nyuuzyou/wikis huggingface dataset:
Distributive Laws of Set Theory/Also known as

== Distributive Laws of Set Theory: Also known as == 
The distributive laws (of set theory) are also known as:
:the distributivity laws (of set theory)
:the distributive properties (of set theory).
== Sources ==
* : Introduction to Set Theory: $1$. Elementary Operations on Sets
* : $\text {(vi)}$
Category:Distributive Laws of Set Theory
Distributivity is Preserved in Induced Structure

== Theorem ==
Let $\struct {T, \oplus, \otimes}$ be an algebraic structure, and let $S$ be a set.
Let $T^S$ denote the set of all mappings from $S$ to $T$.
Let $\struct {T^S, \oplus}$ be the structure on $T^S$ induced by $\oplus$.
Let $\struct {T^S, \otimes}$ be the structure on $T^S$ induced by $\otimes$.
If $\otimes$ is distributive over $\oplus$, then the pointwise operation induced on $T^S$ by $\otimes$ is distributive over the operation induced by $\oplus$.
== Proof ==
Let $\struct {T, \oplus, \otimes}$ be an algebraic structure in which $\otimes$ distributes over $\oplus$.
Let $f, g, h \in T^S$. 
Then:
This shows left distributivity.
The proof for right distributivity is identical.
Category:Pointwise Operations
Category:Distributive Operations
Divergence Operator Distributes over Addition

== Theorem ==
Let $\map {\mathbf V} {x_1, x_2, \ldots, x_n}$ be a vector space of $n$ dimensions.
Let $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ be the standard ordered basis of $\mathbf V$.
Let $\mathbf f$ and $\mathbf g: \mathbf V \to \mathbf V$ be vector-valued functions on $\mathbf V$:
:$\mathbf f := \tuple {\map {f_1} {\mathbf x}, \map {f_2} {\mathbf x}, \ldots, \map {f_n} {\mathbf x} }$
:$\mathbf g := \tuple {\map {g_1} {\mathbf x}, \map {g_2} {\mathbf x}, \ldots, \map {g_n} {\mathbf x} }$
Let $\nabla \cdot \mathbf f$ denote the divergence of $f$.
Then:
:$\nabla \cdot \paren {\mathbf f + \mathbf g} = \nabla \cdot \mathbf f + \nabla \cdot \mathbf g$
== Proof ==
== Sources ==
* : $\S 22$: Miscellaneous Formulas involving $\nabla$: $22.36$
Category:Divergence Operator
Divergence Operator is Invariant under Coordinate Transformation

== Theorem ==
Let $R$ be a region of space in which there exists an vector field $\mathbf V$.
Let $\nabla \cdot \mathbf V$ denote the divergence of $\mathbf V$.
Then $\nabla \cdot \mathbf V$ is invariant under a change of coordinate system on $R$.
== Proof ==
The result follows directly from the physical interpretation of the divergence operator.
== Sources ==
* : Chapter $\text {IV}$: The Operator $\nabla$ and its Uses: $3 a$. The Operation $\nabla \cdot \mathbf V$
Category:Divergence Operator
Divergence Operator on Vector Space is Dot Product of Del Operator

== Theorem ==
Let $R$ be a region of Cartesian $3$-space $\R^3$.
Let $\map {\mathbf V} {x, y, z}$ be a vector field acting over $R$.
Let $\tuple {i, j, k}$ be the standard ordered basis on $\R^3$.
Then
:$\operatorname {div} \mathbf V = \nabla \cdot \mathbf V$
where:
:$\operatorname {div} \mathbf V $ denotes the divergence of $\mathbf V$
:$\nabla$ denotes the del operator.
== Proof ==
We have by definition of divergence of $\mathbf V$:
:$\operatorname {div} \mathbf V = \dfrac {\partial V_x} {\partial x} + \dfrac {\partial V_y} {\partial y} + \dfrac {\partial V_z} {\partial z}$
Now:
== Sources ==
* : Chapter $\text {IV}$: The Operator $\nabla$ and its Uses: $3 a$. The Operation $\nabla \cdot \mathbf V$
Category:Divergence Operator
Divergence Test

== Theorem ==
Let $\sequence {a_n}$ be a sequence in $\R$ which does not converge to $0$.
Then $\ds \sum_{i \mathop = 1}^\infty a_n$ diverges.
== Proof ==
We know that Terms in Convergent Series Converge to Zero.
This is the contrapositive statement of this theorem.
Thus, the theorem holds by Rule of Transposition.
== Also known as ==
Category:Divergence Test
Category:Divergent Series
Category:Convergence Tests
Category:Named Theorems
Category:Proofs by Contraposition
Divergence Theorem for Riemannian Manifold

== Theorem ==
Let $\struct {M, g}$ be a compact Riemannian manifold with boundary $\partial M$.
Let $N$ be the outward-pointing unit normal vector field to $\partial M$.
Let $\hat g$ be the induced metric on $\partial M$.
Let $X$ be a smooth vector field on $M$.
Let $\innerprod \cdot \cdot_g$ be the Riemannian metric.
Let $\rd V_g$ be the Riemannian volume form.
Let $\operatorname {div}$ be the divergence operator.
Then:
:$\ds \int_M \operatorname {div} X \rd V_g = \int_{\partial M} \innerprod X N_g \rd V_{\hat g}$
== Proof ==
== Sources ==
* : $\S 2$: Riemannian Metrics. Problems
Category:Riemannian Geometry
Divergence of Curl is Zero

== Definition ==
Let $\map {\R^3} {x, y, z}$ denote the real Cartesian space of $3$ dimensions.
Let $\mathbf V: \R^3 \to \R^3$ be a vector field on $\R^3$
Then:
:$\map {\operatorname {div} } {\curl \mathbf V} = 0$
where:
:$\curl$ denotes the curl operator
:$\operatorname {div}$ denotes the divergence operator.
== Proof ==
From Curl Operator on Vector Space is Cross Product of Del Operator and Divergence Operator on Vector Space is Dot Product of Del Operator:
where $\nabla$ denotes the del operator.
Hence we are to demonstrate that:
:$\nabla \cdot \paren {\nabla \times \mathbf V} = 0$
Let $\mathbf V$ be expressed as a vector-valued function on $\mathbf V$:
:$\mathbf V := \tuple {\map {V_x} {\mathbf r}, \map {V_y} {\mathbf r}, \map {V_z} {\mathbf r} }$
where $\mathbf r = \tuple {x, y, z}$ is the position vector of an arbitrary point in $R$.
Let $\tuple {\mathbf i, \mathbf j, \mathbf k}$ be the standard ordered basis on $\R^3$.
Hence:
From Clairaut's Theorem:
:$\dfrac {\partial^2 V_z} {\partial x \partial y} = \dfrac {\partial^2 V_z} {\partial y \partial x}$
and the same for the other partial derivatives.
The result follows.
== Sources ==
* : Chapter $\text {V}$: Further Applications of the Operator $\nabla$: $5$. The Operator $\operatorname {div} \curl$: $(5.6)$
* : $\S 22$: Miscellaneous Formulas involving $\nabla$: $22.44$
Category:Divergence Operator
Category:Curl Operator
Divergence of Product of Scalar Field with Curl of Vector Field

== Theorem ==
Let $R$ be a region of space.
Let $U$ be a scalar field over $R$.
Let $\mathbf A = \curl \mathbf B$ be a vector field over $R$ whose vector potential is $\mathbf B$.
Then:
:$\map {\operatorname {div} } {U \curl \mathbf B} = \paren {\curl \mathbf B} \cdot \paren {\grad U}$
where:
:$\operatorname {div}$ denotes the divergence operator
:$\grad$ denotes the gradient operator
:$\curl$ denotes the curl operator.
== Proof ==
== Sources ==
* : Chapter $\text {V}$: Further Applications of the Operator $\nabla$: $8$. Two Useful Divergence Formulae: $(5.10)$
Category:Gradient Operator
Category:Divergence Operator
Category:Curl Operator
Divergence of Product of Scalar Field with Gradient of Scalar Field

== Theorem ==
Let $R$ be a region of space.
Let $U$ and $W$ be scalar fields over $R$.
Then:
:$\map {\operatorname {div} } {U \grad W} = U \nabla^2 W + \paren {\grad U} \cdot \paren {\grad W}$
where:
:$\operatorname {div}$ denotes the divergence operator
:$\grad$ denotes the gradient operator
:$\nabla^2$ denotes the Laplacian.
== Proof ==
== Also presented as ==
This result can also be presented as:
:$\map {\operatorname {div} } {U \grad W} = U \nabla^2 W + \nabla U \cdot \nabla W$
presupposing the implementation of $\grad$ as an operation using the del operator.
== Sources ==
* : Chapter $\text {V}$: Further Applications of the Operator $\nabla$: $8$. Two Useful Divergence Formulae: $(5.9)$
Category:Gradient Operator
Category:Divergence Operator
Divergence of Vector Cross Product

== Definition ==
Let $R$ be a region of space embedded in Cartesian $3$ space $\R^3$.
Let $\mathbf A$ and $\mathbf B$ be vector fields over $R$.
Then:
:$\map {\operatorname {div} } {\mathbf A \times \mathbf B} = \mathbf B \cdot \curl \mathbf A - \mathbf A \cdot \curl \mathbf B$
where:
:$\operatorname {div}$ denotes the divergence operator
:$\curl$ denotes the curl operator
:$\times$ denotes vector cross product
:$\cdot$ denotes dot product.
== Proof ==
From Divergence Operator on Vector Space is Dot Product of Del Operator and Curl Operator on Vector Space is Cross Product of Del Operator:
where $\nabla$ denotes the del operator.
Hence we are to demonstrate that:
:$\nabla \cdot \paren {\mathbf A \times \mathbf B} = \mathbf B \cdot \paren {\nabla \times \mathbf A} - \mathbf A \cdot \paren {\nabla \times \mathbf B}$
Let $\tuple {\mathbf i, \mathbf j, \mathbf k}$ be the standard ordered basis on $\R^3$.
Let $\mathbf A$ and $\mathbf B: \R^3 \to \R^3$ be expressed as vector-valued functions on $\R^3$:
:$\mathbf A := \tuple {\map {A_x} {\mathbf r}, \map {A_y} {\mathbf r}, \map {A_z} {\mathbf r} }$
:$\mathbf B := \tuple {\map {B_x} {\mathbf r}, \map {B_y} {\mathbf r}, \map {B_z} {\mathbf r} }$
where $\mathbf r = \tuple {x, y, z}$ is the position vector of an arbitrary point in $R$.
Then:
== Also presented as ==
This result can also be presented as:
:$\nabla \cdot \paren {\mathbf A \times \mathbf B} = \mathbf B \cdot \paren {\nabla \times \mathbf A} - \mathbf A \cdot \paren {\nabla \times \mathbf B}$
presupposing the implementations of $\operatorname {div}$ and $\curl$ as operations using the del operator.
== Sources ==
* : Chapter $\text {IV}$: The Operator $\nabla$ and its Uses: $6$. Divergence of a Vector Product: $(4.12)$
* : $\S 22$: Miscellaneous Formulas involving $\nabla$: $22.40$
Category:Divergence Operator
Category:Vector Cross Product
Divergent Complex Sequence/Examples/(-1)^n + 1 over n

== Example of Divergent Complex Sequence ==
Let $\sequence {z_n}$ be the complex sequence defined as:
:$z_n \paren {-1}^n + \dfrac i n$
Then $\ds \lim_{n \mathop \to \infty} z_n$ does not exist.
== Proof ==
The real part of $z_n$ is $\paren {-1}^n$.
As can be seen in Divergent Sequence may be Bounded, $\sequence {\paren {-1}^n}$ does not converge to a limit.
== Sources ==
* : $\S 4.2$. Sequences: Example $\text {(iii)}$
Category:Examples of Divergent Sequences
Divergent Complex Sequence/Examples/(2 over 3 + 3i over 4)^n

== Example of Divergent Complex Sequence ==
Let $\sequence {z_n}$ be the complex sequence defined as:
:$z_n = \paren {\dfrac 2 3 + \dfrac {3 i} 4}^n$
Then $\ds \lim_{n \mathop \to \infty} z_n$ does not exist.
== Proof ==
Thus $\cmod {z_n} \to \infty$ and so the limit does not exist.
== Sources ==
* : $\S 4.2$. Sequences: Example $\text {(ii)}$
Category:Examples of Divergent Sequences
Divergent Complex Sequence/Examples/i^n

== Example of Divergent Complex Sequence ==
Let $\sequence {z_n}$ be the complex sequence defined as:
:$z_n = i^n$
Then $\ds \lim_{n \mathop \to \infty} z_n$ does not exist.
== Proof ==
Consider the subsequence $\sequence {x_{n_r} }$ of $\sequence {x_n}$ where $\sequence {n_r}$ be the sequence in $\N$ defined such that $n_r = 2 r$.
Then $\sequence {x_{n_r} }$ is the real sequence defined as:
:$x_{n_r} = \paren {-1}^r$
As can be seen in Divergent Sequence may be Bounded, $\sequence {\paren {-1}^r}$ does not converge to a limit.
== Sources ==
* : $\S 4$. Elementary Functions of a Complex Variable: Exercise $1 \ \text {(ii)}$
Category:Examples of Divergent Sequences
Divergent Function/Examples/Tending to Infinity

== Example of Divergent Function ==
Let $f: \R \to \R$ be such that:
:$\forall H > 0: \exists \delta > 0: \map f x > H$ provided $c < x < c + \delta$
Then (using the language of limits), $\map f x \to +\infty$ as $x \to c^+$.
That is, $f$ is divergent at $c$.
Category:Examples of Divergent Functions
Divergent Function/Examples/Values for Rational Numbers

== Example of Divergent Function ==
Let $f: \R \to \R$ be such that:
:$\map f x = \begin {cases}
0 & : x \in \Q \\
1 & : x \notin \Q
\end {cases}$
Then $x$ converges to neither $0$ nor $1$ and hence is divergent (although, it needs to be noted, not to infinity).
Category:Examples of Divergent Functions
Divergent Real Sequence to Infinity/Examples/(-1)^n times n

== Example of Divergent Real Sequence to Infinity ==
Let $\sequence {a_n}_{n \mathop \ge 1}$ be the real sequence defined as:
:$a_n = \paren {-1} n$
Then $\sequence {a_n}$ is divergent to $\infty$.
However, $\sequence {a_n}$ is neither divergent to $+\infty$ nor divergent to $-\infty$.
== Proof ==
We have that:
:$\size {a_n} = n$
Let $H \in \R_{>0}$ be given.
Then by the Axiom of Archimedes:
:$\exists N \in \N: N > H$
and so for $n \ge N$:
:$n > H$
Thus $\sequence {a_n}$ is divergent to $\infty$.
But $\sequence {a_n}$ cannot be divergent to $+\infty$ because all its odd terms are negative.
Neither can $\sequence {a_n}$ cannot be divergent to $-\infty$ because all its even terms are positive.
Hence the result.
== Sources ==
* : $\S 4$: Convergent Sequences: Exercise $\S 4.29 \ (3) \ \text{(iii)}$
Category:Examples of Divergent Sequences
Divergent Real Sequence to Negative Infinity/Examples/Minus Square Root of n

== Example of Divergent Real Sequence to Negative Infinity ==
Let $\sequence {a_n}_{n \mathop \ge 1}$ be the real sequence defined as:
:$a_n = -\sqrt n$
Then $\sequence {a_n}$ is divergent to $-\infty$.
== Proof ==
Let $H \in \R_{>0}$ be given.
We need to find $N \in \R$ such that:
:$\forall n > N: -\sqrt n < -H$
That is:
:$\forall n > N: \sqrt n > H$
That is:
:$\forall n > N: n > H^2$
We take:
:$N = H^2$
and the result follows.
== Sources ==
* : $\S 4$: Convergent Sequences: Exercise $\S 4.29 \ (3) \ \text{(ii)}$
Category:Examples of Divergent Sequences
Divergent Real Sequence to Positive Infinity/Examples/2^n

== Example of Divergent Real Sequence to Positive Infinity ==
Let $\sequence {a_n}_{n \mathop \ge 1}$ be the real sequence defined as:
:$a_n = 2^n$
Then $\sequence {a_n}$ is divergent to $+\infty$.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : $\S 4$: Convergent Sequences: Exercise $\S 4.29 \ (3) \ \text{(i)}$
Category:Examples of Divergent Sequences
Category:Divergent Real Sequence to Positive Infinity/Examples/2^n
Divergent Real Sequence to Positive Infinity/Examples/2^n/Proof 1

== Example of Divergent Real Sequence to Positive Infinity ==
== Proof ==
Let $H \in \R_{>0}$ be given.
By Boundedness of Nth Powers, $\sequence {a_n}$ is unbounded above.
Hence:
:$\exists N \in \N: 2^N > H$
Then if $n > N$, we have:
:$2^n > 2^N > H$
and the result follows.
== Sources ==
* : $\S 4$: Convergent Sequences: Exercise $\S 4.29 \ (3) \ \text{(i)}$
Category:Divergent Real Sequence to Positive Infinity/Examples/2^n
Divergent Real Sequence to Positive Infinity/Examples/2^n/Proof 2

== Example of Divergent Real Sequence to Positive Infinity ==
== Proof ==
Let $H \in \R_{>0}$ be given.
We have that:
:$2^n > n$
so setting $N = H$:
:$\exists N \in \N: 2^n > H$
for all $n > N$.
== Sources ==
* : $\S 4$: Convergent Sequences: Exercise $\S 4.29 \ (3) \ \text{(i)}$
Category:Divergent Real Sequence to Positive Infinity/Examples/2^n
Divergent Real Sequence to Positive Infinity/Examples/n^alpha

== Example of Divergent Real Sequence to Positive Infinity ==
Let $\alpha \in \Q_{>0}$ be a strictly positive rational number.
Let $\sequence {a_n}_{n \mathop \ge 1}$ be the real sequence defined as:
:$a_n = n^\alpha$
Then $\sequence {a_n}$ is divergent to $+\infty$.
== Proof ==
We are to demonstrate that $n^\alpha \to +\infty$ as $n \to \infty$.
Let $H \in \R_{>0}$ be given.
We need to find $N \in \N$ such that:
:$\forall n > N: n^\alpha > H$
That is:
:$\forall n > N: n > H^{1 / \alpha}$
We choose $N = H^{1 / \alpha}$.
The result follows.
== Sources ==
* : $\S 4$: Convergent Sequences: $\S 4.28$: Example
Category:Examples of Divergent Sequences
Divergent Sequence may be Bounded

== Theorem ==
While every Convergent Sequence is Bounded, it does not follow that every bounded sequence is convergent.
That is, there exist bounded sequences which are divergent.
== Proof 1 ==
== Proof 2 ==
== Also see ==
* Definition:Grandi's Series
* Definition:Oscillating Series
== Sources ==
* : $\S 1.2$: Infinite Series of Constants
Category:Limits of Sequences
Category:Divergent Sequences
Category:Divergent Sequence may be Bounded
Divergent Sequence may be Bounded/Proof 1

== Theorem ==
== Proof ==
Let $\sequence {x_n}$ be the sequence in $\R$ which forms the basis of Grandi's series, defined as:
:$x_n = \paren {-1}^n$
It is clear that $\sequence {x_n}$ is bounded: above by $1$ and below by $-1$.
 $x_n \to l$ as $n \to \infty$.
Let $\epsilon > 0$.
Then $\exists N \in \R: \forall n > N: \size {\paren {-1}^n - l} < \epsilon$.
But there are values of $n > N$ for which $\paren {-1}^n = \pm 1$.
It follows that $\size {1 - l} < \epsilon$ and $\size {-1 - l} < \epsilon$.
From the Triangle Inequality for Real Numbers, we have:
This is a contradiction whenever $\epsilon < 1$.
Thus $\sequence {x_n}$ has no limit and, while definitely bounded, is unmistakably divergent.
== Sources ==
* : $\S 4$: Convergent Sequences: $\S 4.27$: Example
Category:Divergent Sequence may be Bounded
Divergent Sequence may be Bounded/Proof 2

== Theorem ==
== Proof ==
Let $\sequence {x_n}$ be the sequence in $\R$ which forms the basis of Grandi's series, defined as:
:$x_n = \paren {-1}^n$
It is clear that $\sequence {x_n}$ is bounded: above by $1$ and below by $-1$.
Note the following subsequences of $\sequence {x_n}$:
:$(1): \quad \sequence {x_{n_r} }$ where $\sequence {n_r}$ is the integer sequence defined as $n_r = 2 r$
:$(2): \quad \sequence {x_{n_s} }$ where $\sequence {n_s}$ is the integer sequence defined as $n_s = 2 s + 1$.
We have that:
:$\sequence {x_{n_r} }$ is the sequence $1, 1, 1, 1, \ldots$
:$\sequence {x_{n_s} }$ is the sequence $-1, -1, -1, -1, \ldots$
So $\sequence {x_n}$ has two subsequences with different limits.
From Limit of Subsequence equals Limit of Real Sequence, that means $\sequence {x_n}$ can not be convergent.
== Sources ==
* : $\S 5$: Subsequences: $\S 5.3$
* : $\S 5$: Subsequences: Bolzano-Weierstrass Theorem: $\S 5.11$: Example $\text{(i)}$
Category:Divergent Sequence may be Bounded
Divergent Sequence with Finite Number of Terms Deleted is Divergent

== Theorem ==
Let $\left({X, d}\right)$ be a metric space.
Let $\left \langle {x_k} \right \rangle$ be a sequence in $X$.
Let $\left \langle {x_k} \right \rangle$ be divergent.
Let a finite number of terms be deleted from $\left \langle {x_k} \right \rangle$.
Then the resulting subsequence is divergent.
== Proof ==
== Also see == 
* Tail of Convergent Series tends to Zero
Category:Series
Category:Sequences
Category:Convergence Tests
Divergent Series/Examples

== Examples of Divergent Series ==
=== Example: $\dfrac {n + i} {n^2}$ ===
=== Example: $\paren {\dfrac {2 + 3 i} {3 - 2 i} }^n$ ===
=== Example: $\dfrac n {n^2 + i}$ ===
=== Example: $\dfrac {\sin i n} {n^2}$ ===
Category:Examples of Divergent Series
Divergent Series/Examples/((2+3i) over (3-2i))^n

== Example of Divergent Series ==
The complex series defined as:
:$\ds S = \sum_{n \mathop = 1}^\infty \paren {\dfrac {2 + 3 i} {3 - 2 i} }^n$
is divergent.
== Proof ==
Thus the sequence $\sequence {\paren {\dfrac {2 + 3 i} {3 - 2 i} }^n}$ does not tend to zero.
Hence from Terms in Convergent Series Converge to Zero it follows that $S$ is divergent.
== Sources ==
* : $\S 4.3$. Series: Example $\text{(ii)}$
Category:Examples of Divergent Series
Divergent Series/Examples/(n + i) over n^2

== Example of Divergent Series ==
The complex series defined as:
:$\ds S = \sum_{n \mathop = 1}^\infty \dfrac {n + i} {n^2}$
is divergent.
== Proof ==
From Harmonic Series is Divergent, $\ds \sum_{n \mathop = 1}^\infty \dfrac 1 n$ is a divergent series.
The result follows from Convergence of Series of Complex Numbers by Real and Imaginary Part.
== Sources ==
* : $\S 4.3$. Series
Category:Examples of Divergent Series
Divergent Series/Examples/n over n^2 + i

== Example of Divergent Series ==
The complex series defined as:
:$\ds S = \sum_{n \mathop = 1}^\infty \dfrac n {n^2 + i}$
is divergent.
== Proof ==
Then we have:
From Harmonic Series is Divergent, $\ds \sum_{n \mathop = 1}^\infty \dfrac 1 n$ is a divergent series.
The result follows from Convergence of Series of Complex Numbers by Real and Imaginary Part.
== Sources ==
* : $\S 4$. Elementary Functions of a Complex Variable: Exercise $2 \ \text {(ii)}$
Category:Examples of Divergent Series
Divergent Series/Examples/sin i n over n^2

== Example of Divergent Series ==
The complex series defined as:
:$\ds S = \sum_{n \mathop = 1}^\infty \dfrac {\sin i n} {n^2}$
is divergent.
== Proof ==
Hence the result.
== Sources ==
* : $\S 4$. Elementary Functions of a Complex Variable: Exercise $2 \ \text {(v)}$
Category:Examples of Divergent Series
Divided by Positive Element of Field of Quotients

== Theorem ==
Let $\struct {K, +, \circ}$ be the field of quotients of a totally ordered integral domain $\struct {D, +, \circ, \le}$.
Then:
:$\forall z \in K: \exists x, y \in D: z = \dfrac x y, y \in D_{>0}$
== Proof ==
By definition of field of quotients:
:$\forall z \in K: \exists x, y \in D: z = \dfrac x y, y \in D_{\ne 0}$
Suppose $z = x' / y'$ such that $y' \notin D_{>0}$.
Then $y' < 0$ as $D$ is totally ordered.
Then:
If $y' < 0$, then $\paren {-y'} > 0$ from Properties of Ordered Ring $(4)$.
So all we need to do is set $x = -x', y = -y'$ and the result follows.
Category:Fields of Quotients
Category:Integral Domains
Divisibility Test for 7, 11 and 13

== Theorem ==
Mark off the integer $N$ being tested into groups of $3$ digits.
Because of the standard way of presenting integers, this may already be done, for example:
:$N = 22 \, 846 \, 293 \, 462 \, 733 \, 356$
Number the groups of $3$ from the right: 
:$N = \underbrace{22}_6 \, \underbrace{846}_5 \, \underbrace{293}_4 \, \underbrace{462}_3 \, \underbrace{733}_2 \, \underbrace{356}_1$
Considering each group a $3$-digit integer, add the even numbered groups together, and subtract the odd numbered groups:
:$22 - 846 + 293 - 462 + 733 - 356 = -616$
where the sign is irrelevant.
If the result is divisible by $7$, $11$ or $13$, then so is $N$.
In this case:
:$616 = 2^3 \times 7 \times 11$
and so $N$ is divisible by $7$ and $11$ but not $13$.
== Proof ==
Let $N$ be expressed as:
:$N = \ds \sum_{k \mathop = 0}^n a_k 1000^k = a_0 + a_1 1000 + a_2 1000^2 + \cdots + a_n 1000^n$
where $n$ is the number of groups of $3$ digits.
We have that:
:$1000 \equiv -1 \pmod {1001}$
Hence from Congruence of Powers:
:$1000^r \equiv \paren {-1}^r \pmod {1001}$
Thus:
:$N \equiv a_0 + \paren {-1} a_1 + \paren {-1}^2 a_2 + \cdots + \paren {-1}^n a_n \pmod {1001}$
from the definition of Modulo Addition.
Then we note that:
:$1001 = 7 \times 11 \times 13$
and the result follows.
== Sources ==
* : $\text {XI}$. Divisibility: $315$: A Test of Divisibility by $7$, $11$ and $13$
* : $1001$
* : $1001$
Category:Divisibility Tests
Category:7
Category:11
Category:13
Category:1001
Divisibility by 10

== Theorem ==
An integer $N$ expressed in decimal notation is divisible by $10$ the of $N$ is $0$.
That is:
:$N = \sqbrk {a_n \ldots a_2 a_1 a_0}_{10} = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$ is divisible by $10$ 
:
:$a_0 = 0$
== Proof ==
Let $N$ be divisible by $10$.
Then:
== Sources ==
* : $10$
* : $10$
* 
* 
Category:Divisibility Tests
Category:10
Divisibility by 11

== Theorem ==
Let $N \in \N$ be expressed as:
:$N = a_0 + a_1 10 + a_2 10^2 + \cdots + a_n 10^n$
Then $N$ is divisible by $11$ $\ds \sum_{r \mathop = 0}^n \paren {-1}^r a_r$ is divisible by $11$.
That is, a divisibility test for $11$ is achieved by alternately adding and subtracting the digits and taking the result modulo $11$.
== Proof ==
As:
:$10 \equiv -1 \pmod {11}$
we have:
:$10^r \equiv \paren {-1}^r \pmod {11}$
from Congruence of Powers.
Thus:
:$N \equiv a_0 + \paren {-1} a_1 + \paren {-1}^2 a_2 + \cdots + \paren {-1}^n a_n \pmod {11}$
from the definition of Modulo Addition.
The result follows.
== Sources ==
* : Chapter $2$. Equivalence Relations: Exercise $9$
* : $\text {XI}$. Divisibility: $314$: A Test of Divisibility by $11$
* : $\S 2.3$: Congruences: Exercise $11$
* : $11$
* : $11$
* 
* 
Category:Divisibility Tests
Category:11