nyuuzyou/wikis huggingface dataset:
Consecutive Integers whose Sums of Squares of Divisors are Equal

== Theorem ==
The only two consecutive positive integers whose sums of the squares of their divisors are equal are $6$ and $7$.
== Proof ==
The divisors of $6$ are
:$1, 2, 3, 6$
and so the sum of the squares of the divisors of $6$ is:
:$1^2 + 2^2 + 3^2 + 6^2 = 1 + 4 + 9 + 36 = 50$
The divisors of $7$ are
:$1, 7$
and so the sum of the squares of the divisors of $7$ is:
:$1^2 + 7^2 = 1 + 49 = 50$
It remains to be shown that there are no more.
Let $n \ge 7$ be an odd number.
Then both $n - 1$ and $n + 1$ are even.
Denote $\map {\sigma_2} n$ the sum of squares of the divisors of $n$.
We will show that:
:$\map {\sigma_2} {n + 1} > \map {\sigma_2} n$
:$\map {\sigma_2} {n - 1} > \map {\sigma_2} n$ for $n \ge 151$
Since:
Therefore:
and for $n - 1$:
:$\map {\sigma_2} {n - 1} - \map {\sigma_2} n = \dfrac {10 - \pi^2} 8 n^2 - \dfrac 5 2 n + \dfrac {15} 2$
By Solution to Quadratic Equation, the above is greater than zero when:
:$n > \dfrac {\paren {5/2} + \sqrt {\paren {5/2}^2 - 4 \paren {\paren {10 - \pi^2} / 8} \paren {15/2} } } {2 \paren {\paren {10 - \pi^2} / 8} } \approx 150.3$
hence there are no solutions for $\map {\sigma_2} {n - 1} = \map {\sigma_2} n$ for $n \ge 151$.
Our estimate of $\map {\sigma_2} n$ is very rough.
If $n$ is one of the following, we can get sharper estimates:
Suppose $n = p^k$ for a prime $p \ge 3$ and $k \ge 1$.
Then:
We have:
The above is greater than $0$ when $n \ge 17$.
Suppose $n = p q$, where $p, q \ge 3$ are primes and $p \ne q$.
Then:
We have:
The above is greater than $0$ when $n \ge 25$.
Therefore we just need to check the following $n \le 149$:
:$3, 5, 7, 9, 11, 13, 15, 21, 45, 63, 75, 99, 105, 117, 135, 147$
and thus the only pair is $\map {\sigma_2} 6 = \map {\sigma_2} 7 = 50$.
We have also inadvertently proved that $\map {\sigma_2} {2 n} > \map {\sigma_2} {2 n + 1}$ for $n \ge 8$.
== Sources ==
* : $6$
Category:6
Category:7
Consecutive Integers with Same Divisor Sum

== Theorem ==
Let $\sigma_1: \Z_{>0} \to \Z_{>0}$ be the divisor sum function, defined on the strictly positive integers.
The equation:
:$\map {\sigma_1} n = \map {\sigma_1} {n + 1}$
is satisfied by integers in the sequence:
:$14, 206, 957, 1334, 1364, 1634, 2685, 2974, 4364, 14841, 18873, \ldots$
== Examples ==
== Sources ==
* 
* : $14$
Category:Divisor Sum Function
Category:Consecutive Integers with Same Divisor Sum
Consecutive Integers with Same Divisor Sum/Examples

== Examples of Consecutive Integers with Same Divisor Sum == 
Let $\sigma_1: \Z_{>0} \to \Z_{>0}$ denote the divisor sum function.
The following are solutions to the equation:
:$\map {\sigma_1} n = \map {\sigma_1} {n + 1}$
=== $\sigma_1$ of $14$ equals $\sigma_1$ of $15$ ===
=== $\sigma_1$ of $206$ equals $\sigma_1$ of $207$ ===
Category:Consecutive Integers with Same Divisor Sum
Consecutive Integers with Same Divisor Sum/Examples/14

== Example of Consecutive Integers with Same Divisor Sum ==
Let $\sigma_1: \Z_{>0} \to \Z_{>0}$ denote the divisor sum function.
Then:
:$\map {\sigma_1} {14} = \map {\sigma_1} {15} = 24$
== Proof ==
From :
:$\map {\sigma_1} {14} = 24$
From :
:$\map {\sigma_1} {15} = 24$
Hence the result.
== Sources ==
* : $14$
* : $14$
Category:Consecutive Integers with Same Divisor Sum
Category:14
Category:15
Consecutive Integers with Same Divisor Sum/Examples/206

== Example of Consecutive Integers with Same Divisor Sum ==
Let $\sigma_1: \Z_{>0} \to \Z_{>0}$ denote the divisor sum function.
Then:
:$\map {\sigma_1} {206} = \map {\sigma_1} {207} = 312$
== Proof ==
From :
:$\map {\sigma_1} {206} = 312$
From :
:$\map {\sigma_1} {207} = 312$
Hence the result.
Category:Consecutive Integers with Same Divisor Sum
Category:206
Category:207
Consecutive Integers with Same Euler Phi Value

== Theorem ==
Let $\phi: \Z_{>0} \to \Z_{>0}$ be the Euler $\phi$ function, defined on the strictly positive integers.
The equation:
:$\map \phi n = \map \phi {n + 1}$
is satisfied by integers in the sequence:
:$1, 3, 15, 104, 164, 194, 255, 495, 584, 975, 2204, \ldots$
== Examples ==
== Sources ==
* : $104$
Category:Euler Phi Function
Category:Consecutive Integers with Same Euler Phi Value
Consecutive Integers with Same Euler Phi Value/Examples

== Examples of Consecutive Integers with Same Euler Phi Value == 
Let $\phi: \Z_{>0} \to \Z_{>0}$ denote the Euler $\phi$ function: the number of strictly positive integers less than or equal to $n$ which are prime to $n$.
The following are solutions to the equation:
:$\phi \left({n}\right) = \phi \left({n + 1}\right)$
=== $\phi$ of $1$ equals $\phi$ of $2$ ===
=== $\phi$ of $3$ equals $\phi$ of $4$ ===
=== $\phi$ of $15$ equals $\phi$ of $16$ ===
=== $\phi$ of $104$ equals $\phi$ of $105$ ===
=== $\phi$ of $164$ equals $\phi$ of $165$ ===
=== $\phi$ of $194$ equals $\phi$ of $195$ ===
=== $\phi$ of $255$ equals $\phi$ of $256$ ===
Category:Consecutive Integers with Same Euler Phi Value
Consecutive Integers with Same Euler Phi Value/Examples/1

== Example of Consecutive Integers with Same Euler Phi Value ==
Let $\phi: \Z_{>0} \to \Z_{>0}$ denote the Euler $\phi$ function: the number of strictly positive integers less than or equal to $n$ which are prime to $n$.
Then:
:$\map \phi 1 = \map \phi 2 = 1$
== Proof ==
From Euler Phi Function of 1:
:$\map \phi 1 = 1$
From Euler Phi Function of Prime:
:$\map \phi 2 = 2 - 1 = 1$
Hence the result.
Category:Consecutive Integers with Same Euler Phi Value
Consecutive Integers with Same Euler Phi Value/Examples/104

== Example of Consecutive Integers with Same Euler Phi Value ==
Let $\phi: \Z_{>0} \to \Z_{>0}$ denote the Euler $\phi$ function: the number of strictly positive integers less than or equal to $n$ which are prime to $n$.
Then:
:$\map \phi {104} = \map \phi {105} = 48$
== Proof ==
From :
:$\map \phi {104} = 48$
From :
:$\map \phi {105} = 48$
Hence the result.
== Sources ==
* : $104$
Category:Consecutive Integers with Same Euler Phi Value
Category:104
Category:105
Consecutive Integers with Same Euler Phi Value/Examples/15

== Example of Consecutive Integers with Same Euler Phi Value ==
Let $\phi: \Z_{>0} \to \Z_{>0}$ denote the Euler $\phi$ function: the number of strictly positive integers less than or equal to $n$ which are prime to $n$.
Then:
:$\map \phi {15} = \map \phi {16} = 8$
== Proof ==
From :
:$\map \phi {15} = 8$
From the corollary to Euler Phi Function of Prime Power:
:$\map \phi {16} = \map \phi {2^4} = 2^{4 - 1} = 8$
Hence the result.
Category:Consecutive Integers with Same Euler Phi Value
Consecutive Integers with Same Euler Phi Value/Examples/164

== Example of Consecutive Integers with Same Euler Phi Value ==
Let $\phi: \Z_{>0} \to \Z_{>0}$ denote the Euler $\phi$ function: the number of strictly positive integers less than or equal to $n$ which are prime to $n$.
Then:
:$\map \phi {164} = \map \phi {165} = 80$
== Proof ==
From :
:$\map \phi {164} = 80$
From :
:$\map \phi {165} = 80$
Hence the result.
Category:Consecutive Integers with Same Euler Phi Value
Category:164
Category:165
Consecutive Integers with Same Euler Phi Value/Examples/194

== Example of Consecutive Integers with Same Euler Phi Value ==
Let $\phi: \Z_{>0} \to \Z_{>0}$ denote the Euler $\phi$ function: the number of strictly positive integers less than or equal to $n$ which are prime to $n$.
Then:
:$\map \phi {194} = \map \phi {195} = 96$
== Proof ==
From :
:$\map \phi {194} = 96$
From :
:$\map \phi {195} = 96$
Hence the result.
Category:Consecutive Integers with Same Euler Phi Value
Category:194
Category:195
Consecutive Integers with Same Euler Phi Value/Examples/255

== Example of Consecutive Integers with Same Euler Phi Value ==
Let $\phi: \Z_{>0} \to \Z_{>0}$ denote the Euler $\phi$ function: the number of strictly positive integers less than or equal to $n$ which are prime to $n$.
Then:
:$\map \phi {255} = \map \phi {256} = 128$
== Proof ==
From :
:$\map \phi {255} = 128$
From :
:$\map \phi {256} = 128$
Hence the result.
Category:Consecutive Integers with Same Euler Phi Value
Category:255
Category:256
Consecutive Integers with Same Euler Phi Value/Examples/3

== Example of Consecutive Integers with Same Euler Phi Value ==
Let $\phi: \Z_{>0} \to \Z_{>0}$ denote the Euler $\phi$ function: the number of strictly positive integers less than or equal to $n$ which are prime to $n$.
Then:
:$\map \phi 3 = \map \phi 4 = 2$
== Proof ==
From Euler Phi Function of 3:
:$\map \phi 3 = 2$
From the corollary to Euler Phi Function of Prime Power:
:$\map \phi 4 = \map \phi {2^2} = 2^{2 - 1} = 2$
Hence the result.
Category:Consecutive Integers with Same Euler Phi Value
Consecutive Odd Powerful Numbers

== Theorem ==
$25$ and $27$ are the only known pair of consecutive odd powerful numbers.
It is not known whether there are any more.
== Progress ==
We have that:
:$25 = 5^2$
:$27 = 3^3$
and so by definition both are powerful.
== Sources ==
* 
* : $25$
Category:25
Category:27
Category:Powerful Numbers
Category:Open Questions
Consecutive Pairs of Quadratic Residues/Examples

== Examples of Consecutive Pairs of Quadratic Residues ==
=== Consecutive Pairs of Quadratic Residues of $3$ ===
=== Consecutive Pairs of Quadratic Residues of $5$ ===
=== Consecutive Pairs of Quadratic Residues of $7$ ===
=== Consecutive Pairs of Quadratic Residues of $11$ ===
=== Consecutive Pairs of Quadratic Residues of $17$ ===
=== Consecutive Pairs of Quadratic Residues of $29$ ===
=== Consecutive Pairs of Quadratic Residues of $61$ ===
Category:Examples of Quadratic Residues
Consecutive Pairs of Quadratic Residues/Examples/11

== Examples of Consecutive Pairs of Quadratic Residues ==
There are $2$ consecutive pairs of quadratic residues modulo $11$.
This is consistent with the number of such consecutive pairs being $\floor {\dfrac {11} 4}$.
== Proof ==
From Quadratic Residues modulo $11$:
:$\set {1, 3, 4, 5, 9}$ are the quadratic residues modulo $11$
The set of pairs of consecutive quadratic residues modulo $11$ is therefore:
:$\set {\set {3, 4}, \set {4, 5} }$
The result follows.
== Sources ==
* : $\text {3-5}$ The Use of Computers in Number Theory: Exercise $7$
Category:Examples of Quadratic Residues
Category:11
Consecutive Pairs of Quadratic Residues/Examples/17

== Examples of Consecutive Pairs of Quadratic Residues ==
There are $4$ consecutive pairs of quadratic residues modulo $17$.
This is consistent with the number of such consecutive pairs being $\floor {\dfrac {17} 4}$.
== Proof ==
From Quadratic Residues modulo $17$:
:$\set {1, 2, 4, 8, 9, 13, 15, 16}$ are the quadratic residues modulo $17$
The set of pairs of consecutive quadratic residues modulo $17$ is therefore:
:$\set {\set {1, 2}, \set {8, 9}, \set {15, 16}, \set {16, 1} }$
The result follows.
== Sources ==
* : $\text {3-5}$ The Use of Computers in Number Theory: Exercise $7$
Category:Examples of Quadratic Residues
Category:17
Consecutive Pairs of Quadratic Residues/Examples/29

== Examples of Consecutive Pairs of Quadratic Residues ==
There are $7$ consecutive pairs of quadratic residues modulo $29$.
This is consistent with the number of such consecutive pairs being $\floor {\dfrac {29} 4}$.
== Proof ==
From Quadratic Residues modulo $29$:
:$\set {1, 4, 5, 6, 7, 9, 13, 16, 20, 22, 23, 24, 25, 28}$ are the quadratic residues modulo $29$
The set of pairs of consecutive quadratic residues modulo $29$ is therefore:
:$\set {\set {4, 5}, \set {5, 6}, \set {6, 7}, \set {22, 23}, \set {23, 24}, \set {24, 25}, \set {28, 1} }$
The result follows.
== Sources ==
* : $\text {3-5}$ The Use of Computers in Number Theory: Exercise $7$
Category:Examples of Quadratic Residues
Category:29
Consecutive Pairs of Quadratic Residues/Examples/3

== Examples of Consecutive Pairs of Quadratic Residues ==
There are no Consecutive Pairs of Quadratic Residues modulo $3$.
This is consistent with the number of such consecutive pairs being $\floor {\dfrac 3 4}$.
== Proof ==
From Quadratic Residues modulo $3$:
:$1$ is a quadratic residue
:$2$ is not a quadratic residue.
Hence neither is an element of a pair of consecutive quadratic residues.
The result follows.
== Sources ==
* : $\text {3-5}$ The Use of Computers in Number Theory: Exercise $7$
Category:Examples of Quadratic Residues
Category:3
Consecutive Pairs of Quadratic Residues/Examples/5

== Examples of Consecutive Pairs of Quadratic Residues ==
There is $1$ consecutive pair of quadratic residues modulo $5$.
This is consistent with the number of such consecutive pairs being $\floor {\dfrac 5 4}$.
== Proof ==
From Quadratic Residues modulo $5$:
:$4$ and $1$ are quadratic residue
:$2$ and $3$ are not quadratic residues.
In this context $4$ and $1$ are considered a pair of consecutive quadratic residues.
The result follows.
== Sources ==
* : $\text {3-5}$ The Use of Computers in Number Theory: Exercise $7$
Category:Examples of Quadratic Residues
Category:5
Consecutive Pairs of Quadratic Residues/Examples/61

== Examples of Consecutive Pairs of Quadratic Residues ==
There are $15$ consecutive pairs of quadratic residues modulo $61$.
This is consistent with the number of such consecutive pairs being $\floor {\dfrac {61} 4}$.
== Proof ==
From Quadratic Residues modulo $61$:
:$\set {1, 3, 4, 5, 9, 12, 13, 14, 15, 16, 19, 20, 22, 25, 27, 34, 36, 39, 41, 42, 45, 46, 47, 48, 49, 53, 56, 57, 58, 60}$ are the quadratic residues modulo $61$
The set of pairs of consecutive quadratic residues modulo $61$ is therefore:
:$\set {\set {3, 4}, \set {4, 5}, \set {12, 13}, \set {13, 14}, \set {14, 15}, \set {15, 16}, \set {19, 20}, \set {41, 42}, \set {45, 46}, \set {46, 47}, \set {47, 48}, \set {48, 49}, \set {56, 57}, \set {57, 58}, \set {60, 1} }$
The result follows.
== Sources ==
* : $\text {3-5}$ The Use of Computers in Number Theory: Exercise $7$
Category:Examples of Quadratic Residues
Category:61
Consecutive Pairs of Quadratic Residues/Examples/7

== Examples of Consecutive Pairs of Quadratic Residues ==
There is $1$ consecutive pair of quadratic residues modulo $7$.
This is consistent with the number of such consecutive pairs being $\floor {\dfrac 7 4}$.
== Proof ==
From Quadratic Residues modulo $7$:
:$\set {1, 2, 4}$ are the quadratic residues modulo $7$
The only pair of consecutive quadratic residues is therefore $\set {1, 2}$.
The result follows.
== Sources ==
* : $\text {3-5}$ The Use of Computers in Number Theory: Exercise $7$
Category:Examples of Quadratic Residues
Category:7
Consecutive Powerful Numbers

== Theorem ==
The following pairs are of consecutive positive integers both of which are powerful:
:$\left({8, 9}\right), \left({288, 289}\right), \left({675, 676}\right), \left({9800, 9801}\right), \left({332 \, 928, 332 \, 929}\right), \ldots$
== Proof ==
== Sources ==
* : $288$
Category:Powerful Numbers
Consecutive Powerful Numbers/Mistake

== Source Work ==
:
:The Dictionary
::$288$
== Mistake ==
:With $289$, the second pair of consecutive powerful numbers: $288 = 2^5 3^2$ and $289 = 17^2$. The smallest such pair is $8, 9$ and the next two pairs are $9800, 9801$ and $332, 928, \ 332, 929$.
The $3$rd pair is $675, 676$ which are reported in the entry against $675$. It may have been omitted because it is the odd pair out, being the only one known where the smaller element is odd.
== Sources ==
* : $288$
Category:Mistakes/Curious and Interesting Numbers
Consecutive Primes of form 4n+1

== Theorem ==
The sequence of $16$ consecutive prime numbers beginning from $207 \, 622 \, 273$ are all of the form $4 n + 1$.
== Proof ==
Note that the $11 \, 477 \, 481$st prime:
:$207 \, 622 \, 271 = 4 \times 51 \, 905 \, 568 - 1$
and the $11 \, 477 \, 498$th prime:
:$207 \, 622 \, 567 = 4 \times 51 \, 905 \, 642 - 1$
and so are not of the form $4 n + 1$.
== Historical Note ==
== Sources ==
* : $207,622,273$
Category:Prime Numbers
Consecutive Sophie Germain Primes cannot be Pair of Twin Primes

== Theorem ==
Let $p$ and $p + 2$ be twin primes.
Then unless $p = 3$ it is not possible for both $p$ and $p + 2$ to be Sophie Germain primes.
== Proof ==
First it is noted that $3$ and $5$ twin primes which are both Sophie Germain.
Prime numbers greater than $3$ are of the form $6 n - 1$ and $6 n + 1$.
Thus a pair of twin primes is of the form $\left({6 n - 1, 6 n + 1}\right)$.
The result follows from Sophie Germain Prime cannot be 6n+1.
Category:Twin Primes
Category:Sophie Germain Primes
Consecutive Subsets of N

== Theorem ==
Let $\N_k$ denote the initial segment of the natural numbers determined by $k$:
:$\N_k = \left\{{0, 1, 2, 3, \ldots, k - 1}\right\}$
Then:
:$\N_k = \N_{k + 1} \setminus \left\{{k}\right\}$
In particular:
:$\N_{k - 1} = \N_k \setminus \left\{{k - 1}\right\}$
== Proof ==
The result follows as a direct application of Strict Lower Closure of Sum with One.
Category:Natural Numbers
Consecutive Triple of Repeated Digit-Products

== Theorem ==
The triplet of integers $281, 282, 283$ have the property that if their digits are multiplied, and the process repeated on the result until only $1$ digit remains, that final digit is the same for all three, that is, $6$.
There does not exist an set of four consecutive integers which also all end up at the same single digit.
== Proof ==
== Historical Note ==
== Sources ==
* : $281$
Category:281
Category:282
Category:283
Category:Recreational Mathematics
Consecutive Triples with Same Euler Phi value/Examples/5186

== Example of Consecutive Triples with Same Euler Phi value ==
$5186$, $5187$ and $5188$ all have the same Euler $\phi$ value:
:$\map \phi {5186} = \map \phi {5187} = \map \phi {5188} = 2592 = 2^5 \times 3^4$
== Proof ==
From Euler Phi Function of 5186:
:$\map \phi {5186} = 2592$
From Euler Phi Function of 5187:
:$\map \phi {5187} = 2592$
From Euler Phi Function of 5188:
:$\map \phi {5188} = 2592$
== Sources ==
* : $5186$
* : $5186$
Category:Euler Phi Function
Category:5186
Consecutive Triplets not Sum of Pentagonal Numbers

== Theorem ==
The following triplets of consecutive positive integers are such that none is the sum of $3$ pentagonal numbers:
:$\tuple {19, 20, 21}$
:$\tuple {88, 89, 90}$
:$\tuple {99, 100, 101}$
:$\tuple {111, 112, 113}$
== Proof ==
== Sources ==
* 
* : $88$
Category:Pentagonal Numbers
Conservation of Angular Momentum (Lagrangian Mechanics)

== Theorem ==
Let $P$ be a physical system composed of a finite number of particles.
Let $P$ have the action $S$:
:$\ds S = \int_{t_0}^{t_1} L \rd t$
where:
:$L$ is the standard Lagrangian
:$t$ is time.
Let $L$ be invariant rotation around the $z$-axis.
Then the total angular momentum of $P$ along the $z$-axis is conserved.
== Proof ==
By assumption, $S$ is invariant under the following family of transformations:
where $\epsilon \in \R$.
By Noether's Theorem:
:$\nabla_{\dot {\mathbf x} } L \cdot \boldsymbol \psi + \paren {L - \dot {\mathbf x} \cdot \nabla_{\dot {\mathbf x} } L } \phi = C$
where:
and $C$ is an arbitrary constant.
Then it follows that:
:$\ds \sum_i \paren {\dfrac {\partial L} {\partial {\dot x}_i} y_i - \dfrac {\partial L} {\partial {\dot y}_i} x_i} = C$
Since the is the $z$ component of angular momentum of $P$ along the $z$-axis, we conclude that it is conserved.
== Sources ==
* : $\S 4.22$: Conservation Laws
Category:Lagrangian Mechanics
Conservation of Energy

== Theorem ==
Let $P$ be a physical system.
Let it have the action $S$:
:$\ds S = \int_{t_0}^{t_1} L \rd t$
where $L$ is the standard Lagrangian, and $t$ is time.
Suppose $L$ does not depend on time explicitly:
:$\dfrac {\partial L} {\partial t} = 0$
Then the total energy of $P$ is conserved.
== Proof ==
By assumption, $S$ is invariant under the following family of transformations:
:$T = t + \epsilon$
:$\mathbf X = \mathbf x$
By Noether's Theorem:
:$\nabla_{\dot {\mathbf x} } L \cdot \boldsymbol \psi + \paren {L - \dot {\mathbf x} \cdot \nabla_{\dot {\mathbf x} } L} \phi = C$
where $\phi = 1$, $\boldsymbol \psi = \mathbf 0$ and $C$ is an arbitrary constant.
Then it follows that:
Since the last term is the total energy of $P$, we conclude that it is conserved.
== Sources ==
* : $\S 4.22$: Conservation Laws
* 
* 
Category:Conservation Laws
Category:Energy
Category:Lagrangian Mechanics
Category:Physics
Conservation of Momentum

== Theorem ==
Let $P$ be a physical system.
Let it have the action $S$:
:$\ds S = \int_{t_0}^{t_1} L \rd t$
where $L$ is the standard Lagrangian, and $t$ is time.
Suppose $L$ does not depend on one of the coordinates explicitly:
:$\dfrac {\partial L} {\partial x_j} = 0$
Then the total momentum of $P$ along the axis $x_j$ is conserved.
== Proof ==
By assumption, $S$ is invariant under the following family of transformations:
:$T = t$
:$X_j = x_j + \epsilon$
:$X_{i \mathop \ne j} = x_{i \mathop \ne j}$
By Noether's Theorem:
:$\nabla_{\mathbf x} L \cdot \boldsymbol \psi + \paren {L - \dot {\mathbf x} \cdot \nabla_{\dot {\mathbf x} } L} \phi = C$
where $\phi = 0$, $\psi_i = 1$, $\psi_{j \mathop \ne i} = 0$ and $C$ is an arbitrary constant.
Then it follows that:
:$\dfrac {\partial L} {\partial x_j} = C$
Since the last term is the momentum of $P$, we conclude that it is conserved.
== Sources ==
* : $\S 4.22$: Conservation Laws
* 
* 
Category:Conservation Laws
Category:Momentum
Category:Lagrangian Mechanics
Category:Physics
Conservative Vector Field/Examples/Electric Field

== Example of Conservative Vector Field ==
Let $F$ be an electric potential field over a region of space $R$.
Let $F$ give rise to the electric field $\mathbf V$.
Then $\mathbf V$ is a conservative vector field, as it is the gradient of $F$.
== Sources ==
* : Chapter $\text {IV}$: The Operator $\nabla$ and its Uses: $2 a$. The Operation $\nabla S$
Category:Examples of Conservative Vector Fields
Consistency of Logical Formulas has Finite Character

== Theorem ==
Let $P$ be the property of collections of logical formulas defined as:
:$\forall \FF: \map P \FF$ denotes that $\FF$ is consistent.
Then $P$ is of finite character.
That is:
:$\FF$ is a consistent set of formulas every finite subset of $\FF$ is also a consistent set of formulas.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles
Category:Logical Consistency
Category:Finite Character
Category:Consistency of Logical Formulas has Finite Character
Consistency of Logical Formulas has Finite Character/Proof 2

== Theorem ==
== Proof ==
=== Sufficient Condition ===
Let $\FF$ be a consistent set of formulas.
 there exists $\GG \subseteq \FF$ where $\GG$ is finite such that $\GG$ is inconsistent.
But then from Set of Logical Formulas is Inconsistent iff it has Finite Inconsistent Subset, $\FF$ is itself inconsistent.
From this contradiction it follows that $\GG$ must be consistent.
As $\GG$ is arbitrary, it follows that every finite subset of $\FF$ is consistent.
=== Necessary Condition ===
Let every finite subset of $\FF$ is consistent.
 $\FF$ is inconsistent.
Then by Set of Logical Formulas is Inconsistent iff it has Finite Inconsistent Subset there exists a finite subset $\GG \subseteq \FF$ such that $\GG$ is inconsistent.
But $\GG$ is consistent.
From this contradiction it follows that $\FF$ must also be consistent.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles
Category:Consistency of Logical Formulas has Finite Character
Consistent Set of Formulas can be Extended to Maximal Consistent Set

== Theorem ==
Let $\FF$ be a collection of consistent logical formulas.
Then $\FF$ can be extended to (that is, is a subset of) a maximal consistent set of formulas.
== Proof ==
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 5$ Maximal principles
Category:Logical Consistency
Consistent Set of Logical Formulas is Subset of Maximally Consistent Set

== Theorem ==
Let $\LL$ be a formal language used in the field of symbolic logic.
Let $\FF$ be the set of logical formulas of $\LL$.
Let $\FF$ be countable.
Let $S$ be a consistent subset of $\FF$.
Then $S$ is a subset of some maximal consistent set of formulas.
== Proof ==
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 6$ Another approach to maximal principles
Category:Logical Consistency