}$:
:$\forall x \in S: x \circ e = x$
It follows directly from $(2)$ that:
:$\forall x \in S: \exists z \in S: z \circ x = e$
As $\struct {S, \circ}$ is a commutative semigroup, it follows also that:
:$\forall x \in S: x \circ z = e$
Thus every element $x$ of $\struct {S, \circ}$ has an inverse $z$.
All the group axioms are thus seen to be fulfilled, and so $\struct {S, \circ}$ is a group.
== Warning ==
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.16 \ \text {(d)}$
Category:Conditions under which Commutative Semigroup is Group
Category:Group Theory
Category:Commutative Semigroups
Conditions under which Commutative Semigroup is Group/Lemma 1

== Lemma for Conditions under which Commutative Semigroup is Group ==
Suppose the following:
Then:
:If $y \circ x = x = y' \circ x$, then $y = y'$.
== Proof ==
Similarly:
Thus we have:
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.16 \ \text {(a)}$
Category:Conditions under which Commutative Semigroup is Group
Conditions under which Commutative Semigroup is Group/Lemma 2

== Lemma for Conditions under which Commutative Semigroup is Group ==
Suppose the following:
Then:
:If $y \circ x = x$, then $y \circ y = y$.
== Proof ==
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.16 \ \text {(b)}$
Category:Conditions under which Commutative Semigroup is Group
Conditions under which Commutative Semigroup is Group/Lemma 3

== Lemma for Conditions under which Commutative Semigroup is Group ==
Suppose the following:
Then:
:If $y \circ x = x$ and $z \circ w = w$, then $y = z$.
== Proof ==
Then:
Thus:
:$y \circ \paren {w \circ x} = w \circ x = z \circ \paren {w \circ x}$
and $y = z$ follows from Lemma 1.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.16 \ \text {(c)}$
Category:Conditions under which Commutative Semigroup is Group
Conditions under which Commutative Semigroup is Group/Statement of Conditions

== Statement of Conditions under which Commutative Semigroup is Group ==
Let $\struct {S, \circ}$ be a non-empty commutative semigroup.
Let $\struct {S, \circ}$ have the following properties:
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.16$
Category:Conditions under which Commutative Semigroup is Group
Conditions under which Commutative Semigroup is Group/Warning

== Warning concerning Conditions under which Commutative Semigroup is Group ==
Consider an algebraic structure $\struct {S, \circ}$ fulfilling the following conditions:
While it is the case that such an algebraic structure $\struct {S, \circ}$ is a group, if $\struct {S, \circ}$ is a semigroup which is not commutative, this does not necessarily follow.
== Proof ==
Let $S$ be a set with more than $1$ element.
Let $\struct {S, \circ}$ be a semigroup such that $\circ$ is the left operation.
From Structure under Left Operation is Semigroup, $\struct {S, \circ}$ is indeed a semigroup.
From Right Operation is Anticommutative we have that $\circ$ is specifically not commutative.
By definition of the left operation:
:$\forall x \in S: x \circ x = x$
Thus $x$ fulfils the role of $y$ in condition $(1)$:
:$\forall x \in S: \exists y \in S: y \circ x = x$
and so $(1)$ is satisfied.
By definition of the left operation:
:$\forall x, y \in S: y \circ x = y$
That is:
:$\forall x, y \in S: \exists z \in S: z \circ x = y$
where here $z = y$.
Hence $x$ fulfils the role of $y$ and $y$ fulfils the role of $z$ in:
:$\forall x, y \in S: y \circ x = x \implies \exists z \in S: z \circ x = y$
and so $(2)$ is satisfied.
But $\struct {S, \circ}$ is not a group.
== Sources ==
* : Chapter $\text I$: Algebraic Structures: $\S 7$: Semigroups and Groups: Exercise $7.16 \ \text {(e)}$
Category:Conditions under which Commutative Semigroup is Group
Cone on Compact Space is Compact

== Theorem ==
Let $A$ be a compact topological space.
Let $C A$ denote the cone on $A$.
Then, $C A$ is compact.
== Proof ==
By definition of cone:
:$C A = T \ast A$
where:
:$T$ denotes the trivial topological space
:$T \ast A$ denotes the join of $T$ and $A$
By Finite Topological Space is Compact, $T$ is compact.
Therefore, by Join of Compact Spaces is Compact:
:$C A$ is compact.
Category:Compact Spaces
Cones on Homeomorphic Spaces are Homeomorphic

== Theorem ==
Let $X, Y$ be topological spaces.
Suppose:
:$X \sim Y$
denoting that $X$ is homeomorphic to $Y$.
Then:
:$C X \sim C Y$
where $C X$ denotes the cone on $X$.
== Proof ==
Let $T$ be the trivial topological space used in the definition of cone.
We have:
:$T \sim T$ by Homeomorphism Relation is Equivalence
:$X \sim Y$ 
Then, by Joins of Homeomorphic Spaces are Homeomorphic:
:$C X \sim C Y$
Category:Homeomorphisms (Topological Spaces)
Cones or Cylinders are Equal iff Bases are Reciprocally Proportional to Heights

== Theorem ==
== Proof ==
:
Let there be cones and cylinders which are similar.
Let the circles $\map c {ABCD}$ and $\map c {EFGH}$ be their bases.
Let $KL$ and $MN$ be the axes of the cones and cylinders.
Let $L$ and $N$ be the apices of the cones.
Thus:
:let $KL$ be the height of the cone $\map c {ABCDL}$ and the cylinder $AO$
:let $MN$ be the height of the cone $\map c {EFGHM}$ and the cylinder $EP$.
It is to be shown that:
:$\map c {ABCD} : \map c {EFGH} = MN : KL$
That is, the bases of the cones and cylinders are reciprocally proportional to their heights.
Either $LK = MN$ or $LK \ne MN$.
First suppose $LK = MN$.
We have that $AO = EP$.
From :
:$\map c {ABCD} = \map c {EFGH}$
Thus:
:$\map c {ABCD} : \map c {EFGH} = MN : KL$
, suppose $LK \ne MN$.
Suppose $MN > LK$.
Let $QN$ be cut off from $MN$ equal to $KL$.
Let the cylinder $EP$ be cut by the plane $TUS$ through $Q$ parallel to the planes holding the circles $\map c {EFGH}$ and $\map c {RP}$.
Let the cylinder $ES$ be described with the circle $\map c {EFGH}$ as its base and with height $NQ$.
We have that the cylinder $AO$ equals the cylinder $EP$.
Therefore from :
:$AO : ES = EP : ES$
But from :
:$AO : ES = \map c {ABCD} : \map c {EFGH}$
and from :
:$EP : ES = MN : QN$
Therefore from :
:$\map c {ABCD} : \map c {EFGH} = MN : QN$
But:
:$QN = KL$
Therefore:
:$\map c {ABCD} : \map c {EFGH} = MN : KL$
Therefore in the cylinders $AO$ and $EP$, the bases are reciprocally proportional to their heights.
Let the bases of the cylinders $AO$ and $EP$ be reciprocally proportional to their heights:
:$\map c {ABCD} : \map c {EFGH} = MN : KL$
It is to be proved that cylinders $AO$ and $EP$ are equal.
Let $QN$ be cut off from $MN$ equal to $KL$.
Let the cylinder $EP$ be cut by the plane $TUS$ through $Q$ parallel to the planes holding the circles $\map c {EFGH}$ and $\map c {RP}$.
Let the cylinder $ES$ be described with the circle $\map c {EFGH}$ as its base and with height $NQ$.
We have that $KL = QN$.
Thus:
:$\map c {ABCD} : \map c {EFGH} = MN : QN$
But from :
:$\map c {ABCD} : \map c {EFGH} = AO : ES$
and from :
:$MN : QN = EP : ES$
Therefore from :
:$AO : ES = EP : ES$
Therefore from :
:$AO = EP$
== Sources ==
* : Book $\text{XII}$. Propositions
Category:Right Circular Cylinders
Category:Cones
Confidence Interval/Examples/95 percent

== Examples of Confidence Intervals ==
A $95 \%$ confidence interval is a confidence interval whose $\alpha$ parameter is:
:$\alpha = 0 \cdotp 05$
Let $\bar x$ be the mean of a sample of $n$ observations from a normal distribution with unknown mean $\mu$ and known standard deviation $\sigma$.
Then a $95 \%$ confidence interval for $\mu$ is:
:$\closedint {\bar x - \dfrac {1 \cdotp 96 \sigma} {\sqrt n} } {\bar x + \dfrac {1 \cdotp 96 \sigma} {\sqrt n} }$
== Sources ==
* 
* 
Category:Examples of Confidence Intervals
Confidence Level/Examples

== Examples of Confidence Levels ==
Common values of confidence levels are:
:$90 \%$, corresponding to $\alpha = 0 \cdotp 10$
:$95 \%$, corresponding to $\alpha = 0 \cdotp 05$
:$99 \%$, corresponding to $\alpha = 0 \cdotp 01$
:$99 \cdotp 9 \%$, corresponding to $\alpha = 0 \cdotp 001$
== Sources ==
* 
* 
Category:Examples of Confidence Levels
Confocal Conics are Self-Orthogonal

== Theorem ==
The confocal conics defined by:
:$\quad \dfrac {x^2} {a^2} + \dfrac {y^2} {a^2 - c^2} = 1$
forms a family of orthogonal trajectories which is self-orthogonal.
:
== Proof ==
Consider:
:$(1): \quad \dfrac {x^2} {a^2} + \dfrac {y^2} {a^2 - c^2} = 1$
From Equation of Confocal Ellipses: Formulation 2:
:$(1)$ defines an ellipse when $a^2 > c^2$.
From Equation of Confocal Hyperbolas: Formulation 2:
:$(1)$ defines a hyperbola when $a^2 < c^2$.
Thus it is seen that $(1)$ is that of a conic section.
We use the technique of formation of ordinary differential equation by elimination.
Differentiating $x$ gives:
:$\dfrac {2 x} {a^2} + \dfrac {2 y} {a^2 - c^2} \dfrac {\d y} {\d x} = 0$
so
:$\dfrac {\d y} {\d x} = - \dfrac {a^2 - c^2} {a^2} \dfrac x y$
Now we need to eliminate $c$ from the above.
We go back to $(1)$:
Substituting for $a^2 - c^2$:
This is separable, so separate it:
:$\ds \int \frac {\d y} y = -\int \frac {x \rd x} {a^2 - x^2}$
which leads to:
Now $k^2$ is still arbitrary at this point, so set:
:$k^2 = \dfrac {a^2 - c^2} {a^2}$
Substituting this into the above gives us:
:$\dfrac {x^2} {a^2} + \dfrac {y^2} {a^2 - c^2} = 1$
which is $(1)$.
Hence the result by definition of self-orthogonal.
== Sources ==
* : $1$: The Nature of Differential Equations: : Miscellaneous Problems for Chapter $1$: $6$
* 
* 
Category:Orthogonal Trajectories
Category:Conic Sections
Category:Orthogonal Curves
Conformality is Equivalence Relation on Set of Riemannian Metrics

== Theorem ==
Let $M$ be a smooth manifold.
Let $\set g$ be the set of Riemannian metrics on $M$.
Suppose $\set g$ is equipped with the conformality relation.
That is, suppose that for any pair of Riemannian metrics there is a conformal transformation.
Then conformality relation is equivalence relation.
== Proof ==
== Sources ==
* : $\S 3$: Model Riemannian Manifolds. Euclidean Spaces
Category:Riemannian Geometry
Category:Examples of Equivalence Relations
Confusion of Bound Variables

== Mistake ==
Let $\mathbf A$ be the WFF of predicate logic:
:$\forall x: ( \exists y: x < y )$
Suppose we wished to substitute $y$ for $x$.
If we paid no heed to whether $y$ were free for $x$, we would obtain:
:$\forall y: ( \exists y: y < y )$
This is plainly false for the natural numbers, but $\forall x: \exists y: x < y$ is true (just take $y = x + 1$).
This problem is called confusion of bound variables.
== Resolution ==
In formal systems with the language of predicate calculus, substitution is restricted to substitution for free occurrences.
Since $x$ is in the WFF $(\exists y: x < y)$, it is by definition not the case that $y$ is free for $x$.
Thus this substitution is not asserted to preserve truth.
Category:Fallacies and Mistakes
Congruence (Number Theory)/Examples

== Examples of Congruence ==
=== Congruence Modulo $1$ ===
=== Congruence Modulo $2 \pi$ as Angular Measurement ===
=== Example: $8 \equiv -1 \pmod 3$ ===
=== Example: $42 \equiv 18 \pmod 8$ ===
=== Example: $365 \equiv 1 \pmod 7$ ===
=== Example: $1 \cdotp 6 \equiv 0 \cdotp 6 \pmod 1$ ===
=== Example: $5 \cdotp 74 \equiv -3 \cdotp 26 \pmod 3$ ===
Category:Examples of Congruence (Number Theory)
Congruence (Number Theory)/Examples/Modulo 1

== Example of Congruence Modulo an Integer ==
Let $x \equiv y \pmod 1$ be defined as congruence on the real numbers modulo $1$:
:$\forall x, y \in \R: x \equiv y \pmod 1 \iff \exists k \in \Z: x - y = k$
That is, if their difference $x - y$ is an integer.
The equivalence classes of this equivalence relation are of the form:
:$\eqclass x 1 = \set {\dotsc, x - 2, x - 1, x, x + 1, x + 2, \dotsc}$
Each equivalence class has exactly one representative in the half-open real interval:
:$\hointr 0 1 = \set {x \in \R: 0 \le x < 1}$
== Also see ==
* Definition:Modulo 1
== Sources ==
* : Chapter $2$. Equivalence Relations: Exercise $1 \ \text {(i)}$
* : Chapter $2$: 'And you do addition?': $\S 2.3$: Relations: Example $2.3.6$
Category:Examples of Congruence (Number Theory)
Congruence (Number Theory)/Examples/Modulo 2 pi as Angular Measurement

== Example of Congruence Modulo an Integer ==
Let $\RR$ denote the relation on the real numbers $\R$ defined as:
:$\forall x, y \in \R: \tuple {x, y} \in \RR \iff \text {$x$ and $y$}$ measure the same angle in radians
Then $\RR$ is the congruence relation modulo $2 \pi$.
The equivalence classes of this equivalence relation are of the form:
:$\eqclass \theta {2 \pi} = \set {\theta + 2 k \pi: k \in \Z}$
:
Hence for example:
:$\eqclass 0 {2 \pi} = \set {2 k \pi: k \in \Z}$
and:
:$\eqclass {\dfrac \pi 2} {2 \pi} = \set {\dfrac {\paren {4 k + 1} \pi} 2: k \in \Z}$
Each equivalence class has exactly one representative in the half-open real interval:
:$\hointr 0 {2 \pi} = \set {x \in \R: 0 \le x < 2 \pi}$
and have a one-to-one correspondence with the points on the circumference of a circle.
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Relations
Category:Examples of Congruence (Number Theory)
Congruence (Number Theory)/Integers/Examples/Modulo 1

== Example of Congruence Modulo an Integer ==
Let $x \equiv y \pmod 1$ be defined on the integers as congruence modulo $1$:
:$\forall x, y \in \Z: x \equiv y \pmod 1 \iff \exists k \in \Z: x - y = k$
That is, if their difference $x - y$ is an integer.
The equivalence classes of this equivalence relation is the set of integers:
:$\eqclass x 1 = \Z$
== Sources ==
* : $\S 2.5$. Congruence of integers: Example $38$
Category:Examples of Congruence Modulo Integer
Congruence (Number Theory)/Integers/Examples/Modulo 2

== Example of Congruence Modulo an Integer ==
Let $x \equiv y \pmod 2$ be defined on the integers as congruence modulo $2$:
:$\forall x, y \in \Z: x \equiv y \pmod 2 \iff \exists k \in \Z: x - y = 2 k$
That is, if their difference $x - y$ is an even integer.
The equivalence classes of this equivalence relation are of the form:
:$\eqclass x 2 = \set {\dotsc, x - 4, x - 2, x, x + 2, x + 4, \dotsc}$
which are:
Each equivalence class has exactly one representative in the set $\set {0, 1}$.
== Sources ==
* : Chapter $2$. Equivalence Relations: Exercise $5 \ \text{(i)}$
* : Chapter $2$: 'And you do addition?': $\S 2.3$: Relations: Ponderable $2.3.4$
Category:Examples of Congruence Modulo Integer
Congruence (Number Theory)/Integers/Examples/Modulo 3

== Example of Congruence Modulo an Integer ==
Let $x \mathrel \RR y$ be the equivalence relation defined on the integers as congruence modulo $3$:
:$x \mathrel \RR y \iff x \equiv y \pmod 3$
defined as:
:$\forall x, y \in \Z: x \equiv y \pmod 3 \iff \exists k \in \Z: x - y = 3 k$
That is, if their difference $x - y$ is a multiple of $3$.
The equivalence classes of this equivalence relation are of the form:
:$\eqclass x 3 = \set {\dotsc, x - 6, x - 3, x, x + 3, x + 6, \dotsc}$
which are:
Thus the partition of $\Z$ induced by $\RR$ is:
:$\Bbb S = \set {\eqclass 0 3, \eqclass 1 3, \eqclass 2 3}$
Each equivalence class has exactly one representative in the set $\set {0, 1, 2}$.
So, for example, $\eqclass {17} 3 = \set {\dotsc -1, 2, 5, \dotsc, 14, 17, 20, 23, \dotsc}$
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Relations
Category:Examples of Congruence Modulo Integer
Congruence (Number Theory) is Congruence Relation

== Theorem ==
Congruence modulo $m$ is a congruence relation on $\struct {\Z, +}$.
== Proof ==
Suppose $a \equiv b \bmod m$ and $c \equiv d \bmod m$.
Then by the definition of congruence there exists $k, k' \in \Z$ such that:
:$\paren {a - b} = k m$
:$\paren {c - d} = k' m$
Hence:
:$\paren {a - b} + \paren {c - d} = k m + k' m$
Using the properties of the integers:
:$\paren {a + c} - \paren {b + d} = m \paren {k + k'}$
Hence $\paren {a + c} \equiv \paren {b + d} \bmod m$ and congruence modulo $m$ is a congruence relation.
== Sources ==
* : $\text{I}$: Groups: $\S 1$: Semigroups, Monoids and Groups
Category:Integers