nyuuzyou/wikis huggingface dataset:
Compound Statement/Examples/Atmospheric Testing

== Example of Compound Statement ==
:Atmospheric testing of nuclear weapons will be discontinued or this planet will become uninhabitable.
is a compound statement whose substatements are:
:Atmospheric testing of nuclear weapons will be discontinued
:This planet will become uninhabitable.
== Sources ==
* : $2$ Arguments Containing Compound Statements: $2.1$: Simple and Compound Statements
Category:Examples of Statements
Compound Statement/Examples/Barometer

== Example of Compound Statement ==
:If the barometer falls then either it will rain or it will snow
is a compound statement whose substatements are:
:The barometer falls
:It will rain
:It will snow.
== Sources ==
* : $\S 1$: Informal statement calculus: $\S 1.1$: Statements and connectives
Category:Examples of Statements
Compton Wavelength of Electron

== Theorem ==
The Compton wavelength of the electron is:
=== Symbol ===
== Proof ==
By definition, the Compton wavelength $\lambda_\E$ of an electron is given as:
:$\lambda_\E = \dfrac h {m_\E c}$
where:
:$m_\E$ denotes the mass of the electron
:$h$ denotes Planck's constant
:$c$ denotes the speed of light.
Then we have:
== Sources ==
* : $2$. Physical Constants and Conversion Factors: Table $2.3$ Adjusted Values of Constants
::which gives the mantissa as $2 \cdotp 426 \, 309 \, 6$ with an uncertainty of $\pm 74$ corresponding to the $2$ least significant figures
Category:Compton Wavelength
Category:Electrons
Compton Wavelength of Proton

== Theorem ==
The Compton wavelength of the proton is:
=== Symbol ===
== Proof ==
By definition, the Compton wavelength $\lambda_{\mathrm p}$ of a proton is given as:
:$\lambda_{\mathrm p} = \dfrac h {m_{\mathrm p} c}$
where:
:$m_{\mathrm p}$ denotes the mass of the proton
:$h$ denotes Planck's constant
:$c$ denotes the speed of light.
Then we have:
== Sources ==
* : $2$. Physical Constants and Conversion Factors: Table $2.3$ Adjusted Values of Constants
::which gives the mantissa as $1 \cdotp 321 \, 440 \, 9$ with an uncertainty of $\pm 90$ corresponding to the $2$ least significant figures
Category:Compton Wavelength
Category:Protons
Computable Rational Sequence is Computable Real Sequence

== Theorem ==
Let $\sequence {x_n}$ be a computable rational sequence.
Then, $\sequence {x_n}$ is a computable real sequence.
== Proof ==
By Computable Real Sequence iff Limits of Computable Rational Sequences, it suffices to show that there exist:
:A computable rational sequence $\sequence {a_k}$
:Total recursive functions $\phi, \psi : \N^2 \to \N$
such that:
:$\forall m, p \in \N: \forall n \ge \map \psi {m, p}: \size {a_{\map \phi {m, n}} - x_m} < \dfrac 1 {p + 1}$
We will choose:
:$\forall k \in \N: a_k = x_k$
:$\map \phi {m, n} = m$
:$\map \psi {m, p} = 0$
$\phi$ and $\psi$ are trivially primitive recursive, and thus total recursive by Primitive Recursive Function is Total Recursive Function.
As $\sequence {a_k} = \sequence {x_k}$, it is computable by assumption.
Finally, we have that, for any $m, n, p \in \N$:
Category:Computability Theory
Computable Real Sequence iff Limits of Computable Rational Sequences/Corollary

== Corollary to Computable Real Sequence iff Limits of Computable Rational Sequences ==
Let $\sequence {x_m}$ be a real sequence.
Then:
:$\sequence {x_m}$ is a computable real sequence
 there exists a computable rational sequence $\sequence {a_k}$ such that, for all $m, p \in \N$:
:$\size {a_{\map \pi {m, p}} - x_m} < \dfrac 1 {p + 1}$
where $\pi$ is the Cantor pairing function.
== Proof ==
By Computable Real Sequence iff Limits of Computable Rational Sequences, it suffices to show that there exist:
* A computable rational sequence $\sequence {b_k}$
* Total recursive functions $\phi, \psi : \N^2 \to \N$
such that:
:$\forall m, p \in \N: \forall n \ge \map \psi {m, p}: \size {b_{\map \phi {m, n}} - x_m} < \dfrac 1 {p + 1}$
 there exists a computable rational sequence $\sequence {a_k}$ such that, for all $m, p \in \N$:
:$\size {a_{\map \pi {m, p}} - x_m} < \dfrac 1 {p + 1}$
where $\pi$ is the Cantor pairing function.
=== Necessary Condition ===
Let:
* $\sequence {b_k}$ be a computable rational sequence
* $\phi, \psi : \N^2 \to \N$ be total recursive functions
satisfying:
:$\forall m, p \in \N: \forall n \ge \map \psi {m, p}: \size {b_{\map \phi {m, n}} - x_m} < \dfrac 1 {p + 1}$
Define $\chi : \N \to \N$ as:
:$\map \chi k = \map \phi {\map {\pi_1} k, \map \psi {\map {\pi_1} k, \map {\pi_2} k} }$
where $\pi_1, \pi_2 : \N \to \N$ are the projections on the Cantor pairing function.
We have that $\chi$ is total recursive by:
* Inverse of Cantor Pairing Function is Primitive Recursive
* Primitive Recursive Function is Total Recursive Function
Let $\sequence {a_k}$ be defined as:
:$a_k = b_{\map \chi k}$
By Computable Subsequence of Computable Rational Sequence is Computable:
:$\sequence {a_k}$ is a computable rational sequence.
Let $m, p \in \N$ be arbitrary.
We have:
=== Sufficient Condition ===
Let $\sequence {a_k}$ be a computable rational sequence such that, for all $m, p \in \N$:
:$\size {a_{\map \pi {m, p}} - x_m} < \dfrac 1 {p + 1}$
where $\pi$ is the Cantor pairing function.
Let $\sequence {b_k}$ be defined as:
:$b_k = a_k$
Let $\phi, \psi : \N \to \N$ be defined as:
:$\map \phi {m, n} = \map \pi {m, n}$
:$\map \psi {m, p} = p$
Both $\phi$ and $\psi$ are total recursive by:
* Cantor Pairing Function is Primitive Recursive
* Primitive Recursive Function is Total Recursive Function
Let $m, n, p \in \N$ satisfy:
:$n \ge \map \psi {m, p}$
We have:
Category:Computability Theory
Computable Subsequence of Computable Rational Sequence is Computable

== Theorem ==
Let $\sequence {x_n}_{n \in \N}$ be a computable rational sequence.
Let $\phi : \N \to \N$ be a total recursive function.
Then:
:$\sequence {x_{\map \phi n}}_{n \in \N}$
is a computable rational sequence.
=== Corollary ===
== Proof ==
By definition of computable rational sequence, there exist total recursive $N, D : \N \to \N$ such that, for every $n \in \N$:
:$x_n = \dfrac {k_n} {\map D n + 1}$
where $\map N n$ codes the integer $k_n$.
We define total recursive $N', D' : \N \to \N$ as:
:$\map {N'} n = \map N {\map \phi n}$
:$\map {D'} n = \map N {\map \phi n}$
Now, we just need to show that, for every $n \in \N$:
:$x_{\map \phi n} = \dfrac {k'_n} {\map {D'} n + 1}
where $\map {N'} n$ codes the integer $k'_n$.
We have:
Category:Computability Theory
Computable Subsequence of Computable Rational Sequence is Computable/Corollary

== Theorem ==
Let $\sequence {x_k}$ be a computable rational sequence.
Let $\phi : \N^2 \to \N$ be a total recursive function.
Then, there exists a computable rational sequence $\sequence {y_k}$ such that, for all $n, m \in \N$:
:$y_{\map \pi {n, m}} = x_{\map \phi {n, m}}$
where $\pi : \N^2 \to \N$ is the Cantor pairing function.
== Proof ==
Let $\psi : \N^2 \to \N$ be defined as:
:$\map \psi k = \map \phi {\map {\pi_1} k, \map {\pi_2} k}$
where $\pi_1, \pi_2 : \N \to \N$ are the projections on the Cantor pairing function.
By Inverse of Cantor Pairing Function is Primitive Recursive, we have that $\psi$ is total recursive.
Therefore, by Computable Subsequence of Computable Rational Sequence is Computable:
:$\sequence {x_{\map \psi k}}$
is a computable rational sequence.
It remains to be shown that:
:$y_k = x_{\map \psi k}$
for all $k \in \N$.
We have:
Category:Computability Theory
Concatenation (Formal Systems)/Examples/Arbitrary Example 1

== Example of Concatenation in the context of Formal Systems ==
Let $a, b, c \in \AA$.
Then concatenating $b$ to $a$ results in the word $ab$, and concatenating $c$ gives the word $abc$.
Concatenating $b$ then gives $abcb$.
Concatenating $aba$ to $aabcba$ gives the word $aabcbaaba$.
Category:Examples of Concatenation (Formal Systems)
Concatenation (Formal Systems)/Examples/Arbitrary Example 2

== Example of Concatenation in the context of Formal Systems ==
Let $A$ be the string $\text {dog}$.
Let $B$ be the string $\text {house}$.
Then concatenating $B$ to $A$ results in the string $\text {doghouse}$.
== Sources ==
* : Chapter $1$: Preliminaries: $1.1$ Strings, Alphabets and Languages
Category:Examples of Concatenation (Formal Systems)
Concatenation of Contours is Contour

== Theorem ==
Let $C$ and $D$ be contours in the complex plane.
That is, $C$ is a finite sequence of directed smooth curves $C_1, \ldots, C_n$.
Let $C_k$ be parameterized by the smooth path $\gamma_k: \closedint {a_k} {b_k} \to \C$ for all $k \in \set {1, \ldots, n}$.
Similarly, $D$ is a finite sequence of directed smooth curves $D_1, \ldots, D_m$.
Let $D_j$ be parameterized by the smooth path $\sigma_j: \closedint {c_j} {d_j} \to \C$ for all $j \in \set {1, \ldots, m}$.
Suppose $\map {\gamma_n} {b_n} = \map {\sigma_1} {c_1}$.
Then the finite sequence:
:$C_1, \ldots, C_n, D_1, \ldots, D_m$
defines a contour.
== Proof ==
By definition of contour, each $C_k$ and $D_j$ is a directed smooth curve for all $k \in \set {1, \ldots, n}, j \in \set {1, \ldots, m}$.
By definition of contour:
:$\map {\gamma_k} {b_k} = \map {\gamma_{k + 1} } {a_{k + 1} }$
and:
:$\map {\sigma_j} {d_j} = \map {\sigma_{j + 1} } {c_{j + 1} }$
for all $k \in \set {1, \ldots, n - 1}$ and $j \in \set {1, \ldots, m - 1}$.
By assumption:
:$\map {\gamma_n} {b_n} = \map {\sigma_1} {c_1}$
Hence, $C_1, \ldots, C_n, D_1, \ldots, D_m$ defines a contour.
== Sources ==
* $\S 2.2$
Category:Complex Contour Integrals
Concatenation of URM Programs is Associative

== Theorem ==
Let $P, Q, R$ be one-variable URM programs.
Then the concatenated URM programs $P * \paren {Q * R}$ and $\paren {P * Q} * R$ are the same.
== Proof ==
For ease of analysis, let us assume that:
:Each of $P$ and $Q$ have already had the appropriate Clear Registers Program $\map Z {2, \map \rho P}$ and $\map Z {2, \map \rho Q}$ appended to them
:Each of $P$ and $Q$ have already had the appropriate amendments made to their exit jumps so as to lead to the first line of the appropriate Clear Registers Program. (Note that if the latter is a null URM program, this will be the line immediately following the end of the program).
Thus we can discuss $P$ and $Q$ without reference to these technical details.
When concatenating programs, the only amendments made to the programs themselves are to the contents of the Jump.
So it is clear that the number, type and order of the basic instructions of both programs are the same.
All we have to consider are the amendments to the Jumps.
Suppose the lengths of $P$ and $Q$ are as follows:
:$\map \lambda P = t$
:$\map \lambda Q = s$
First we look at the Jumps of $R$.
When we form $\paren {Q * R}$, every Jump of $R$ of the form $\map J {m, n, q}$ is replaced by $\map J {m, n, q + s}$.
Then when forming $P * \paren {Q * R}$ the Jump $\map J {m, n, q + s}$ is replaced by $\map J {m, n, q + s + t}$.
On the other hand, when we form $\paren {P * Q} * R$, every Jump of $R$ of the form $\map J {m, n, q}$ is replaced directly by $\map J {m, n, q + \paren {s + t} }$.
Thus the Jumps of $R$ have been replaced by the same instructions in each of $\paren {P * Q} * R$ and $P * \paren {Q * R}$.
Now we look at the Jumps of $Q$.
When forming $\paren {Q * R}$, we have already ensured that every Jump of $Q$ of the form $\map J {m, n, q}$ where $q > s$ has been replaced by $\map J {m, n, s + 1}$.
So when forming $P * \paren {Q * R}$ the Jump $\map J {m, n, s + 1}$ is replaced by $\map J {m, n, s + t + 1}$.
On the other hand, when forming $\paren {P * Q}$, every Jump of $Q$ of the form $\map J {m, n, q}$ is replaced by $\map J {m, n, q + t}$.
We have that $\paren {P * Q}$ has $s + t$ instructions. When $q > s$, we take account of the fact that $q + t > s + t$.
So in $\paren {P * Q} * R$, Jumps of the form $\map J {m, n, s + 1}$ are replaced by $\map J {m, n, s + t + 1}$.
This agrees with the corresponding Jumps in $P * \paren {Q * R}$.
Similarly, we can show that all other Jumps of $Q$, and all Jumps of $P$, are amended in exactly the same way in each of $\paren {P * Q} * R$ and $P * \paren {Q * R}\right)$.
Hence the result.
Category:URM Programs
Concave Real Function is Continuous

== Theorem ==
Let $f$ be a real function which is concave on the open interval $\openint a b$.
Then $f$ is continuous on $\openint a b$.
== Proof ==
From Concave Real Function is Left-Hand and Right-Hand Differentiable, $f$ is left-hand and right-hand differentiable on $\openint a b$.
From Left-Hand and Right-Hand Differentiable Function is Continuous, $f$ is continuous on $\openint a b$.
== Also see ==
* Convex Real Function is Continuous
Category:Concave Real Functions
Category:Continuous Real Functions
Concave Real Function is Left-Hand and Right-Hand Differentiable

== Theorem ==
Let $f$ be a real function which is concave on the open interval $\openint a b$.
Then the left-hand derivative $\map {f'_-} x$ and right-hand derivative $\map {f'_+} x$ both exist for all $x \in \openint a b$.
== Proof ==
Let $f$ be concave on $\openint a b$.
Then by definition of concavity:
:$\forall x_1, x_2, x_3 \in \openint a b: x_1 < x_2 < x_3: \dfrac {\map f {x_2} - \map f {x_1} } {x_2 - x_1} \ge \dfrac {\map f {x_3} - \map f {x_1} } {x_3 - x_1}$
Let $0 < h_1 < h_2$.
Substitute $x_1 = x$, $x_2 = x + h_1$, $x_3 = x + h_2$. Then:
:$\dfrac {\map f {x + h_1} - \map f x} {h_1} \ge \dfrac {\map f {x + h_2} - \map f x} {h_2}$
Hence the function $\map F h = \dfrac {\map f {x + h} - \map f x} h$ decreases in some $\closedint 0 \delta$.
Thus from Limit of Decreasing Function it follows that both $\ds \lim_{h \mathop \to 0^+} \map F h = \map {f'_+} x$ and $\ds \lim_{h \mathop \to 0^-} \map F h = \map {f'_-} x$ exist.
== Also see ==
* Convex Real Function is Left-Hand and Right-Hand Differentiable
Category:Concave Real Functions
Concentration of Lipschitz Functionals of Normal Vector

== Theorem ==
Let $X_1, \dots, X_n$ be continuous random variables with a standard normal distribution $\Gaussian 0 1$.
Let $F: \R^n \to \R$ be L-Lipschitz.
Then:
:$\map \Pr {\size {\map F X - \expect {\map F X} } > L u} \le \map \exp {-\dfrac {u^2} {2 C} }$
where:
:$\Pr$ denotes the probability function
:$\expect {\, \cdot \,}$ denotes expectation.
== Proof (The Pisier-Maurey approach) ==
The goal will be to control bound the Laplace transform:
:$\expect {\map \exp {\lambda \paren {\map f X - \expect {\map f X} } } } \le \map \exp {C \dfrac {\paren {\lambda L}^2} 2}$
Then by Chernoff's bound:
:$\map \Pr {\size {\map F X - \expect {\map F X} } > L u} \le \inf_{\lambda \mathop \ge 0} \map \exp {-\lambda L u + C \dfrac {\paren {\lambda L}^2} 2}$
is minimized for $\lambda = \dfrac u {L C}$.
Consider another normal $n$-vector $Y$ independent of $X$.
Then by Jensen's inequality:
:$\expect {\map \exp {\lambda(\map f X - \expect {\map f X}) } } \leq \expect {\map \exp {\lambda(\map f X - \map f Y) } }$
Next we will estimate the difference $\map f X - \map f Y$.
Consider:
:$X_\theta := \map \cos {\dfrac \pi 2 \theta} X + \map \sin {\dfrac \pi 2 \theta} Y$
Then:
:$\dfrac \pi 2 Z_\theta := \dfrac \partial {\partial \theta} X_\theta = -\dfrac \pi 2 \map \sin {\dfrac \pi 2 \theta} X + \dfrac \pi 2 \map \cos {\dfrac \pi 2 \theta} Y$
Because $\cov {\dfrac \pi 2 Z_\theta, X_\theta} = 0$ we have that $\dfrac \pi 2 Z_\theta, X_\theta$ are independent. 
By FTC we have:
:$\ds \map f X - \map f Y = \int_0^1 \dfrac \partial {\partial \theta} \map f {X_\theta} \rd \theta = \int_0^1 \dfrac \pi 2 Z_\theta \cdot \nabla \map f {X_\theta} \rd \theta$
By Jensen's we find the bound:
:$\ds \int_0^1 \expect {\map \exp {\lambda \paren {\dfrac \pi 2 Z_\theta \cdot \nabla \map f {X_\theta} } } } \rd \theta$
For $v \in \R^n$ we have:
:$\expect {\map \exp {\lambda \paren {\dfrac \pi 2 Z_\theta \cdot v} } } = \map \exp {\dfrac {\size v^2} 2}$
Therefore, by independence of $\dfrac \pi 2 Z_\theta, X_\theta$ we find:
:$\ds \int_0^1 \expect {\map \exp {\dfrac {\paren {\lambda \pi/2}^2} 2 \size {\nabla \map f {X_\theta} }^2} } \rd \theta$
By the Lipschitz bound we find the bound:
:$\map \exp {\dfrac {\paren {\lambda \pi/2}^2} 2 L^2}$
Concentric Circles do not Intersect

== Theorem ==
If two circles are concentric, they share no points on their circumferences.
Alternatively, this can be worded:
:If two concentric circles share one point on their circumferences, then they share them all (that is, they are the same circle).
== Proof ==
This follows directly from:
:Intersecting Circles have Different Centers
:Touching Circles have Different Centers.
Category:Concentric Circles
Conclusion of Valid Categorical Syllogism is Negative iff one Premise is Negative

== Theorem ==
The conclusion of a valid categorical syllogism is negative one of the premises is also negative.
== Proof ==
== Also see ==
* Rules of Quality
* No Valid Categorical Syllogism contains two Negative Premises
== Historical Note ==
== Sources ==
* : Chapter $4$: The Predicate Calculus $2$: $4$ The Syllogism: Exercises: $\text{B}$: Rules of Quality: $1$
Category:Rules of Quality
Concomitant Variable/Examples/Insecticide

== Example of Concomitant Variable ==
Consider an experiment to test the effect of various insecticides at reducing a pest on fruit trees.
An appropriate concomitant variable would be a measure of the level of infestation on each experimental unit immediately before applying the insecticides.
Hence the technique of analysis of covariance adjusts for the effects of differing initial infestation on the response to the insecticides.
== Sources ==
* 
* 
Category:Examples of Concomitant Variables
Condition Number/Examples/Coefficient Matrix

== Example of Condition Number ==
Let $\mathbf A$ be the coefficient matrix of a system of simultaneous linear equations.
A condition number for $\mathbf A$ is:
:the norm of $\mathbf A$ multiplied by the norm of its inverse $\mathbf A^{-1}$
If the system of simultaneous linear equations is ill-conditioned, $\mathbf A^{-1}$ will have relatively large elements and the condition number will likewise be large.
== Sources ==
* 
* 
Category:Examples of Condition Number
Condition for 3 Points in Plane to be Collinear

== Theorem ==
Let $A = \tuple {x_1, y_1}, B = \tuple {x_2, y_2}, C = \tuple {x_3, y_3}$ be points in the Cartesian plane.
Then:
:$A$, $B$ and $C$ are collinear
 the determinant:
:$\begin {vmatrix}
x_1 & y_1 & 1 \\
x_2 & y_2 & 1 \\
x_3 & y_3 & 1 \\
\end {vmatrix}$
equals zero.
== Proof ==
We have that:
:$A$, $B$ and $C$ are collinear
:
:the area of $\triangle ABC = 0$
:
:$\dfrac 1 2 \size {\paren {\begin {vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end {vmatrix} } } = 0$ (from Area of Triangle in Determinant Form)
:
:$\begin {vmatrix} x_1 & y_1 & 1 \\ x_2 & y_2 & 1 \\ x_3 & y_3 & 1 \\ \end {vmatrix} = 0$
== Sources ==
* : Chapter $\text I$. Coordinates: $9$. Collinear points
* : $\S 10$: Formulas from Plane Analytic Geometry: Area of Triangle with Vertices at $\tuple {x_1, y_1}$, $\tuple {x_2, y_2}$, $\tuple {x_3, y_3}$: $10.10$
* : $\S 8$: Formulas from Plane Analytic Geometry: Area of Triangle with Vertices at $\tuple {x_1, y_1}$, $\tuple {x_2, y_2}$, $\tuple {x_3, y_3}$: $8.10.$
Category:Collinear Points
Condition for 3 over n producing 3 Egyptian Fractions using Greedy Algorithm when 2 Sufficient

== Theorem ==
Consider proper fractions of the form $\dfrac 3 n$ expressed in canonical form.
Let Fibonacci's Greedy Algorithm be used to generate a sequence $S$ of Egyptian fractions for $\dfrac 3 n$.
Then $S$ consists of $3$ terms, where $2$ would be sufficient the following conditions hold:
:$n \equiv 1 \pmod 6$
:$\exists d: d \divides n$ and $d \equiv 2 \pmod 3$
=== Corollary ===
The smallest $n$ for which $S$ consists of $3$ terms, where $2$ would be sufficient, is $25$.
== Proof ==
By Upper Limit of Number of Unit Fractions to express Proper Fraction from Greedy Algorithm, $S$ consists of no more than $3$ terms.
Suppose $n$ has our desired property.
Since $\dfrac 3 n$ is proper, $n \ge 4$.
Since $\dfrac 3 n$ is in canonical form, $3 \nmid n$.
We also have that $S$ consists of at least $2$ terms.
Consider the case $n = 3 k - 1$.
:$\dfrac 3 n = \dfrac 1 k + \dfrac 1 {k \paren {3 k - 1} } \quad$ as $\ceiling {\dfrac n 3} = \ceiling {k - \dfrac 1 3} = k$
Fibonacci's Greedy Algorithm produces $2$ terms only.
hence it must be the case that $n = 3 k - 2$.
We have:
:$\dfrac 3 n = \dfrac 1 k + \dfrac 2 {k \paren {3 k - 2} } \quad$ as $\ceiling {\dfrac n 3} = \ceiling {k - \dfrac 2 3} = k$
If $k$ is even, $\dfrac 1 {\paren {k / 2} \paren {3 k - 2} }$ is an Egyptian fraction.
Then Fibonacci's Greedy Algorithm would produce $2$ terms only.
hence it must be the case that $k$ is odd.
This happens $n$ is odd.
We have shown that for Fibonacci's Greedy Algorithm to produce $3$ terms, $n$ must be odd and $n = 3 k - 2$.
By Chinese Remainder Theorem, these conditions can be merged into:
:$n \equiv 1 \pmod 6$
We need find when Fibonacci's Greedy Algorithm gives minimum terms.
Write:
Since $\dfrac 3 n$ is in canonical form, $x y \divides n$.
By Divisor of Product, we can find $p, q \in \N$ such that:
:$p \divides x$, $q \divides y$
:$p q = n$
Rewrite:
If $\exists d: d \divides n$ and $d \equiv 2 \pmod 3$, set $q = 1$ and $a = \dfrac {d + 1} 3$.
Then $p = n$ and $b = \dfrac {a n} {3 a - 1} = \dfrac {n \paren {d + 1} } {3 d}$, which is a solution:
:$\dfrac 3 n = \dfrac 3 {n \paren {d + 1} } + \dfrac {3 d} {n \paren {d + 1} }$
Now suppose $d$ does not exist.
Then any divisor of $n$ is equivalent to $1 \pmod 3$.
Hence $q \equiv 1 \pmod 3$.
Thus $3 a - q \equiv 2 \pmod 3$.
Let $r = \gcd \set {3 a - q, a}$.
We have:
:$a \divides b \paren {3 a - q} = b r \paren {\dfrac {3 a - q} r}$
Thus:
:$\dfrac a r \divides b \paren {\dfrac {3 a - q} r}$
By Integers Divided by GCD are Coprime:
:$\dfrac a r \perp \dfrac {3 a - q} r$
Finally, by Euclid's Lemma:
:$\dfrac a r \divides b$
Hence $\dfrac {3 a - q} r, \dfrac {b r} a \in \Z$, and $p = \paren {\dfrac {3 a - q} r} \paren {\dfrac {b r} a}$.
Thus we also have:
:$\dfrac {3 a - q} r \divides p \divides n$
:$r = \gcd \set {q, a} \divides q \divides n$
Hence:
:$\dfrac {3 a - q} r \equiv 1 \pmod 3$
:$r \equiv 1 \pmod 3$
Taking their product:
which is a contradiction.
Therefore $n$ cannot be expressed as the sum of $2$ Egyptian fractions.
Hence the result.
Category:Fibonacci's Greedy Algorithm
Category:Egyptian Fractions
Condition for 4 Points to be Coplanar

== Theorem ==
Let:
be distinct points in Cartesian $3$-space.
Then $p_1$, $p_2$, $p_3$ and $p_4$ are coplanar :
:$\begin {vmatrix} x_1 & y_1 & z_1 & 1 \\ x_2 & y_2 & z_2 & 1 \\ x_3 & y_3 & z_3 & 1 \\ x_4 & y_4 & z_4 & 1 \end {vmatrix} = 0$
where the construct is evaluated as a determinant.
== Proof ==
== Sources ==
* 
* 
Category:Coplanar Points