nyuuzyou/wikis huggingface dataset:
Double Orthocomplement of Closed Linear Subspace

== Theorem ==
Let $H$ be a Hilbert space.
Let $A \subseteq H$ be a closed linear subspace of $H$.
Then:
:$\paren {A^\perp}^\perp = A$
== Proof ==
Let $I : H \to H$ be the identity operator (viz., $I h = h$).
Also, let $P : H \to A$ be the orthogonal projection.
Then $I - P : H \to A^\perp$ is the Orthogonal Projection onto Orthocomplement.
By Kernel of Orthogonal Projection:
:$\map \ker {I - P} = \paren {A^\perp}^\perp$
From Orthogonal Projection is Projection:
:$h \in P \sqbrk H \implies h = P h$
where $P \sqbrk H$ denotes the image of $H$ under $P$.
Also:
:$0 = \paren {I - P} h \iff h = P h$ 
Therefore,:
:$\map \ker {I - P} = P \sqbrk H$
Finally, from Range of Orthogonal Projection:
:$P \sqbrk H = A$
To conclude:
:$\paren {A^\perp}^\perp = \map \ker {I - P} = P \sqbrk H = A$
== Sources ==
* : $\text{I}$ Hilbert Spaces: $\S 2.$ Orthogonality: Corollary $2.9$
Category:Hilbert Spaces
Double Pointed Countable Complement Topology fulfils no Separation Axioms

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Let $T \times D$ be the double pointed topology on $T$.
Then $T \times D$ is not a $T_0$ space, $T_1$ space, $T_2$ space, $T_3$ space, $T_4$ space or $T_5$ space.
== Proof ==
From Separation Axioms on Double Pointed Topology, we have that:
:$T \times D$ is not a $T_0$ space.
:$T \times D$ is not a $T_1$ space.
:$T \times D$ is not a $T_2$ space.
Also from Separation Axioms on Double Pointed Topology, we have that:
:$T \times D$ is a $T_3$ space $T$ is a $T_3$ space
:$T \times D$ is a $T_4$ space $T$ is a $T_4$ space
:$T \times D$ is a $T_5$ space $T$ is a $T_5$ space
But from Countable Complement Space is not $T_3$, $T_4$ or $T_5$:
:$T$ is not a $T_3$ space
:$T$ is not a $T_4$ space
:$T$ is not a $T_5$ space
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $21$. Double Pointed Countable Complement Topology: $6$
Category:Separation Axioms
Category:Double Pointed Topologies
Category:Countable Complement Topology
Double Pointed Countable Complement Topology is Weakly Countably Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a countable complement topology on an uncountable set $S$.
Let $T \times D$ be the double pointed topology on $T$.
Then $T \times D$ is weakly countably compact.
== Proof ==
By definition, $T$ is weakly countably compact every infinite subset of $S$ has a limit point in $S$.
Let $D = \set {0, 1}$.
Let $\tuple {p, 0}$ belong to some infinite $A \subseteq S$.
Then its twin $\tuple {p, 1}$ is a limit point of $A$.
Hence the result by definition of weakly countably compact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $21$. Double Pointed Countable Complement Topology: $7$
Category:Weakly Countably Compact Spaces
Category:Double Pointed Topologies
Category:Countable Complement Topology
Double Pointed Discrete Real Number Space is Weakly Countably Compact

== Theorem ==
Let $T_\R = \struct {\R, \tau_\R}$ be the (uncountable) discrete space on the set of real numbers.
Let $T_D = \struct {D, \tau_D}$ be the indiscrete topology on the doubleton $D = \set {a, b}$.
Let $T = T_\R \times T_D$ be the double pointed (uncountable) discrete space which is the product space of $T_\R$ and $T_D$.
Then $T$ is weakly countably compact. 
== Proof ==
We have that $T$ is a partition topology, whose basis $\PP$ is defined as:
:$\PP = \set {\set {\tuple {s, a}, \tuple {s, b} }: s \in \R}$
Let $A \subseteq \R \times D$ such that $A$ is infinite.
Let $x \in A$.
Let $U$ be the union of sets of the form $\set {\tuple {s, a}, \tuple {s, b} }$, and hence open in $T$.
Now if $x \in U$, it will be an element in some $\set {\tuple {s, a}, \tuple {s, b} }$.
So there will exist $y \in U$ which will also be an element in that $\set {\tuple {s, a}, \tuple {s, b} }$.
So, by definition, $x$ is a limit point of $A$.
So, by definition, $T$ is weakly countably compact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $5$. Partition Topology: $7$
Category:Weakly Countably Compact Spaces
Category:Partition Topology
Category:Double Pointed Topologies
Category:Discrete Topology
Double Pointed Discrete Real Number Space is not Lindelöf

== Theorem ==
Let $T_\R = \struct {\R, \tau_\R}$ be the (uncountable) discrete space on the set of real numbers.
Let $T_D = \struct {D, \tau_D}$ be the indiscrete topology on the doubleton $D = \set {a, b}$.
Let $T = T_\R \times T_D$ be the double pointed (uncountable) discrete space which is the product space of $T_\R$ and $T_D$.
Then $T$ is not a Lindelöf space. 
== Proof ==
We have that $T$ is a partition topology, whose basis $\PP$ is defined as:
:$\PP = \set {\set {\paren {s, a}, \paren {s, b} }: s \in \R}$
We have that $\PP$ is an open cover of $T$.
But $\PP$ has no countable subcover.
Hence the result, by definition of Lindelöf space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $5$. Partition Topology: $7$
Category:Lindelöf Spaces
Category:Partition Topology
Category:Double Pointed Topologies
Category:Discrete Topology
Double Pointed Finite Complement Topology fulfils no Separation Axioms

== Theorem ==
Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Let $T \times D$ be the double pointed topology on $T$.
Then $T \times D$ is not a $T_0$ space, $T_1$ space, $T_2$ space, $T_3$ space, $T_4$ space or $T_5$ space.
== Proof ==
From Separation Axioms on Double Pointed Topology, we have that:
:$T \times D$ is not a $T_0$ space.
:$T \times D$ is not a $T_1$ space.
:$T \times D$ is not a $T_2$ space.
Also from Separation Axioms on Double Pointed Topology, we have that:
:$T \times D$ is a $T_3$ space $T$ is a $T_3$ space
:$T \times D$ is a $T_4$ space $T$ is a $T_4$ space
:$T \times D$ is a $T_5$ space $T$ is a $T_5$ space
But from Finite Complement Space is not $T_3$, $T_4$ or $T_5$:
:$T$ is not a $T_3$ space
:$T$ is not a $T_4$ space
:$T$ is not a $T_5$ space
Hence the result.
== Sources ==
* : Part $\text {II}$: Counterexamples: $18 \text { - } 19$. Finite Complement Topology: $7$
Category:Separation Axioms
Category:Double Pointed Topologies
Category:Finite Complement Topology
Double Pointed Finite Complement Topology is Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a finite complement topology on an infinite set $S$.
Let $T \times D$ be the double pointed topology on $T$.
Then $T \times D$ is compact.
== Proof ==
The Finite Complement Topology is Compact.
Also, a Finite Topological Space is Compact.
So $D$, defined as the indiscrete space on a doubleton, is also compact.
The result follows directly from Tychonoff's Theorem.
== Sources ==
* : Part $\text {II}$: Counterexamples: $18 \text { - } 19$. Finite Complement Topology: $7$
Category:Compact Spaces
Category:Double Pointed Topologies
Category:Finite Complement Topology
Double Pointed Fortissimo Space is Lindelöf

== Theorem ==
Let $T = \struct {S, \tau}$ be a Fortissimo space.
Let $T \times D$ be the double pointed topology on $T$.
Then $T \times D$ is a Lindelöf space.
== Proof ==
Let $D = \set {0, 1}$.
Let $\CC$ be an open cover of $T \times D$.
Then $\exists A \times B \in \CC$ such that $\tuple {p, 0} \in A \times B$.
We must have $\relcomp S A$ is countable and $B = D$.
Hence $\relcomp S A \times D$, a product of countable sets, must be countable.
So $A \times D$, together with an open neighborhood of each of the elements of $\relcomp S A \times D$, is a countable subcover of $\CC$.
Hence the result by definition of Lindelöf space.
== Sources ==
* : Part $\text {II}$: Counterexamples: $25$. Fortissimo Space: $4$
Category:Double Pointed Topologies
Category:Fortissimo Space
Category:Lindelöf Spaces
Double Pointed Fortissimo Space is Weakly Countably Compact

== Theorem ==
Let $T = \struct {S, \tau_p}$ be a Fortissimo space.
Let $T \times D$ be the double pointed topology on $T$.
Then $T \times D$ is weakly countably compact.
== Proof ==
Let $D = \set {0, 1}$.
Let $\tuple {p, 0}$ belong to some infinite $A \subseteq S$.
Then its twin $\tuple {p, 1}$ is a limit point of $A$.
Hence the result by definition of weakly countably compact.
== Sources ==
* : Part $\text {II}$: Counterexamples: $25$. Fortissimo Space: $4$
Category:Double Pointed Topologies
Category:Fortissimo Space
Category:Weakly Countably Compact Spaces
Double Pointed Fortissimo Space is not Pseudocompact

== Theorem ==
Let $T = \struct {S, \tau}$ be a Fortissimo space.
Let $T \times D$ be the double pointed topology on $T$.
Then $T \times D$ is not pseudocompact.
== Proof ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $25$. Fortissimo Space: $4$
Category:Double Pointed Topologies
Category:Fortissimo Space
Category:Pseudocompact Spaces
Double Pointed Fortissimo Space is not Sigma-Compact

== Theorem ==
Let $T = \struct {S, \tau}$ be a Fortissimo space.
Let $T \times D$ be the double pointed topology on $T$.
Then $T \times D$ is not $\sigma$-compact.
== Proof ==
== Sources ==
* : Part $\text {II}$: Counterexamples: $25$. Fortissimo Space: $4$
Category:Double Pointed Topologies
Category:Fortissimo Space
Category:Sigma-Compact Spaces
Double Pointed Topology is not T0

== Theorem ==
Let $T_1 = \struct {S, \tau_S}$ be a topological space.
Let $D = \struct {A, \set {\O, A} }$ be the indiscrete topology on an arbitrary doubleton $A = \set {a, b}$.
Let $T = \struct {T_1 \times D, \tau}$ be the double pointed topological space on $T_1$.
Then $T$ is not a $T_0$ (Kolmogorov) space.
== Proof 1 ==
== Proof 2 ==
== Sources ==
* : Notes: Part $1$: Basic Definitions: Section $2$. Separation Axioms: $1$
Category:Double Pointed Topologies
Category:T0 Spaces
Category:Double Pointed Topology is not T0
Double Pointed Topology is not T0/Proof 1

== Theorem ==
== Proof ==
By definition, the double pointed topology $\tau$ on $T_1$ is the product topology on $T_1 \times D$.
Let $x \in S$, and consider the point $\tuple {x, a} \in S \times A$.
Then:
:$\forall U \in \tau: \tuple {x, a} \in U \implies \tuple {x, b} \in U$
:$\forall U \in \tau: \tuple {x, b} \in U \implies \tuple {x, a} \in U$
as $D$ is an indiscrete space.
Hence the result, by definition of $T_0$ (Kolmogorov) space.
Category:Double Pointed Topology is not T0
Double Pointed Topology is not T0/Proof 2

== Theorem ==
== Proof ==
By definition, the double pointed topology $\tau$ on $T_1$ is the product topology on $T_1 \times D$.
By definition, $D$ is the indiscrete space on a doubleton.
 $T$ is a $T_0$ (Kolmogorov) space.
Then from Product Space is $T_0$ iff Factor Spaces are $T_0$ it follows that $D$ is also a $T_0$ (Kolmogorov) space.
But from Indiscrete Non-Singleton Space is not $T_0$, $D$ is not a $T_0$ (Kolmogorov) space.
The result follows by Proof by Contradiction.
Category:Double Pointed Topology is not T0
Double Reductio ad Absurdum

== Proof Structure ==
Let $N$ be a real number.
Let $P$ be the proposition:
:$\map P x = N$
that is, that a certain number $x$ is equal to $N$.
The Double Reductio ad Absurdum is an argument in the form:
:$(1): \quad$ Suppose $x < N$.
Then it is possible to derive a contradiction.
Therefore $x \not < N$.
:$(2): \quad$ Suppose $x > N$.
Then it is possible to derive a contradiction.
Therefore $x \not > N$.
So, because $x \not < N$ and $x \not > N$, it follows that:
:$x = N$
* 
*
Double Root is Root of First Derivative of Equation but not Second Derivative

== Theorem ==
Let $\EE$ be the equation:
:$\map f x = 0$
where $f: \R \to \R$ is a real function.
Let $\xi$ be a double root of $\EE$.
Then:
:$\xi$ is a root of the equation $\map {f'} x = 0$
:$\xi$ is not a root of the equation $\map {f' '} x = 0$
where:
:$f': \R \to \R$ is the first derivative of $f$ $x$
:$f' ': \R \to \R$ is the second derivative of $f$ $x$.
== Proof ==
== Sources ==
* 
* 
Category:Double Roots
Category:Differential Calculus
Double Root of Polynomial is Root of Derivative

== Theorem ==
Let $R$ be a commutative ring with unity.
Let $f \in R \sqbrk X$ be a polynomial.
Let $a \in R$ be a root of $f$ with multiplicity at least $2$.
Let $f'$ denote the formal derivative of $f$.
Then $a$ is a root of $f'$.
== Proof ==
Because $a$ has multiplicity at least $2$, we can write:
:$\map f X = \paren {X - a}^2 \map g X$
with $\map g X \in R \sqbrk X$.
From Formal Derivative of Polynomials Satisfies Leibniz's Rule:
:$\map {f'} X = 2 \paren {X - a} \map g X + \paren {X - a}^2 \map {g'} X$
and thus:
:$\map {f'} a = 0$
Category:Polynomial Theory
Double Superinduction Principle

== Theorem ==
Let $M$ be a class.
Let $g: M \to M$ be a mapping on $M$.
Let $M$ be a minimally superinductive class under $g$.
Let $\RR$ be a relation on $M$ which satisfies:
Then $\map \RR {x, y}$ holds for all $x, y \in M$.
== Proof ==
The proof proceeds by Proof by Superinduction.
Let an element $x$ of $M$ be defined as:
:left normal with respect to $\RR$ $\map \RR {x, y}$ for all $y \in M$
:right normal with respect to $\RR$ $\map \RR {y, x}$ for all $y \in M$.
Let the hypothesis be assumed.
First we demonstrate a lemma:
=== Lemma ===
We now show by Proof by Superinduction that every $x \in M$ is right normal with respect to $\RR$.
It then follows from the lemma that $\map \RR {x, y}$ for all $x, y \in M$.
=== Basis for the Induction ===
$\map P \O$ is the case:
From condition $\text D_1$ of the definition of $\RR$, we have immediately that:
:$\map \RR {x, \O}$
for all $x \in M$.
That is, that $\O$ is right normal with respect to $\RR$.
Thus $\map P \O$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P x$ is true, where $x \in M$, then it logically follows that $\map P {\map g x}$ is true.
So this is the induction hypothesis:
:$x$ is right normal with respect to $\RR$
from which it is to be shown that:
:$\map g x$ is right normal with respect to $\RR$.
=== Induction Step ===
This is the induction step:
Let $x \in M$ be right normal with respect to $\RR$:
:$\forall y \in M: \map \RR {y, x}$
By the lemma we have that $x$ is left normal with respect to $\RR$.
That is:
:$\forall y \in M: \map \RR {x, y}$
Thus by condition $\text D_2$ of the definition of $\RR$:
:$\forall y \in M: \map \RR {y, \map g x}$
That is, $\map g x$ is right normal with respect to $\RR$.
So $\map P x \implies \map P {\map g x}$ for all $x \in M$.
=== Closure under Chain Unions ===
This is the closure under chain unions:
Now suppose that $C$ is a chain of right normal elements with respect to $\RR$.
Let $x \in M$ be arbitrary.
Then by $\text D_3$ we have that $\ds \map \RR {x, \bigcup C}$ holds.
That is, $\ds \bigcup C$ is right normal with respect to $\RR$.
It follows by Proof by Superinduction that:
:$\forall x \in M$: $x$ is right normal with respect to $\RR$.
The result follows.
== Also known as ==
The Double Superinduction Principle can also be referred to as the Principle of Double Superinduction.
Some authors refer to this principle by its initials: D.S.P.
== Also see ==
* Double Induction Principle
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 2$ Superinduction and double superinduction: Theorem $2.1$
Category:Mathematical Induction
Category:Named Theorems
Category:Proof Techniques
Category:Double Superinduction Principle
Category:Closure under Chain Unions
Double Superinduction Principle/Lemma

== Lemma for Double Superinduction Principle ==
Let $M$ be a minimally superinductive class under a mapping $g$.
Let $\RR$ be a relation on $M$ which satisfies:
Let $x$ be a right normal element of $M$ with respect to $\RR$.
Then $x$ is also a left normal element of $M$ with respect to $\RR$.
== Proof ==
The proof proceeds by Proof by Superinduction.
Let $x \in M$ be right normal with respect to $\RR$.
Let $\map P y$ be the proposition:
:$\map \RR {x, y}$ holds.
=== Basis for the Induction ===
By condition $\text D_1$ of the definition of $\RR$:
:$\map \RR {x, \O}$
for all $x \in M$.
Thus $\map P \O$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that if $\map P y$ is true, where $x \in M$, then it logically follows that $\map P {\map g y}$ is true.
So this is the induction hypothesis:
:$\map \RR {x, y}$ holds
from which it is to be shown that:
:$\map \RR {x, \map g y}$ holds
=== Induction Step ===
This is the induction step:
Let $\map \RR {x, y}$ hold.
As $x$ is right normal with respect to $\RR$:
:$\map \RR {y, x}$ holds.
Thus by condition $\text D_2$ of the definition of $\RR$:
:$\map \RR {x, \map g y}$ holds.
So $\map P x \implies \map P {\map g x}$ and it follows by Proof by Superinduction that $\map \RR {x, y}$ holds for every $x, y \in M$.
=== Closure under Chain Unions ===
This is the closure under chain unions:
Now suppose that $C$ is a chain of elements of $M$ such that $\map \RR {x, y}$ holds for all $y \in C$.
Then by $\text D_3$ we have $\map \RR {x, \bigcup C}$.
Therefore:
:$\forall x \in M$: if $x$ is a right normal element of $M$ with respect to $\RR$, then $x$ is a left normal element of $M$ with respect to $\RR$.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text I$ -- Superinduction and Well Ordering: $\S 2$ Superinduction and double superinduction: Theorem $2.1$
Category:Double Superinduction Principle
Double of Antiperiod is Period

== Theorem ==
Let $f: \R \to \R$ be a real antiperiodic function with an anti-period of $A$.
Then $f$ is also periodic with a period of $2A$.
== Proof ==
Let $L_f$ be the set of all periodic elements of $f$.
By Periodic Element is Multiple of Antiperiod and Absolute Value of Real Number is not less than Divisors:
:$\forall p \in L_f: A \divides p \land A \le p$
Suppose there is a $p \in L_f$ such that $p = A$. 
Then: 
which contradicts Constant Function has no Period.
Therefore $\forall p \in L_f: A \divides p \land A < p$.
The smallest number $r$ such that $A < r \land A \divides r$ is $2A$.
But by Double of Antiperiodic Element is Periodic, this is a periodic element of $f$.
Hence the result.
Category:Antiperiodic Functions
Double of Antiperiodic Element is Periodic

== Theorem ==
Let $f: \R \to \R$ be a real function.
Let $L \in \R_{>0}$ be an anti-periodic element of $f$.
Then $2 L$ is a periodic element of $f$.
In other words, every anti-periodic function is also periodic.
== Proof ==
By Non-Zero Real Numbers Closed under Multiplication we have that $2 L \in \R_{>0}$.
Then:
Category:Antiperiodic Functions
Doubleton Class can be Formed from Two Sets

== Theorem ==
Let $V$ be a basic universe.
Let $a, b \in V$ be sets.
Then the doubleton class $\set {a, b}$ can be formed, which is a subclass of $V$.
== Proof ==
Using the , let $A$ be the class defined as:
:$A := \set {x: x \in V \land \paren {x = a \lor x = b} }$
That is:
:$A = \set {a, b}$
By the , $\set {a, b}$ is the only such class which has $a$ and $b$ as elements.
== Also see ==
* Definition:Doubleton Class
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 4$ The pairing axiom
Category:Doubleton Classes
Doubleton Class of Equal Sets is Singleton Class

== Theorem ==
Let $V$ be a basic universe.
Let $a, b \in V$ be sets.
Consider the doubleton class $\set {a, b}$.
Let $a = b$.
Then:
:$\set {a, b} = \set a$
where $\set a$ denotes the singleton class of $a$.
== Proof ==
Let $A = \set {a, b}$
The existence of $A$ is shown in Doubleton Class can be Formed from Two Sets:
:$A := \set {x: \paren {x = a \lor x = b} }$
Let $a = b$.
Then:
== Sources ==
* : Chapter $2$: Some Basics of Class-Set Theory: $\S 4$ The pairing axiom
Category:Doubleton Classes
Doubleton of Elements is Subset

== Theorem ==
Let $S$ be a set.
Let $\set {x,y}$ be the doubleton of distinct $x$ and $y$.
Then:
:$x, y \in S \iff \set {x,y} \subseteq S$
== Proof ==
=== Necessary Condition ===
Let $x, y \in S$.
From Singleton of Element is Subset:
:$\set x \subseteq S$
:$\set y \subseteq S$
From Union of Subsets is Subset:
:$\set x \cup \set y \subseteq S$
From Union of Disjoint Singletons is Doubleton:
: $\set x \cup \set y = \set {x, y}$
Hence:
:$\set {x,y} \subseteq S$
=== Sufficient Condition ===
Let $\set {x,y} \subseteq S$.
From the definition of a subset:
:$x \in \set {x,y} \implies x \in S$
:$y \in \set {x,y} \implies y \in S$
Category:Subsets
Category:Doubletons
Doubleton of Sets can be Derived using Axiom of Abstraction

== Theorem ==
Let $a$ and $b$ be sets.
By application of the , the set $\set {a, b}$ can be formed.
Hence the doubleton $\set {a, b}$ can be derived as a valid object in Frege set theory.
== Proof ==
Let $P$ be the property defined as:
:$\forall x: \map P x := \paren {x = a \lor x = b}$
where $\lor$ is the disjunction operator.
Hence, using the , we form the set:
:$\set {a, b} := \set {x: x = a \lor x = b}$
== Sources ==
* : Chapter $1$: General Background: $\S 7$ Frege set theory
Category:Doubletons
Category:Axiom of Abstraction
Doubling the Cube

== Classic Problem ==
Doubling the cube is the exercise to construct a cube whose volume is double that of a given cube.
== Solution ==
Let $C$ be the cube in question.
Let $x$ be the length of one of the edges of $C$.
 demonstrated that the problem is equivalent to finding line segments of length $p$ and $q$ such that:
:$\dfrac x p = \dfrac p q = \dfrac q {2 x}$
from which:
:$2 x^3 = p^3$
and so:
:$\dfrac p x = \sqrt [3] 2$
Hence to find a cube whose volume is double that of $C$ is equivalent to finding the Cube Root of 2.
But Doubling the Cube by Compass and Straightedge Construction is Impossible.
Various techniques have been devised which use more elaborate apparatus than just the straightedge and compass:
=== Archytas Curve ===
=== Intersection of Parabolas ===
=== Intersection of Parabola and Hyperbola ===
=== Conchoid of Nicomedes ===
=== Cissoid of Diocles ===
== Fallacious Proofs ==
== Also see ==
* Doubling the Cube by Compass and Straightedge Construction is Impossible
* Trisecting the Angle
* Squaring the Circle
== Also known as ==
== Historical Note ==
== Sources ==
* : Chapter $\text{II}$: Modern Minds in Ancient Bodies
* : $1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
* 
* : $1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
* 
* 
* 
* 
* 
* 
* 
* 
Category:Doubling the Cube
Category:Classic Problems
Category:Euclidean Geometry
Category:Solid Geometry
Doubling the Cube/Also known as

== Doubling the Cube ==
The problem of Doubling the Cube is known as the Delian Problem, after the location () of the altar whose dimensions were under the question.
Some sources refer to the problem of Doubling the Cube as Duplicating the Cube.
However, the position taken by is that duplication can also mean making an exact copy of, which could cause misunderstanding.
== Sources ==
* 
* 
* 
* 
* 
* 
Category:Doubling the Cube
Doubling the Cube/Archytas Curve

== Theorem ==
The Archytas curve can be used for Doubling the Cube.
== Proof ==
== Historical Note ==
== Sources ==
* : $1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
* : $1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
Category:Doubling the Cube
Doubling the Cube/Archytas Curve/Historical Note

== Historical Note on using Archytas curve for Doubling the Cube ==
The use of the Archytas curve for Doubling the Cube was devised by .
== Sources ==
* : $1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
* : $1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
Category:Historical Notes
Doubling the Cube/Cissoid of Diocles

== Theorem ==
The problem of Doubling the Cube can be solved by using a cissoid of Diocles.
== Proof ==
== Historical Note ==
== Sources ==
* : $1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
* : $1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
Category:Doubling the Cube
Doubling the Cube/Cissoid of Diocles/Historical Note

== Historical Note on Doubling the Cube by Cissoid of Diocles ==
The technique for doubling the cube by using a Cissoid of Diocles was devised by .
== Sources ==
* : $1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
* : $1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
Category:Historical Notes
Doubling the Cube/Conchoid of Nicomedes

== Theorem ==
The problem of Doubling the Cube can be solved by using a conchoid of Nicomedes.
== Proof ==
== Historical Note ==
== Sources ==
* : $1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
* : $1 \cdotp 25992 \, 10498 \, 94873 \, 16476 \ldots$
Category:Doubling the Cube