nyuuzyou/wikis huggingface dataset:
Principal Ideal of Commutative Ring

== Theorem ==
Let $\struct {R, +, \circ}$ be a commutative ring with unity.
Let $a \in R$.
Let $Ra$ be the principal left ideal of $R$ generated by $a$.
Let $aR$ be the principal right ideal of $R$ generated by $a$.
Let $\ideal a$ be the principal ideal of $R$ generated by $a$.
Then $Ra = \ideal a = aR$.
== Proof ==
By definition of principal left ideal:
:$Ra = \set{r \circ a: r \in R}$
By definition of commutative ring with unity and center of ring:
:$a$ is in the center of $R$
From Principal Ideal from Element in Center of Ring:
:$\ideal a = R \circ a = \set{r \circ a: r \in R}$
Hence:
:$Ra = \ideal a$
We have:
Category:Ideal Theory
Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal

== Theorem ==
Let $\struct {D, +, \circ}$ be a principal ideal domain.
Let $\ideal p$ be the principal ideal of $D$ generated by $p$.
Then $p$ is irreducible $\ideal p$ is a maximal ideal of $D$.
== Proof ==
=== Necessary Condition ===
=== Sufficient Condition ===
Category:Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal
Category:Principal Ideals of Rings
Category:Maximal Ideals of Rings
Category:Principal Ideal Domains
Category:Factorization
Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal/Forward Implication

== Theorem ==
Let $\struct {D, +, \circ}$ be a principal ideal domain.
Let $p$ be an irreducible element of $D$.
Let $\ideal p$ be the principal ideal of $D$ generated by $p$.
Then $\ideal p$ is a maximal ideal of $D$.
== Proof ==
Let $p$ be irreducible in $D$.
Let $U_D$ be the group of units of $D$.
By definition, an irreducible element is not a unit.
So from Principal Ideals in Integral Domain:
:$\ideal p \subset D$
Suppose the principal ideal $\ideal p$ is not maximal.
Then there exists an ideal $K$ of $D$ such that:
:$\ideal p \subset K \subset R$
Because $D$ is a principal ideal domain:
:$\exists x \in R: K = \ideal x$
Thus:
:$\ideal p \subset \ideal x \subset D$
Because $\ideal p \subset \ideal x$:
:$x \divides p$
by Principal Ideals in Integral Domain.
That is:
:$\exists t \in D: p = t \circ x$
But $p$ is irreducible in $D$, so $x \in U_D$ or $t \in U_D$.
That is, either $x$ is a unit or $x$ is an associate of $p$.
But since $K \subset D$:
:$\ideal x \ne D$ so $x \notin U_D$
by Principal Ideals in Integral Domain.
Also, since $\ideal p \subset \ideal x$:
:$\ideal p \ne \ideal x$
so $x$ is not an associate of $p$, by Principal Ideals in Integral Domain.
This contradiction shows that $\ideal p$ is a maximal ideal of $D$.
== Sources ==
* : $\S 63.2$ Construction of fields as factor rings
Category:Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal
Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal/Reverse Implication

== Theorem ==
Let $\left({D, +, \circ}\right)$ be a principal ideal domain.
Let $\left({p}\right)$ be the principal ideal of $D$ generated by $p$.
Let $\left({p}\right)$ be a maximal ideal of $D$.
Then $p$ is irreducible.
== Proof ==
Let $\left({p}\right)$ be a maximal ideal of $D$.
Let $p = f g$ be any factorization of $p$.
We must show that one of $f, g$ is a unit.
Suppose that neither of $f, g$ is a unit.
First it will be shown that:
:$\left({p}\right) \subsetneqq \left({f}\right)$
Let $x \in \left({p}\right)$.
That is:
:$\exists q \in D: x = p q$
Then:
:$x = f g q \in \left({f}\right)$
so:
:$\left({p}\right) \subseteq \left({f}\right)$
Now suppose $f \in \left({p}\right)$.
Then:
:$\exists r \in D: f = r p$
and so from $p = f g$ above:
:$f = r g f$
Therefore:
:$r g = 1$
and so $g$ is a unit.
This is a contradiction.
Thus:
:$f \notin \left({p}\right)$
and clearly:
:$f \in \left({f}\right)$
so:
:$\left({p}\right) \subsetneqq \left({f}\right)$
as claimed.
Therefore, since $\left({p}\right)$ is maximal, we must have:
:$\left({f}\right) = D$
But we assumed that $f$ is not a unit.
So there is no $h \in D$ such that $f h = 1$.
Therefore:
:$1 \notin \left({f}\right) = \left\{{f h: h \in D}\right\}$
and:
:$\left({f}\right) \subsetneqq D$
This is a contradiction.
Therefore at least one of $f, g$ must be a unit.
This completes the proof.
Category:Principal Ideal of Principal Ideal Domain is of Irreducible Element iff Maximal
Principal Ideal of Ring/Examples/Integer Multiples of 3

== Examples of Principal Ideals of Rings ==
Let $\struct {\Z, + \times}$ denote the ring of integers.
Let $3 \Z$ denote the set of integer multiples of $3$.
Then $\struct {3 \Z, +, \times}$, where $+$ and $\times$ denote integer addition and integer multiplication respectively, forms the principal ideal of $\Z$ generated by $3$.
== Proof ==
== Sources ==
* 
* 
Category:Examples of Principal Ideals of Rings
Principal Ideals in Integral Domain

== Theorem ==
Let $\struct {D, +, \circ}$ be an integral domain.
Let $U_D$ be the group of units of $D$.
Let $\ideal x$ be the principal ideal of $D$ generated by $x$.
Let $x, y \in \struct {D, +, \circ}$.
Then:
=== Element in Integral Domain is Divisor iff Principal Ideal is Superset ===
=== Element in Integral Domain is Unit iff Principal Ideal is Whole Domain ===
=== Elements in Integral Domain are Associates iff Principal Ideals are Equal ===
Category:Integral Domains
Category:Principal Ideals of Rings
Category:Factorization
Principal Ideals of Integers

== Theorem ==
Let $J$ be a non-zero ideal of $\Z$.
Then $J = \ideal b$ where $b$ is the smallest strictly positive integer belonging to $J$.
== Proof ==
It follows from Ring of Integers is Principal Ideal Domain that $J$ is a principal ideal.
Let $c \in J, c \ne 0$.
Then $-c \in J$ and by Natural Numbers are Non-Negative Integers, exactly one of them is strictly positive.
Thus $J$ does actually contain strictly positive elements, so that's a start.
Let $b$ be the smallest strictly positive element of $J$.
This exists because Natural Numbers are Non-Negative Integers and the Well-Ordering Principle.
By definition of a principal ideal, we have $\ideal b \subseteq J$ as $b \in J$.
We need to show that $J \subseteq \ideal b$.
So, let $a \in J$.
By the Division Theorem, $\exists q, r: a = b q + r, 0 \le r < b$.
As $a, b \in J$, then so does $r = a - b q$.
So, by the definition of $b$, it follows that $r = 0$.
Thus $a = b q \in \ideal b$.
== Sources ==
* : Chapter $\text {IV}$: Rings and Fields: $24$. The Division Algorithm: Theorem $24.3$
* : $\S 59.2$ Principal ideals in a commutative ring with a one
Category:Ideal Theory
Principal Left Ideal is Left Ideal

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity.
Let $a \in R$.
Let $Ra$ be the principal left ideal of $R$ generated by $a$.
Then $Ra$ is an left ideal of $R$.
== Proof ==
We establish that $Ra$ is an left ideal of $R$, by verifying the conditions of Test for Left Ideal.
$Ra \ne \O$, as $1_R \circ a = a \in Ra$.
	
Let $x, y \in Ra$.
Then:
Let $s \in Ra, x \in R$.
Thus by Test for Left Ideal, $Ra$ is a left ideal of $R$.
== Also see ==
* Principal Right Ideal is Right Ideal where it is shown that a principal right ideal is a right ideal
Category:Ideal Theory
Principal Right Ideal is Right Ideal

== Theorem ==
Let $\struct {R, +, \circ}$ be a ring with unity.
Let $a \in R$.
Let $aR$ be the principal right ideal of $R$ generated by $a$.
Then $aR$ is an right ideal of $R$.
== Proof ==
We establish that $aR$ is an right ideal of $R$, by verifying the conditions of Test for Right Ideal.
$aR \ne \O$, as $a \circ 1_R = a \in aR$.
	
Let $x, y \in ar$.
Then:
Let $s \in aR, x \in R$.
Thus by Test for Right Ideal, $aR$ is a right ideal of $R$.
Category:Ideal Theory
Principal Ultrafilter is All Sets Containing Cluster Point

== Theorem ==
Let $S$ be a set.
Let $\powerset S$ denote the power set of $S$.
Let $\FF \subset \powerset S$ be a principal ultrafilter on $S$.
Let its cluster point be $x$.
Then $\FF$ is the set of all subsets $T$ of $S$ such that $x \in T$.
== Proof ==
We have that the cluster point of $\FF$ is $x$.
 there is some $A \subseteq S$ such that $x \in A$ and $A \notin \FF$.
Then, by definition of relative complement:
:$x \notin \relcomp S A$
By definition of cluster point, $x$ is in every element of $\FF$.
But as $x \notin \relcomp S A$, it follows that:
:$\relcomp S A \notin \FF$
This contradicts $\FF$ being an ultrafilter:
:for every $A \subseteq S$, either $A \in \FF$ or $\relcomp S A \in \FF$
Therefore, by Proof by Contradiction, every $A \subseteq S$ such that $x \in A$ is in $\FF$.
Thus, $A \subseteq S$ is in $\FF$ precisely when $x \in A$.
== Sources ==
* : Part $\text I$: Basic Definitions: Section $1$: General Introduction: Filters
Category:Filter Theory
Principal Value of One over x is Distribution

== Theorem ==
Let $\phi \in \map \DD \R$ be a test function.
Let $T : \map \DD \R \to \C$ be a mapping such that:
:$\forall \phi \in \map \DD \R : \map T \phi = \PV \frac {\map \phi x} x \rd x : = \lim_{\epsilon \mathop \to 0} \int_{\size x \mathop > \epsilon} \frac {\map \phi x} x \rd x$
where $\PV$ denotes the Cauchy principal value.
Then $T$ is a distribution.
== Proof ==
Let $\phi \in \map \DD \R$ be a test function with a support on $\closedint {-a} a$.
Then:
Furthermore:
=== Existence of the limit ===
Since $\phi \in \map \DD \R$, the integral exists for any $\epsilon$.
Hence, the limit exists.
Thus, we can rewrite $T$ as:
:$\ds \map T \phi = \int_{-a}^a \int_0^1 \dfrac {\d \map \phi {t x} } {\map \d {t x} } \rd t \rd x$
=== Linearity ===
Follows from Riemann Integral Operator is Linear Mapping.
=== Continuity ===
By Convergent Sequence Minus Limit, we can shift the sequence to set its limit to zero.
Let $\mathbf 0 : \R \to 0$ be a zero mapping.
Let $\sequence {\phi_n}_{n \mathop \in \N}$ be a sequence with support on $\closedint {-a} a$ such that it converges to $\mathbf 0$:
:$\phi_n \stackrel \DD {\longrightarrow} \mathbf 0$
Then:
Take the limit $n \to \infty$.
Then:
:$\map T {\mathbf 0} = 0$
Hence, by definition, $\PV \frac 1 x$ is a distribution.
== Sources ==
* : Chapter $\S 6.1$: A glimpse of distribution theory. Test functions, distributions, and examples
Category:Examples of Distributions
Category:Functional Analysis
Principle E

== Theorem ==
Let $S$ be a set of sets.
Let $S$ be closed under chain unions.
Then every element $b$ of $S$ is a subset of some element of $S$ that has no immediate extension in $S$.
== Principle E and Axiom of Choice ==
=== Axiom of Choice implies Principle E ===
=== Principle E implies Axiom of Choice ===
== Linguistic Note ==
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 6$ Another approach to maximal principles
Category:Class Extensions
Category:Principle E
Category:Closure under Chain Unions
Principle E is Equivalent to Kuratowski's Lemma

== Theorem ==
Kuratowski's Lemma is equivalent to Principle E.
== Proof ==
Recall Kuratowski's Lemma:
Recall Principle E:
=== Kuratowski's Lemma implies Principle E ===
Let Kuratowski's Lemma hold.
Let $x$ be a maximal element under the subset relation.
Then $x$ has no immediate extension which is not $x$ itself.
That is, Principle E holds directly.
=== Principle E implies Kuratowski's Lemma ===
Let Principle E hold.
Let $S$ be swelled.
That is, every subset of every element of $S$ is an element of $S$.
Then an element $x$ of $S$ has no immediate extension in $S$ $x$ is a maximal element of $S$ under the subset relation.
Thus Principle E implies that for every swelled set $S$ which is closed under chain unions, every $b \in S$ is a subset of some maximal element of $S$.
By Class of Finite Character is Swelled, a set of finite character is swelled.
By Class of Finite Character is Closed under Chain Unions a set of finite character is closed under chain unions.
Thus Principle E implies Tukey's Lemma.
From Maximal Principles are Equivalent, we have that Tukey's Lemma is equivalent to Kuratowski's Lemma.
== Sources ==
* : Chapter $4$: Superinduction, Well Ordering and Choice: Part $\text {II}$ -- Maximal principles: $\S 6$ Another approach to maximal principles: Theorem $6.1$
Category:Principle E
Category:Kuratowski's Lemma
Principle of Commutation/Formulation 1

== Theorem ==
:$p \implies \paren {q \implies r} \dashv \vdash q \implies \paren {p \implies r}$
This can of course be expressed as two separate theorems:
=== Forward Implication ===
=== Reverse Implication ===
== Proof 1 ==
== Proof 2 ==
Category:Principle of Commutation
Principle of Commutation/Formulation 1/Proof 2

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|ccccc||ccccc|} \hline
p & \implies & (q & \implies & r) & q & \implies & (p & \implies & r) \\
\hline
F & T & F & T & F & F & T & F & T & F \\
F & T & F & T & T & F & T & F & T & T \\
F & T & T & F & F & T & T & F & T & F \\
F & T & T & T & T & T & T & F & T & T \\
T & T & F & T & F & F & T & T & F & F \\
T & T & F & T & T & F & T & T & T & T \\
T & F & T & F & F & T & F & T & F & F \\
T & T & T & T & T & T & T & T & T & T \\
\hline
\end{array}$
Category:Principle of Commutation
Category:Truth Table Proofs
Principle of Commutation/Formulation 2

== Theorem ==
:$\vdash \paren {p \implies \paren {q \implies r} } \iff \paren {q \implies \paren {p \implies r} }$
This can of course be expressed as two separate theorems:
=== Forward Implication ===
=== Reverse Implication ===
== Proof ==
=== Proof of Forward Implication ===
=== Proof of Reverse Implication ===
== Sources ==
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Theorem $\text T 107$
Category:Principle of Commutation
Principle of Composition/Formulation 2

== Theorem ==
:$\paren {\paren {p \implies r} \lor \paren {q \implies r} } \iff \paren {\paren {p \land q} \implies r}$
== Proof ==
== Sources ==
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Theorem $\text{T60}$
Category:Principle of Composition
Principle of Condensation of Singularities

== Theorem ==
Let $\struct {X, \norm {\,\cdot\,}_X}$ be a Banach space.
Let $\struct {Y, \norm {\,\cdot\,}_Y}$ be a normed vector space.
Let $\family {T_\alpha: X \to Y}_{\alpha \mathop \in A}$ be an $A$-indexed family of bounded linear transformations from $X$ to $Y$ such that: 
:$\ds \sup_{\alpha \in A} \norm {T_\alpha} = \infty$
Then: 
:$\ds \sup_{\alpha \in A} \norm {T_\alpha x} = \infty$ for some $x \in X$.
== Proof ==
 that: 
:$\ds \sup_{\alpha \in A} \norm {T_\alpha x} < \infty$ for each $x \in X$.
From the Banach-Steinhaus Theorem, we have: 
:$\ds \sup_{\alpha \in A} \norm {T_\alpha} < \infty$.
contradicting that: 
:$\ds \sup_{\alpha \in A} \norm {T_\alpha} = \infty$
So, we have: 
:$\ds \sup_{\alpha \in A} \norm {T_\alpha x} = \infty$ for some $x \in X$.
== Sources ==
* $22.2$: The Principle of Uniform Boundedness
Category:Linear Transformations on Banach Spaces
Principle of Conservation of Angular Momentum

== Theorem ==
Newton's Laws of Motion imply the conservation of angular momentum in systems of masses in which no external force is acting.
== Proof ==
We start by stating Newton's Third Law of Motion in all its detail.
We consider a collection of massive bodies denoted by the subscripts $1$ to $N$.
These bodies interact with each other and exert forces on each other and these forces occur in equal and opposite pairs.
The force $F_{i j}$ exerted by body $i$ on body $j$ is related to the force exerted by body $j$ on body $i$ by:
:$(1): \quad \vec {F_{i j}} = -\vec {F_{j i} }$
The final part of Newton's Third Law of Motion is that these equal and opposite forces act through the line that connects the two bodies in question.
This can be stated thus:
:$(2): \quad \vec {F_{i j}} = a_{i j} \paren {\vec {r_j} - \vec {r_i} }$
where:
:$\vec{r_i}$ is the position of body $i$
:$a_{i j} $ is the magnitude of the force.
As a consequence of $(1)$:
:$a_{j i} = a_{i j}$
Let the total torque $\vec {\tau_{\operatorname {total} } }$ on the system be measured about an origin located at $\vec {r_0}$.
Thus:
We have that there is no external force.
Thus the second term disappears, and:
In summary, in a system of masses in which there is no external force, the total torque on the system is equal to $0$.
This is because the pair of torque between two bodies must cancel out.
Since the rate of change of angular momentum is proportional to the torque, the angular momentum of a system is conserved when no external force is applied.
== Also known as ==
== Examples ==
== Sources ==
* 
* 
* 
* 
Category:Principle of Conservation of Angular Momentum
Category:Angular Momentum
Category:Classical Mechanics
Principle of Conservation of Angular Momentum/Examples/Cloud of Gas

== Examples of Use of Principle of Conservation of Angular Momentum ==
Let $\CC$ be a rotating cloud of gas in space.
Suppose $\CC$ is contracting under its own gravitation.
Then it rotates with a greater angular velocity as it contracts, so as to preserve its angular momentum.
== Sources ==
* 
* 
Category:Principle of Conservation of Angular Momentum
Principle of Conservation of Energy

== Physical Law ==
Let $B$ be a body in a force field.
Let $B$ be subject only to forces within that force field.
Let $P$ be the potential energy of $B$.
Let $K$ be the kinetic energy of $B$.
Then:
:$P + K = \text{constant}$
== Historical Note ==
== Sources ==
* : $1$: The Nature of Differential Equations: $\S 5$: Falling Bodies and Other Rate Problems
* 
* 
* 
* 
* 
* 
* 
* 
Category:Energy
Category:Potential Energy
Category:Kinetic Energy
Principle of Counting

== Theorem ==
Let $T$ be a set such that $T \sim \N_n$.
Then:
:$\forall m \in \N: n \ne m \implies T \nsim \N_m$
== Proof ==
The result follows directly from Equality of Natural Numbers and the fact that set equivalence is transitive.
== Motivation ==
Suppose we have a set of objects which we want to count.
That is, we want to know how many objects there are in our set.
What we are doing is trying to match up our set with one of the sets $\N_n$.
That's what we (effectively) do when we count. We pick a set $\N_n$ and compare it in size to see whether it matches the one we're counting. If it doesn't, we pick another one and try again.
In fact, what we do is test them in order. We give each $n \in \N$ a name, and we learn what order they go in and which name matches which $n$. Then we separate the set we are counting into two subsets: the ones counted and the ones still to be counted. As we move each object one by one from the latter into the former, we recite the name of the element of $n$ which follows the previous one we have just counted.
When there are no more elements in the ones still to be counted subset, the $n$ we have reached is the name of the $\N_n$ which is equivalent to the ones counted subset.
== Also see ==
This is (confusingly) not to be confused with the Fundamental Principle of Counting, otherwise known as the counting principle.
== Sources ==
* : $\S 2.4$: Definition $2.4$
* : Chapter $\text {III}$: The Natural Numbers: $\S 17$: Finite Sets: Theorem $17.3$
Category:Natural Numbers
Category:Named Theorems
Principle of Definition by Structural Induction

== Theorem ==
Let $\LL$ be a formal language.
Let the formal grammar of $\LL$ be a bottom-up grammar with unique parsability.
A definition $\map D \phi$ (in the metalanguage of $\LL$) for all well-formed formulas $\phi$ of $\LL$ is uniquely specified by:
:$(1): \quad$ A definition $\map D a$ for each letter $a$ of $\LL$
:$(2): \quad$ For each rule of formation for $\LL$, a definition $\map D \phi$ of the resultant WFF $\phi$ in terms of the constituent WFFs' definitions $\map D {\phi_1}, \ldots, \map D {\phi_n}$.
== Proof ==
We apply the Principle of Structural Induction on the following statement $\map P \phi$:
:The definition is specified for $\phi$
The given hypotheses verify the conditions for the Principle of Structural Induction.
It follows that $\map D \phi$ is specified for each WFF $\phi$.
Moreover, by virtue of the unique parsability, it is ensured that each WFF $\phi$ has a unique definition $\map D \phi$.
Category:Formal Languages
Principle of Dilemma

== Theorem ==
=== Formulation 1 ===
=== Formulation 2 ===
== Also known as ==
This proof structure is also referred to as proof by cases but it is usual to reserve that name for a more general concept.
== Also see ==
* Proof by Cases
Category:Principle of Dilemma
Category:Conditional
Category:Conjunction
Category:Logical Negation
Category:Proof Techniques
Principle of Dilemma/Formulation 1

== Theorem ==
If both a premise and its negation imply the conclusion, then the conclusion must be true independently of any premises.
:$\paren {p \implies q} \land \paren {\neg p \implies q} \dashv \vdash q$
This can be expressed as two separate theorems:
=== Forward Implication ===
=== Reverse Implication ===
== Proof ==
== Also see ==
* Proof by Cases
Category:Principle of Dilemma
Principle of Dilemma/Formulation 1/Forward Implication

== Theorem ==
:$\paren {p \implies q} \land \paren {\neg p \implies q} \vdash q$
== Proof 1 ==
== Proof 2 ==
== Proof 3 ==
== Sources ==
* : $\text{I}$: 'NOT' and 'IF': $\S 3$: Exercises, Group $\text{I}: \ 15$
* : $\S 1.12$: Valid Arguments
Category:Principle of Dilemma
Principle of Dilemma/Formulation 1/Forward Implication/Proof 3

== Theorem ==
== Proof ==
From the Constructive Dilemma we have:
:$p \implies q, r \implies s \vdash p \lor r \implies q \lor s$
from which, changing the names of letters strategically:
:$p \implies q, \neg p \implies q \vdash p \lor \neg p \implies q \lor q$
From Law of Excluded Middle, we have:
:$\vdash p \lor \neg p$
From the Rule of Idempotence we have:
:$q \lor q \vdash q$
and the result follows by Hypothetical Syllogism.
Category:Principle of Dilemma
Principle of Dilemma/Formulation 1/Proof

== Theorem ==
== Proof ==
We apply the Method of Truth Tables.
As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.
$\begin{array}{|cccccccc||c|} \hline
(p & \implies & q) & \land & (\neg & p & \implies & q) & q \\
\hline
F & T & F & F & T & F & F & F & F \\
F & T & T & T & T & F & T & T & T \\
T & F & F & F & F & T & T & F & F \\
T & T & T & T & F & T & T & T & T \\
\hline
\end{array}$
Category:Truth Table Proofs
Category:Principle of Dilemma
Principle of Duality in Space

== Theorem ==
Let $P$ be a theorem of projective geometry proven using the propositions of incidence.
Let $Q$ be the statement created from $P$ by interchanging:
:$(1) \quad$ the terms point and plane
:$(2) \quad$ the terms lie on and intersect at
and so on.
Then $Q$ is also a theorem of projective geometry.
== Proof ==
== Sources ==
* : Chapter $1$: Introduction: The Propositions of Incidence: $1.2$: The projective method: The principle of duality
Category:Duality Principles
Category:Projective Geometry
Principle of Duality in the Plane

== Theorem ==
Let $P$ be a theorem of projective geometry proven using the propositions of incidence.
Let $Q$ be the statement created from $P$ by interchanging:
:$(1): \quad$ the terms point and (straight) line
:$(2): \quad$ the terms collinear (of points) and concurrent (of lines)
:$(3): \quad$ the terms lie on and intersect at
and so on.
Then $Q$ is also a theorem of projective geometry.
== Proof ==
== Sources ==
* : Chapter $1$: Introduction: The Propositions of Incidence: $1.2$: The projective method: The principle of duality
Category:Duality Principles
Category:Projective Geometry