nyuuzyou/wikis huggingface dataset:
Fibonacci Number plus Constant in terms of Fibonacci Numbers

== Theorem ==
Let $c$ be a number.
Let $\sequence {b_n}$ be the sequence defined as:
:$b_n = \begin{cases}
0 & : n = 0 \\
1 & : n = 1 \\
b_{n - 2} + b_{n - 1} + c & : n > 1
\end{cases}$
Then $\sequence {b_n}$ can be expressed in Fibonacci numbers as:
:$b_n = c F_{n - 1} + \paren {c + 1} F_n - c$
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 0}$, let $\map P n$ be the proposition:
:$b_n = c F_{n - 1} + \paren {c + 1} F_n - c$
=== Basis for the Induction ===
$\map P 0$ is the case:
Thus $\map P 0$ is seen to hold.
$\map P 1$ is the case:
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P {k - 1}$ and $\map P k$ are true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$b_{k - 1} = c F_{k - 2} + \paren {c + 1} F_{k - 1} - c$
:$b_k = c F_{k - 1} + \paren {c + 1} F_k - c$
from which it is to be shown that:
:$b_{k + 1} = c F_k + \paren {c + 1} F_{k + 1} - c$
=== Induction Step ===
This is the induction step:
So $\map P {k - 1} \land \map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{\ge 0}: b_n = c F_{n - 1} + \paren {c + 1} F_n - c$
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $13 \ \text{(b)}$
Category:Fibonacci Numbers
Fibonacci Number whose Index is Plus or Minus Prime p is Multiple of p

== Theorem ==
Let $p$ be a prime number distinct from $5$.
Let $F_n$ denote the $n$th Fibonacci number.
Then either $F_{p - 1}$ or $F_{p + 1}$ (but not both) is a multiple of $p$.
== Proof ==
First we consider the edge cases:
For $p = 2$ we have that:
:$F_{2 + 1} = F_3 = 2$ which is a multiple of $2$.
:$F_{2 - 1} = F_1 = 1$ which is not a multiple of $2$.
For $p = 3$ we have that:
:$F_{3 + 1} = F_4 = 3$ which is a multiple of $3$.
:$F_{3 - 1} = F_2 = 1$ which is not a multiple of $3$.
Let $p$ be an odd prime.
From Cassini's Identity:
:$(1): \quad F_{p - 1} F_{p + 1} - {F_p}^2 = \left({-1}\right)^p = -1$
Then:
Hence either $F_{p - 1}$ or $F_{p + 1}$ is a multiple of $p$.
But as $F_{p + 1} = F_p + F_{p - 1}$, only one of them can be.
Finally, note that if $p = 5$:
:$F_{5 + 1} = F_6 = 8$ which is not a multiple of $5$.
:$F_{5 - 1} = F_4 = 3$ which is not a multiple of $5$.
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $27$
Category:Fibonacci Numbers
Fibonacci Number with Negative Index

== Theorem ==
Let $F_n$ be the $n$th Fibonacci number.
Then:
:$\forall n \in \Z_{> 0} : F_{-n} = \paren {-1}^{n + 1} F_n$
== Proof ==
From the initial definition of Fibonacci numbers, we have:
:$F_0 = 0, F_1 = 1, F_2 = 1, F_3 = 2, F_4 = 3$
By definition of the extension of the Fibonacci numbers to negative integers:
:$F_n = F_{n + 2} - F_{n - 1}$
The proof proceeds by induction.
For all $n \in \N_{>0}$, let $\map P n$ be the proposition:
:$F_{-n} = \paren {-1}^n F_n$
=== Basis for the Induction ===
$\map P 1$ is the case:
So $\map P 1$ is seen to hold.
$\map P 2$ is the case:
So $\map P 2$ is seen to hold.
This is our basis for the induction.
=== Induction Hypothesis ===
Now we need to show that, if $\map P k$ and $\map P {k - 1}$ are true, where $k > 1$, then it logically follows that $\map P {k + 1}$ is true.
So this is our induction hypothesis:
:$F_{-\paren {k - 1} } = \paren {-1}^k F_{k - 1}$
:$F_{-k} = \paren {-1}^{k + 1} F_k$
Then we need to show:
:$F_{-\paren {k + 1} } = \paren {-1}^{k + 2} F_{k + 1}$
=== Induction Step ===
This is our induction step:
So $\map P k \land \map P {k - 1} \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{>0} : F_{-n} = \paren {-1}^{n + 1} F_n$
== Sources ==
* 
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $8$
Category:Fibonacci Numbers
Category:Proofs by Induction
Fibonacci Number with Prime Index 2n+1 is Congruent to 5^n Modulo p

== Theorem ==
Let $p = 2 n + 1$ be an odd prime.
Then:
:$F_p \equiv 5^n \pmod p$
== Proof ==
From Fibonacci Number by Power of 2:
:$2^p F_p = 2 \ds \sum_{k \mathop \in \Z} \dbinom p {2 k + 1} 5^k$
From Binomial Coefficient of Prime:
:$\forall k \in \Z: 0 < k < p: \dbinom p k \equiv 0 \pmod p$
and from Fermat's Little Theorem: Corollary 1:
:$2^p \equiv 2 \pmod p$
Hence:
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $26$
Category:Fibonacci Numbers
Category:Binomial Coefficients
Fibonacci Number with Prime Index is not necessarily Prime

== Theorem ==
Let $p \in \Z_{>0}$ be a prime number.
Let $F_p$ be the $p$th Fibonacci number.
Then $F_p$ is not itself necessarily prime.
== Proof ==
Proof by Counterexample:
:$F_{19} = 4181 = 37 \times 113$
== Also see ==
* Prime Fibonacci Number has Prime Index except for 3
== Sources ==
* : $5$
* : $5$
Category:Fibonacci Numbers
Category:Prime Numbers
Fibonacci Numbers/Examples/F1000

== Example of Fibonacci Number ==
The Fibonacci number $F_{1000}$ is a number with $209$ decimal digits beginning with $4$.
== Proof ==
By the corollary to the Euler-Binet Formula:
:$F_{1000} \approx \dfrac {\phi^{1000} } {\sqrt 5}$
From Number of Digits in Number, the number of decimal digits $m$ in $F_{1000}$ is given by:
:$m = \floor {\log_{10} F_{1000} } + 1$
Thus, by calculation:
:$m = \floor {208 \cdotp 64} + 1 = 209$
and the first digit can be obtained by evaluating $10^{0 \cdotp 64} \approx 4 \cdotp 36$.
Hence the result.
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $2$
Category:Fibonacci Numbers/Examples
Fibonacci Numbers/Examples/F173

== Example of Fibonacci Number ==
The Fibonacci number $F_{173}$ is:
:$638 \, 817 \, 435 \, 613 \, 190 \, 763 \, 972 \, 389 \, 505 \, 493$
It can be factorized as:
:$6 \, 260 \, 874 \, 567 \, 713 \times 102 \, 033 \, 258 \, 884 \, 875 \, 061$
== Historical Note ==
Category:Fibonacci Numbers/Examples
Category:638,817,435,613,190,763,972,389,505,493
Fibonacci Numbers which equal the Square of their Index

== Theorem ==
The only Fibonacci numbers which equal the square of their index are:
== Proof ==
By definition of the Fibonacci numbers:
Then it is observed that $F_{12} = 144$.
After that, for $n > 12$, we have that $F_n > n^2$.
== Also see ==
* Square Fibonacci Number: after $144$, there are no more square Fibonacci numbers at all.
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $5$
Category:Fibonacci Numbers
Fibonacci Numbers which equal their Index

== Theorem ==
The only Fibonacci numbers which equal their index are:
== Proof ==
By definition of the Fibonacci numbers:
Then it is observed that $F_5 = 5$.
After that, for $n > 5$, we have that $F_n > n$.
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $4$
Category:Fibonacci Numbers
Fibonacci Prime has Prime Index except for 3

== Theorem ==
Let $F_n$ denote the $n$th Fibonacci number.
Let $F_n$ be a prime number.
Then, apart from $F_4 = 3$, $n$ is a prime number.
== Proof ==
Let $F_n$ be a prime number.
 $n$ is a composite number greater than $4$.
Then $n = r s$ for some $1 < r, s < n$.
Except for the case where $n = 4$, at least one of $r$ and $s$ is greater than $2$.
From Divisibility of Fibonacci Number:
:$F_r \divides F_n$
and:
:$F_s \divides F_n$
where $\divides$ denotes divisibility.
When $k > 2$ we have that $F_k > 1$.
Thus when $n$ is composite such that $n > 4$, $F_n$ has at least one proper divisor.
That is, $F_n$ is not prime.
Thus by Proof by Contradiction, $n$ cannot be composite.
The exception is when $n = 4$, as noted above in which case its only proper divisor is $2$.
But $F_2 = 1$, and so divisibility by $F_2$ does not preclude primality.
Indeed, $F_4 = 3$, which is prime.
== Also see ==
* Fibonacci Number with Prime Index is not necessarily Prime, demonstrating the converse does not hold.
== Sources ==
* : $5$
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $7$
* : $5$
Category:Fibonacci Numbers
Category:Prime Numbers
Category:Fibonacci Primes
Fibonacci String Begins with ba

== Theorem ==
Let $S_n$ be a Fibonacci string of length $n$.
Then for $n \ge 3$, $S_n$ begins with $\text {ba}$.
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 3}$, let $\map P n$ be the proposition:
:$S_n$ begins with $\text {ba}$
We note in passing that $S_1 = \text a$ and $S_2 = \text b$, so neither of these begin with $\text{ba}$.
=== Basis for the Induction ===
$\map P 3$ is the case:
:$S_3 = \text {ba}$
Thus $\map P 3$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 3$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$S_k$ begins with $\text {ba}$
from which it is to be shown that:
:$S_{k + 1}$ begins with $\text {ba}$
=== Induction Step ===
This is the induction step:
By definition of Fibonacci string:
:$S_{k + 1} = S_k S_{k - 1}$
concatenated.
So $S_{k + 1}$ begins with $S_k$.
By the induction hypothesis, $S_k$ begins with $\text {ba}$.
Thus $S_{k + 1}$ likewise begins with $\text {ba}$.
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:for all $n \in \Z$ such that $n \ge 3$, $S_n$ begins with $\text {ba}$.
Category:Fibonacci Strings
Fibonacci String Ends with ab or ba

== Theorem ==
Let $S_n$ be a Fibonacci string of length $n$.
Then for $n \ge 3$, $S_n$ ends either with $\text{ba}$ or with $\text{ab}$.
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{\ge 3}$, let $P \left({n}\right)$ be the proposition:
:$S_n$ ends either with $\text{ba}$ or with $\text{ab}$
We note in passing that $S_1 = \text a$ and $S_2 = \text b$, so neither of these ends either with $\text{ba}$ or with $\text{ab}$.
=== Basis for the Induction ===
$P \left({3}\right)$ is the case:
:$S_3 = \text{ba}$
Thus $P \left({3}\right)$ is seen to hold.
$P \left({4}\right)$ is the case:
:$S_4 = \text{bab}$
Thus $P \left({4}\right)$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $P \left({k}\right)$ is true, where $k \ge 3$, then it logically follows that $P \left({k + 1}\right)$ is true.
So this is the induction hypothesis:
:$S_k$ ends either with $\text{ba}$ or with $\text{ab}$
and:
:$S_{k - 1}$ ends either with $\text{ba}$ or with $\text{ab}$
from which it is to be shown that:
:$S_{k + 1}$ ends either with $\text{ba}$ or with $\text{ab}$
=== Induction Step ===
This is the induction step:
By definition of Fibonacci string:
:$S_{k + 1} = S_k S_{k - 1}$
concatenated.
So $S_{k + 1}$ end with $S_{k - 1}$.
By the induction hypothesis, $S_k$ ends either with $\text{ba}$ or with $\text{ab}$.
Thus $S_{k + 1}$ likewise ends either with $\text{ba}$ or with $\text{ab}$.
So $P \left({k}\right) \implies P \left({k + 1}\right)$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:for all $n \in \Z$ such that $n \ge 3$, $S_n$ends either with $\text{ba}$ or with $\text{ab}$.
Category:Fibonacci Strings
Fibonomial Coefficient is Integer

== Theorem ==
Let $\dbinom n k_\FF$ be a Fibonomial coefficient.
Then $\dbinom n k_\FF$ is an integer.
== Proof ==
Recurrence Relation for Fibonomial Coefficients gives:
:$\dbinom n k_\FF = F_{k - 1} \dbinom {n - 1} k_\FF + F_{n - k + 1} \dbinom {n - 1} {k - 1}_\FF$
where $F_{k - 1}$ etc. denote Fibonacci numbers.
It follows that each Fibonomial coefficient is the sum of integers, and so an integer.
== Sources ==
* : $\S 1.2.8$: Fibonacci Numbers: Exercise $29 \ \text{(b)}$
Category:Fibonomial Coefficients
Field Adjoined Algebraic Elements is Algebraic

== Definition ==
Let $L / K$ be a field extension and $S \subseteq L$ a subset.
If each $x \in S$ is algebraic over $K$ then $\map K S$ is algebraic over $K$.
== Proof ==
Let $S \subseteq L$ be arbitrary, and $x \in \map K S$.
By Field Adjoined Set $x \in \map K S$ $x \in \map K {\alpha_1, \ldots, \alpha_n}$ for some $\alpha_1, \ldots, \alpha_n \in S$.
We have that $\map K {\alpha_1, \ldots, \alpha_n} / K$ is finite by Finitely Generated Algebraic Extension is Finite.
Moreover a Finite Field Extension is Algebraic.
Therefore $x$ is algebraic over $K$.
Category:Field Extensions
Field Adjoined Set

== Theorem ==
Let $F$ be a field.
Let $S \subseteq F$ be a subset of $F$.
Let $K \le F$ be a subfield of $F$.
The subring $K \sqbrk S$ of $F$ generated by $K \cup S$ is the set of all finite linear combinations of powers of elements of $S$ with coefficients in $K$.
The subfield $\map K S$ of $F$ generated by $K \cup S$ is the set of all $x y^{-1} \in F$ with $a, b \in K \sqbrk S$, $b \ne 0$.
$\map K S$ is isomorphic to the field of quotients $Q$ of $K \sqbrk S$.
=== Corollary ===
== Proof ==
Let $\set {X_s: s \in S}$ be a family of indeterminates indexed by $S$.
Let $\phi$ be the Evaluation Homomorphism such that $\phi \sqbrk {X_s} = s$.
By Ring Homomorphism Preserves Subrings, $\Img \phi$ is a subring of $F$
This subring contains $\phi \sqbrk {X_s} = s$ for each $s \in S$ and $\map \phi k = k$ for all $k \in K$.
Moreover since it is obtained by evaluating polynomials on $S$ it is the set of all finite linear combinations of powers elements of $S$ with coefficients in $K$ as claimed.
All these linear combinations must belong to any subring of $F$ that contains $K$ and $S$ (otherwise it is not closed), so $\Img \phi$ is the smallest such subring.
By Universal Property for Field of Quotients, the inclusion $K \sqbrk S \to F$ extends uniquely to a homomorphism $\psi : Q \to F$, given by $\map \psi {a / b} = a b^{-1}$.
We have:
:$\Img \psi = \set {a b^{-1} \in F: a, b \in K \sqbrk S, \ b \ne 0} \simeq Q$
The isomorphism comes from the fact that a field homomorphism is injective and the First Ring Isomorphism Theorem.
Thus $\Img \phi$ is a subfield of $F$ containing $K \sqbrk S$ and $K \cup S$.
Any subfield of $F$ containing $K$ and $S$ must contain $K \cup S$, $K \sqbrk S$ and all $a b^{-1}$ with $a, b \in K \sqbrk S$ (otherwise it would not be closed).
Therefore $Q \simeq \Img \phi$ is the smallest such subfield.
Category:Field Extensions
Field Adjoined Set/Corollary

== Corollary to Field Adjoined Set ==
Let $F$ be a field.
Let $S \subseteq F$ be a subset of $F$.
Let $K \le F$ be a subfield of $F$.
Let $A = K \sqbrk {X_1, \ldots, X_n}$ be the ring of polynomial functions in $n$ indeterminates over $K$. 
Let $B = \map K {X_1, \ldots, X_n}$ be the field of rational functions in $n$ indeterminates over $K$. 
Let $\alpha_1, \ldots, \alpha_n \in F$.
Then:
:$(1): \quad x \in K \sqbrk {\alpha_1, \ldots, \alpha_n} \iff x = \map f {\alpha_1, \ldots, \alpha_n}$ for some $f \in A$
:$(2): \quad x \in \map K {\alpha_1, \ldots, \alpha_n} \iff x = \map f {\alpha_1, \ldots, \alpha_n}$ for some $f \in B$
:$(3): \quad x \in K \sqbrk S \iff x \in K \sqbrk {\alpha_1, \ldots, \alpha_n}$ for some $\alpha_1, \ldots, \alpha_n \in S$
:$(4): \quad x \in \map K S \iff x \in \map K {\alpha_1, \ldots, \alpha_n}$ for some $\alpha_1, \ldots, \alpha_n \in S$
== Proof ==
Category:Field Extensions
Field Contains at least 2 Elements

== Theorem ==
Let $\struct {F, +, \times}$ be a field.
Then $E$ contains at least $2$ elements.
== Proof ==
By definition, $\struct {F, +, \times}$ is an algebraic structure such that:
:$\struct {F, +}$ is an abelian group whose identity element is $0 \in F$
:$\struct {F^*, \times}$ is also an abelian group, where $F^* = F \setminus \set 0$.
From Group is not Empty, $\struct {F^*, \times}$ cannot be the empty set.
So there are at least $2$ elements in $F$: 
:the identity element $0$ of $\struct {F, +}$
:the identity element $1$ of $\struct {F^*, \times}$.
Category:Field Theory
Field Extension/Examples/Complex Numbers over Reals

== Examples of Field Extensions ==
The complex numbers $\C$ forms a finite field extension over the real numbers $\R$ of degree $2$.
== Proof ==
It follows from Real Numbers form Subfield of Complex Numbers that $\C$ is an extension of $\R$.
From Vector Space over Division Subring: Real Numbers in Complex Numbers we have that the dimension of the vector space on $\C$ over $\R$ is $2$.
== Sources ==
* : Chapter $8$: Field Extensions: $\S 36$. The Degree of a Field Extension: Example $69$
Category:Examples of Field Extensions
Field Extension/Examples/Real Numbers of Type Rational a plus b root 2

== Examples of Field Extensions ==
Let $\Q \sqbrk {\sqrt 2}$ denote the set:
:$\Q \sqbrk {\sqrt 2} := \set {a + b \sqrt 2: a, b \in \Q}$
that is, all numbers of the form $a + b \sqrt 2$ where $a$ and $b$ are rational numbers.
Then $\Q \sqbrk {\sqrt 2}$ forms a finite field extension over the rational numbers $\Q$ of degree $2$.
== Proof ==
From Real Numbers of Type Rational a plus b root 2 form Field, $\Q \sqbrk {\sqrt 2}$ forms a field.
From Rational Numbers form Field, $\Q$ is also a field.
We have that $\Q \subseteq \Q \sqbrk {\sqrt 2}$, as:
:$\Q = \set {x \in \Q \sqbrk {\sqrt 2}: b = 0}$
Thus $\Q \sqbrk {\sqrt 2}$ is a field extension of $\Q$.
Thus $\Q \sqbrk {\sqrt 2}$ can be considered as a vector space over $\Q$.
Then we have that $\set {1, \sqrt 2}$ forms a basis of $\Q \sqbrk {\sqrt 2}$.
Hence $\Q \sqbrk {\sqrt 2}$ forms a finite field extension over the rational numbers $\Q$ of degree $2$.
== Sources ==
* : Chapter $8$: Field Extensions: $\S 36$. The Degree of a Field Extension: Example $70$
Category:Examples of Field Extensions
Field Generated by Macroscopic Charge Density

== Theorem ==
Let $B$ be a body of matter.
Let $P$ be a point inside $B$ whose position vector is $\mathbf r$.
The electric field at $P$ generated by the macroscopic charge density within $B$ is given by:
:$\ds \map {\mathbf E} {\mathbf r} = \dfrac 1 {4 \pi \varepsilon_0} \int_{\text {all space} } \dfrac {\paren {\mathbf r - \mathbf r'} \map \rho {\mathbf r'} } {\size {\mathbf r - \mathbf r'}^3} \rd V'$
where:
:$\d V'$ is an infinitesimal volume element
:$\mathbf r'$ is the position vector of $\d V'$
:$\map \rho {\mathbf r'}$ is the macroscopic charge density of the macroscopic electric field at $\mathbf r'$
:$\varepsilon_0$ denotes the vacuum permittivity.
== Proof ==
From Electric Field Strength from Assemblage of Point Charges, the electric field strength caused by an assemblage of point charges $q_1, q_2, \ldots, q_n$ is given by:
:$\ds \map {\mathbf E} {\mathbf r} = \dfrac 1 {4 \pi \epsilon_0} \sum_i \dfrac {\paren {\mathbf r - \mathbf r_i} q_i} {\size {\mathbf r - \mathbf r_i}^3}$
where $\mathbf r_1, \mathbf r_2, \ldots, \mathbf r_n$ are the position vectors of $q_1, q_2, \ldots, q_n$ respectively.
We apply the same principle to the macroscopic charge density and convert the summation into a definite integral, as follows:
Consider a volume element $\d V'$ which is smaller than the scale used for a macroscopic electric field, but still large enough to contain many atoms.
:
The electric field strength caused by $\d V'$ is:
:$\map {\mathbf E_{\text {atomic} } } {\mathbf r'} = \dfrac 1 {4 \pi \epsilon_0} \dfrac {\paren {\mathbf r - \mathbf r'} \rd q} {\size {\mathbf r - \mathbf r'}^3}$
where $\d q$ is the electric charge on $\d V'$.
By definition of macroscopic charge density:
:$\map \rho {\mathbf r'} = \dfrac {\d q} {\d V'}$
and so:
:$\map {\mathbf E} {\mathbf r'} = \dfrac 1 {4 \pi \epsilon_0} \dfrac {\paren {\mathbf r - \mathbf r'} \map \rho {\mathbf r'} } {\size {\mathbf r - \mathbf r'}^3} \rd V'$
Integrating over all space:
:$\ds \map {\mathbf E} {\mathbf r} = \dfrac 1 {4 \pi \varepsilon_0} \int_{\text {all space} } \dfrac {\paren {\mathbf r - \mathbf r'} \map \rho {\mathbf r'} } {\size {\mathbf r - \mathbf r'}^3} \rd V'$
Hence the result.
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.3$ Electric Fields in Matter: $1.3.3$ The macroscopic electric field
Category:Macroscopic Charge Density
Field Generated by Surface Charge Density

== Theorem ==
Let $B$ be a body of matter.
Let $P$ be a point inside $B$ whose position vector is $\mathbf r$.
The electric field at $P$ generated by the surface charge density over $B$ is given by:
:$\ds \map {\mathbf E} {\mathbf r} = \dfrac 1 {4 \pi \varepsilon_0} \int_{\text {all surfaces} } \dfrac {\paren {\mathbf r - \mathbf r'} \map \sigma {\mathbf r'} } {\size {\mathbf r - \mathbf r'}^3} \rd S'$
where:
:$\d S'$ is an infinitesimal area element
:$\mathbf r'$ is the position vector of $\d S'$
:$\map \sigma {\mathbf r'}$ is the surface charge density at $\mathbf r'$
:$\varepsilon_0$ denotes the vacuum permittivity.
== Proof ==
From Electric Field Strength from Assemblage of Point Charges, the electric field strength caused by an assemblage of point charges $q_1, q_2, \ldots, q_n$ is given by:
:$\ds \map {\mathbf E} {\mathbf r} = \dfrac 1 {4 \pi \epsilon_0} \sum_i \dfrac {\paren {\mathbf r - \mathbf r_i} q_i} {\size {\mathbf r - \mathbf r_i}^3}$
where $\mathbf r_1, \mathbf r_2, \ldots, \mathbf r_n$ are the position vectors of $q_1, q_2, \ldots, q_n$ respectively.
We apply the same principle to the surface charge density and convert the summation into a definite integral, as follows:
Consider an area element $\d S'$.
The electric field strength caused by $\d S'$ is:
:$\map {\mathbf E} {\mathbf r'} = \dfrac 1 {4 \pi \epsilon_0} \dfrac {\paren {\mathbf r - \mathbf r'} \rd q} {\size {\mathbf r - \mathbf r'}^3}$
where $\d q$ is the electric charge on $\d S'$.
By definition of surface charge density:
:$\map \sigma {\mathbf r'} = \dfrac {\d q} {\d S'}$
and so:
:$\map {\mathbf E} {\mathbf r'} = \dfrac 1 {4 \pi \epsilon_0} \dfrac {\paren {\mathbf r - \mathbf r'} \map \sigma {\mathbf r'} } {\size {\mathbf r - \mathbf r'}^3} \rd S'$
Integrating over all surfaces of $B$:
:$\ds \map {\mathbf E} {\mathbf r} = \dfrac 1 {4 \pi \varepsilon_0} \int_{\text {all surfaces} } \dfrac {\paren {\mathbf r - \mathbf r'} \map \rho {\mathbf r'} } {\size {\mathbf r - \mathbf r'}^3} \rd S'$
Hence the result.
== Examples ==
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.3$ Electric Fields in Matter: $1.3.3$ The macroscopic electric field
Category:Field Generated by Surface Charge Density
Category:Surface Charge Density
Field Generated by Surface Charge Density/Examples/Arbitrary Rectangular Area

== Example of Use of Field Generated by Surface Charge Density ==
Consider a rectangular surface $S$ embedded in the $x$-$y$ plane in a Cartesian $3$-space.
Let the corners of $S$ be at $x = \pm a$ and $y = \pm b$.
The electric field at $P$ generated by the surface charge density over $S$ is given by:
:$\ds \map {\mathbf E} {\mathbf r} = \dfrac 1 {4 \pi \varepsilon_0} \int_{x \mathop = -a}^a \int_{y \mathop = -b}^b \dfrac {\paren {\mathbf r - \mathbf r'} \map \sigma {\mathbf r'} } {\size {\mathbf r - \mathbf r'}^3} \rd y' \rd x'$
where:
:$\d y' \rd x'$ is an infinitesimal area element of $S$
:$\mathbf r'$ is the position vector of $\d y' \rd x'$
:$\map \sigma {\mathbf r'}$ is the surface charge density at $\mathbf r'$
:$\varepsilon_0$ denotes the vacuum permittivity.
:
== Sources ==
* : Chapter $1$: Force and energy in electrostatics: $1.3$ Electric Fields in Matter: $1.3.3$ The macroscopic electric field
Category:Field Generated by Surface Charge Density
Field Homomorphism Preserves Product Inverses

== Theorem ==
Let $\phi: \struct {F_1, +_1, \times_1} \to \struct {F_2, +_2, \times_2}$ be a field homomorphism.
Then:
:$\forall x \in F_1^*: \map \phi {x^{-1} } = \map \phi x^{-1}$
== Proof ==
By definition, if $\struct {F_1, +_1, \times_1}$ and $\struct {F_2, +_2, \times_2}$ are fields then $\struct {F_1^*, \times_1}$ and $\struct {F_2^*, \times_2}$ are groups.
Again by definition:
:the product inverse of $x$ in $\struct {F_1, +_1, \times_1}$ for $\times_1$ is the product inverse of $x$ in $\struct {F_1^*, \times_1}$
:the product inverse of $x$ in $\struct {F_2, +_2, \times_2}$ for $\times_2$ is the product inverse of $x$ in $\struct {F_2^*, \times_2}$
The result follows from Group Homomorphism Preserves Inverses.
== Sources ==
* : Chapter $\text {I}$: Elementary Definitions: $\S 3$. Homomorphisms
Category:Field Homomorphisms