nyuuzyou/wikis huggingface dataset:
Uniqueness of Polynomial Ring in One Variable

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $\struct {R \sqbrk X, \iota, X}$ and $\struct {R \sqbrk Y, \kappa, Y}$ be polynomial rings in one variable over $R$.
Then there exists a unique ring homomorphism $f: R \sqbrk X \to R \sqbrk Y$ such that:
:$f \circ \iota = \kappa$
:$\map f X = Y$
and it is an isomorphism.
== Outline of proof ==
Using the universal property, we construct ring homomorphisms in both directions, and apply uniqueness to their compositions to show that they are mutual inverses.
== Proof ==
The existence and uniqueness of $f$ follows from the universal property.
Likewise, there is a unique ring homomorphism $g: R \sqbrk Y \to R \sqbrk X$ such that:
:$g \circ \kappa = \iota$
:$\map g Y = X$
and a unique ring homomorphism $h: R \sqbrk X \to R \sqbrk X$ such that:
:$h \circ \iota = \iota$
:$\map X = X$
By uniqueness and Identity Mapping is Ring Homomorphism, $h = \operatorname{id}$ is the identity mapping on $R \sqbrk X$.
Again by uniqueness and Composition of Ring Homomorphisms is Ring Homomorphism, $g\circ f = \operatorname{id}_{R \sqbrk X}$.
By symmetry, $f \circ g = \operatorname{id}_{R \sqbrk Y}$.
Thus $f$ is an isomorphism.
Category:Polynomial Theory
Uniqueness of Positive Root of Positive Real Number/Negative Exponent

== Theorem ==
Let $x \in \R$ be a real number such that $x > 0$.
Let $n \in \Z$ be an integer such that $n < 0$.
Then there is at most one $y \in \R: y \ge 0$ such that $y^n = x$.
== Proof ==
Let $m = -n$.
Let $g$ be the real function defined on $\hointr 0 \to$ defined by:
:$\map g y = y^m$
From the definition of power:
:$\map g y = \dfrac 1 {\map f y}$
Hence $\map g y$ is strictly decreasing.
From Strictly Monotone Mapping with Totally Ordered Domain is Injective:
:there is at most one $y \in \R: y \ge 0$ such that $y^n = x$.
Category:Uniqueness of Positive Root of Positive Real Number
Uniqueness of Positive Root of Positive Real Number/Positive Exponent

== Theorem ==
Let $x \in \R$ be a real number such that $x > 0$.
Let $n \in \Z$ be an integer such that $n > 0$.
Then there is at most one $y \in \R: y \ge 0$ such that $y^n = x$.
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
Category:Uniqueness of Positive Root of Positive Real Number
Uniqueness of Positive Root of Positive Real Number/Positive Exponent/Proof 1

== Theorem ==
== Proof ==
Let the real function $f: \hointr 0 \to \to \hointr 0 \to$ be defined as:
:$\map f y = y^n$
First let $n > 0$.
By Identity Mapping is Order Isomorphism, the identity function $I_\R$ on $\hointr 0 \to$ is strictly increasing.
We have that:
:$\map f y = \paren {\map {I_\R} y}^n$
By Product of Positive Strictly Increasing Mappings is Strictly Increasing, $f$ is strictly increasing on $\hointr 0 \to$.
From Strictly Monotone Mapping with Totally Ordered Domain is Injective:
:there is at most one $y \in \R: y \ge 0$ such that $y^n = x$.
Category:Uniqueness of Positive Root of Positive Real Number
Uniqueness of Positive Root of Positive Real Number/Positive Exponent/Proof 2

== Theorem ==
== Proof ==
We have that:
:$0 < y_1 < y_2 \implies y_1^n < y_2^n$
so there exists at most one $y \in \R: y \ge 0$ such that $y^n = x$.
== Sources ==
* : Chapter $1$: The Real and Complex Number Systems: Real Numbers: $1.37$. Theorem
Category:Uniqueness of Positive Root of Positive Real Number
Uniqueness of Positive Root of Positive Real Number/Positive Exponent/Proof 3

== Theorem ==
== Proof ==
To prove uniqueness, we must show that:
:$y_1^n = x = y_2^n$ implies $y_1 = y_2$
 that $y_1 \ne y_2$.
Then $y_1 < y_2$ or $y_2 < y_1$.
, assume that $y_1 < y_2$.
We will show by induction that $y_1^n < y_2^n$, contradicting the assumption that $y_1^n = x = y_2^n$.
=== Basis for the Induction ===
By assumption:
:$y_1^1 < y_2^1$
This is our basis for the induction.
=== Induction Hypothesis ===
We need to show that $y_1^n < y_2^n$ implies $y_1^{n + 1} < y_2^{n + 1}$.
So this is our induction hypothesis:
:$y_1^n < y_2^n$
=== Induction Step ===
By the induction hypothesis, $y_1^n < y_2^n$.
By assumption, both $y_1$ and $y_2$ are positive, giving the following choices:
:$y_1^n = 0 \quad \text{and} \quad y_2^n > 0$
:$y_1^n > 0 \quad \text{and} \quad y_2^n > 0$
The first case violates our assumption that $y_1^n = x = y_2^n$.
Assume the second case:
:$y_1^n > 0 \quad y_2^n > 0$
Then by the Real Number Axioms and our assumption that $y_1 < y_2$, the following hold:
:$y_1^n \cdot y_1 < y_2^n \cdot y_1$
:$y_2^n \cdot y_1 < y_2^n \cdot y_2$
This gives:
:$y_1^{n + 1} < y_2^{n + 1}$
completing the proof by induction.
Thus $y_1^n \ne y_2^n$, contradicting our assumption.
Therefore, for strictly positive $n \in \Z$ and $x \in \R$, there is a unique positive $y \in \R$ such that $y^n=x$.
Category:Uniqueness of Positive Root of Positive Real Number
Uniqueness of Product Measures

== Theorem ==
Let $\struct {X, \Sigma_1, \mu}$ and $\struct {Y, \Sigma_2, \nu}$ be measure spaces.
Let $\GG_1$ and $\GG_2$ be generators for $\Sigma_1$ and $\Sigma_2$, respectively.
Suppose that $\GG_1$ and $\GG_2$ are closed under intersection.
Suppose further that there are exhausting sequences $\sequence {G_{1, n} }_{n \mathop \in \N}$ and $\sequence {G_{2, n} }_{n \mathop \in \N}$ in $\GG_1$ and $\GG_2$, respectively, such that:
:$\forall n \in \N: \map \mu {G_{1, n} } < \infty$
:$\forall n \in \N: \map \nu {G_{2, n} } < \infty$
Then there is at most one measure $\rho$ on the product space $\struct {X \times Y, \Sigma_1 \otimes \Sigma_2}$ such that:
:$\forall G_1 \in \GG_1, G_2 \in \GG_2: \map \rho {G_1 \times G_2} = \map \mu {G_1} \, \map \nu {G_2}$
That is, there can be at most one product measure on $\struct {X \times Y, \Sigma_1 \otimes \Sigma_2}$.
== Proof ==
== Sources ==
* : $13.4$
Category:Product Measure
Uniqueness of Real z such that x Choose n+1 Equals y Choose n+1 Plus z Choose n

== Theorem ==
Let $n \in \Z_{\ge 0}$ be a positive integer.
Let $x, y \in \R$ be real numbers which satisfy:
:$n \le y \le x \le y + 1$
Then there exists a unique real number $z$ such that:
:$\dbinom x {n + 1} = \dbinom y {n + 1} + \dbinom z n$
where $n - 1 \le z \le y$.
== Proof ==
We have:
== Sources ==
* : Problem $13.31 \ \text {(a)}$
* : $\S 1.2.6$: Binomial Coefficients: Exercise $66$
Category:Binomial Coefficients
Uniqueness of Representing Objects

== Theorem ==
Let $C$ be a locally small category.
Let $\mathbf{Set}$ be the category of sets.
Let $F : \mathbf C \to \mathbf{Set}$ be a covariant functor.
Let $(A, \eta)$ and $(B, \xi)$ be representations of $F$.
Then there exists a unique isomorphism $f : A \to B$ such that $\eta \circ h^f = \xi$, where:
:$h^f$ is the precomposition natural transformation
:$\circ$ denotes vertical composition of natural transformations
== Proof ==
Category:Category Theory
Unit-Speed Admissible Curve is Critical Point of Riemannian Length iff Geodesic

== Theorem ==
Let $\struct {M, g}$ be a Riemannian manifold.
Let $\gamma$ be a unit-speed admissible curve.
Let $L_g$ the Riemannian length of some admissible curve.
Then $\gamma$ is the critical point of $L_g$ $\gamma$ is geodesic.
== Proof ==
== Sources ==
* : $\S 6$: Geodesics and Distance. Minimizing Curves Are Geodesics
Category:Riemannian Manifolds
Category:Curves
Category:Geodesics
Unit-Speed Minimizing Curve is Geodesic

== Theorem ==
Let $\struct {M, g}$ be a Riemannian manifold.
Let $I = \closedint a b$ be a closed real interval.
Let $\gamma : I \to M$ be an admissible curve.
Suppose $\gamma$ is parametrized so that $\gamma$ is a unit-speed curve.
Suppose $\gamma$ is minimizing.
Then $\gamma$ is a geodesic.
== Proof ==
== Sources ==
* : $\S 6$: Geodesics and Distance. Minimizing Curves Are Geodesics
Category:Riemannian Manifolds
Category:Curves
Category:Geodesics
Unit Cylinder as Surface of Revolution

== Theorem ==
Let $\struct {\R^3, d}$ be the Euclidean space.
Let $S_C \subseteq \R^3$ be the surface of revolution.
Let $C$ be a straight line in the open upper half-plane.
Let the smooth local parametrization of $C$ be:
:$\map \gamma t = \tuple {t, 1}$
Then the induced metric on $S_C$ is:
:$g = d t^2 + d \theta^2$
== Proof ==
We have that:
:$\map {\gamma'} t = \tuple {1, 0}$
Hence, $\map \gamma t$ is a unit-speed curve.
By the corollary of the induced metric on the surface of revolution:
:$g = d t^2 + d \theta^2$
== Sources ==
* : $\S 2$: Riemannian Metrics. Methods for Constructing Riemannian Metrics
Category:Surfaces of Revolution
Category:Solid Geometry
Category:Induced Metrics
Unit Ideal iff Radical is Unit Ideal

== Theorem ==
Let $A$ be a commutative ring with unity.
Let $\mathfrak a$ be an ideal of $A$.
Then:
:$ \mathfrak a = A \iff \map \Rad {\mathfrak a} = A$
where:
:$A$ is called the unit ideal of $A$
:$\map \Rad {\mathfrak a}$ denotes the radical of $\mathfrak a$
== Proof ==
Category:Radical of Ideals
Unit Ideal is Principal Ideal Generated by Unity

== Theorem ==
Let $A$ be a commutative ring with unity.
Then:
:$A = \ideal 1$
where:
:$A$ is called the unit ideal of $A$
:$\ideal 1$ denotes the principal ideal generated by the unity of $A$
== Proof ==
$\ideal 1 \subseteq A$ is clear by definition of principal ideal.
To see $A \subseteq \ideal 1$, let $a \in A$ be an arbitrary element.
Then:
Category:Commutative Algebra
Category:Ideal Theory
Unit Interval is Path-Connected in Real Numbers

== Theorem ==
Let $\R$ be the real number line with the usual (Euclidean} metric.
The closed unit interval $\mathbf I = \closedint 0 1$ is a path-connected metric subspace of $\R$.
== Proof ==
Follows directly from Subset of Real Numbers is Path-Connected iff Interval.
== Sources ==
* : Chapter $\text{III}$: Metric Spaces: Path-Connectedness
Category:Connected Spaces
Category:Real Intervals
Unit Matrix is Identity for Matrix Multiplication

== Theorem ==
Let $R$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\map {\MM_R} n$ denote the metric space of square matrices of order $n$ over $R$.
Let $\mathbf I_n$ denote the unit matrix of order $n$:
Then:
:$\forall \mathbf A \in \map {\MM_R} n: \mathbf A \mathbf I_n = \mathbf A = \mathbf I_n \mathbf A$
That is, the unit matrix $\mathbf I_n$ is the identity element for (conventional) matrix multiplication over $\map {\MM_R} n$.
== Proof ==
=== Lemma: Left Identity ===
=== Lemma: Right Identity ===
Thus:
$\mathbf A \mathbf I_n = \mathbf A = \mathbf I_n \mathbf A$
Hence, by definition, $\mathbf I_n$ is an identity element for (conventional) matrix multiplication over $\map {\MM_R} n$.
That $\mathbf I_n$ is the identity element follows from Identity is Unique.
== Sources ==
* : Chapter $1$: Definitions and Examples: Example $1.7$
* : Part $\text I$: Matrices and vector spaces: $1$ Matrices: $1.2$ Addition and multiplication of matrices: $10$
Category:Conventional Matrix Multiplication
Category:Unit Matrices
Category:Unit Matrix is Identity for Matrix Multiplication
Unit Matrix is Identity for Matrix Multiplication/Left

== Theorem ==
Let $R$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $m, n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\map {\MM_R} {m, n}$ denote the $m \times n$ metric space over $R$.
Let $I_m$ denote the unit matrix of order $m$.
Then:
:$\forall \mathbf A \in \map {\MM_R} {m, n}: \mathbf I_m \mathbf A = \mathbf A$
== Proof ==
Let $\sqbrk a_{m n} \in \map {\MM_R} {m, n}$.
Let $\sqbrk b_{m n} = \mathbf I_m \sqbrk a_{m n}$.
Then:
Thus $\sqbrk b_{m n} = \sqbrk a_{m n}$ and $\mathbf I_m$ is shown to be a left identity.
Category:Unit Matrix is Identity for Matrix Multiplication
Unit Matrix is Identity for Matrix Multiplication/Right

== Theorem ==
Let $R$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $m, n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\map {\MM_R} {m, n}$ denote the $m \times n$ metric space over $R$.
Let $I_n$ denote the unit matrix of order $n$.
Then:
:$\forall \mathbf A \in \map {\MM_R} {m, n}: \mathbf A \mathbf I_n = \mathbf A$
== Proof ==
Let $\sqbrk a_{m n} \in \map {\MM_R} {m, n}$.
Let $\sqbrk b_{m n} = \sqbrk a_{m n} \mathbf I_n$.
Then:
Thus $\sqbrk b_{m n} = \sqbrk a_{m n}$ and $\mathbf I_n$ is shown to be a right identity.
Category:Unit Matrix is Identity for Matrix Multiplication
Unit Matrix is Orthogonal

== Theorem ==
The unit matrix $\mathbf I_n$ of order $n$ is orthogonal.
== Proof ==
By Unit Matrix is its own Inverse the inverse $I_n^{-1}$ of $I_n$ is $I_n$.
By definition a unit matrix is a diagonal matrix.
Hence by Diagonal Matrix is Symmetric:
:$I_n = I_n^\intercal$
where $I_n^\intercal$ is the transpose of $I_n$.
Thus:
:$I_n^{-1} = I_n^\intercal$
and the result follows by definition of orthogonal.
Category:Unit Matrices
Category:Orthogonal Matrices
Unit Matrix is Proper Orthogonal

== Theorem ==
The unit matrix $\mathbf I_n$ of order $n$ is proper orthogonal.
== Proof ==
By Unit Matrix is Orthogonal, $\mathbf I_n$ is an orthogonal matrix.
By Determinant of Unit Matrix, the determinant of $\mathbf I_n$ is equal to $1$.
The result follows by definition of proper orthogonal.
Category:Unit Matrices
Category:Orthogonal Matrices
Unit Matrix is Unity of Ring of Square Matrices

== Theorem ==
Let $R$ be a ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Let $\struct {\map {\MM_R} n, +, \times}$ denote the ring of square matrices of order $n$ over $R$.
The unit matrix over $R$:
:$\mathbf I_n = \begin {pmatrix} 1_R & 0_R & 0_R & \cdots & 0_R \\ 0_R & 1_R & 0_R & \cdots & 0_R \\ 0_R & 0_R & 1_R & \cdots & 0_R \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0_R & 0_R & 0_R & \cdots & 1_R \end {pmatrix}$
is the identity element of $\struct {\map {\MM_R} n, +, \times}$.
== Proof ==
In Unit Matrix is Identity for Matrix Multiplication, it is demonstrated that:
:$\forall \mathbf A \in \map {\MM_R} n: \mathbf A \mathbf I_n = \mathbf A = \mathbf I_n \mathbf A$ 
Hence the result, by definition of identity element
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 29$. Matrices
Category:Rings of Square Matrices
Category:Unit Matrices
Unit Matrix is its own Inverse

== Theorem ==
The inverse of the unit matrix $\mathbf I_n$ of order $n$ is $\mathbf I_n$.
That is, a unit matrix it its own inverse.
== Proof ==
By definition, a unit matrix is a diagonal matrix.
From Inverse of Diagonal Matrix, the inverse of a diagonal matrix:
:$\mathbf D = \begin{bmatrix}
a_{11} & 0 & \cdots & 0 \\
0 & a_{22} & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & a_{nn} \\
\end{bmatrix}$
is the diagonal matrix:
: $\mathbf D^{-1} = \begin{bmatrix}
\dfrac 1 {a_{11}} & 0 & \cdots & 0 \\
0 & \dfrac 1 {a_{22}} & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \dfrac 1 {a_{nn}} \\
\end{bmatrix}$
When $\mathbf D$ is the unit matrix $\mathbf I_n$, all elements $a_{kk}$ are equal to $1$:
:$\mathbf I_n = \begin{bmatrix}
1 & 0 & \cdots & 0 \\
0 & 1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & 1 \\
\end{bmatrix}$
Hence:
:$\mathbf I_n^{-1} = \begin{bmatrix}
\dfrac 1 1 & 0 & \cdots & 0 \\
0 & \dfrac 1 1 & \cdots & 0 \\
\vdots & \vdots & \ddots & \vdots \\
0 & 0 & \cdots & \dfrac 1 1 \\
\end{bmatrix}$
Hence the result.
Category:Unit Matrices
Category:Inverse Matrices
Unit Sphere as Surface of Revolution

== Theorem ==
Let $\struct {\R^3, d}$ be the Euclidean space.
Let $S_C \subseteq \R^3$ be the surface of revolution.
Let $C$ be a semi-circle defined by $x^2 + y^2 = 1$ in the open upper half-plane.
Let the smooth local parametrization of $C$ be:
:$\map \gamma \phi = \tuple {\cos \phi, \sin \phi}$
where $\phi \in \openint 0 \pi$.
Then the induced metric on $S_C$ is:
:$g = \d \phi^2 + \sin^2 \phi \rd \theta^2$
== Proof ==
By Smooth Local Parametrization of Surface of Revolution, the smooth local parametrization of $S_C$ can be written as:
:$\map X {\phi, \theta} = \tuple {\sin \phi \cos \theta, \sin \phi \sin \theta, \cos \phi}$
By Induced Metric on Surface of Revolution:
This is a metric of a unit sphere with the points on the $x$-axis removed.
== Sources ==
* : $\S 2$: Riemannian Metrics. Methods for Constructing Riemannian Metrics
Category:Surfaces of Revolution
Category:Solid Geometry
Category:Induced Metrics
Unit Sphere is Closed/Normed Vector Space

== Theorem ==
Let $M = \struct {X, \norm {\, \cdot \,} }$ be a normed vector space.
Let $\Bbb S := \set {x \in X : \norm x = 1}$ be a unit sphere in $M$.
Then $\Bbb S$ is closed in $M$.
== Proof ==
Let $\map {B_1} 0 = \set {x \in X : \norm x < 1}$ be an open ball.
Let $\map { {B_1}^-} 0 = \set {x \in X : \norm x \le 1}$ be a closed ball.
Then:
:$\ds X = \Bbb S \bigcup \relcomp X {\Bbb S}$
where
:$\ds \relcomp X {\Bbb S} = \map {B_1} 0 \bigcup \paren {X \setminus \map { {B_1}^-} 0}$
is the relative complement of $\Bbb S$ in $X$.
We have that Closed Ball is Closed in Normed Vector Space.
By definition, $X \setminus \map { {B_1}^-} 0$ is open.
Furthermore, Open Ball is Open Set in Normed Vector Space.
By Union of Open Sets of Normed Vector Space is Open, $\relcomp X {\Bbb S}$ is open.
By definition, $\Bbb S$ is closed.
== Sources ==
* : $\S 1.3$: Normed and Banach spaces. Topology of normed spaces
Category:Closed Sets (Normed Vector Spaces)
Unit Tangent Bundle is Connected iff Manifold is Connected

== Theorem ==
Let $\struct {M,g}$ be a Riemannian manifold of dimension $n > 1$.
Let $UTM$ be the unit tangent bundle of $M$.
Then $UTM$ is connected $M$ is connected.
== Proof ==
== Sources ==
* : $\S 2$. Riemannian Metrics. Definitions
Category:Tangent Bundles
Category:Riemannian Geometry
Unit Vector in Direction of Vector

== Theorem ==
Let $\mathbf v$ be a vector quantity.
The unit vector $\mathbf {\hat v}$ in the direction of $\mathbf v$ is:
:$\mathbf {\hat v} = \dfrac {\mathbf v} {\norm {\mathbf v} }$
where $\norm {\mathbf v}$ is the magnitude of $\mathbf v$.
== Proof ==
From Vector Quantity as Scalar Product of Unit Vector Quantity:
:$\mathbf v = \norm {\mathbf v} \mathbf {\hat v}$
whence the result.
== Also presented as ==
This result is often seen presented as:
:$\mathbf {\hat v} = \dfrac {\mathbf v} v$
as in this context $v$ is usually understood as being the magnitude of $\mathbf v$.
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.1$ Electric Charge
* 
Category:Unit Vectors
Unit Vector in terms of Direction Cosines

== Theorem ==
Let $\mathbf r$ be a vector quantity embedded in a Cartesian $3$-space.
Let $\mathbf i$, $\mathbf j$ and $\mathbf k$ be the unit vectors in the positive directions of the $x$-axis, $y$-axis and $z$-axis respectively.
Let $\cos \alpha$, $\cos \beta$ and $\cos \gamma$ be the direction cosines of $\mathbf r$ with respect to the $x$-axis, $y$-axis and $z$-axis respectively.
Let $\mathbf {\hat r}$ denote the unit vector in the direction of $\mathbf r$.
Then:
:$\mathbf {\hat r} = \paren {\cos \alpha} \mathbf i + \paren {\cos \beta} \mathbf j + \paren {\cos \gamma} \mathbf k$
== Proof ==
From Components of Vector in terms of Direction Cosines:
:$(1): \quad \mathbf r = r \cos \alpha \mathbf i + r \cos \beta \mathbf j + r \cos \gamma \mathbf k$
where $r$ denotes the magnitude of $\mathbf r$, that is:
:$r := \size {\mathbf r}$
By Unit Vector in Direction of Vector:
:$\mathbf {\hat r} = \dfrac {\mathbf r} r$
The result follows by multiplication of both sides of $(1)$ by $\dfrac 1 r$.
== Sources ==
* : Chapter $\text I$. Addition and Subtraction of Vectors. Centroids: Components of a Vector: $7$. The unit vectors $\mathbf i$, $\mathbf j$, $\mathbf k$
Category:Direction Cosines
Category:Unit Vectors
Unit Vectors in Complex Plane which are Vertices of Equilateral Triangle

== Theorem ==
Let $\epsilon_1, \epsilon_2, \epsilon_3$ be complex numbers embedded in the complex plane such that:
:$\epsilon_1, \epsilon_2, \epsilon_3$ all have modulus $1$
:$\epsilon_1 + \epsilon_2 + \epsilon_3 = 0$
Then:
:$\paren {\dfrac {\epsilon_2} {\epsilon_1} }^3 = \paren {\dfrac {\epsilon_3} {\epsilon_2} }^2 = \paren {\dfrac {\epsilon_1} {\epsilon_3} }^2 = 1$
== Proof ==
We have that:
Thus by Geometrical Interpretation of Complex Subtraction, $\epsilon_1$, $\epsilon_2$ and $\epsilon_3$ form the sides of a triangle.
As the modulus of each of $\epsilon_1$, $\epsilon_2$ and $\epsilon_3$ equals $1$, $\triangle \epsilon_1 \epsilon_2 \epsilon_3$ is equilateral.
:
By Complex Multiplication as Geometrical Transformation:
The same analysis can be done to the other two pairs of sides of $\triangle \epsilon_1 \epsilon_2 \epsilon_3$.
Hence the result.
== Sources ==
* : $\S 3$. Roots of Unity: Example $2$.
Category:Geometry of Complex Plane
Category:Equilateral Triangles
Unit Vectors of Cartesian 3-Space form Basis

== Theorem ==
Consider the Cartesian $3$-space $C$.
Let $\mathbf i$, $\mathbf j$ and $\mathbf k$ denote the unit vectors in the positive directions of the $x$-axis, $y$-axis and $z$-axis respectively.
Then $\set {\mathbf i, \mathbf j, \mathbf k}$ forms a basis for $C$.
== Proof ==
== Sources ==
* : Chapter $1$ Vector Analysis $1.1$ Definitions, Elementary Approach
Category:Unit Vectors
Unit n-Cube under Chebyshev Distance is Subspace of Real Vector Space

== Theorem ==
Let $n \in \N$.
Let $I^n$ denote the unit $n$-cube:
:$I^n = \closedint 0 1^n$
that is, the Cartesian product of $n$ instances of the closed real interval $\set {x \in \R: 0 \le x \le 1}$.
Let $d_c: I^n \times I^n \to \R$ be defined as:
:$\ds \map {d_c} {x, y} = \max_{i \mathop = 1}^n \set {\size {x_i - y_i} }$
where $x = \tuple {x_1, x_2, \ldots, x_n}, y = \tuple {x_1, x_2, \ldots, x_n} \in I^n$.
Then $\struct {I_n, d_c}$ is a metric subspace of $\struct {\R^n, d_\infty}$, where $d_\infty$ is the Chebyshev distance on the real vector space $\R_n$.
== Proof ==
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 7$: Subspaces and Equivalence of Metric Spaces: Example $2$
Category:Metric Subspaces
Unit n-Sphere under Euclidean Metric is Metric Subspace of Euclidean Real Vector Space

== Theorem ==
Let $\Bbb S^n$ be the unit $n$-sphere.
Let $d_S: \Bbb S^n \times \Bbb S^n \to \R$ be the real-valued function defined as:
:$\ds \forall x, y \in \Bbb S^n: \map {d_S} {x, y} = \sqrt {\sum_{i \mathop = 1}^{n + 1} \paren {x_i - y_i}^2}$
where $x = \tuple {x_1, x_2, \ldots, x_{n + 1} }, y = \tuple {y_1, y_2, \ldots, y_{n + 1} }$.
Then $\struct {\Bbb S^n, d_S}$ is a metric subspace of $\struct {\R^{n + 1}, d}$, where $d$ is the Euclidean metric on the real vector space $\R^{n + 1}$.
== Proof ==
The metric given is the Euclidean metric restricted to the subset $\Bbb S^n$ of the real vector space $\R^{n + 1}$.
The result follows from Subspace of Metric Space is Metric Space.
== Sources ==
* : Chapter $2$: Metric Spaces: $\S 7$: Subspaces and Equivalence of Metric Spaces: Example $3$
Category:Metric Subspaces
Unit of Integral Domain divides all Elements

== Theorem ==
Let $\struct {D, +, \circ}$ be an integral domain whose unity is $1_D$.
Let $\struct {U_D, \circ}$ be the group of units of $\struct {D, +, \circ}$.
Then:
:$\forall x \in D: \forall u \in U_D: u \divides x$
That is, every unit of $D$ is a divisor of every element of $D$.
== Proof ==
== Sources ==
* : $\S 62.1$ Factorization in an integral domain: $\text{(ii)}$
Category:Integral Domains
Unit of Measurement/Examples/Mass

== Examples of Units of Measurement ==
The physical quantity mass can be defined by specifying how two bodies are compared using a balance.
The mass of one of the bodies can be taken as a standard, and given a name, for example kilogram or pound.
Then the mass of any other body can be expressed in terms of this unit of measurement.
== Sources ==
* 
* 
Category:Examples of Units of Measurement
Unit of Ring is not Zero Divisor

== Theorem ==
Let $\struct {R, +, \circ}$ be a non-null ring with unity whose zero is $0_R$ and whose unity is $1_R$.
Let $x$ be a unit of $\struct {R, +, \circ}$.
Then $x$ is neither a left zero divisor nor a right zero divisor of $\struct {R, +, \circ}$.
== Proof ==
 $x$ is either a left zero divisor or a right zero divisor of $\struct {R, +, \circ}$.
, suppose $x$ is a left zero divisor of $\struct {R, +, \circ}$.
That is:
:$x \circ y = 0_R$
for some $y \in R \setminus \set {0_R}$.
Then:
This contradicts the deduction that $y \ne 0_R$.
Thus by Proof by Contradiction $x$ is neither a left zero divisor nor a right zero divisor of $\struct {R, +, \circ}$.
Category:Rings with Unity
Category:Units of Rings
Category:Zero Divisors