nyuuzyou/wikis huggingface dataset:
Green's Theorem

== Theorem ==
Let $\Gamma$ be a positively oriented piecewise smooth Jordan curve in $\R^2$.
Let $U = \Int \Gamma$, that is, the interior of $\Gamma$.
Let $A$ and $B$ be functions of $\tuple {x, y}$ defined on an open region containing $U$ and have continuous partial derivatives in such a set.
Then:
:$\ds \oint_\Gamma \paren {A \rd x + B \rd y} = \iint_U \paren {\frac {\partial B} {\partial x} - \frac {\partial A} {\partial y} } \rd x \rd y$
where the is a contour integral.
== Proof ==
It suffices to demonstrate the theorem for rectangular regions in the $x y$-plane.
The Riemann-sum nature of the double integral will then guarantee the proof of the theorem for arbitrary regions, because a Riemann-sum is technically a summation of the areas of arbitrarily small rectangles. 
As the proof is for a rectangle, the proof will work for arbitrary regions, which can be approximated by collections of ever smaller rectangles.
Let $R = \set {\tuple {x, y}: a \le x \le b, c \le y \le d}$ be a rectangular region.
Let the boundary $C$ of $R$ be oriented counterclockwise.
We break the boundary into $4$ pieces:
:$C_1$, which runs from $\tuple {a, c}$ to $\tuple {b, c}$
:$C_2$, which runs from $\tuple {b, c}$ to $\tuple {b, d}$
:$C_3$, which runs from $\tuple {b, d}$ to $\tuple {a, d}$
:$C_4$, which runs from $\tuple {a, d}$ to $\tuple {a, c}$
Then:
We note that $y$ is constant along $C_1$ and $C_3$.
So:
:$\ds \int_{C_1} B \rd y = \int_{C_3} B \rd y = 0$
Hence:
A similar argument demonstrates that:
:$\ds \iint_R \frac {\partial A} {\partial y} \rd x \rd y = -\oint_C A \d x$
and hence:
:$\ds \oint_C \paren {A \rd x + B \rd y} = \iint_R \paren {\frac {\partial B} {\partial x} - \frac {\partial A} {\partial y} } \rd x \rd y$
== Also see ==
* Gauss-Ostrogradsky Theorem
* Stokes' Theorem
== Sources ==
* 
* 
* 
Category:Green's Theorem
Category:Integral Calculus
Greenwood's Conjecture

== Famous False Conjecture ==
Let $p$ be a prime number.
Let $T_p$ be the $p$th triangular number.
The conjecture states that:
:If $T_p$ is even, then $T_p + 1$ is prime
:If $T_p$ is odd, then $T_p - 2$ is prime.
== Refutation ==
We have that $T_{31} = 496$, but $497 = 7 \times 71$ and so is not prime.
== Historical Note ==
== Sources ==
* : $496$
* : $496$
Category:Famous False Conjectures
Category:Triangular Numbers
Category:Prime Numbers
Gregorian Calendar/Mistake

== Source Work ==
:
:The Dictionary
::$3333$
== Mistake ==
:The Gregorian calendar is approximately $1$ day ahead every $3333$ years.
== Correction ==
Using a value of $365 \cdotp 2422$ days per tropical year, it can be established that the Gregorian calendar starts one day later relative to the tropical year after $3333$ years (to the nearest day).
Using the more precise value of $365 \cdotp 24219 \, 878$ days in the tropical year, it is shown in Discrepancy between Gregorian Year and Tropical Year that the number of Gregorian years after which there is a day's discrepancy is in fact $3319$ (to the nearest day).
In of $1997$, uses the more precise $365 \cdotp 24219 \, 878$, to obtain an approximate value of $3320$ such years.
The section $3333$ has been dropped from of $1997$.
== Sources ==
* : $3333$
Category:Mistakes/Curious and Interesting Numbers
Gregory Series

== Theorem ==
For $-\dfrac \pi 4 \le \theta \le \dfrac \pi 4$:
:$\theta = \tan \theta - \dfrac 1 3 \tan^3 \theta + \dfrac 1 5 \tan^5 \theta - \ldots$
This is called the Gregory series.
== Proof ==
By the root test the radius of convergence is $-\dfrac \pi 4 \le \theta \le \dfrac \pi 4$.
== Also presented as ==
== Also known as ==
== Also see ==
* Leibniz's Formula for Pi
== Historical Note ==
== Sources ==
* 
* 
Category:Gregory Series
Category:Tangent Function
Gregory Series/Also presented as

== Gregory Series: Also presented as == 
The Gregory series can also be seen in the form:
:$\tan^{-1} \theta = \theta - \dfrac {\theta^3} 3 + \dfrac {\theta^5} 5 - \ldots$
which is valid for $-1 \le x \le 1$.
== Sources ==
* 
* 
Category:Gregory Series
Grelling-Nelson Paradox

== Paradox ==
Define an adjective to be autological if it is true when applied to itself.
For instance, the word English is autological, as it is a word in English.
The word multisyllabic is also autological, as it contains multiple syllables.
Define an adjective to be heterological if it is not true when applied to itself.
For instance, the word long is heterological, as it is not a long word.
The word monosyllabic is also heterological, as it does not contain only one syllable.
All adjectives must either be autological or heterological, as they either apply to themselves or they don't.
Autological and heterological are thus each defined as the negation of the other.
The Grelling-Nelson paradox arises when trying to apply one of these adjectives to the word heterological.
If heterological is autological, then it does apply to itself.
But then by the definition of heterological, it does not apply to itself, making it heterological.
If heterological is heterological, then it does not apply to itself.
But then by the definition of heterological, it is not the case that heterological does not apply to itself, and so it does apply to itself, making it autological.
We thus have:
:heterological is autological $\iff$ heterological is heterological
for the contradictory predicates heterological and autological.
This paradox is closely related to :
:$\not \exists x: \forall y: \paren {\map \RR {x, y} \iff \neg \map \RR {y, y} }$
Define $x$ to be the predicate heterological, and define the relation $\map \RR {x, y}$ to mean:
:the predicate $x$ applies to $y$
The sentence:
:$\map \RR {x, y} \iff \neg \map \RR {y, y}$
is thus interpreted:
:$y$ is heterological $y$ does not apply to itself.
The assertion that the predicate heterological exists and is defined as such for all $y$ is then equivalent to:
:$\forall y: \paren {\map \RR {x, y} \iff \neg \map \RR {y, y} }$
This leads to the same contradiction as -- when we instantiate $y$ as $x$, we obtain:
:$\map \RR {x, x} \iff \neg \map \RR {x, x}$
== Also known as ==
== Historical Note ==
== Sources ==
* 
* 
Category:Grelling-Nelson Paradox
Category:Antinomies
Category:Semantic Paradoxes
Grimm's Conjecture

== Conjecture ==
Let $n$ and $k$ be positive integers.
Let $n+1, n+2, \dots, n+k$ be composite numbers.
Then there exists a finite sequence $\left\langle{p_i}\right\rangle$ of $k$ distinct primes such that $p_i$ divides $n + i$ for each natural number $i$ such that $1 \le i \le k$.
== Sources ==
* 
Category:Number Theory
Category:Unproven Hypotheses
Grothendieck Universe is Closed under Binary Cartesian Product

== Theorem ==
Let $\mathbb U$ be a Grothendieck universe.
Let $u, v \in \mathbb U$.
Let $u \times v$ be the binary cartesian product of $u$ and $v$ realized as a set of ordered pairs in Kuratowski formalization.
Then $u \times v \in \mathbb U$.
== Proof ==
From Binary Cartesian Product in Kuratowski Formalization contained in Power Set of Power Set of Union:
:$u \times v \subseteq \powerset {\powerset {u \cup v} }$ 
Then:
Category:Grothendieck Universes
Grothendieck Universe is Closed under Binary Union

== Theorem ==
Let $\mathbb U$ be a Grothendieck universe.
Let $u, v \in \mathbb U$.
Then $u \cup v \in \mathbb U$.
== Proof ==
If $\mathbb U = \O$, the claim is true.
Assume $\mathbb U \ne \O$.
By Nonempty Grothendieck Universe contains Von Neumann Natural Numbers, every von Neumann natural number is an element of $\mathbb U$.
In particular:
:$2 = \set {\O, \set \O} \in \mathbb U$
Using $2$ as an indexing set, we remember that $0 = \O$ and $1 = \set \O$, and define:
This sets up the structures needed to exploit Grothendieck Universe: Axiom $(4)$ below.
Then:
Category:Grothendieck Universes
Grothendieck Universe is Closed under Mappings

== Theorem ==
Let $\mathbb U$ be a Grothendieck universe.
Let $u, v \in \mathbb U$.
Let $f: u \to v$ be a mapping realized as a relation consisting of ordered pairs in Kuratowski formalization.
Then $f \in \mathbb U$.
== Proof ==
Let $u \times v$ be the finite cartesian product of $u$ and $v$ realized as a set of ordered pairs in Kuratowski formalization.
By definition of mapping, we have $f \subseteq u \times v$.
Then:
Category:Grothendieck Universes
Group/Examples

== Examples of Groups ==
=== Example: $\dfrac {x + y} {1 + x y}$ ===
=== Example: $x + y + 2$ over $\R$ ===
=== Example: $x + y + x y$ over $\R \setminus \set {-1}$ ===
=== Example: $x^{-1} = 1 - x$ ===
=== Example: Group of Linear Functions ===
=== Example: Operation Induced by Self-Inverse and Cancellable Elements ===
=== Example: $\tuple {a c, a d + b}$ on $\R_{\ge 0} \times \R$ ===
Category:Examples of Groups
Group/Examples/Linear Functions

== Theorem ==
Let $G$ be the set of all real functions $\theta_{a, b}: \R \to \R$ defined as:
:$\forall x \in \R: \map {\theta_{a, b} } x = a x + b$
where $a, b \in \R$ such that $a \ne 0$.
The algebraic structure $\struct {G, \circ}$, where $\circ$ denotes composition of mappings, is a group.
$\struct {G, \circ}$ is specifically non-abelian.
== Proof ==
We verify the group axioms, in the following order (for convenience):
=== ===
Let $\theta_{a, b}$ and $\theta_{c, d}$ be elements of $G$.
From Composition of Linear Real Functions:
:$\theta_{c, d} \circ \theta_{a, b} = \theta_{a c, b c + d}$
As $a \ne 0$ and $c \ne 0$ it follows that $a c \ne 0$.
Thus $\theta_{a c, b c + d} \in G$
This proves closure of $\circ$.
=== ===
We have that Composition of Mappings is Associative.
=== ===
We assert that $\theta_{1, 0}$ is the identity with respect to $\circ$.
So indeed $\theta_{1, 0}$ is the identity for $\circ$.
=== ===
For any element $\theta_{a, b} \in S$, we claim that $\theta_{1 / a, \, -b / a}$ is the inverse for $\theta_{a, b}$.
From Inverse of Linear Function on Real Numbers:
:$\theta^{-1}_{a, b} = \theta_{1 / a, \, -b / a}$
Verifying this:
and:
We conclude that $\theta_{1 / a, \, -b / a}$ is indeed the inverse for $\theta_{a, b}$.
It follows that $\struct {G, \circ}$ is indeed a group.
=== Non-Abelian ===
Let $\theta_{a, b}$ and $\theta_{c, d}$ be elements of $G$.
From Condition for Composition of Linear Real Functions to be Commutative, it is not true in general that $\theta_{c, d} \circ \theta_{a, b} = \theta_{a, b} \circ \theta_{c, d}$.
Thus $\struct {G, \circ}$ is non-abelian by definition.
== Sources ==
* : Chapter $4$. Groups: Exercise $15$
Category:Examples of Groups
Group/Examples/Self-Inverse and Cancellable Elements

== Theorem ==
Let $S$ be a set with an operation which assigns to each $\tuple {a, b} \in S \times S$ an element $a \ast b \in S$ such that:
:$(1): \quad \exists e \in S: a \ast b = e \iff a = b$
:$(2): \quad \forall a, b, c \in S: \paren {a \ast c} \ast \paren {b \ast c} = a \ast b$
Then $\struct {S, \circ}$ is a group, where $\circ$ is defined as $a \circ b = a \ast \paren {e \ast b}$.
== Proof ==
We verify the group axioms, in the following order (for convenience):
=== ===
Let $a, b \in S$. Then from the definition of $\ast$, we have $a \ast b \in S$, and hence also $a \circ b = a \ast \paren {e \ast b} \in S$.
This proves closure of $\circ$.
=== ===
We assert that $e$ is the identity with respect to $\circ$.
We verify this as follows (let $a \in S$):
It follows that $a \circ e = e \circ a = a \ast e$.
It remains to prove the following identity:
:$(3): \quad a \ast e = a$
By property $(1)$, it suffices to prove $a \ast \paren {a \ast e} = e$:
So indeed $e$ is the identity for $\circ$.
=== ===
For any element $a \in S$, we claim that $e \ast a$ is the inverse for $a$.
This is verified as follows:
We conclude that $e \ast a$ is indeed the inverse for $a$.
=== ===
Lastly, we prove that $\circ$ is associative, i.e.:
:$a \circ \paren {b \circ c} = \paren {a \circ b} \circ c$
To this end, we observe the following property of $\ast$:
Subsequently, we compute:
This somewhat cumbersome calculation shows that we indeed have associativity.
It follows that $\struct {S, \circ}$ is indeed a group.
== Also see ==
* $B$-Algebra, where it can be seen that the operation $*$ as defined here is related to that in the definition of a $B$-Algebra.
== Sources ==
* : Chapter $2$: The Definition of Group Structure: $\S 28 \delta$
Category:Examples of Groups
Group/Examples/ac, ad+b on Positive Reals by Reals

== Example of Group ==
Let $\R$ denote the set of real numbers.
Let $\R_{\ge 0}$ denote the set of positive real numbers.
Let $S = \R_{\ge 0} \times \R$ denote the Cartesian product of $\R_{\ge 0}$ and $\R$.
Let $\oplus: S \to S$ be the operation on $S$ defined as:
:$\forall \tuple {a, b}, \tuple {c, d} \in S: \tuple {a, b} \oplus \tuple {c, d} := \tuple {a c, a d + b}$
Then the algebraic structure $\struct {S, \oplus}$ is a group.
== Proof ==
Taking the group axioms in turn:
=== ===
Let $\tuple {a, b}, \tuple {c, d} \in S$ be arbitrary.
We have that:
Then from Real Numbers form Field we haveL
:$a d + b \in \R$
Thus $\tuple {a c, a d + b} \in \R_{\ge 0} \times \R$ and so $\struct {S, \oplus}$ is closed.
=== ===
Let $\tuple {a, b}, \tuple {c, d}, \tuple {e, f} \in S$ be arbitrary.
Thus $\paren {\tuple {a, b} \oplus \tuple {c, d} } \oplus \tuple {e, f} = \tuple {a, b} \oplus \paren {\tuple {c, d} \oplus \tuple {e, f} }$ and so $\struct {S, \oplus}$ is associative.
=== ===
We have:
Thus $\tuple {1, 0}$ is the identity element of $\struct {S, \oplus}$.
=== ===
We have that $\tuple {1, 0}$ is the identity element of $\struct {S, \oplus}$.
Hence we need to find $\tuple {c, d} \in S$ such that $\tuple {a, b} \oplus \tuple {c, d} = \tuple {1, 0}$.
Hence:
and:
Thus every element $\tuple {a, b}$ of $\struct {S, \oplus}$ has an inverse $\tuple {\dfrac 1 a, -\dfrac b a}$.
All the group axioms are thus seen to be fulfilled, and so $\struct {S, \oplus}$ is a group.
== Sources ==
* : Chapter $\text {II}$: New Structures from Old: $\S 8$: Compositions Induced on Subsets: Exercise $8.7 \ \text {(a)}$
Category:Examples of Groups
Group/Examples/inv x = 1 - x

== Theorem ==
Let $S = \set {x \in \R: 0 < x < 1}$.
Then an operation $\circ$ can be found such that $\struct {S, \circ}$ is a group such that the inverse of $x \in S$ is $1 - x$.
== Proof ==
Define $f: \openint 0 1 \to \R$ by:
:$\map f x := \map \ln {\dfrac {1 - x} x}$
Let us show that $f$ is a bijection by constructing an inverse mapping $g: \R \to \openint 0 1$:
:$\map g z := \dfrac 1 {1 + \exp z}$
=== Lemma 1 ===
=== Lemma 2 ===
Thus $f$ is a bijection.
Let $\struct {\R, +}$ be the additive group on $\R$.
Now define $\circ := +_f$ to be the operation induced on $\openint 0 1$ by $f$ and $+$:
:$x \circ y := \map {f^{-1} } {\map f x + \map f y}$
Let us determine the behaviour of $\circ$ more explicitly:
We see that $\circ$ is commutative.
Let $x = \dfrac 1 2$, $\dfrac {1 - x} x = 1$ so that:
so that $\dfrac 1 2$ is the identity element for $\circ$.
Furthermore, putting $y = 1 - x$, the following obtains:
establishing $1 - x$ to be the inverse of $x$, as desired.
That $\circ$ in fact determines a group on $\openint 0 1$ follows from Pullback of Group is Group.
== Sources ==
* : Chapter $2$: The Definition of Group Structure: $\S 26 \mu$
Category:Examples of Groups
Category:Group/Examples/inv x = 1 - x
Group/Examples/inv x = 1 - x/Lemma 1

== Lemma for Group Example: $x^{-1} = 1 - x$ ==
Define $f: \openint 0 1 \to \R$ by:
:$\map f x := \map \ln {\dfrac {1 - x} x}$
and $g: \R \to \openint 0 1$:
:$\map g z := \dfrac 1 {1 + \exp z}$
Then:
:$\map {f \circ g} x = x$
== Proof ==
Category:Group/Examples/inv x = 1 - x
Group/Examples/inv x = 1 - x/Lemma 2

== Lemma for Group Example: $x^{-1} = 1 - x$ ==
Define $f: \openint 0 1 \to \R$ by:
:$\map f x := \map \ln {\dfrac {1 - x} x}$
and $g: \R \to \openint 0 1$:
:$\map g z := \dfrac 1 {1 + \exp z}$
Then:
:$\map {g \circ f} x = x$
== Proof ==
Category:Group/Examples/inv x = 1 - x
Group/Examples/x+y+2 over Reals

== Example of Group ==
Let $\circ: \R \times \R$ be the operation defined on the real numbers $\R$ as:
:$\forall x, y \in \R: x \circ y := x + y + 2$
Then $\struct {\R, \circ}$ is a group whose identity is $-2$.
== Proof ==
Taking the group axioms in turn:
=== ===
$\forall x, y \in \R: x + y + 2 \in \R$
Thus $x \circ y \in \R$ and so $\struct {\R, \circ}$ is closed.
=== ===
Thus $\circ$ is associative.
=== ===
Let $y$ be such that $x \circ y = x$.
Then:
Then it is noted that:
Thus $-2$ is the identity element of $\struct {\R, \circ}$.
=== ===
We have that $-2$ is the identity element of $\struct {\R, \circ}$.
So:
Then it is noted that:
Thus every element of $\struct {\R, \circ}$ has an inverse $-x - 4$.
All the group axioms are thus seen to be fulfilled, and so $\struct {\R, \circ}$ is a group.
== Sources ==
* : Chapter $1$: Definitions and Examples: Exercise $1 \ \text{(b)}$
Category:Examples of Groups
Group/Examples/x+y+xy over Reals less -1

== Example of Group ==
Let $\circ: \R \times \R$ be the operation defined on the real numbers $\R$ as:
:$\forall x, y \in \R: x \circ y := x + y + x y$
Let:
:$\R' := \R \setminus \set {-1}$
that is, the set of real numbers without $-1$.
Then $\struct {\R', \circ}$ is a group whose identity is $0$.
== Proof ==
From Monoid Example: Operation Defined as $x + y + x y$ on Real Numbers, $\struct {\R, \circ}$ is a monoid whose identity is $0$.
Let $x \in \R$.
We have that:
demonstrating that $-1$ is a zero for $\circ$ on $\R$.
Thus for $\struct {S, \circ}$ to be a group, where $S \subseteq \R$, it is necessary for $-1 \notin S$.
It has been established that $0$ is the identity of $\struct {\R, \circ}$.
Let $y \in \R'$.
Then:
Thus the inverse of $x$ is $-\dfrac x {1 + x}$, which exists for all $x$ except for $x = -1$.
Suppose $x, y \in \R \setminus \set {-1}$ such that $x \circ y = -1$.
Then:
So we have that $\struct {\R', \circ}$ is closed.
Thus all the group axioms are fulfilled, and so $\struct {\R', \circ}$ is a group.
== Sources ==
* : Chapter $4$. Groups: Exercise $11$
Category:Examples of Groups
Group/Examples/x+y over 1+xy

== Theorem ==
Let $G = \set {x \in \R: -1 < x < 1}$ be the set of all real numbers whose absolute value is less than $1$.
Let $\circ: G \times G \to \R$ be the binary operation defined as:
:$\forall x, y \in G: x \circ y = \dfrac {x + y} {1 + x y}$
The algebraic structure $\struct {G, \circ}$ is a group.
=== Isomorphic to Real Numbers ===
== Proof ==
Let $-1 < x, y, z < 1$.
We check the group axioms in turn:
=== ===
Thus $\circ$ has been shown to be associative.
=== ===
Similarly, putting $y = 0$ we find $x \circ y = x$.
So $0$ is the identity.
=== ===
Similarly, putting $x = -y$ gives us $\paren {-y} \circ y = 0$.
So each $x$ has an inverse $-x$.
=== ===
First note that:
:$-1 < x, y < 1 \implies x y > -1 \implies 1 + x y > 0$
Next:
Finally:
Thus:
:$-1 < x, y < 1 \implies -1 < x \circ y < 1$
and we see that in this range, $\circ$ is closed.
Thus the given set and operation form a group.
== Sources ==
* : Chapter $2$: The Definition of Group Structure: $\S 26 \lambda$
Category:Examples of Groups
Category:Examples of Groups/x+y over 1+xy
Group/Examples/x+y over 1+xy/Isomorphic to Real Numbers

== Theorem ==
Let $G = \set {x \in \R: -1 < x < 1}$ be the set of all real numbers whose absolute value is less than $1$.
Let $\circ: G \times G \to G$ be the binary operation defined as:
:$\forall x, y \in G: x \circ y = \dfrac {x + y} {1 + x y}$
It has been established in Group Examples: $\dfrac {x + y} {1 + x y}$ that $\struct {G, \circ}$ forms a group.
$\struct {G, \circ}$ is isomorphic to the additive group of real numbers $\struct {\R, +}$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $2$: Group Homomorphism and Isomorphism: $\S 62 \epsilon$
Category:Examples of Groups/x+y over 1+xy
Group/Examples/x+y over 1+xy/Isomorphic to Real Numbers/Proof 1

== Theorem ==
Let $G := \set {x \in \R: -1 < x < 1}$ be the set of all real numbers whose absolute value is less than $1$.
Let $\circ: G \times G \to G$ be the binary operation defined as:
:$\forall x, y \in G: x \circ y = \dfrac {x + y} {1 + x y}$
== Proof ==
To prove $G$ is isomorphic to $\struct {\R, +}$, we need to find a bijective homorphism $\phi: \openint {-1} 1 \to \R$:
:$\forall x, y \in G: \map \phi {x \circ y} = \map \phi x + \map \phi y$
From Group Examples: $\dfrac {x + y} {1 + x y}$:
:the identity element of $G$ is $0$
:the inverse of $x$ in $G$ is $-x$.
This also holds for $\struct {\R, +}$.
This hints at the structure of a possible isomorphism, namely:
:$(1): \quad$ that it is an odd function
:$(2): \quad$ that it passes through $0$
:$(3): \quad$ that it is defined on the open interval $\openint {-1} 1$.
One such function is the inverse hyperbolic arctangent function $\tanh^{-1}$, which is indeed defined on $\openint {-1} 1$ and has the above properties.
Hence:
This demonstrates the homorphism between $\struct {G, \circ}$ and $\struct {\R, +}$.
We have that Real Inverse Hyperbolic Tangent Function is Strictly Increasing over $\openint {-1} 1$.
Hence from Strictly Monotone Real Function is Bijective, $\tanh^{-1}: \openint {-1} 1 \to \R$ is a bijection.
Hence the result that $\struct {G, \circ} \cong \struct {\R, +}$.
== Also see ==
* Hyperbolic Tangent of Sum
== Sources ==
* https://math.stackexchange.com/questions/934974/proving-a-defined-group-g-is-isomorphic-to-mathbbr
Category:Examples of Groups/x+y over 1+xy
Group/Examples/x+y over 1+xy/Isomorphic to Real Numbers/Proof 2

== Theorem ==
Let $G := \set {x \in \R: -1 < x < 1}$ be the set of all real numbers whose absolute value is less than $1$.
Let $\circ: G \times G \to G$ be the binary operation defined as:
:$\forall x, y \in G: x \circ y = \dfrac {x + y} {1 + x y}$
== Proof ==
To prove $G$ is isomorphic to $\struct {\R, +}$, it is sufficient to find a bijective homorphism $\phi: \to \R \to G$:
:$\forall x, y \in G: \map \phi {x + y} = \map \phi x \circ \map \phi y$
From Group Examples: $\dfrac {x + y} {1 + x y}$:
:the identity element of $G$ is $0$
:the inverse of $x$ in $G$ is $-x$.
This also holds for $\struct {\R, +}$.
This hints at the structure of a possible such $\phi$:
:$(1): \quad$ that it is an odd function
:$(2): \quad$ that it passes through $0$
:$(3): \quad$ that its image is the open interval $\openint {-1} 1$.
One such function is the hyperbolic tangent function $\tanh$, which indeed has the above properties on $\R$.
Then we have that:
This demonstrates the homorphism between $\struct {\R, +}$ and $\struct {G, \circ}$.
We have that Real Hyperbolic Tangent Function is Strictly Increasing over $\R$.
Hence from Strictly Monotone Real Function is Bijective, $\tanh: \R \to \openint {-1} 1$ is a bijection.
Hence the result that $\struct {\R, +} \cong \struct {G, \circ}$.
Category:Examples of Groups/x+y over 1+xy
Group/Non-Group Examples/Arbitrary Order 4

== Example of Algebraic Structure which is not a Group ==
Let $S = \set {1, 2, 3, 4}$.
Consider the algebraic structure $\struct {S, \circ}$ given by the Cayley table:
:$\begin{array}{r|rrrr}
\circ & 1 & 2 & 3 & 4 \\
\hline
1 & 1 & 2 & 3 & 4
\\
2 & 2 & 4 & 3 & 1
\\
3 & 3 & 2 & 4 & 3
\\
4 & 4 & 3 & 1 & 2
\\
\end{array}$
Then $\struct {S, \circ}$ is not a group.
== Proof ==
It is immediately seen that $\struct {S, \circ}$ violates the Latin Square Property.
For example, both the column headed $3$ and the row headed $3$ have $2$ instances of $3$ in them.
== Sources ==
* : Chapter $\text{II}$: Groups: Exercise $\text{B vi}$
Category:Examples of Groups
Group Action/Examples/Cyclic Group on Polygon

== Example of Group Action ==
Consider the cyclic group $C_n$ defined as $\gen g$ whose identity is $e$.
Let $P_n$ be a regular $n$-sided polygon.
Then $C_n$ acts on on $P_n$ by the mapping for which, for each vertex $x$ of $P_n$:
:$e x = x$
:$g^k x$ is the vertex obtained when $P_n$ is rotated through $\dfrac {2 \pi k} n$ radians about the center of $P_n$.
== Sources ==
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Category:Examples of Group Actions