nyuuzyou/wikis huggingface dataset:
Construction of Square on Given Straight Line

== Theorem ==
== Proof ==
:
Let $AB$ be the given line segment.
Construct $AC$ perpendicular to $AB$.
By Construction of Equal Straight Lines from Unequal, place $D$ on $AC$ so $AD = AB$.
Construct $DE$ parallel to $AB$, and construct $BE$ parallel to $AD$.
So $ADEB$ is a parallelogram.
So $AB = DE$ and $AD = BE$ from Opposite Sides and Angles of Parallelogram are Equal.
So the parallelogram is equilateral.
Also, we have that $AD$ falls on the parallels $AB$ and $DE$.
So from Parallelism implies Supplementary Interior Angles, we have that $\angle BAD + \angle ADE$ equals two right angles.
But as $\angle BAD$ is right, so is $\angle ADE$.
And, from Opposite Sides and Angles of Parallelogram are Equal, so are $\angle ABE$ and $\angle BED$.
So $ADEB$ is equilateral and equiangular, and therefore, by definition, a square.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Squares
Construction of Straight Line Perpendicular to Plane from point not on Plane

== Theorem ==
== Proof ==
:
Let $A$ be the given elevated point.
Let the given plane be identified as the plane of reference.
It is required that a straight line be drawn perpendicular to the plane of reference.
Let any arbitrary straight line $BC$ be drawn in the plane of reference.
From :
:Let $AD$ be drawn from $A$ perpendicular to $BC$.
If $AD$ is perpendicular to the plane of reference, then the construction is complete.
Otherwise, let $DE$ be drawn from $D$ perpendicular to $BC$ and in the plane of reference.
From :
:Let $AF$ be drawn from $A$ perpendicular to $DE$.
From :
:Let $GH$ be drawn from the point $F$ parallel to $BC$.
We have that $BC$ is perpendicular to each of $DA$ and $DE$.
Therefore from :
:$BC$ is perpendicular to the plane through $ED$ and $DA$.
We have that $GH$ is parallel to $BC$.
But from :
:$GH$ is also perpendicular to the plane through $ED$ and $DA$.
Therefore by :
:$GH$ is perpendicular to all the straight lines which meet it and are in the plane through $ED$ and $DA$.
But $AF$ meets $GH$ and is in the plane through $ED$ and $DA$.
Therefore $GH$ is perpendicular to $FA$.
So $FA$ is perpendicular to $GH$.
But $FA$ is also perpendicular to $DE$.
Therefore $FA$ is perpendicular to both $GH$ and $DE$.
Therefore from :
:$AF$ is perpendicular to the plane through $DE$ and $GH$.
But the plane through $DE$ and $GH$ is the plane of reference.
Therefore $AF$ is perpendicular to the plane of reference.
Thus $AF$ is the required line.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Planes
Construction of Straight Line Perpendicular to Plane from point on Plane

== Theorem ==
== Proof ==
:
Let the given plane be identified as the plane of reference.
Let $A$ be the point from which the perpendicular is to be constructed.
Let $B$ be any elevated point.
From :
:let $BC$ be constructed from $B$ perpendicular to the plane of reference.
From :
:let $AD$ be drawn from $A$ parallel to $BC$.
We have that $AD$ and $BC$ are two parallel straight lines of which one of them $BC$ is perpendicular to the plane of reference.
From :
:$AD$ is also perpendicular to the plane of reference.
Therefore $AD$ is the required perpendicular.
== Sources ==
* : Book $\text{XI}$. Propositions
Category:Planes
Category:Perpendiculars
Construction of Tangent from Point to Circle/Proof 1

== Theorem ==
== Proof ==
:
Let $A$ be the given point and let $BCD$ be the given circle.
It is required that a straight line be drawn from $A$ to $BCD$.
Let the center $E$ of $BCD$ be found.
Join $AE$ and draw the circle $AFG$ with center $E$ and radius $AE$.
From $D$ let $DF$ be drawn perpendicular to $EA$.
Join $EF$ and let $B$ be the point at which $EF$ joins the circle $BCD$.
Join $AB$.
Then $AB$ is the required tangent to $BCD$.
=== Proof of Construction ===
Since $E$ is the center of the circles $BCD$ and $AFG$, $EA = EF$ and $ED = EB$.
Therefore the two sides $AE, EB$ equal the two sides $FE, ED$, and they contain a common angle at $E$.
So by Triangle Side-Angle-Side Congruence, $\triangle DEF = \triangle BEA$.
Therefore $\angle EDF = \angle EBA$.
But $\angle EDF$ is a right angle, so $\angle EBA$ is also a right angle.
But $EB$ is a radius of circle $BCD$.
It follows from the porism to Line at Right Angles to Diameter of Circle that $AB$ is tangent to $BCD$.
== Sources ==
* : Book $\text{III}$. Propositions
Category:Construction of Tangent from Point to Circle
Construction of Tangent from Point to Circle/Proof 2

== Theorem ==
== Proof ==
:
Let $BCD$ with center $A$ be the circle, and let $E$ be the exterior point from which a tangent is to be drawn.
Bisect $AE$ at $F$.
Then draw a circle $AEG$ whose center is $F$ and whose radius is $AF$.
The point $G$ is where $AEG$ intersects $BCD$.
The line $EG$ is the required tangent.
=== Proof of Construction ===
By the method of construction, $AE$ is the diameter of $AEG$.
By Thales' Theorem $\angle AGE$ is a right angle.
But $AG$ is a radius of $BCD$.
The result follows from Radius at Right Angle to Tangent.
== Historical Note ==
An easier solution is of course possible as soon as we know $(\text{III}$. $31)$ that the angle in a semicircle is a right angle; for we have only to describe a circle on $AE$ as diameter, and this circle cuts the given circle in the two points of contact.
:: -- : Footnote to Book $\text{III}$ Proposition $19$
== Sources ==
* : Book $\text{III}$. Propositions
Category:Construction of Tangent from Point to Circle
Construction of Third Apotome

== Theorem ==
== Proof ==
:
Let $A$ be a rational straight line.
Let $E$, $BC$ and $CD$ be numbers set out which do not have pairwise between them the ratio that a square number has to another square number.
But let $CB : BD$ be the ratio that a square number has to a square number.
Using , let it be contrived that:
:$E : BC = A^2 : FG^2$
and:
:$BC : CD = FG^2 : GH^2$
We have that:
:$E : BC = A^2 : FG^2$.
From :
:$A^2$ is commensurable with $FG^2$.
But $A^2$ is rational.
Therefore $FG^2$ is rational.
Therefore $FG$ is rational.
We have that $E : BC$ is not the ratio that a square number has to another square number.
Therefore $A^2 : FG^2$ is not the ratio that a square number has to another square number.
Therefore from :
:$A$ is incommensurable in length with $FG$.
We have that:
:$BC : CD = FG^2 : GH^2$
From :
:$FG^2$ is commensurable with $GH^2$.
But $FG^2$ is rational.
Therefore $GH^2$ is rational.
Therefore $GH$ is rational.
We have that $BC : CD$ is not the ratio that a square number has to another square number.
Therefore $FG^2 : GH^2$ is not the ratio that a square number has to another square number.
Therefore from :
:$FG$ is incommensurable in length with $GH$.
Both $FG$ and $GH$ are rational.
Therefore $FG$ and $GH$ rational straight lines which are commensurable in square only.
Therefore $FH$ is an apotome.
It remains to be shown that $FH$ is a third apotome.
We have that:
:$E : BC = A^2 : FG^2$.
and:
:$BC : CD = FG^2 : GH^2$
Therefore from :
:$E : CD = A^2 : HG^2$
But $E : CD$ is not the ratio that a square number has to another square number.
Therefore neither is $A^2 : GH^2$ the ratio that a square number has to another square number.
Therefore from :
:$A$ is incommensurable in length with $GH$.
Therefore neither $FG$ and $GH$ is incommensurable in length with the rational straight line $A$.
Now let $K^2 = FG^2 - GH^2$.
We have that:
:$BC : CD = FG^2 GH^2$
Therefore by :
:$BC : BD = FG^2 : K^2$
But $BC$ has to $BD$ the ratio that a square number has to another square number.
Therefore $FG^2$ has to $K^2$ the ratio that a square number has to another square number.
Therefore by :
:$FG$ is incommensurable in length with $K$.
So:
:$FG^2$ is greater than $GH^2$ by the square on a straight line which is commensurable in length with the rational straight line $A$.
Therefore, by definition, $FH$ is a third apotome.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of Third Binomial Straight Line

== Theorem ==
== Proof ==
:
Let $AC$ and $CB$ be straight lines constructed such that $AB = AC + CB$ is itself a straight line.
Using , let:
: $AB : BC = m^2 : n^2$
where $m$ and $n$ are numbers such that $m^2 - n^2$ is not square.
Let $D$ be a number which is not square.
Let $D$ have to neither $AB$ nor $AC$ the ratio that a square number has to another square number.
Let $E$ be a rational straight line.
Using , let:
:$D : AB = E^2 : FG^2$
where $FG$ is a straight line.
From :
:$E^2$ is commensurable with $FG^2$.
We have that $E$ is rational.
Therefore $FG$ is also rational.
Since:
: $D$ does not have to $AB$ the ratio that a square number has to another square number
then:
: $E^2$ does not have to $FG^2$ the ratio that a square number has to another square number.
From :
:$E$ is incommensurable in length with $FG$.
Using , let:
:$BA : AC = FG^2 : GH^2$
where $GH$ is a straight line constructed such that $FH = FG + GH$ is itself a straight line.
From :
:$FG^2$ is commensurable with $GH^2$.
We have that $FG$ is rational.
Therefore $GH$ is also rational.
Since:
: $BA$ does not have to $AC$ the ratio that a square number has to another square number
then:
: $FG^2$ does not have to $GH^2$ the ratio that a square number has to another square number.
From :
:$FG$ is incommensurable in length with $GH$.
Therefore $FG$ and $GH$ are rational straight lines which are commensurable in square only.
Therefore by definition $FH$ is a binomial.
We have that:
:$D : AB = E^2 : FG^2$
and
:$BA : AC = FG^2 : GH^2$
From :
:$D : AC = E^2 : GH^2$
But $D$ does not have to $AC$ the ratio that a square number has to another square number.
Therefore, neither does $E$ have to $GH$ the ratio that a square number has to another square number.
Therefore from :
:$E$ is incommensurable in length with $GH$.
We have that:
:$BA : AC = FG^2 : GH^2$
Therefore:
: $FG^2 > GH^2$
Let:
: $FG^2 = GH^2 + K^2$
for some $K$.
From :
:$AB : BC = FG : K$
But: 
: $AB : BC = m^2 : n^2$
where $m$ and $n$ are numbers.
Therefore:
:$FG : K = m^2 : n^2$
Therefore from :
:$FG$ is commensurable in length with $K$.
Therefore $FG^2$ is greater than $GH^2$ by the square on a straight line which is commensurable in length with $FG$.
But $FG$ and $GH$ are rational straight lines which are commensurable in square only.
Also, neither $FG$ nor $GH$ is commensurable in length with $E$.
Therefore $FH$ is a third binomial straight line.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of Third Proportional Straight Line

== Theorem ==
Given any two straight lines of given length $a$ and $b$, it is possible to construct a third straight line of length $c$ such that $a : b = b : c$.
== Construction ==
Let $AB, AC$ be the two given straight lines.
Let them be placed to contain any angle.
Let $AB$ be produced to $D$, and $AC$ be produced to $E$.
Let $BD$ be constructed equal to $AC$.
Join $BC$ and construct $DE$ parallel to $BC$.
Then $CE$ is the required third proportional line.
== Proof ==
:
We have that:
:$BC \parallel DE$
So from Parallel Transversal Theorem:
:$AB : BD = AC : CE$
But $BD = AC$ and so:
:$AB : AC = AC : CE$
as required.
== Sources ==
* : Book $\text{VI}$. Propositions
Category:Lines
Construction of Transitive Closure of Relation

== Theorem ==
Let $\RR$ be a relation.
Let $\RR^+$ be the relation which is constructed from $\RR$ as follows:
:$(1): \quad$ If $\tuple {a, b} \in \RR$, then $\tuple {a, b} \in \RR^+$
:$(2): \quad$ If $\tuple {a, b} \in \RR^+$ and $\tuple {b, c} \in \RR$, then $\tuple {a, c} \in \RR^+$
:$(3): \quad$ Nothing is in $\RR^+$ unless it so follows from $(1)$ and $(2)$.
Then $\RR^+$ is the transitive closure of $\RR$.
== Proof ==
Let $\tuple {x, y} \in \R^+$ from rules $(1)$ and $(2)$.
Then either:
:$\tuple {x, y}$ belongs there because $\RR \subseteq \RR^+$
or:
:$\tuple {x, y}$ belongs there, because if it were not then $\RR^+$ would not be transitive.
It remains to be shown that $\RR^+$ is in fact transitive.
Because of the method of construction, it follows that $\RR^+$ is the smallest transitive relation on $S$ which contains $\RR$ as a subset.
== Sources ==
* : Chapter $1$: Preliminaries: $1.5$ Relations: Closures of Relations
Category:Transitive Closures
Construction of Triangle from Given Lengths

== Theorem ==
Given three straight lines such that the sum of the lengths of any two of the lines is greater than the length of the third line, it is possible to construct a triangle having the lengths of these lines as its side lengths.
== Construction ==
:
Let the three straight lines from which we are going to construct the triangle be $a$, $b$, and $c$.
Let $D$ and $E$ be any distinct points.
Construct $DE$ and extend it beyond $E$.
We cut off a length $DF$ on $DE$ equal to $a$.
Similarly, we cut off $FG = b$ and $GH = c$ on $DE$.
Now we can construct a circle centered at $F$ with radius $DF$.
Similarly, we can construct a circle centered at $G$ with radius $GH$.
Call one of the intersections of the two circles $K$.
, let this be the top one in the accompanying diagram.
Finally, we can construct the segment $FK$.
$\triangle FGK$ is the required triangle.
== Proof ==
Since $F$ is the center of the circle with radius $FD$, it follows from that $DF = KF$, so $a = KF$ by Euclid's first common notion.
Since $G$ is the center of the circle with radius $GH$, it follows from that $GH = GK$, so $c = GK$ by Euclid's first common notion.
$FG = b$ by construction.
Therefore the lines $FK$, $FG$, and $GK$ are, respectively, equal to the lines $a$, $b$, and $c$, so $\triangle FGK$ is indeed the required triangle.
Note that the condition required of the lengths of the segments is the inequality shown in Proposition $20$: Sum of Two Sides of Triangle Greater than Third Side. Thus, this is a necessary condition for the construction of a triangle.
When first wrote the proof of this proposition in , he neglected to prove that the two circles described in the construction actually do intersect, just as he did in Proposition $1$: Construction of Equilateral Triangle.
== Sources ==
* : Book $\text{I}$. Propositions
Category:Triangles
Construction of that which produces Medial Whole with Medial Area is Unique

== Theorem ==
== Proof ==
:
Let $AB$ be a straight line which produces with a medial area a medial whole.
Let $BC$ be added to $AB$ such that:
: $AC$ and $CB$ are incommensurable in square
: $AC^2 + CB^2$ is medial
: $2 \cdot AC \cdot CB$ is a medial rectangle
: $2 \cdot AC \cdot CB$ is a incommensurable with $AC^2 + CB^2$.
It is to be proved that no other straight line can be added to $AB$ which fulfils these conditions.
Suppose $BD$, different from $BC$, can be added to $AB$ such that:
:$AD$ and $DB$ are incommensurable in square
:$AD^2 + DB^2$ is medial
: $2 \cdot AD \cdot DB$ is a medial rectangle
: $2 \cdot AD \cdot DB$ is a incommensurable with $AD^2 + DB^2$.
Let $EF$ be a rational straight line.
Let $EG = AC^2 + CB^2$ be applied to $EF$, producing $EM$ as breadth.
Let $HG = 2 \cdot AC \cdot CB$ be subtracted from $EG$ producing $HM$ as breadth.
Therefore from :
:the remainder $EL$ equals $AB^2$.
Thus $AB$ equals the "side" of $EL$.
Let $EI = AD^2 + DB^2$ be applied to $EF$, producing $EN$ as breadth.
But $EL = AB^2$.
Therefore the remainder $HI$ equals $2 \cdot AD \cdot DB$.
Therefore $AC^2 + CB^2$ are medial.
Also $AC^2 + CB^2 = EG$.
Therefore $EG$ is medial.
But $EG$ is applied to the rational straight line $EF$, producing $EM$ as breadth.
Therefore from :
:$EM$ is rational and incommensurable in length with $EF$.
We have that $2 \cdot AC \cdot CB$ is a medial rectangle.
But $2 \cdot AC \cdot CB = HG$.
Therefore $HG$ is medial.
Also, $HG$ has been applied to the rational straight line $EF$, producing $EM$ as breadth.
Therefore by :
:$HM$ is rational and incommensurable in length with $EF$.
We have that:
:$AC^2 + CB^2$ incommensurable with $2 \cdot AC \cdot CB$.
We also have that:
:$EG = AC^2 + CB^2$
and:
:$GH = 2 \cdot AC \cdot CB$
Therefore:
:$EG$ is incommensurable with $HG$.
But from :
:$EG : HG = EM : HM$
Therefore from :
:$EM$ is incommensurable in length with $MH$.
But both $EM$ and $MH$ are rational straight lines.
Therefore $EM$ and $MH$ are rational straight lines which are commensurable in square only.
Therefore $EH$ is an apotome, and $HM$ an annex to it.
Similarly it can be shown that $HN$ is also an annex to it.
Therefore we have different straight lines which are annexes to an apotome which are commensurable in square only to the whole.
From , this is impossible.
The result follows.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Construction of that which produces Medial Whole with Rational Area is Unique

== Theorem ==
== Proof ==
:
Let $AB$ be a straight line which produces with a rational area a medial whole.
Let $BC$ be added to $AB$ such that:
: $AC$ and $CB$ are incommensurable in square
: $AC^2 + CB^2$ is medial
: $2 \cdot AC \cdot CB$ is a rational] rectangle.
It is to be proved that no other straight line can be added to $AB$ which fulfils these conditions.
Suppose $BD$, different from $BC$, can be added to $AB$ such that:
:$AD$ and $DB$ are incommensurable in square
:$AD^2 + DB^2$ is [[Definition:Medial Area|medial]
:$2 \cdot AD \cdot DB$ is a rational] rectangle.
From :
:$AD^2 + DB^2 - 2 \cdot AD \cdot DB = AC^2 + CB^2 - 2 \cdot AC \cdot CB = AB^2$
Therefore:
:$AD^2 + DB^2 - AC^2 + CB^2 = 2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$
But $2 \cdot AD \cdot DB$ and $2 \cdot AC \cdot CB$ are both rational.
Therefore $2 \cdot AD \cdot DB - 2 \cdot AC \cdot CB$ is rational.
Therefore $AD^2 + DB^2 - AC^2 + CB^2$ is rational.
But from:
:
and:
::
both $AD^2 + DB^2$ and $AC^2 + CB^2$ is medial.
By this cannot happen.
Hence the result.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Constructive Dilemma/Formulation 1

== Theorem ==
== Proof 1 ==
== Proof by Truth Table ==
== Sources ==
* : Chapter $1$: The Propositional Calculus $1$: $3$ Conjunction and Disjunction: Exercise $1 \ \text{(h)}$
* : $\S 1.2.1$: Rules for natural deduction: Exercises $1.4: \ 2 \ \text{(d)}$
Category:Constructive Dilemma
Constructive Dilemma/Formulation 1/Proof by Truth Table

== Theorem ==
== Proof ==
We apply the Method of Truth Tables to the proposition.
As can be seen for all boolean interpretations by inspection, where the truth value under the main connective on the is $\T$, that under the one on the is also $\T$:
:$\begin{array}{|ccccccc||ccccccc|} \hline
(p & \implies & q) & \land & (r & \implies & s) & (p & \lor & r) & \implies & (q & \lor & s) \\
\hline
\F & \T & \F & \T & \F & \T & \F & \F & \F & \F & \T & \F & \F & \F \\
\F & \T & \F & \T & \F & \T & \T & \F & \F & \F & \T & \F & \T & \T \\
\F & \T & \F & \F & \T & \F & \F & \F & \T & \T & \F & \F & \F & \F \\
\F & \T & \F & \T & \T & \T & \T & \F & \T & \T & \T & \F & \T & \T \\
\F & \T & \T & \T & \F & \T & \F & \F & \F & \F & \T & \T & \T & \F \\
\F & \T & \T & \T & \F & \T & \T & \F & \F & \F & \T & \T & \T & \T \\
\F & \T & \T & \F & \T & \F & \F & \F & \T & \T & \T & \T & \T & \F \\
\F & \T & \T & \T & \T & \T & \T & \F & \T & \T & \T & \T & \T & \T \\
\T & \F & \F & \F & \F & \T & \F & \T & \T & \F & \F & \F & \F & \F \\
\T & \F & \F & \F & \F & \T & \T & \T & \T & \F & \T & \F & \T & \T \\
\T & \F & \F & \F & \T & \F & \F & \T & \T & \T & \F & \F & \F & \F \\
\T & \F & \F & \F & \T & \T & \T & \T & \T & \T & \T & \F & \T & \T \\
\T & \T & \T & \T & \F & \T & \F & \T & \T & \F & \T & \T & \T & \F \\
\T & \T & \T & \T & \F & \T & \T & \T & \T & \F & \T & \T & \T & \T \\
\T & \T & \T & \F & \T & \F & \F & \T & \T & \T & \T & \T & \T & \F \\
\T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T & \T \\
\hline
\end{array}$
Hence the result.
Note that the two formulas are not equivalent, as the relevant columns do not match exactly.
Category:Truth Table Proofs
Category:Constructive Dilemma
Constructive Dilemma/Formulation 2

== Theorem ==
== Proof ==
== Sources ==
* : $2$ Arguments Containing Compound Statements: $2.3$: Argument Forms and Truth Tables: Exercise $\text{I} \ \mathbf k.$
* : $3$: The Method of Deduction: $3.1$: Formal Proof of Validity: Rules of Inference: $5.$
* : $\S \text{II}$: The Logic of Statements $(2): \ 1$: Decision procedures and proofs: $3$
Category:Constructive Dilemma
Constructive Dilemma/Formulation 3

== Theorem ==
: $\vdash \paren {\paren {p \lor r} \land \paren {p \implies q} \land \paren {r \implies s} } \implies \paren {q \lor s}$
== Proof ==
== Sources ==
* : $\text{II}$: 'AND', 'OR', 'IF AND ONLY IF': $\S 5$: Theorem $\text{T48}$
Category:Constructive Dilemma
Constructive Dilemma for Join Semilattices

== Theorem ==
Let $\struct {S, \vee, \preceq}$ be a join semilattice.
Let $a, b, c, d \in S$.
Let $a \preceq b$.
Let $c \preceq d$.
Then $\paren {a \vee c} \preceq \paren {b \vee d}$.
== Proof ==
By Join Semilattice is Ordered Structure, $\preceq$ is compatible with $\vee$.
By the definition of ordering, $\preceq$ is transitive.
Thus the theorem holds by Operating on Transitive Relationships Compatible with Operation.
Category:Lattice Theory
Constructive Proof/Examples/Nonconstructive Proof

== Example of Nonconstructive Proof ==
By way of illustration, this is the shape of a proof which is not constructive:
:It is false that every number $n$ lacks the property $P$.
:Therefore there exists at least one number $n_0$ that has property $P$.
Unless an example of such a $n_0$ can be constructed, such an argument is not allowed in a constructive proof.
== Sources ==
* 
Category:Examples of Constructive Proofs
Content of Cayley-Menger Determinant

== Theorem ==
Let $V_j$ denote the Cayley-Menger determinant for a $j$-simplex $S$ in $\R^n$.
Let the vertices of $S$ be $v_1, v_2, \ldots, v_{j + 1}$.
Let $B = \sqbrk {\beta_{i j} }$ denote the order $j + 1$ matrix given by:
:$\beta_{ij} = {\size {v_i - v_j}_2}^2$
where $\size {v_i - v_j}_2$ is the vector $2$-norm of the vector $v_i - v_j$.
Then the content $V_j$ is given by:
:$\map {V_j^2} S = \dfrac {\paren {-1}^{j + 1} } {2^j \paren {j!}^2} \det C $
where $C$ is the order $j + 1$ matrix obtained from $B$ by bordering $B$ with a top row $\tuple {0, 1, \ldots, 1}$ and a left column $\tuple {0, 1, \ldots, 1}^\intercal$. 
== Proof ==
== Also see ==
* Tartaglia's Formula
* Heron's Formula
== Sources ==
* 
Category:Linear Algebra
Category:Distance Geometry
Category:Examples of Determinants
Category:Cayley-Menger Determinant
Content of Finite Set of Rectangles not Less than Covered Subrectangles

== Theorem ==
Let $S$ be a finite set of closed $n$-rectangles.
Let $\RR$ be a closed $n$-rectangle that contains every rectangle in $S$.
Let:
:$P = \tuple {P_1, \dotsc, P_n}$
be a subdivision of $\RR$ such that for every:
:$\ds R = \prod_{i \mathop = 1}^n \closedint {a_i} {b_i} \in S$
we have that:
:$a_i, b_i \in P_i$
That is, the endpoints $a_i$ and $b_i$ are elements of the finite subdivision $P_i$.
Let $Q$ be the set of subrectangles of $P$ that are contained in some rectangle in $S$.
Then:
:$\ds \sum_{R \mathop \in S} \map V R \ge \sum_{R \mathop \in Q} \map V R$
where $\map V R$ denotes the content of the rectangle $R$.
== Proof ==
Content of Monic Polynomial

== Theorem ==
Let $f$ be a polynomial with rational coefficients.
Let $\cont f$ be the content of $f$.
If $f$ is monic, then $\cont f = \dfrac 1 n$ for some integer $n$.
== Proof ==
Since $f$ is monic, it can be written as:
:$f = X^r + \cdots + a_1 X + a_0$
Let $n = \inf \set {n \in \N : n f \in \Z \sqbrk X}$.
Let $d = \cont {n f}$.
Then by definition of content:
:$d = \gcd \set {n, n a_{r - 1}, \ldots, n a_1, n a_0}$
Therefore, by definition of GCD, $d$ divides $n$.
So say $n = k d$ with $k \in \Z$.
Then: 
:$\cont f = \dfrac d {k d} = \dfrac 1 k$
as required.
Category:Content of Polynomial
Category:Monic Polynomials
Content of Polynomial in Dedekind Domain is Multiplicative

== Theorem ==
Let $R$ be a Dedekind domain.
Let $f, g \in R \sqbrk X$ be polynomials.
Let $\cont f$ denote the content of $f$.
Then $\cont {f g} = \cont f \cont g$ is the product of $\cont f$ and $\cont g$.
== Proof ==
Category:Content of Polynomial
Category:Gauss's Lemma (Polynomial Theory)