nyuuzyou/wikis huggingface dataset:
Triple with Sum and Product Equal

== Theorem ==
For $a, b, c \in \Z$, $a \le b \le c$, the solutions to the equation:
:$a + b + c = a b c$
are:
:$\tuple {1, 2, 3}$
:$\tuple {-3, -2, -1}$
and the trivial solution set:
:$\set {\tuple {-z, 0, z}: z \in \N}$
== Proof ==
Suppose one of $a, b, c$ is zero.
Then $a b c = 0 = a + b + c$.
The remaining two numbers sum to $0$, giving the solution set:
:$\set {\tuple {-z, 0, z}: z \in \N}$
Suppose $a < 0$ and $0 < b \le c$.
Then $a b c \le a < a + b + c$.
Hence equality never happens.
Similarly, for $a \le b < 0$ and $c > 0$:
:$a b c \ge c > a + b + c$
Hence equality never happens.
Now it remains the case $0 < a \le b \le c$.
Suppose $a \ge 2$.
Then:
hence $a = 1$.
Suppose $b \ge 3$.
Then:
hence $b = 1$ or $b = 2$.
For $b = 1$:
:$c = 1 + 1 + c$
which is a contradiction.
For $b = 2$:
:$2 c = 1 + 2 + c$
giving $c = 3$.
Therefore $\tuple {1, 2, 3}$ is the only solution for strictly positive integer values of $a, b, c$.
Similarly, for $a \le b \le c < 0$, we have $0 < -c \le -b \le -a$.
Also:
and we have already show that $-c = 1, -b = 2, -a = 3$.
Therefore we have $\tuple {a, b, c} = \tuple {-3, -2, -1}$.
We have considered all possible signs of $a, b, c$.
Hence the result.
Category:6
Triples of Consecutive Sphenic Numbers

== Theorem ==
The sequence of triplets of consecutive sphenic numbers starts:
:$\tuple {1309, 1310, 1311}, \tuple {1885, 1886, 1887}, \tuple {2013, 2014, 2015}, \ldots$
== Proof ==
Note that there cannot be quadruplets of such numbers, since one of the quadruplets must be divisible by $4$, making it non-sphenic.
We have:
hence each number above is sphenic.
== Sources ==
* : $1309$
Category:Sphenic Numbers
Triples with Same Sum and Same Product/Mistake

== Source Work ==
:
:$\mathbf D$: Diophantine Equations
::$\mathbf {D 16}$: Triples with Same Sum and Same Product
== Mistake ==
:It may be of interest to ask for the smallest sums or products with each multiplicity. For example, for $4$ triples, finds the smallest common sum to be $118$ ... and the smallest common product to be $25200$ ...
== Correction ==
In the article cited by , in fact does no such thing. Instead, he raises the question for $5$ such triples.
== Also see ==
* Smallest Number which is Sum of 4 Triples with Equal Products
* Smallest Integer which is Product of 4 Triples all with Same Sum
== Sources ==
* 
* 
* : $\mathbf {D 16}$: Triples with the same sum and same product
Category:Mistakes/Unsolved Problems in Number Theory
Triplet in Arithmetic Sequence with equal Divisor Sum

== Theorem ==
The smallest triple of integers in arithmetic sequence which have the same divisor sum is:
:$\map {\sigma_1} {267} = \map {\sigma_1} {295} = \map {\sigma_1} {323} = 360$
== Proof ==
We have that:
demonstrating that $267, 295, 323$ are in arithmetic sequence with common difference $28$.
Then:
== Sources ==
* : $267$
Category:Divisor Sum Function
Category:Arithmetic Sequences
Category:267
Category:295
Category:323
Triplets of Products of Two Distinct Primes

== Theorem ==
The following triplets of consecutive positive integers are the smallest in which each number is the product of $2$ distinct prime numbers:
:$33, 34, 35$
:$85, 86, 87$
:$93, 94, 95$
:$141, 142, 143$
:$201, 202, 203$
:$213, 214, 215$
:$217, 218, 219$
== Proof ==
Taking each triplet in turn:
It is noted that the triplet:
:$121, 122, 123$
while consisting of semiprimes, not all of these are the product of $2$ distinct prime numbers, as $121 = 11^2$.
== Sources ==
* : $33$
:::but beware a mistake: $85, 86, 87$ is omitted.
Category:Semiprimes
Triplets of Products of Two Distinct Primes/Mistake

== Source Work ==
:
:The Dictionary
::$33$
== Mistake ==
:The triplet $33, 34, 35$ is the smallest in which each number is the product of $2$ distinct primes. ... The next such triplets are: $93, 94, 95$; $141, 142, 143$; $201, 202, 203$; $213, 214, 215$; $217, 218, 219$; ...
The next such triplet is in fact $85, 86, 87$, which is omitted from the above list.
== Sources ==
* : $33$
Category:Mistakes/Curious and Interesting Numbers
Trisecting the Angle

== Classic Problem ==
Trisecting the angle is dividing a given angle into three equal parts.
== Solution ==
Trisecting the Angle by Compass and Straightedge Construction is Impossible.
Various techniques were devised which used more elaborate apparatus than just the straightedge and compass:
=== Archimedean Spiral ===
=== Neusis Construction ===
=== By Parabola ===
=== By Hyperbola ===
=== Quadratrix of Hippias ===
=== Cissoid of Diocles ===
=== Conchoid of Nicomedes ===
=== Trisectrix of Maclaurin ===
== Fallacious Proofs ==
== Also see ==
* Squaring the Circle
* Doubling the Cube
== Historical Note ==
== Sources ==
* : Chapter $\text{II}$: Modern Minds in Ancient Bodies
* : $3$
* : $3$
* 
* 
* 
* 
Category:Trisecting the Angle
Category:Compass and Straightedge Constructions
Category:Euclidean Geometry
Category:Classic Problems
Trisecting the Angle/Archimedean Spiral

== Theorem ==
Let $\alpha$ be an angle which is to be trisected.
This can be achieved by means of an Archimedean spiral.
== Construction ==
In the figure, the blue line is an Archimedean spiral.
Let $\angle AOB$ be the angle to be trisected, where $OB$ is on the polar axis.
Trisect the line segment $OA$ such that $OC = \dfrac 1 3 OA$.
Draw a circle with $O$ as the center and $OC$ as the radius.
Let $D$ be the intersection of the circle and the spiral.
Then:
:$\angle DOB = \dfrac 1 3 \angle AOB$
== Proof ==
Let the equation of the Archimedean spiral be $r = a \theta$.
Then:
== Also see ==
* Trisection of Angle by Neusis Construction
== Historical Note ==
Category:Trisecting the Angle
Category:Archimedean Spiral
Trisecting the Angle/Cissoid of Diocles

== Theorem ==
Let $\alpha$ be an angle which is to be trisected.
This can be achieved by means of a cissoid of Diocles.
== Construction ==
== Proof ==
== Also see ==
* Trisection of Angle by Neusis Construction
== Historical Note ==
== Sources ==
* : $3$
Category:Cissoid of Diocles
Category:Trisecting the Angle
Trisecting the Angle/Conchoid of Nicomedes

== Theorem ==
Let $\alpha$ be an angle which is to be trisected.
This can be achieved by means of a conchoid of Nicomedes.
== Construction ==
== Proof ==
== Also see ==
* Trisection of Angle by Neusis Construction
== Historical Note ==
== Sources ==
* : $3$
* : $3$
Category:Conchoid of Nicomedes
Category:Trisecting the Angle
Trisecting the Angle/Fallacious Proof/Edw. J. Goodwin

== Fallacious Proof of Trisecting the Angle ==
:The trisection of a right line taken as the chord of any arc of a circle trisects the angle of the arc.
:::: -- 
== Resolution ==
Basically this says $\sin x = x$.
Make the angle large.
Draw a $135 \degrees$ arc, so that trisection is obviously at $45 \degrees$ and $90 \degrees$.
Observe that the middle piece of the chord is too short for the chord to be trisected.
== Sources ==
* 
Category:Trisecting the Angle
Category:Fallacies
Trisecting the Angle/Historical Note

== Historical Note on Trisecting the Angle ==
The exercise to trisect of the general angle using a compass and straightedge construction, was an exercise that the ancient Greeks failed to succeed in.
This was one of three such problems: the other two being Squaring the Circle and Doubling the Cube.
There are several techniques available that use other tools, but these were considered unacceptably vulgar to the followers of .
== Sources ==
* : Chapter $\text{II}$: Modern Minds in Ancient Bodies
* : $3$
* : $3$
* 
* 
Category:Historical Notes
Trisecting the Angle/Hyperbola

== Theorem ==
Let $\alpha$ be an angle which is to be trisected.
This can be achieved by means of a hyperbola.
However, the points on the hyperbola that are required for this construction cannot be found by using only a straightedge and compass.
== Construction ==
== Proof ==
== Also see ==
* Trisection of Angle by Parabola
* Trisection of Angle by Neusis Construction
* Trisection of Angle by Archimedean Spiral
== Historical Note ==
== Sources ==
* : $3$
* : $3$
Category:Trisecting the Angle
Category:Hyperbolas
Trisecting the Angle/Neusis Construction

== Theorem ==
Let $\alpha$ be an angle which is to be trisected.
This can be achieved by means of a neusis construction.
== Construction ==
Let there be a neusis ruler $S$ marked by two points $p$ and $q$.
:
Let the vertex of $\alpha$ be labelled $B$.
Let $A$ be the point of one of the branches of $\alpha$ such that the distance $AB$ equals $pq$.
Let the other branch of $\alpha$ be extended past $B$ through $D$.
Let a circle be drawn with center at $B$ with radius $AB$.
Let the straightedge $S$ be placed so that:
:$S$ passes over $A$
:$p$ lies on the circle at $C$
:$q$ lies on the straight line $BD$ at $D$.
Then the angle $3 \angle CBD = \alpha$
Thus $\alpha$ has been trisected.
== Proof ==
We have that $\angle BCD + \angle ACB$ make a straight angle.
As $CD = AB$ by construction, $CD = BC$ by definition of radius of circle.
Thus $\triangle BCD$ is isosceles.
By Isosceles Triangle has Two Equal Angles:
:$\angle CBD = \angle CDB$
From Sum of Angles of Triangle equals Two Right Angles:
:$\angle BCD + 2 \angle CBD$ equals two right angles.
Thus:
:$2 \angle CBD = \angle ACB$
Similarly, by Isosceles Triangle has Two Equal Angles:
:$\angle ACB = \angle CAB$
and again from Sum of Angles of Triangle equals Two Right Angles:
:$\angle ABC + 2 \angle ACB$ equals two right angles.
and so:
:$\angle ABC + 4 \angle CBD$ equals two right angles.
But $\alpha + \angle ABC + \angle CBD$ make a straight angle.
Thus:
:$\alpha + \angle ABC + \angle CBD = \angle ABC + 4 \angle CBD$
and so:
:$\alpha = 3 \angle CBD$
== Also see ==
* Trisection of Angle by Archimedean Spiral
== Sources ==
* : Chapter $2$: The Logic of Shape: Problems for the Greeks
Category:Trisecting the Angle
Category:Neusis Constructions
Trisecting the Angle/Parabola

== Theorem ==
Let $\alpha$ be an angle which is to be trisected.
This can be achieved by means of a parabola.
However, the points on the parabola that are required for this construction cannot be found by using only a straightedge and compass.
== Construction ==
Let $\angle POQ$ be the angle which is to be trisected.
:
Let the parabola $\PP$ be constructed whose equation is $y = 2 x^2$.
Construct the circle $\CC_1$ whose center is at $O$ and whose radius is $1$.
By Equation of Circle, this has the equation:
:$x^2 + y^2 = 1$
Let $\CC$ intersect $OQ$ at $A$.
Let $AB$ be constructed parallel to the $x$-axis to intersect the $y$-axis at $B$.
Let $AB$ be bisected at $C$.
Let $CE$ be constructed perpendicular to $AB$.
Let $DE$ be tangent to the circle $\CC_1$ at the $y$-axis
Hence $E$ is the intersection of $CE$ and $DE$.
Construct the circle $\CC_2$ whose center is at $E$ which passes through $O$.
Let $F$ be the point at which the circle $\CC_2$ intersects the parabola $\PP$.
Let $FG$ be dropped perpendicular to the $x$-axis.
Let $FG$ intersect the circle $\CC_1$ at $H$.
The angle $\angle POH$ is the required trisection of $\angle POQ$.
== Proof ==
First, notice that because $A$ lies on $\CC_1$:
:$A = \tuple {\cos \angle POQ, \sin \angle POQ}$
This means:
:$B = \tuple {0, \sin \angle POQ}$
Because $C$ is the midpoint of $AB$:
:$C = \tuple {\dfrac {\cos \angle POQ} 2, \sin \angle POQ}$
Because $D$ lies on $\CC_1$:
:$D = \tuple {0, 1}$
and so:
:$E = \tuple {\dfrac {\cos \angle POQ} 2, 1}$
From Equation of Circle, $C_2$ has the equation:
:$\paren {x - \dfrac {\cos \angle POQ} 2}^2 + \paren {y -1}^2 = \dfrac {\cos^2 \angle POQ} 4 + 1$
Because $F$ lies on both $\CC_2$ and $\PP$, we can solve for the $x$-coordinate of $F$:
Thus $x = 0$ or $4 x^3 - 3 x - \cos \angle POQ = 0$.
This confirms what is obvious by the construction, namely that $\CC_2$ and $\PP$ intersect at the origin.
We can also see that the solution we are after must lie in the first quadrant, where $x \ne 0$.
So there must be one positive $x$ such that:
:$4 x^3 - 3 x = \cos \angle POQ$
By the Triple Angle Formula for Cosine $\cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta$, it is clear that one solution that works is:
:$x = \map \cos {\dfrac {\angle POQ} 3}$
Because, by construction:
:there is a vertical line through $F$ and $H$
:$H$ lies on the unit circle $\CC_1$
it follows that:
:$H = \tuple {\cos \dfrac {\angle POQ} 3, \sin \dfrac {\angle POQ} 3}$
Therefore:
:$\angle POH = \dfrac {\angle POQ} 3$
== Also see ==
* Trisection of Angle by Neusis Construction
* Trisection of Angle by Archimedean Spiral
== Historical Note ==
== Sources ==
* : $3$
* : $3$
Category:Trisecting the Angle
Category:Parabolas
Trisecting the Angle/Quadratrix of Hippias

== Theorem ==
Let $\alpha$ be an angle which is to be trisected.
This can be achieved by means of a quadratrix of Hippias.
== Construction ==
== Proof ==
== Also see ==
* Trisection of Angle by Neusis Construction
== Historical Note ==
== Sources ==
* : $3$
* : $3$
Category:Quadratrix of Hippias
Category:Trisecting the Angle
Trisecting the Angle/Trisectrix of Maclaurin

== Theorem ==
Let $\alpha$ be an angle which is to be trisected.
This can be achieved by means of a trisectrix of Maclaurin.
== Construction ==
:
In the figure, the blue line is a trisectrix of Maclaurin.
Let us denote the trisectrix of Maclaurin by $\TT$.
Let $3 \theta$ be the angle to be trisected.
Position the vertex of $3 \theta$ at the point $\tuple {2 a, 0}$ with one arm along the $x$-axis.
Let the other arm of $3 \theta$ intersect the loop of $\TT$ at $P'$.
Let $OP'$ be constructed, where $O$ denotes the origin of the Cartesian plane in which $\TT$ is embedded.
Then:
:$\angle POP' = \theta$
and hence $3 \theta$ has been trisected.
== Proof ==
== Sources ==
* 
* 
Category:Trisecting the Angle
Category:Trisectrix of Maclaurin
Trisecting the Angle by Compass and Straightedge Construction is Impossible

== Theorem ==
There is no compass and straightedge construction for the trisection of the general angle.
== Proof ==
Let $OA$ and $OB$ intersect at $O$.
It will be shown that there is no general method using a compass and straightedge construction to construct $OC$ such that $\angle AOB = 3 \times \angle AOC$.
It is sufficient to demonstrate that this is impossible for one specific angle.
Hence we choose $\angle AOB = 60 \degrees$.
Let $A$ and $B$ be points on the unit circle whose center is at $\tuple {0, 0}$.
Let $A$ lie on the $x$-axis.
Thus:
:$O$ is the point $\tuple {0, 0}$
:$A$ is the point $\tuple {1, 0}$
:$B$ is the point $\tuple {\cos 60 \degrees, \sin 60 \degrees}$
These all belong to $\Q \sqbrk {\sqrt 3}$.
trisection of $AOB$ is equivalent to constructing the point $\tuple {\cos 20 \degrees, \sin 20 \degrees}$.
From Triple Angle Formula for Cosine:
:$\cos 3 \theta = 4 \cos^3 \theta - 3 \cos \theta$
so:
:$8 \cos^3 20 \degrees - 6 \cos 20 \degrees = 2 \cos 60 \degrees = 1$
Thus $\cos 20 \degrees$ is a root of the polynomial:
:$8 x^3 = 6 x - 1$
which by Irreducible Polynomial: $8 x^3 - 6 x - 1$ in Rationals is irreducible over $\Q$.
Thus $\cos 20 \degrees$ is algebraic over $\Q$ with degree $3$.
Thus by Algebraic Element of Degree 3 is not Element of Field Extension of Degree Power of 2, $\cos 20 \degrees$ is not an element of any extension of $\Q$ of degree $2^m$.
The result follows from Point in Plane is Constructible iff Coordinates in Extension of Degree Power of 2.
== Sources ==
* : Chapter $\text{II}$: Modern Minds in Ancient Bodies
* : Chapter $8$: Field Extensions: $\S 40$. Construction with Ruler and Compasses: Theorem $76$
* 
* 
* 
* 
* : Chapter $2$: The Logic of Shape: Problems for the Greeks
Category:Trisecting the Angle
Category:Compass and Straightedge Constructions
Trisectrix of Maclaurin has Asymptote

== Theorem ==
Let $\TT$ be the trisectrix of Maclaurin embedded in a Cartesian plane with the equation:
:$x^3 + x y^2 + a y^2 - 3 a x^2 = 0$
Then $\TT$ has one asymptote, which is the straight line $x = -a$.
== Proof ==
== Sources ==
* 
* 
Category:Trisectrix of Maclaurin
Trisectrix of Maclaurin has Loop

== Theorem ==
Let $\TT$ be the trisectrix of Maclaurin embedded in a Cartesian plane with the equation:
:$x^3 + x y^2 + a y^2 - 3 a x^2 = 0$
Then $\TT$ has a loop.
== Proof ==
== Sources ==
* 
* 
Category:Trisectrix of Maclaurin
Trisectrix of Maclaurin is Symmetrical about X-Axis

== Theorem ==
Let $\TT$ be the trisectrix of Maclaurin embedded in a Cartesian plane with the equation:
:$x^3 + x y^2 + a y^2 - 3 a x^2 = 0$
Then $\TT$ is symmetrical about the $x$-axis.
== Proof ==
== Sources ==
* 
* 
Category:Trisectrix of Maclaurin
Trivial Estimate for Cyclotomic Polynomials

== Theorem ==
Let $n \ge 1$ be a natural number.
Let $\Phi_n$ be the $n$th cyclotomic polynomial.
Let $\phi$ be the Euler totient function.
Let $z \in \C$ be a complex number.
Then:
:$\size {\size z - 1}^{\map \phi n} \le \size {\map {\Phi_n} z} \le \paren {\size z + 1}^{\map \phi n}$
where:
:the first inequality becomes an equality only if:
:::$n = 1$ and $z \in \R_{\ge 0}$
::or:
:::$n = 2$ and $z \in \R_{\le 0}$
:the second inequality becomes an equality only if:
:::$n = 1$ and $z \in \R_{\le 0}$
::or:
:::$n = 2$ and $z \in \R_{\ge 0}$
== Proof ==
Category:Cyclotomic Polynomials
Trivial Gradation is Gradation

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring.
Let $\struct {M, e, \cdot}$ be a monoid.
Let
:$\ds R = \bigoplus_{m \mathop \in M} R_m$
be the trivial $M$-gradation on $R$.
This is a gradation on $R$.
== Proof ==
We are required to show that:
:$\forall x \in R_m, y \in R_n: x \circ y \in R_{m \cdot n}$
First suppose that $m = n = e$ are both the identity.
In this case, $R_m = R_n = R$.
Since by definition, $R$ is closed under $\circ$, it follows that
:$\forall x \in R, y \in R: x \circ y \in R$
as required.
Now suppose that either $m \ne e$ or $n \ne e$.
After possibly exchanging $m$ and $n$, we may as well assume that $n \ne e$.
In particular, $R_n = \mathbf 0$ is the zero ring.
So if $y \in R_n$, then $y = 0$.
Therefore, for every $x \in R_m$, by Ring Product with Zero, we must have 
:$x \circ y = x \circ 0 = 0$
Since $R_{m \cdot n}$ is an abelian group it must by definition contain $0$.
Therefore $x \circ y \in R_{m \cdot n}$ as required.
Category:Ring Theory
Trivial Group Action is Group Action

== Theorem ==
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $S$ be a non-empty set.
Let $*: G \times S \to S$ be the trivial group action:
:$\forall \tuple {g, s} \in G \times S: g * s = s$
Then $*$ is indeed a group action.
== Proof ==
The group action axioms are investigated in turn.
Let $g_1, g_2 \in G$ and $s \in S$.
Thus:
demonstrating that holds.
Then:
demonstrating that holds.
Category:Group Actions
Category:Trivial Group Action
Trivial Group is Abelian

== Theorem ==
The trivial group is an abelian group.
== Proof ==
From Trivial Group is Cyclic Group, it is shown that the algebraic structure $\struct {\set e, \circ}$ such that $e \circ e = e$ is a cyclic group.
The result follows from Cyclic Group is Abelian.
Category:Trivial Group
Category:Abelian Groups
Trivial Group is Cyclic Group

== Theorem ==
The trivial group is a cyclic group.
== Proof ==
In Trivial Group is Group it is shown that the algebraic structure $\struct {\set e, \circ}$ such that $e \circ e = e$ is in fact a group.
It remains to be shown that it is cyclic.
In order for $G$ to be a cyclic group, every element $x$ of $G$ has to be expressible in the form $x = g^n$ for some $g \in G$ and some $n \in \Z$.
In this case, for every integer $n$, every element of $G$ can be expressed in the form $e^n$.
Thus $G$ is trivially a cyclic group.
== Sources ==
* : $\S 5.4$. Cyclic groups: Example $100$
* : $\S 1.2$
Category:Trivial Group
Category:Cyclic Groups
Trivial Group is Group

== Theorem ==
The trivial group is a group.
== Proof ==
Let $G = \struct {\set e, \circ}$ be an algebraic structure.
=== ===
For $G$ to be a group, it must be closed.
So it must be the case that:
:$\forall e \in G: e \circ e = e$
=== ===
$\circ$ is associative:
:$e \circ \paren {e \circ e} = e = \paren {e \circ e} \circ e$
trivially.
=== ===
$e$ is the identity:
:$\forall e \in G: e \circ e = e$
=== ===
Every element of $G$ (all one of them) has an inverse:
This follows from the fact that the Identity is Self-Inverse, and the only element of $G$ is indeed the identity:
:$e \circ e = e \implies e^{-1} = e$
== Sources ==
* : $\S 4.5$. Examples of groups: Example $78$
* : $\S 1.3$: Example $10$
* : Chapter $2$: The Definition of Group Structure: $\S 26$
Category:Trivial Group
Trivial Group is Initial Object

== Theorem ==
Let $\mathbf{Grp}$ be the category of groups.
Let $1 = \set e$ be the trivial group.
Then $1$ is an initial object of $\mathbf{Grp}$.
== Proof ==
Let $\struct {G, \circ}$ be a group with identity $e_G$.
By Group Homomorphism Preserves Identity, any hypothetical group homomorphism $\phi: 1 \to G$ must satisfy:
:$\map \phi e = e_G$
Let us define the mapping $\phi$ in this way.
By Equality of Mappings, only one such mapping $1 \to G$ can exist, establishing uniqueness.
Now to verify that $\phi$ is actually a group homomorphism.
Since $1$ has only the element $e$, this is verified by:
where we used $*$ to denote the group operation on $1$.
Thus $\phi$ is a group homomorphism.
We have thus established that there is a unique morphism $1 \to G$ in $\mathbf{Grp}$ for all $G$.
That is, $1$ is an initial object in $\mathbf{Grp}$.
== Sources ==
* : $\S 2.2$: Example $2.11$: $3$
Category:Category of Groups
Category:Trivial Group
Trivial Group is Smallest Group

== Theorem ==
Let $G = \struct {\set e, \circ}$ be a trivial group.
Then $G$ is the smallest group possible, in that there exists no set with lower cardinality which is the underlying set of a group.
== Proof ==
From Trivial Group is Group, we have that there does exist a group of cardinality $1$.
From Group is not Empty, there can be no group of smaller order.
== Sources ==
* : $1$
Category:Trivial Group
Trivial Group is Terminal Object of Category of Groups

== Theorem ==
Let $\mathbf {Grp}$ be the category of groups.
Let $\set e$ be the trivial group.
Then $\set e$ is a terminal object of $\mathbf {Grp}$.
== Proof ==
Let $\struct {G, \circ}$ be any group.
By Singleton is Terminal Object of Category of Sets, there is precisely one mapping:
:$!: G \to \set e$
defined by:
:$\forall g \in G: ! (g) = e$
By definition, any group homomorphism is also a mapping.
Hence, there is at most one morphism $\struct {G, \circ} \to \set e$ in $\mathbf {Grp}$.
Now to verify that the mapping $!$ is a group homomorphism.
For any $g, h \in G$, we have (using $*$ for the group operation on $\set e$):
That is, $!$ is a group homomorphism.
Thus for all groups $\struct {G, \circ}$, there is a unique group homomorphism $!: G \to \set e$.
That is, $\set e$ is a terminal object of $\mathbf {Grp}$.
== Sources ==
* : $\S 2.2$: Example $2.11$: $3$
Category:Category of Groups
Category:Trivial Group
Trivial Module is Module

== Theorem ==
Let $\struct {G, +_G}$ be an abelian group whose identity is $e_G$.
Let $\struct {R, +_R, \circ_R}$ be a ring.
Let $\struct {G, +_G, \circ}_R$ be the trivial $R$-module, such that:
:$\forall \lambda \in R: \forall x \in G: \lambda \circ x = e_G$
Then $\struct {G, +_G, \circ}_R$ is a module.
== Proof ==
Checking the module axioms in turn:
:: $\quad \lambda \circ \paren {x +_G y} = e_G = e_G +_G e_G = \paren {\lambda \circ x} +_G \paren {\lambda \circ y}$
:: $\quad \paren {\lambda +_R \mu} \circ x = e_G = e_G +_G e_G = \paren {\lambda \circ x} +_G \paren {\mu \circ x}$
:: $\quad \paren {\lambda \times_R \mu} \circ x = e_G = \lambda \circ e_G = \lambda \circ \paren {\mu \circ x}$
Thus the trivial module is indeed a module.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 26$. Vector Spaces and Modules: Example $26.6$
Category:Module Theory