nyuuzyou/wikis huggingface dataset:
Graph of Real Function in Cartesian Plane intersects Vertical at One Point

== Theorem ==
Let $f: \R \to \R$ be a real function.
Let its graph be embedded in the Cartesian plane $\CC$:
:
Every vertical line through a point $a$ in the domain of $f$ intersects the graph of $f$ at exactly one point $P = \tuple {a, \map f a}$.
== Proof ==
From Equation of Vertical Line, a vertical line in $\CC$ through the point $\tuple {a, 0}$ on the $x$-axis has an equation $x = a$.
A real function is by definition a mapping.
Hence:
:$\forall a_1, a_2 \in \Dom f: a_1 = a_2 \implies \map f {a_1} = \map f {a_2}$
where $\Dom f$ denotes the domain of $f$.
Thus for each $a \in \Dom f$ there exists exactly one ordered pair $\tuple {a, y}$ such that $y = \map f a$.
That is, there is exactly one point on $x = a$ which is also on the graph of $f$.
The result follows.
== Sources ==
* : Chapter $\text{I}$: Sets and Functions: Functions
Category:Real Functions
Category:Graphs of Mappings
Graph of Real Injection in Coordinate Plane intersects Horizontal Line at most Once

== Theorem ==
Let $f: \R \to \R$ be a real function which is injective.
Let its graph be embedded in the Cartesian plane $\CC$:
:
Let $\LL$ be a horizontal line through a point $b$ in the codomain of $f$.
Then $\LL$ intersects the graph of $f$ on at most one point $P = \tuple {a, b}$ where $b = \map f a$.
== Proof ==
From Equation of Horizontal Line, a horizontal line in $\CC$ through the point $\tuple {0, b}$ on the $y$-axis has an equation $y = b$.
By hypothesis, $f$ is a injection.
Hence:
:$\forall a_1, a_2 \in \Dom f: \map f {a_1} = \map f {a_2} \implies a_1 = a_2$
where $\Dom f$ denotes the domain of $f$.
Thus for each $b \in \Cdm f$ there exists exactly one ordered pair $\tuple {a, b}$ such that $b = \map f a$.
That is, there is exactly one point on $y = b$ which is also on the graph of $f$.
The result follows.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $6 \ \text {(b)}$
Category:Real Functions
Category:Injections
Category:Graphs of Mappings
Graph of Real Surjection in Coordinate Plane intersects Every Horizontal Line

== Theorem ==
Let $f: \R \to \R$ be a real function which is surjective.
Let its graph be embedded in the Cartesian plane $\CC$:
:
Every horizontal line through a point $b$ in the codomain of $f$ intersects the graph of $f$ on at least one point $P = \tuple {a, b}$ where $b = \map f a$.
== Proof ==
From Equation of Horizontal Line, a horizontal line in $\CC$ through the point $\tuple {0, b}$ on the $y$-axis has an equation $y = b$.
By hypothesis, $f$ is a surjection.
Hence:
:$\forall b \in \R: \exists a \in \R: b = \map f a$
Thus for each $b \in \R$ there exists at least one ordered pair $\tuple {a, b}$ such that $b = \map f a$.
That is, there exists at least one point on $y = b$ which is also on the graph of $f$.
The result follows.
== Sources ==
* : $\S 5$. Induced mappings; composition; injections; surjections; bijections: Exercise $6 \ \text {(a)}$
Category:Real Functions
Category:Surjections
Category:Graphs of Mappings
Graph of y equals a x^n on Logarithmic Scales

== Theorem ==
Let $a$ and $n$ be real constants.
Let the equation $y = a x^n$ be a plotted on logarithmic scales.
Then the graph of $y = a x^n$ is a straight line.
== Proof ==
== Sources ==
* 
* 
* 
Category:Logarithmic Scales
Category:Logarithmic Graph Paper
Graph with Even Vertices Partitions into Cycles

== Theorem ==
Let $G = \struct {V, E}$ be a graph whose vertices are all even.
Then its edge set $E$ can be partitioned into cycles, no two of which share an edge.
The converse also holds: a graph which can be partitioned into cycles must have all its vertices even.
== Proof ==
Let $G = \struct {V, E}$ be a graph whose vertices are all even.
If there is more than one vertex in $G$, then each vertex must have degree greater than $0$.
Begin at any vertex $u$.
Since the graph is connected (if the graph is not connected then the argument will be applied to separate components), there must be an edge $\set {u, u_1}$ for some vertex $u_1 \ne u$.
Since $u_1$ has even degree greater than $0$, there is an edge $\set {u_1, u_2}$.
These two edges make a trail from $u$ to $u_2$.
Continue this trail, leaving each vertex on an edge that was not previously used, until we reach a vertex $v$ that we have met before.
(Note: $v$ may or may not be the same vertex as $u$. It does not matter either way.)
The edges of the trail between the two occurrences of $v$ must form a cycle.
Call the cycle formed by this process $C_1$.
If $C_1$ covers all the edges of $G$, the proof is complete.
Otherwise, remove the edges forming $C_1$ from the graph, leaving the graph $G_1$.
All the vertices in $G_1$ are still even.
So pick some vertex $u'$ in $G_1$.
Repeat the same process as before, starting with an edge $\set {u', u'_1}$.
By the same argument, we can generate a new cycle $C_2$, which has no edges in common with $C_1$.
If $C_2$ covers all the rest of the edges of $G$, then we are done.
Otherwise, remove the edges forming $C_2$ from the graph, leaving the graph $G_2$, which again contains only even vertices.
We continue in this way until we have used up all the edges of $G$.
By this time we have a number of cycles, $C_1, C_2, \ldots, C_k$ which between them contain all the edges of $G$ but no two of them have an edge in common.
The converse is trivial:
All the vertices of a cycle by definition are even.
So if we were to take all these cycles and add them to the graph in turn, each vertex would increase its order by an even number.
== Sources ==
* : $\S 3.1$: The Königsberg Bridge Problem: An Introduction to Eulerian Graphs: Problem $15$
Category:Graph Theory
Graphical Solution to Simultaneous Equations/Examples/Single Equation

== Example of Graphical Solution to Simultaneous Equations ==
It is possible to apply the technique of a graphical solution to simultaneous equations to a single equation by setting one of the simultaneous equations to $y = \map f x$, and the other to $y = 0$.
== Sources ==
* 
* 
Category:Examples of Graphical Solutions to Simultaneous Equations
Grassmann's Identity

== Theorem ==
Let $K$ be a division ring.
Let $\struct {G, +_G, \circ}_K$ be a $K$-vector space.
Let $M$ and $N$ be finite-dimensional subspaces of $G$.
Then the sum $M + N$ and intersection $M \cap N$ are finite-dimensional, and:
:$\map \dim {M + N} + \map \dim {M \cap N} = \map \dim M + \map \dim N$
== Proof 1 ==
== Proof 2 ==
== Also known as ==
This result can also be referred to as Grassmann's formula.
Category:Linear Algebra
Category:Grassmann's Identity
Grassmann's Identity/Proof 1

== Theorem ==
== Proof ==
=== Outline ===
Starting from a basis of $M \cap N$, we complete it to a basis of $M$ and one of $N$.
We then verify that the union of these basis is a basis of $M + N$.
=== Proof ===
First, suppose $M \subseteq N$ or $N \subseteq M$.
Then the assertion is clear.
Assume that $M \cap N$ is a proper subspace of both $M$ and $N$.
Let $B$ be a basis of $M \cap N$.
By Dimension of Proper Subspace is Less Than its Superspace this is finite-dimensional.
By Results concerning Generators and Bases of Vector Spaces, there exist non-empty sets $C$ and $D$ disjoint from $B$ such that:
:$B \cup C$ is a basis of $M$
:$B \cup D$ is a basis of $N$.
The space generated by $B \cup C \cup D$ contains both $M$ and $N$.
Hence it contains $M + N$.
But as $B \cup C \cup D \subseteq M \cup N$, the space it generates is contained in $M + N$.
Therefore $B \cup C \cup D$ is a generator for $M + N$.
If $d$ is a linear combination of $D$ and also of $B \cup C$, then $d \in M \cap N$.
So $d$ is a linear combination of $B$, and consequently $d = 0$ as $B \cup D$ is linearly independent and $D$ is disjoint from $B$.
In particular, $D$ is disjoint from $B \cup C$.
Next we show that $B \cup C \cup D$ is linearly independent and hence a basis of $M + N$.
Let $\sequence {b_m}$ and $\sequence {d_p}$ be sequences of distinct vectors such that $B \cup C = \set {b_1, \ldots, b_m}$ and $D = \set {d_1, \ldots, d_p}$.
Let $\ds \sum_{j \mathop = 1}^m \lambda_j b_j + \sum_{k \mathop = 1}^p \mu_k d_k = 0$.
Then:
:$\ds \sum_{k \mathop = 1}^p \mu_k d_k = - \sum_{j \mathop = 1}^m \lambda_j b_j$
Hence $\ds \sum_{k \mathop = 1}^p \mu_k d_k$ is a linear combination of $D$ and also of $B \cup C$.
By the preceding:
:$\ds \sum_{k \mathop = 1}^p \mu_k d_k = 0$
Hence $\mu_k = 0$ for all $k \in \closedint 1 p$.
Thus:
:$\ds \sum_{j \mathop = 1}^m \lambda_j b_j = 0$
and therefore $\lambda_j = 0$ for all $j \in \closedint 1 m$.
Therefore $B \cup C \cup D$ is linearly independent.
Thus we have:
Hence the result.
== Sources ==
* : Chapter $\text {V}$: Vector Spaces: $\S 27$. Subspaces and Bases: Theorem $27.15$
Category:Grassmann's Identity
Gravity at Earth's Surface

== Physical Law ==
The acceleration due to gravity at the surface of Earth is approximately given by:
:$g = 9.8 \ \mathrm N \ \mathrm {kg}^{-1}$
Note that this is equivalent to 
:$g = 9.8 \ \mathrm M \ \mathrm s^{-2}$
as an acceleration can be defined as the force per unit mass from Newton's Second Law of Motion.
== Proof ==
From Acceleration Due to Gravity, we have that:
:$g = \dfrac {G M} {r^2}$
where:
:$g$ is the acceleration on the body caused by the gravitational field of Earth
:$M$ is the mass of Earth, approximately $5.9736 \times 10^{24}\,\mathrm {kg}$
:$G$ is the universal gravitational constant, approximately $6.674 \times 10^{-11} \, \mathrm N \, \mathrm m^2 \, \mathrm {kg}^{-2}$
:$r$ is the radius of Earth, approximately $6.371 \times 10^{6}\,\mathrm m$.
Hence the result, by direct numerical calculation.
Note that the value of $r$ is significantly variable, depending on where you are on Earth.
For fuller detail, see at .
== Sources ==
* : $\text {I}$: Motion, Sound and Heat: Chapter $2$: Falling Bodies: Free Fall
Category:Gravity
Grazing Cows

== Problem ==
Let:
:$a$ cows graze $b$ fields in $c$ days
:$a'$ cows graze $b'$ fields in $c'$ days
:$a' '$ cows graze $b' '$ fields in $c' '$ days.
What is the relationship between the $9$ magnitudes $a$ to $c' '$?
== Solution ==
Suppose that:
:each field initially contains the same quantity of grass: $M$
:the daily growth in each field is the same: $m$
:each cow consumes the same amount of grass per day: $Q$.
Then:
Consider the matrix:
:$A = \begin {pmatrix} b & b c & c a \\ b' & b' c' & c' a' \\ b' ' & b' ' c' ' & c' ' a' ' \end {pmatrix}$
We have:
:$A \tuple {M, m, -Q}^T = \mathbf 0$
Given that each of $M$, $m$ and $Q$ are not zero, by Matrix is Singular iff Product with Non-Zero Vector is Zero, $A$ is singular.
It follows that $\det A = 0$.
After algebra, we have:
:$b' ' c c' \paren {a b' - b a'} + c' ' b' ' \paren {b c' a' - b' c a} + c' ' a' ' b b' \paren {c - c'} = 0$
== Historical Note ==
== Sources ==
* 
* 
* : Prince Rupert's Cube: $128$
Category:Classic Problems
Greater Angle of Triangle Subtended by Greater Side

== Theorem ==
== Proof ==
:
Let $\triangle ABC$ be a triangle such that $\angle ABC$ is greater than $\angle BCA$.
Suppose $AC$ is not greater than $AB$.
If $AC$ were equal to $AB$, then by Isosceles Triangle has Two Equal Angles, $\angle ABC = \angle BCA$, but they're not so it isn't.
If $AC$ were less than $AB$, then by Greater Side of Triangle Subtends Greater Angle it would follow that $\angle ABC$ is less than $\angle BCA$, but it's not so it isn't.
So $AC$ must be greater than $AB$
Hence the result.
== Sources ==
* : Book $\text{I}$. Propositions
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.17$
Category:Triangles
Greater Side of Triangle Subtends Greater Angle

== Theorem ==
== Proof ==
:
Let $\triangle ABC$ be a triangle such that $AC$ is greater than $AB$.
Let $AD$ be made equal to $AB$.
Let $BD$ be joined.
Then $\angle ADB$ is an exterior angle of the triangle $\triangle BCD$.
Therefore $\angle ADB$ is greater than $\angle ACB$.
As $AD = AB$, the triangle $\triangle ABD$ is isosceles.
From Isosceles Triangle has Two Equal Angles, $\angle ADB = \angle ABD$.
Therefore $\angle ABD$ is greater than $\angle ACB$.
Therefore, as $\angle ABC = \angle ABD + \angle DBC$, it follows that $\angle ABC$ is greater than $\angle ACB$.
Hence the result.
== Sources ==
* : Book $\text{I}$. Propositions
* : Chapter $1$: Triangles and Polygons: Theorems and Corollaries $1.16$
Category:Triangles
Greatest Area of Quadrilateral with Sides in Arithmetic Sequence

== Theorem ==
Let $Q$ be a quadrilateral whose sides $a$, $b$, $c$ and $d$ are in arithmetic sequence.
Let $\AA$ be the area of $Q$.
Let $Q$ be such that $\AA$ is the greatest area possible for one with sides $a$, $b$, $c$ and $d$.
Then:
:$\AA = \sqrt {a b c d}$
== Proof ==
We are given that $\AA$ is the greatest possible for a quadrilateral whose sides are $a$, $b$, $c$ and $d$.
From Area of Quadrilateral with Given Sides is Greatest when Quadrilateral is Cyclic, $Q$ is cyclic.
Hence $\AA$ can be found using Brahmagupta's Formula.
Let $s$ denote the semiperimeter of $Q$:
:$s = \dfrac {a + b + c + d} 2$
We are given that $a$, $b$, $c$ and $d$ are in arithmetic sequence.
, that means there exists $k$ such that:
where $k$ is the common difference.
Then:
and so:
Category:Cyclic Quadrilaterals
Greatest Common Divisor divides Lowest Common Multiple

== Theorem ==
Let $a, b \in \Z$ such that $a b \ne 0$.
Then:
:$\gcd \set {a, b} \divides \lcm \set {a, b}$
where:
:$\lcm$ denotes lowest common multiple
:$\gcd$ denotes greatest common divisor.
:$\divides$ denotes divisibility.
== Proof ==
We have that:
:$\gcd \set {a, b} \divides a$
and:
:$a \divides \lcm \set {a, b}$
The result follows from Divisor Relation is Transitive.
== Sources ==
* : $\text {2-2}$ Divisibility: Exercise $9$
Category:Greatest Common Divisor
Category:Lowest Common Multiple
Greatest Common Divisor is Associative

== Theorem ==
Let $a, b, c \in \Z$.
Then:
:$\gcd \set {a, \gcd \set {b, c} } = \gcd \set {\gcd \set {a, b}, c}$
where $\gcd$ denotes the greatest common divisor.
== Proof ==
Follows directly from GCD from Prime Decomposition and Min Operation is Associative.
== Also see ==
* Lowest Common Multiple is Associative
Category:Greatest Common Divisor
Category:Examples of Associative Operations‎
Greatest Common Divisor is at least 1

== Theorem ==
Let $a, b \in \Z$ be integers.
The greatest common divisor of $a$ and $b$ is at least $1$:
:$\forall a, b \in \Z_{\ne 0}: \gcd \set {a, b} \ge 1$
== Proof ==
From One Divides all Integers:
:$\forall a, b \in \Z: 1 \divides a \land 1 \divides b$
and so:
:$1 \le \gcd \set {a, b}$
as required.
Category:Greatest Common Divisor
Greatest Common Divisor of Integers/Examples

== Examples of Greatest Common Divisors of Integers ==
=== $12$ and $-8$ ===
=== $-12$ and $30$ ===
=== $-5$ and $5$ ===
=== $8$ and $17$ ===
=== $-8$ and $-36$ ===
=== $n$ and $0$ ===
=== $20$, $70$ and $80$ ===
Category:Examples of Greatest Common Divisors
Greatest Common Divisor of Integers/Examples/-12 and 30

== Example of Greatest Common Divisor of Integers ==
The greatest common divisor of $-12$ and $30$ is:
:$\gcd \set {-12, 30} = 6$
== Proof ==
The strictly positive divisors of $-12$ are:
:$\set {x \in \Z_{>0}: x \divides \paren {-12} } = \set {1, 2, 3, 4, 6, 12}$
The strictly positive divisors of $30$ are:
:$\set {x \in \Z_{>0}: x \divides 30} = \set {1, 2, 3, 5, 6, 10, 15, 30}$
Of these, the common divisors are:
:$\set {x \in \Z_{>0}: x \divides \paren {-12} \land x \divides 30} = \set {1, 2, 3, 6}$
The greatest of these is $6$.
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Example $2 \text{-} 1$
Category:Examples of Greatest Common Divisors
Greatest Common Divisor of Integers/Examples/-5 and 5

== Example of Greatest Common Divisor of Integers ==
The greatest common divisor of $-5$ and $5$ is:
:$\gcd \set {-5, 5} = 5$
== Proof 1 ==
== Proof 2 ==
Category:Examples of Greatest Common Divisors
Category:Greatest Common Divisor of Integers/Examples/-5 and 5
Greatest Common Divisor of Integers/Examples/-5 and 5/Proof 1

== Example of Greatest Common Divisor of Integers ==
== Proof ==
The strictly positive divisors of $-5$ are:
:$\set {x \in \Z_{>0}: x \divides \paren {-5} } = \set {1, 5}$
The strictly positive divisors of $5$ are:
:$\set {x \in \Z_{>0}: x \divides 5 } = \set {1, 5}$
Of these, the common divisors are:
:$\set {x \in \Z_{>0}: x \divides \paren {-5} \land x \divides 5 } = \set {1, 5}$
The greatest of these is $5$.
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Example $2 \text{-} 1$
Category:Greatest Common Divisor of Integers/Examples/-5 and 5
Greatest Common Divisor of Integers/Examples/-8 and -36

== Example of Greatest Common Divisor of Integers ==
The greatest common divisor of $-8$ and $-36$ is:
:$\gcd \set {-8, -36} = 4$
== Proof ==
The strictly positive divisors of $-8$ are:
:$\set {x \in \Z_{>0}: x \divides \paren {-8} } = \set {1, 2, 4, 8}$
The strictly positive divisors of $-36$ are:
:$\set {x \in \Z_{>0}: x \divides \paren {-36} } = \set {1, 2, 3, 4, 6, 9, 12, 18, 36}$
Of these, the common divisors are:
:$\set {x \in \Z_{>0}: x \divides \paren {-8} \land x \divides \paren {-36} } = \set {1, 2, 4}$
The greatest of these is $4$.
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Example $2 \text{-} 1$
Category:Examples of Greatest Common Divisors
Greatest Common Divisor of Integers/Examples/12 and -8

== Example of Greatest Common Divisor of Integers ==
The greatest common divisor of $12$ and $-8$ is:
:$\gcd \set {12, -8} = 4$
== Proof ==
The strictly positive divisors of $12$ are:
:$\set {x \in \Z_{>0}: x \divides 12} = \set {1, 2, 3, 4, 6, 12}$
The strictly positive divisors of $-8$ are:
:$\set {x \in \Z_{>0}: x \divides \paren {-8} } = \set {1, 2, 4, 8}$
Of these, the common divisors are:
:$\set {x \in \Z_{>0}: x \divides 12 \land x \divides \paren {-8} } = \set {1, 2, 4}$
The greatest of these is $4$.
== Sources ==
* : $\text {2-2}$ Divisibility: Example $\text {2-5}$
Category:Examples of Greatest Common Divisors
Greatest Common Divisor of Integers/Examples/20, 70 and 80

== Example of Greatest Common Divisor of Integers ==
The greatest common divisor of $20$, $70$ and $80$ is:
:$\gcd \set {20, 70, 80} = 10$
== Proof ==
The strictly positive divisors of $20$ are:
:$\set {x \in \Z_{>0}: x \divides 20} = \set {1, 2, 4, 5, 10, 20}$
The strictly positive divisors of $70$ are:
:$\set {x \in \Z_{>0}: x \divides 70} = \set {1, 2, 5, 7, 10, 14, 35, 70}$
The strictly positive divisors of $80$ are:
:$\set {x \in \Z_{>0}: x \divides 80} = \set {1, 2, 4, 5, 8, 10, 16, 20, 40, 80}$
Of these, the common divisors are:
:$\set {x \in \Z_{>0}: x \divides 20 \land x \divides 70 \land x \divides 80} = \set {1, 2, 5, 10}$
The greatest of these is $10$.
== Sources ==
* 
* 
Category:Examples of Greatest Common Divisors
Greatest Common Divisor of Integers/Examples/8 and 17

== Example of Greatest Common Divisor of Integers ==
The greatest common divisor of $8$ and $17$ is:
:$\gcd \set {8, 17} = 1$
That is, $8$ and $17$ are coprime.
== Proof ==
The strictly positive divisors of $8$ are:
:$\set {x \in \Z_{>0}: x \divides 8} = \set {1, 2, 4, 8}$
The strictly positive divisors of $17$ are:
:$\set {x \in \Z_{>0}: x \divides 17} = \set {1, 17}$
It is seen that there is only one strictly positive common divisor of $8$ and $17$, and that is $1$.
Hence by definition $8$ and $17$ are coprime.
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.2$ The Greatest Common Divisor: Example $2 \text{-} 1$
Category:Examples of Greatest Common Divisors
Greatest Common Divisor of Integers/Examples/n and 0

== Example of Greatest Common Divisor of Integers ==
Let $n \in \Z_{>0}$.
The greatest common divisor of $n$ and $0$ is:
:$\gcd \set {n, 0} = n$
== Proof ==
The strictly positive divisors of $0$ are:
:$\set {x \in \Z_{>0}: x \divides 0} = \Z_{>0}$
Category:Examples of Greatest Common Divisors
Greatest Common Divisor of Set of Integers/Examples/39, 42, 54

== Example of Greatest Common Divisor of Set of Integers ==
Let $S = \set {39, 42, 54}$.
The greatest common divisor of $S$ is:
:$\map \gcd S = 3$
== Proof ==
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.3$ The Euclidean Algorithm
Category:Examples of Greatest Common Divisors
Greatest Common Divisor of Set of Integers/Examples/49, 210, 350

== Example of Greatest Common Divisor of Set of Integers ==
Let $S = \set {49, 210, 350}$.
The greatest common divisor of $S$ is:
:$\map \gcd S = 7$
== Proof ==
== Sources ==
* : Chapter $2$: Divisibility Theory in the Integers: $2.3$ The Euclidean Algorithm
Category:Examples of Greatest Common Divisors
Greatest Common Divisor of Three Numbers

== Theorem ==
== Proof ==
Let $A, B, C$ be the three given (natural) numbers not prime to one another.
:
Let the GCD $D$ of $A, B$ be found, by the Euclidean Algorithm.
Then $D$ either divides or does not divide $C$.
First, suppose $D$ divides $C$.
Then as it also divides $A$ and $B$, it is a common divisor of $A, B$ and $C$.
Suppose $D$ is not the GCD of $A, B, C$.
Then some number $E$ is a common divisor of $A, B$ and $C$ such that $E > D$.
But then $E$ is a common divisor of $A$ and $B$ and so from the porism to Euclid's Algorithm $E$ also divides $D$.
That means $E \le D$ which contradicts $E > D$.
So $D$ is the GCD of $A, B, C$ after all.
Next, suppose $D$ is not a divisor of $C$.
Since $A, B, C$ are not prime to one another, some number will divide them.
That number which divides $A, B, C$ will also divide $A, B$ and so from the porism to Euclid's Algorithm will also divide $D$.
But it also divides $C$.
So some number divides both $D$ and $C$, and so $C$ and $D$ are not prime to one another.
Let the GCD of $C$ and $D$ be $E$, by the Euclidean Algorithm.
Since $E$ divides $D$ and $D$ divides $A, B$, then $E$ divides $A, B$.
But $E$ divides $C$, and so $E$ divides $A, B, C$.
So $E$ is a common divisor of $A, B$ and $C$.
Now suppose $E$ is not the GCD of $A, B, C$.
Then some number $F$ such that $F > E$ divides $A, B, C$.
It follows that $F$ divides $A$ and $B$, and so from the porism to Euclid's Algorithm will also divide $D$, the GCD of $A, B$.
So $F$ divides $D$ and $F$ also divides $C$.
So $F$ is a common divisor of $D$ and $C$.
So by the porism to Euclid's Algorithm, $F$ divides the GCD of $C, D$, which is $E$.
But this contradicts the supposition that $F > E$.
So no such $F$ can exist, and so $E$ is the GCD of $A, B, C$.
== Sources ==
* : Book $\text{VII}$. Propositions
Category:Number Theory
Greatest Common Divisors in Principal Ideal Domain are Associates

== Theorem ==
Let $\struct {D, +, \circ}$ be a principal ideal domain.
Let $S = \set {a_1, a_2, \dotsc, a_n}$ be a set of non-zero elements of $D$.
Let $y_1$ and $y_2$ be greatest common divisors of $S$.
Then $y_1$ and $y_2$ are associates.
== Proof ==
From Finite Set of Elements in Principal Ideal Domain has GCD we have that at least one such greatest common divisor exists.
So, let $y_1$ and $y_2$ be greatest common divisors of $S$.
Then:
Thus we have:
:$y_1 \divides y_2$ and $y_2 \divides y_1$
where $\divides$ denotes divisibility.
Hence the result, by definition of associates.
== Sources ==
* : $\S 62.4$ Factorization in an integral domain: $\text{(ii)}$
Category:Principal Ideal Domains
Category:Greatest Common Divisor
Category:Associates