nyuuzyou/wikis huggingface dataset:
Construction of Inverse Completion

== Theorem ==
This page consists of a series of linked theorems, each of which builds towards one result.
To access the proofs for the individual theorems, click on the links which form the titles of each major section.
== Initial Definitions ==
Let $\struct {S, \circ}$ be a commutative semigroup which has cancellable elements.
Let $\struct {C, \circ {\restriction_C} } \subseteq \struct {S, \circ}$ be the subsemigroup of cancellable elements of $\struct {S, \circ}$, where $\circ {\restriction_C}$ denotes the restriction of $\circ$ to $C$.
== Cartesian Product with Cancellable Elements ==
== Congruence Relation ==
=== Natural Number Difference ===
== Quotient Structure ==
== Properties of Quotient Structure ==
== Notes ==
Elements of $T'$ are equivalence classes of ordered pairs $\tuple {x, y}$ where $x \in S, y \in C$.
Each of the elements of $\eqclass {\tuple {x, y} } \boxtimes$ is such that $x \circ y^{-1}$ have the same value, where $y^{-1} \in C^{-1}$.
Hence is it a natural progression to define an operation $\odot$, say, such that:
:$x \odot y \equiv x \circ y^{-1}$
In the context of the integers, this operation is minus, hence:
:$x - y \equiv x + \paren {-y}$
In the context of the rational numbers, this operation is division, hence:
:$\dfrac x y \equiv x \times y^{-1}$
Each element of $S$, and hence in $C$, is identified (via the isomorphism $\psi$) with one of these equivalence classes.
If $S$ is a monoid, then it has an identity $e_S$, say, which is in $C$.
Hence:
:$\forall x \in C: \map \psi x = \eqclass {\tuple {x, e_S} } \boxtimes$
In particular:
:$\map \psi {e_S} = \eqclass {\tuple {e_S, e_S} } \boxtimes$
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers
Category:Construction of Inverse Completion
Category:Semigroups
Category:Cartesian Product
Category:Inverse Completions
Category:Quotient Mappings
Category:Equivalence Relations
Construction of Inverse Completion/Cartesian Product with Cancellable Elements

== Theorem ==
Let $\struct {S, \circ}$ be a commutative semigroup which has cancellable elements.
Let $\struct {C, \circ {\restriction_C} } \subseteq \struct {S, \circ}$ be the subsemigroup of cancellable elements of $\struct {S, \circ}$, where $\circ {\restriction_C}$ denotes the restriction of $\circ$ to $C$.
Let $\struct {S \times C, \oplus}$ be the external direct product of $\struct {S, \circ}$ and $\struct {C, \circ {\restriction_C} }$, where $\oplus$ is the operation on $S \times C$ induced by $\circ$ on $S$ and $\circ {\restriction_C}$ on $C$.
That is:
:$\forall \tuple {x, y}, \tuple {u, v} \in S \times C: \tuple {x, y} \oplus \tuple {u, v} = \tuple {x \circ u, y \mathop {\circ {\restriction_C} } v}$
Then $\struct {S \times C, \oplus}$ is a commutative semigroup.
== Proof ==
By Cancellable Elements of Semigroup form Subsemigroup, $\struct {C, \circ {\restriction_C} }$ is a subsemigroup of $\struct {S, \circ}$, where $\circ {\restriction_C}$ is the restriction of $\circ$ to $C$.
By Restriction of Commutative Operation is Commutative, as $\struct {C, \circ {\restriction_C} }$ is a substructure of a commutative structure, it is also commutative.
From:
:the external direct product preserves the nature of semigroups
:the external direct product preserves commutativity
we see that $\struct ({S \times C, \oplus}$ is a commutative semigroup.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers
Category:Construction of Inverse Completion
Construction of Inverse Completion/Congruence Relation

== Theorem ==
The cross-relation $\boxtimes$ is a congruence relation on $\struct {S \times C, \oplus}$.
=== Members of Equivalence Classes ===
=== Equivalence Class of Equal Elements ===
== Proof ==
From Semigroup is Subsemigroup of Itself, $\struct {S, \circ}$ is a subsemigroup of $\struct {S, \circ}$.
Also from Semigroup is Subsemigroup of Itself, $\struct {C, \circ {\restriction_C} }$ is a subsemigroup of $\struct {C, \circ {\restriction_C} }$.
The result follows from Cross-Relation is Congruence Relation.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers
Category:Congruence Relations
Category:Construction of Inverse Completion
Construction of Inverse Completion/Equivalence Relation

== Theorem ==
The cross-relation $\boxtimes$ is an equivalence relation on $\struct {S \times C, \oplus}$.
== Proof ==
From Semigroup is Subsemigroup of Itself, $\struct {S, \circ}$ is a subsemigroup of $\struct {S, \circ}$.
Also from Semigroup is Subsemigroup of Itself, $\struct {C, \circ {\restriction_C} }$ is a subsemigroup of $\struct {C, \circ {\restriction_C} }$.
The result follows from Cross-Relation is Equivalence Relation.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers
Category:Construction of Inverse Completion
Construction of Inverse Completion/Equivalence Relation/Equivalence Class of Equal Elements

== Theorem ==
Then:
:$\forall c, d \in C: \tuple {c, c} \boxtimes \tuple {d, d}$
== Proof ==
From Semigroup is Subsemigroup of Itself, $\struct {S, \circ}$ is a subsemigroup of $\struct {S, \circ}$.
Also from Semigroup is Subsemigroup of Itself, $\struct {C, \circ {\restriction_C} }$ is a subsemigroup of $\struct {C, \circ {\restriction_C} }$.
The result follows from Equivalence Class of Equal Elements of Cross-Relation.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers
Category:Construction of Inverse Completion
Construction of Inverse Completion/Equivalence Relation/Members of Equivalence Classes

== Theorem ==
Then:
$\forall x, y \in S, a, b \in C:$
:$(1): \quad \tuple {x \circ a, a} \boxtimes \tuple {y \circ b, b} \iff x = y$
:$(2): \quad \eqclass {\tuple {x \circ a, y \circ a} } \boxtimes = \eqclass {\tuple {x, y} } \boxtimes$
where $\eqclass {\tuple {x, y} } \boxtimes$ is the equivalence class of $\tuple {x, y}$ under $\boxtimes$.
== Proof ==
From Cross-Relation is Equivalence Relation we have that $\boxtimes$ is an equivalence relation.
Hence the equivalence class of $\tuple {x, y}$ under $\boxtimes$ is defined for all $\tuple {x, y} \in S \times C$.
From Semigroup is Subsemigroup of Itself, $\struct {S, \circ}$ is a subsemigroup of $\struct {S, \circ}$.
Also from Semigroup is Subsemigroup of Itself, $\struct {C, \circ {\restriction_C} }$ is a subsemigroup of $\struct {C, \circ {\restriction_C} }$.
The result follows from Elements of Cross-Relation Equivalence Class.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers
Category:Construction of Inverse Completion
Construction of Inverse Completion/Generator for Quotient Structure

== Theorem ==
Let $\struct {S, \circ}$ be a commutative semigroup which has cancellable elements.
Let $C \subseteq S$ be the set of cancellable elements of $S$.
Let $\struct {S \times C, \oplus}$ be the external direct product of $\struct {S, \circ}$ and $\struct {C, \circ {\restriction_C} }$, where:
:$\circ {\restriction_C}$ is the restriction of $\circ$ to $C \times C$
and:
:$\oplus$ is the operation on $S \times C$ induced by $\circ$ on $S$ and $\circ \restriction_C$ on $C$.
Let $\boxtimes$ be the congruence relation $\boxtimes$ defined on $\struct {S \times C, \oplus}$ by:
:$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$
Let the quotient structure defined by $\boxtimes$ be:
:$\struct {T', \oplus'} := \struct {\dfrac {S \times C} \boxtimes, \oplus_\boxtimes}$
where $\oplus_\boxtimes$ is the operation induced on $\dfrac {S \times C} \boxtimes$ by $\oplus$.
$T' = S' \cup \paren {C'}^{-1}$ is a generator for the semigroup $T'$.
== Proof ==
Let $\tuple {x, y} \in S \times C$. Then:
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers
Category:Construction of Inverse Completion
Construction of Inverse Completion/Identity of Quotient Structure

== Theorem ==
Let $c \in C$ be arbitrary.
Then:
:$\eqclass {\tuple {c, c} } \boxtimes$
is the identity of $T'$.
== Proof ==
Hence the result, by definition of identity element.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers
Category:Construction of Inverse Completion
Construction of Inverse Completion/Image of Quotient Mapping is Subsemigroup

== Theorem ==
Let the mapping $\psi: S \to T'$ be defined as:
:$\forall x \in S: \map \psi x = \eqclass {\tuple {x \circ a, a} } \boxtimes$
Let $S'$ be the image $\psi \sqbrk S$ of $S$.
Then $\struct {S', \oplus'}$ is a subsemigroup of $\struct {T', \oplus'}$.
== Proof ==
We have that $S'$ is the image $\psi \sqbrk S$ of $S$.
For $\struct {S', \oplus'}$ to be a subsemigroup of $\struct {T', \oplus'}$, by Subsemigroup Closure Test we need to show that $\struct {S', \oplus'}$ is closed.
Let $x, y \in S'$.
Then $x = \map \phi {x'}, y = \map \phi {y'}$ for some $x', y' \in S$.
But as $\phi$ is an isomorphism, it obeys the morphism property.
So $x \oplus' y = \map \phi {x'} \oplus' \map \phi {y'} = \map \phi {x' \circ y'}$.
Hence $x \oplus' y$ is the image of $x' \circ y' \in S$ and hence $x \oplus' y \in S'$.
Thus by the Subsemigroup Closure Test, $\struct {S', \oplus'}$ is a subsemigroup of $\struct {T', \oplus'}$
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers
Category:Construction of Inverse Completion
Construction of Inverse Completion/Invertible Elements in Quotient Structure

== Theorem ==
Every cancellable element of $S'$ is invertible in $T'$.
== Proof ==
From Identity of Quotient Structure, $\struct {T', \oplus'}$ has an identity, and it is $\eqclass {\tuple {c, c} } \boxtimes$ for any $c \in C$.
Call this identity $e_{T'}$.
Let the mapping $\psi: S \to T'$ be defined as:
:$\forall x \in S: \map \psi x = \eqclass {\tuple {x \circ a, a} } \boxtimes$
From Image of Cancellable Elements in Quotient Mapping:
:$C' = \psi \sqbrk C$
So:
The inverse of $x'$ is $\eqclass {\tuple {a, a \circ x} } \boxtimes$, as follows:
thus showing that the inverse of $\eqclass {\tuple {x \circ a, a} } \boxtimes$ is $\eqclass {\tuple {a, a \circ x} } \boxtimes$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers
Category:Construction of Inverse Completion
Construction of Inverse Completion/Natural Number Difference

== Natural Number Difference in Inverse Completion ==
In the context of the natural numbers, the difference is defined as:
:$n - m = p \iff m + p = n$
from which it can be seen that the above congruence can be understood as:
:$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 + y_2 = x_2 + y_1 \iff x_1 - y_1 = x_2 - y_2$
Thus this congruence defines an equivalence between pairs of elements which have the same difference.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers
Category:Inverse Completions
Construction of Inverse Completion/Properties of Quotient Structure

== Theorem ==
=== Identity of Quotient Structure ===
=== Invertible Elements in Quotient Structure ===
=== Generator for Quotient Structure ===
=== Quotient Structure is Inverse Completion ===
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers
Category:Construction of Inverse Completion
Construction of Inverse Completion/Quotient Mapping/Image of Cancellable Elements

== Theorem ==
Let the mapping $\psi: S \to T'$ be defined as:
:$\forall x \in S: \map \psi x = \eqclass {\tuple {x \circ a, a} } \boxtimes$
Let $S'$ be the image $\psi \sqbrk S$ of $S$.
The set $C'$ of cancellable elements of the semigroup $S'$ is $\psi \sqbrk C$.
== Proof ==
We have Morphism Property Preserves Cancellability.
Thus:
:$c \in C \implies \map \psi c \in C'$
So by Image of Subset under Mapping is Subset of Image:
:$\psi \sqbrk C \subseteq C'$
From above, $\psi$ is an isomorphism.
Hence, also from Morphism Property Preserves Cancellability:
:$c' \in C' \implies \map {\psi^{-1} } {c'} \in C$
So by Preimage of Subset is Subset of Preimage:
:$\psi^{-1} \sqbrk {C'} \subseteq C$
Hence by definition of set equality:
:$\psi \sqbrk C = C'$
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers
Category:Construction of Inverse Completion
Construction of Inverse Completion/Quotient Mapping is Injective

== Theorem ==
Let the mapping $\psi: S \to T'$ be defined as:
:$\forall x \in S: \map \psi x = \eqclass {\tuple {x \circ a, a} } \boxtimes$
Then $\psi: S \to T'$ is an injection, and does not depend on the particular element $a$ chosen.
== Proof ==
The result follows by the definition of injection.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers
Category:Examples of Injections
Category:Construction of Inverse Completion
Construction of Inverse Completion/Quotient Mapping is Monomorphism

== Theorem ==
Let the mapping $\psi: S \to T'$ be defined as:
:$\forall x \in S: \map \psi x = \eqclass {\tuple {x \circ a, a} } \boxtimes$
The mapping $\psi: S \to T'$ is a monomorphism.
== Proof ==
We have that this quotient mapping $\psi: S \to T'$ is an injection.
Let $x, y \in S$. Then:
So $\map \psi {x \circ y} = \map \psi x \oplus' \map \psi y$, and the morphism property is proven.
Thus $\psi$ is an injective homomorphism, and so by definition a monomorphism.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers
Category:Construction of Inverse Completion
Category:Monomorphisms (Abstract Algebra)
Construction of Inverse Completion/Quotient Mapping to Image is Isomorphism

== Theorem ==
Let the mapping $\psi: S \to T'$ be defined as:
:$\forall x \in S: \map \psi x = \eqclass {\tuple {x \circ a, a} } \boxtimes$
Let $S'$ be the image $\psi \sqbrk S$ of $S$.
Then $\psi$ is an isomorphism from $S$ onto $S'$.
== Proof ==
From Quotient Mapping is Monomorphism, $\psi: \struct {S, \circ} \to \struct {S', \oplus'}$ is a monomorphism.
Therefore by definition:
:$\psi$ is a homomorphism
:$\psi$ is an injection.
Because $S'$ is the image of $\psi \sqbrk S$, by Surjection by Restriction of Codomain $\psi$ is a surjection.
Therefore by definition $\psi: S \to S'$ is a bijection.
A bijective homomorphism is an isomorphism.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers
Category:Construction of Inverse Completion
Category:Isomorphisms (Abstract Algebra)
Construction of Inverse Completion/Quotient Structure

== Theorem ==
This cross-relation is a congruence relation on $S \times C$.
Let the quotient structure defined by $\boxtimes$ be:
:$\struct {T', \oplus'} := \struct {\dfrac {S \times C} \boxtimes, \oplus_\boxtimes}$
where $\oplus_\boxtimes$ is the operation induced on $\dfrac {S \times C} \boxtimes$ by $\oplus$.
=== Quotient Structure is Commutative Semigroup ===
=== Quotient Mapping is Injective ===
=== Quotient Mapping is Monomorphism ===
=== Image of Quotient Mapping is Subsemigroup ===
=== Quotient Mapping to Image is Isomorphism ===
=== Image of Cancellable Elements in Quotient Mapping ===
== Proof ==
From the defined equivalence relation, we have that:
:$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$
is a congruence relation on $\struct {S \times C, \oplus}$.
From the definition of the members of the equivalence classes:
:$(1) \quad \forall x, y \in S, a, b \in C: \tuple {x \circ a, a} \boxtimes \tuple {y \circ b, b} \iff x = y$
:$(2) \quad \forall x, y \in S, a, b \in C: \eqclass {\tuple {x \circ a, y \circ a} } \boxtimes = \eqclass {\tuple {x, y} } \boxtimes$
From the definition of the equivalence class of equal elements:
:$(3) \quad \forall c, d \in C: \tuple {c, c} \boxtimes \tuple {d, d}$
where $\eqclass {\tuple {x, y} } \boxtimes$ is the equivalence class of $\tuple {x, y}$ under $\boxtimes$.
Hence we are justified in asserting the existence of the quotient structure:
:$\struct {T', \oplus'} = \struct {\dfrac {S \times C} \boxtimes, \oplus_\boxtimes}$
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers
Category:Construction of Inverse Completion
Construction of Inverse Completion/Quotient Structure is Commutative Semigroup

== Theorem ==
Then:
:$\struct {T', \oplus'}$ is a commutative semigroup.
== Proof ==
The quotient epimorphism from $\struct {S \times C, \oplus}$ onto $\struct {T', \oplus'}$ is given by:
:$q_\boxtimes: \struct {S \times C, \oplus} \to \struct {T', \oplus'}: \map {q_\boxtimes} {x, y} = \eqclass {\tuple {x, y} } \boxtimes$
where, by definition:
By Morphism Property Preserves Closure, as $\oplus$ is closed, then so is $\oplus'$.
By Epimorphism Preserves Associativity, as $\oplus$ is associative, then so is $\oplus'$.
By Epimorphism Preserves Commutativity, as $\oplus$ is commutative, then so is $\oplus'$.
Thus $\struct {T', \oplus'}$ is closed, associative and commutative, and therefore a commutative semigroup.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers
Category:Construction of Inverse Completion
Construction of Inverse Completion/Quotient Structure is Inverse Completion

== Theorem ==
Let $\struct {S, \circ}$ be a commutative semigroup which has cancellable elements.
Let $C \subseteq S$ be the set of cancellable elements of $S$.
Let $\struct {S \times C, \oplus}$ be the external direct product of $\struct {S, \circ}$ and $\struct {C, \circ {\restriction_C} }$, where:
:$\circ {\restriction_C}$ is the restriction of $\circ$ to $C \times C$
and:
:$\oplus$ is the operation on $S \times C$ induced by $\circ$ on $S$ and $\circ {\restriction_C}$ on $C$.
Let $\boxtimes$ be the congruence relation $\boxtimes$ defined on $\struct {S \times C, \oplus}$ by:
:$\tuple {x_1, y_1} \boxtimes \tuple {x_2, y_2} \iff x_1 \circ y_2 = x_2 \circ y_1$
Let the quotient structure defined by $\boxtimes$ be:
:$\struct {T', \oplus'} := \struct {\dfrac {S \times C} \boxtimes, \oplus_\boxtimes}$
where $\oplus_\boxtimes$ is the operation induced on $\dfrac {S \times C} \boxtimes$ by $\oplus$.
$T'$ is an inverse completion of its subsemigroup $S'$.
== Proof ==
Every cancellable element of $S'$ is invertible in $T'$, from Invertible Elements in Quotient Structure.
$T' = S' \cup \paren {C'}^{-1}$ is a generator for the semigroup $T'$, from Generator for Quotient Structure.
Hence the result, by definition of inverse completion
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $\S 20$: The Integers
Category:Construction of Inverse Completion
Construction of Isosceles Triangle whose Base Angle is Twice Apex

== Theorem ==
It is possible to construct an isosceles triangle such that each of the angles at the base is twice that at the apex.
== Construction ==
:
Let $AB$ be a straight line.
Construct $C$ on $AB$ such that $AB \cdot BC = AC^2$.
Construct the circle whose center is at $A$ and whose radius is $AB$.
Fit $BD$ into this circle as a chord such that $BD = AC$.
Join $AD$.
Then $\triangle ABD$ is the required isosceles triangle, as $\angle ABD = \angle BAD = 2 \angle BDA$.
== Proof ==
Join $CD$.
Circumscribe circle $ACD$ about $\triangle ACD$.
As $AC = BD$ we have that $AB \cdot BC = BD^2$.
We have that $B$ is outside the circle $ACD$.
From the converse of the Tangent Secant Theorem it follows that $BD$ is tangent to circle $ACD$.
Then from the Tangent-Chord Theorem:
:$\angle BDC = \angle DAC$
Add $\angle CDA$ to both:
:$\angle CDA + \angle BDC = \angle BDA = \angle CDA + \angle DAC$.
But from Sum of Angles of Triangle Equals Two Right Angles we have that:
:$(1) \quad \angle BCD = \angle CDA + \angle DAC$
So $\angle BDA = \angle BCD$.
But since $AD = AB$, from Isosceles Triangle has Two Equal Angles $\angle BDA = \angle CBD$.
So $\angle BDA = \angle BCD = \angle CBD$.
Since $\angle DBC = \angle BCD$, from Triangle with Two Equal Angles is Isosceles we have $BD = DC$.
But by hypothesis $BD = CA$ and so $CA = CD$.
So from Isosceles Triangle has Two Equal Angles $\angle CDA = \angle DAC$.
So $\angle CDA + \angle DAC = 2 \angle DAC$.
But from $(1)$ we have that $\angle BCD = \angle CDA + \angle DAC$.
So $\angle BCD = 2 \angle CAD = 2 \angle BAD$.
But $\angle BCD = \angle BDA = \angle DBA$.
So $\angle ABD = \angle BAD = 2 \angle BDA$.
== Sources ==
* : Book $\text{IV}$. Propositions
Category:Circles
Category:Isosceles Triangles
Construction of Lattice Point in Cartesian Plane

== Theorem ==
Let $\CC$ be a Cartesian plane.
Let $P = \tuple {a, b}$ be a lattice point in $\CC$.
Then $P$ is constructible using a compass and straightedge construction.
== Proof ==
Let $O$ denote the point $\tuple {0, 0}$.
Let $A$ denote the point $\tuple {1, 0}$.
The $x$-axis is identified with the straight line through $O$ and $A$.
The $y$-axis is constructed as the line perpendicular to $OA$ through $O$.
From Construction of Integer Multiple of Line Segment, the point $\tuple {a, 0}$ is constructed.
Drawing a circle whose center is at $O$ and whose radius is $OA$ the point $A'$ is constructed on the $y$-axis where $OA' = OA$.
Thus $A'$ is the point $\tuple {0, 1}$.
From Construction of Integer Multiple of Line Segment, the point $\tuple {0, b}$ is constructed.
Using Construction of Parallel Line, a straight line is drawn through $\tuple {a, 0}$ parallel to the $y$-axis.
Using Construction of Parallel Line, a straight line is drawn through $\tuple {0, b}$ parallel to the $x$-axis.
By definition of Cartesian plane, their intersection is at $\tuple {a, b}$, which is the required point $P$.
== Sources ==
* : Chapter $8$: Field Extensions: $\S 40$. Construction with Ruler and Compasses
Category:Analytic Geometry
Category:Compass and Straightedge Constructions
Construction of Mean Proportional

== Theorem ==
Given any two straight lines of length $a$ and $b$ it is possible to find a straight line of length $c$ such that $a : c = c : b$.
== Construction ==
Let $AB$ and $BC$ be the two given straight lines.
We require to find a mean proportional to $AB, BC$.
Let $AB$ and $BC$ be placed in a straight line and let the semicircle $ADC$ be placed on $AC$. 
Let $BD$ be drawn perpendicular to $AC$.
Then $BD$ is the required mean proportional.
== Proof ==
:
From Relative Sizes of Angles in Segments, $\angle ADC$ is a right angle.
So from the porism to Perpendicular in Right-Angled Triangle makes two Similar Triangles, $DB$ is the mean proportional between $AB$ and $BC$.
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Lines
Construction of Medial Straight Lines Commensurable in Square Only containing Medial Rectangle whose Square Differences Commensurable with Greater

== Theorem ==
== Proof ==
Let $A$, $B$ and $C$ be rational straight lines which are commensurable in square only.
Let $A^2 > C^2$ such that:
:$A^2 = C^2 + \rho^2$
where $\rho$ is a straight line which is commensurable in length with $A$.
This can be done using Construction of Rational Straight Lines Commensurable in Square Only whose Square Differences Commensurable with Greater.
Let $D$ be a straight line such that $D^2 = A B$.
From Medial is Irrational, $A B$ is medial.
Therefore $D^2$ is medial.
So by definition $D$ is medial.
Let $D E = B C$.
Since:
:$A B : B C = A : C$
and:
:$D^2 = A B$
and:
:$D E = B C$
then:
:$A : C = D^2 : D E$
So as:
:$D^2 : D E = D : E$
it follows that:
:$A : C = D : E$
But $A$ is commensurable in square only with $C$.
Therefore by Commensurability of Elements of Proportional Magnitudes:
:$D$ is commensurable in square only with $E$.
We have that $D$ is medial.
Therefore by Straight Line Commensurable with Medial Straight Line is Medial, $E$ is medial.
We have that:
:$A : C = D : E$
and:
:$A^2 = C^2 + \rho^2$
where $\rho$ is a straight line which is commensurable in length with $A$.
Therefore Commensurability of Squares on Proportional Straight Lines:
:$D^2 = E^2 + \sigma^2$
where $\sigma$ is a straight line which is commensurable in length with $D$.
We have that:
: $B C = D E$
while from Medial is Irrational:
:$B C$ is medial.
Therefore $D E$ is also medial.
Therefore two medial straight lines $D$ and $E$ have been found which are commensurable in square only and which contain a medial rectangle.
Further, $D^2$ is greater than $E^2$ by the square on a straight line commensurable in length with $D$.
Similarly it can be proved that the square on $D$ exceeds the square on $E$ by the square on a straight line incommensurable in length with $D$, when the square on $A$ is greater than the square on $C$ by the square on a straight line incommensurable in length with $A$.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory