nyuuzyou/wikis huggingface dataset:
Square of Sum/Geometric Proof

== Theorem ==
== Proof ==
:
Let the straight line $AB$ be cut at random at $C$.
Construct the square $ADEB$ on $AB$ and join $DB$.
Construct $CF$ parallel to $AD$ through $C$.
Construct $HK$ parallel to $AB$ through $G$.
From Parallelism implies Equal Alternate Angles, $\angle CGB = \angle ADB$.
We have that $BA = AD$.
So from Isosceles Triangle has Two Equal Angles:
:$\angle ADB = \angle ABD$
So $\angle CGB = \angle CBG$, and so from Triangle with Two Equal Angles is Isosceles, $BC = CG$.
From Opposite Sides and Angles of Parallelogram are Equal, we have $CB = GK$ and $CG = KB$ and so $CGKB$ is equilateral.
Now, since $CG \parallel BK$, from Parallelism implies Supplementary Interior Angles we have that $\angle KBC$ and $\angle GCB$ are supplementary.
But as $\angle KBC$ is a right angle, $\angle BCG$ is also a right angle.
So from Opposite Sides and Angles of Parallelogram are Equal, $\angle CGK$ and $\angle GKB$ are also right angles.
So $CGKB$, described on $CB$, is right-angled.
As $CGKB$ is also equilateral, by definition it is square.
For the same reason, $HGDF$ is also square, and described on $HG$, which equals $AB$.
So $HGDF$ and $CGKB$ are the squares on $AC$ and $CB$.
Now we have from Complements of Parallelograms are Equal that $\Box ACGH = \Box GKEF$.
But $\Box ACGH$ is the rectangle contained by $AC$ and $CB$, as $CB = GC$.
So $\Box GKEF$ is also equal to the rectangle contained by $AC$ and $CB$.
But the squares $HGFD$ and $CBKG$ are equal to the squares on $AC$ and $CB$.
So the four areas $HGFD$, $CBKG$, $ACGH$ and $GKEF$ are equal to:
:the squares on $AC$ and $CB$
and:
:twice the rectangle contained by $AC$ and $CB$.
But $HGFD$, $CBKG$, $ACGH$ and $GKEF$ are also equal to the square on $AB$.
Hence the result.
== Sources ==
* : Book $\text{II}$. Propositions
Category:Squares
Category:Rectangles
Category:Square of Sum
Square of Sum less Square/Geometric Proof

== Theorem ==
== Proof ==
Let $AB$ be bisected at $C$, and let $BD$ be added to it in a straight line.
Then the rectangle contained by $AD$ and $BD$ together with the square on $BC$ equals the square on $CD$.
:
The proof is as follows.
Construct the square $CEFD$ on $CD$, and join $DE$.
Construct $BG$ parallel to $CE$ through $B$, and let $BG$ cross $DE$ at $H$.
Construct $KM$ parallel to $AD$ through $H$.
Construct $AK$ parallel to $DF$ through $A$.
As $AC = CB$, from Parallelograms with Equal Base and Same Height have Equal Area we have that $\Box ACLK = \Box CBHL$.
From Complements of Parallelograms are Equal, $\Box CBHL = \Box HMFG$.
So $\Box ACLK = \Box HMFG$.
Add $\Box CDML$ to each.
So the whole of $\Box ADMK$ is equal to the gnomon $CDFGHL$.
But $\Box ADMK$ is the rectangle contained by $AD$ and $BD$, because $BD = DM$.
So the gnomon $CDFGHL$ is equal in area to the rectangle contained by $AD$ and $BD$.
Now $\Box LHGE$ is equal to the square on $BC$.
Add $\Box LHGE$ to each of the gnomon $CDFGHL$ and $\Box ADMK$.
Then the gnomon $CDFGHL$ together with $\Box LHGE$ equals the rectangle contained by $AD$ and $BD$ and the square on $BC$.
But the gnomon $CDFGHL$ together with $\Box LHGE$ is the square $CDFE$.
Hence the result.
== Sources ==
* : Book $\text{II}$. Propositions
Category:Square of Sum less Square
Square of Sum of Vectors

== Theorem ==
Let $\mathbf a$ and $\mathbf b$ be vector quantities.
Then:
:$\paren {\mathbf a + \mathbf b}^2 = \mathbf a^2 + 2 \mathbf a \cdot \mathbf b + \mathbf b^2$
where:
:$\mathbf a \cdot \mathbf b$ denotes dot product
:$\mathbf a^2$ denotes the square of $\mathbf a$, that is: $\mathbf a \cdot \mathbf a$.
== Proof ==
== Also see ==
* Dot Product of Sum with Difference of Vectors
* Square of Sum
== Sources ==
* : Introduction: Vector Notation and Formulae: Products of Vectors
Category:Vector Addition
Category:Dot Product
Square of Sum with Double/Algebraic Proof 2

== Theorem ==
== Proof ==
A direct application of the Binomial Theorem:
:$\ds \forall n \in \Z_{\ge 0}: \paren {x + y}^n = \sum_{k \mathop = 0}^n \binom n k x^{n - k} y^k$
putting $n = 2, x = a, y = 2 b$.
Category:Square of Sum with Double
Square of Sum with Double/Geometric Proof

== Theorem ==
== Proof ==
That is: $4 \paren {a + b} b + a^2 = \paren {a + 2 b}^2$.
:
Let the straight line $AB$ be cut at random at $C$.
Then four times the rectangle contained by $AB$ and $BC$ together with the square on $AC$ equals the square on $AB$ and $BC$ as a single straight line.
The proof is as follows.
Produce $AB$ to $D$ where $BD = BC$.
Construct the square $AEFD$ on $AD$ and join $DE$.
Draw the given figure above, in the same manner as in Square of Sum.
By Parallelograms with Equal Base and Same Height have Equal Area:
: $\Box CBKG = \Box BDNK$
: $\Box KGQR = \Box KNPR$
From Complements of Parallelograms are Equal:
: $\Box CBKG = \Box KNPR$
and so:
: $\Box BDNK = \Box KGQR$
So all of them are equal:
: $\Box CBKG = \Box BDNK = \Box KGQR = \Box KNPR$
So the four of them together equal $4 \Box CBKG$.
Similarly, by Parallelograms with Equal Base and Same Height have Equal Area:
: $\Box ACGM = \Box MGQO$
: $\Box QRLH = \Box RPFL$
From Complements of Parallelograms are Equal:
:$\Box MGQO = \Box QRLH$
and:
: $\Box ACGM = \Box RPFL$
So all of them are equal:
: $\Box ACGM = \Box MGQO = \Box QRLH = \Box RPFL$
So the four of them together equal $4 \Box ACGM$.
Adding all these eight areas together, we see that the gnomon $STU$ equals $4 \Box CBKG + 4 \Box ACGM = 4 \Box ABKM$.
But $\Box ABKM$ is the rectangle contained by $AB$ and $BC$, as $BC = BK$.
So four times the rectangle contained by $AB$ and $BC$ equals the area of the gnomon $STU$.
Now we add $\Box QOEH$ to each.
Note that $\Box QOEH$ equals the square on $AC$.
So four times the rectangle contained by $AB$ and $BC$, together with the square on $AC$, equals the gnomon $STU$ together with $\Box QOEH$.
But the gnomon $STU$ together with $\Box QOEH$ form the whole square $ADFE$, which is the square on $AB$ and $BC$ as a single straight line.
Hence the result.
== Sources ==
* : Book $\text{II}$. Propositions
Category:Square of Sum with Double
Square of Triangular Number equals Sum of Sequence of Cubes

== Theorem ==
:$\ds \sum_{i \mathop = 1}^n i^3 = {T_n}^2$
where $T_n$ denotes the $n$th triangular number.
== Proof 1 ==
== Proof 2 ==
Category:Triangular Numbers
Category:Cube Numbers
Category:Square of Triangular Number equals Sum of Sequence of Cubes
Square of Triangular Number equals Sum of Sequence of Cubes/Proof 2

== Theorem ==
== Proof ==
The proof proceeds by induction.
For all $n \in \Z_{> 0}$, let $\map P n$ be the proposition:
:$\ds \sum_{i \mathop = 1}^n i^3 = {T_n}^2$
=== Basis for the Induction ===
$\map P 1$ is the case:
Thus $\map P 1$ is seen to hold.
This is the basis for the induction.
=== Induction Hypothesis ===
Now it needs to be shown that, if $\map P k$ is true, where $k \ge 2$, then it logically follows that $\map P {k + 1}$ is true.
So this is the induction hypothesis:
:$\ds \sum_{i \mathop = 1}^k i^3 = {T_k}^2$
from which it is to be shown that:
:$\ds \sum_{i \mathop = 1}^{k + 1} i^3 = {T_{k + 1} }^2$
=== Induction Step ===
This is the induction step:
So $\map P k \implies \map P {k + 1}$ and the result follows by the Principle of Mathematical Induction.
Therefore:
:$\forall n \in \Z_{> 0}: \ds \sum_{i \mathop = 1}^n i^3 = {T_n}^2$
== Sources ==
* : $15$
* : $15$
Category:Square of Triangular Number equals Sum of Sequence of Cubes
Square of V Operator on Hilbert Space

== Theorem ==
Let $\struct {\HH, \innerprod \cdot \cdot}$ be a Hilbert space over $\C$.
Let $\struct {\HH \times \HH, \innerprod \cdot \cdot_{\HH \times \HH} }$ be the Hilbert space direct sum of $\HH$ with itself.
Define $V : \HH \times \HH \to \HH \times \HH$ by:
:$\map V {x, y} = \tuple {-y, x}$
for each $\tuple {x, y} \in \HH \times \HH$.
Then $V^2 = -I$. 
== Proof ==
For each $x, y \in \HH$ we have:
== Sources ==
* : $13.7$: Graphs
Category:V Operators on Hilbert Spaces
Square of Vandermonde Matrix

== Theorem ==
The square of the Vandermonde matrix of order $n$:
: $\mathbf V = \begin{bmatrix}
 x_1 & x_2 & \cdots & x_n \\
 x_1^2 & x_2^2 & \cdots & x_n^2 \\
\vdots & \vdots & \ddots & \vdots \\
 x_1^n & x_2^n & \cdots & x_n^n
\end{bmatrix}$
is symmetrical in $x_1, \ldots, x_n$.
== Proof ==
Category:Vandermonde Matrices
Square of Vector Cross Product

== Theorem ==
Let $\mathbf a$ and $\mathbf b$ be vectors in a vector space $\mathbf V$ of $3$ dimensions.
Let $\mathbf a \times \mathbf b$ denote the vector cross product of $\mathbf a$ with $\mathbf b$.
Then:
:$\paren {\mathbf a \times \mathbf b}^2 = \mathbf a^2 \mathbf b^2 - \paren {\mathbf a \cdot \mathbf b}^2$
where:
:$\paren {\mathbf a \times \mathbf b}^2$ denotes the square of $\mathbf a \times \mathbf b$
:$\mathbf a \cdot \mathbf b$ denotes the dot product of $\mathbf a \times \mathbf b$.
== Proof 1 ==
== Proof 2 ==
The following proof is valid in the context of a Cartesian $3$-space:
Category:Square of Vector Cross Product
Category:Vector Cross Product
Category:Dot Product
Square of Vector Quantity in Coordinate Form

== Theorem ==
Let $\mathbf a$ be a vector in a vector space $\mathbf V$ of $n$ dimensions:
$\ds \mathbf a = \sum_{k \mathop = 1}^n a_k \mathbf e_k$
where $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ is the standard ordered basis of $\mathbf V$.
Then:
:$\ds \mathbf a^2 = \sum_{k \mathop = 1}^n a_k^2$
where $\mathbf a^2$ denotes the square of $\mathbf a$.
== Proof ==
By definition of square of $\mathbf a$:
:$\mathbf a^2 = \mathbf a \cdot \mathbf a$
By definition of dot product:
:$\ds \mathbf a \cdot \mathbf a = a_1 a_1 + a_2 a_2 + \cdots + a_n a_n = \sum_{k \mathop = 1}^n a_k^2$
== Sources ==
* : Introduction: Vector Notation and Formulae: Products of Vectors
Category:Dot Product
Square of n such that 2n-1 is Composite is not Sum of Square and Prime

== Theorem ==
Let $n^2$ be a square such that $2 n - 1$ is composite.
Then $n^2$ cannot be expressed as the sum of a square and a prime.
== Proof ==
The case where $n = 1$ is trivial, as there are no prime numbers less than $1$.
Let $n, m \in \Z$ be integers such that $n > 1$.
Let $n^2 = m^2 + p$ where $p$ is prime.
Then:
So if $2 n - 1$ is composite, there exists no prime $p$ such that $n^2 = m^2 + p$.
Further, if such a $p$ does exist, then $m = n - 1$, and so:
:$n^2 = \paren {n - 1}^2 + p$
Category:Numbers not Sum of Square and Prime
Square on Binomial Straight Line applied to Rational Straight Line

== Theorem ==
=== Lemma ===
== Proof ==
:
Let $AB$ be a binomial straight line divided into its terms at $C$.
Let $AC > CB$.
Let $DE$ be a rational straight line.
Let $DEFG$ equal to $AB^2$ be applied to $DE$ producing $DG$ as its breadth.
It is to be demonstrated that $DG$ is a first binomial straight line.
From :
:$AB^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$
Let the rectangle $DH$ be applied to $DE$ such that $DH = AC^2$.
Let the rectangle $KL$ be applied to $DE$ such that $KL = BC^2$.
Then the rectangle $MF$ is equal to $2 \cdot AC \cdot CB$.
Let $MG$ be bisected at $N$.
Let $NO$ be drawn parallel to $ML$ (or $GF$, which is the same thing).
Therefore each of the rectangles $MO$ and $NF$ equals $AC \cdot CB$.
We have that $AB$ is a binomial which has been divided into its terms at $C$.
Therefore, by definition, $AC$ and $CB$ are rational straight lines which are commensurable in square only.
Therefore $AC^2$ and $CB^2$ are commensurable and rational.
Thus by :
:$AC^2 + CB^2$ is a rational area.
But $AC^2 + CB^2 = DL$.
Therefore $DL$ is a rational area.
We have that $DL$ is applied to the rational straight line $DE$.
Therefore from :
:$DM$ is rational and commensurable in length with $DE$.
We have that $AC$ and $CB$ are rational straight lines which are commensurable in square only.
So from :
:$2 \cdot AC \cdot CB = MF$ is medial.
We have that $MF$ is applied to the rational straight line $ML$.
Therefore from :
:$MG$ is also rational and incommensurable in length with $DE$.
But $MD$ is also rational and commensurable in length with $DE$.
Therefore from :
:$DM$ is incommensurable in length with $MG$.
But $DM$ and $MG$ are both rational straight lines which are commensurable in square only.
Therefore, by definition, $DG$ is binomial.
It remains to be proved that $DG$ is first binomial.
From :
:$AC \cdot CB$ is a mean proportional between $AC^2$ and $CB^2$.
Therefore $MO$ is a mean proportional between $DH$ and $KL$.
Therefore by :
:$DK : MN = MN : MK$
Therefore from :
:$DK \cdot KM = MN^2$
We have that $AC^2$ is commensurable with $CB^2$.
Therefore $DH$ is commensurable with $KL$.
So from:
:
and:
:
it follows that:
:$DK$ is commensurable in length with $KM$.
From :
:$AC^2 + CB^2 > 2 \cdot AC \cdot CB$
Therefore:
:$DL > MF$
and so from :
:$DM > MG$
And:
:$DK \cdot KM = MN^2 = \dfrac {MG^2} 4$
and:
:$DK$ is commensurable in length with $KM$.
Thus from :
:$DM^2$ is greater than $MG^2$ by the square on a straight line commensurable in length with $DM$.
We also have that:
:$DM$ and $MG$ are rational
and:
:the greater term $DM$ is commensurable in length with the rational straight line $DE$.
Therefore by definition $DG$ is a first binomial straight line.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Square on Binomial Straight Line applied to Rational Straight Line/Lemma

== Lemma to Square on Binomial Straight Line applied to Rational Straight Line ==
:$\forall x, y \in \R: x \ne y \implies x^2 + y^2 > 2 x y$
== Proof ==
:
Let $AB$ be a straight line.
Let $AB$ be cut into unequal parts at $C$ such that $AC > CB$.
Let $AB$ be bisected at $D$.
From :
:$AC \cdot CB + CD^2 = AD^2$
So:
:$AC \cdot CB < AD^2$
and so:
:$2 \cdot AC \cdot CB < 2 \cdot AD^2$
But from :
:$AC^2 + CB^2 = 2 \cdot AD^2 + DC^2$
Therefore:
:$AC^2 + CB^2 > 2 \cdot AC \cdot CB$
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Square on First Apotome of Medial Straight Line applied to Rational Straight Line

== Theorem ==
== Proof ==
:
Let $AB$ be a first apotome of a medial straight line.
Let $CD$ be a rational straight line.
Let the rectangle $CE$ be applied to $CD$ equal to $AB^2$ producing $CF$ as breadth.
It is to be demonstrated that $CF$ is a second apotome.
Let $BG$ be the annex to $AB$.
Therefore, by definition, $AG$ and $GB$ are medial straight lines which are commensurable in square only which contain a rational rectangle.
Let the rectangle $CH$ be applied to $CD$ equal to the square on $AG$, producing $CK$ as breadth.
Let the rectangle $KL$ be applied to $CD$ equal to the square on $BG$, producing $KM$ as breadth.
Then the whole $CL$ is equal to the squares on $AG$ and $GB$.
From:
:
and:
:
it follows that:
:$CL$ is medial.
We have that $CL$ is applied to the rational straight line $CD$ producing $CM$ as breadth.
Therefore by :
:$CM$ is rational and incommensurable in length with $CD$.
We have that:
:$CL = AG^2 + GB^2$
and:
:$AB^2 = CE$
Therefore by :
:$2 \cdot AG \cdot GB = FL$
But $2 \cdot AG \cdot GB$ is rational.
Therefore $FL$ is rational.
Also $FL$ is applied to the rational straight line $FE$, producing $FM$ as breadth.
Therefore from :
:$FM$ is rational and commensurable in length with $CD$.
We have that:
:$AG^2 + GB^2$ is medial
while:
:$2 \cdot AG \cdot GB$ is rational.
Therefore $CL$ is incommensurable with $FL$.
But from :
:$CL : FL = CM : FM$
Therefore by :
:$CM$ is incommensurable in length with $FM$.
But both $CM$ and $FM$ are rational.
Therefore $CM$ and $FM$ are rational straight lines which are commensurable in square only.
Therefore, by definition, $CF$ is an apotome.
It remains to be shown that $CF$ is a second apotome.
Let $FM$ be bisected at the point $N$.
Let $NO$ be drawn through $N$ parallel to $CD$.
Therefore each of the rectangles $FO$ and $LN$ is equal to the rectangle contained by $AB$ and $GB$.
We have that:
:the rectangle contained by $AG$ and $GB$ is a mean proportional between the squares on $AG$ and $GB$.
and:
:$CH = AG^2$
and:
:$KL = BG^2$
and:
:$NL = AG \cdot GB$
Therefore $NL$ is a mean proportional between $CH$ and $KL$.
Therefore:
:$CH : NL = NL : KL$
But we also have:
:$CH : NL = CK : NM$
and from :
:$NL : KL = NM : KM$
Therefore by :
:$CK \cdot KM = NM^2 = \dfrac {FM^2} 4$
We have that:
:$CM$ and $MF$ are unequal straight lines
and:
:the rectangle $CK \cdot KM$ has been applied to $CM$ equal to $\dfrac {FM^2} 4$ and deficient by a square figure
while:
:$CK$ is commensurable with $KM$.
Therefore from :
:$CM^2$ is greater than $MF^2$ by the square on a straight line which is commensurable in length with $CM$.
Also the annex $FM$ is commensurable in length with the rational straight line $CD$.
Therefore, by definition, $CF$ is a second apotome.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Square on First Bimedial Straight Line applied to Rational Straight Line

== Theorem ==
== Proof ==
:
Let $AB$ be a first bimedial straight line divided into its medials at $C$.
Let $AC > CB$.
Let $DE$ be a rational straight line.
Let $DEFG$ equal to $AB^2$ be applied to $DE$ producing $DG$ as its breadth.
It is to be demonstrated that $DG$ is a second binomial straight line.
From :
:$AB^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$
Let the rectangle $DH$ be applied to $DE$ such that $DH = AC^2$.
Let the rectangle $KL$ be applied to $DE$ such that $KL = BC^2$.
Then the rectangle $MF$ is equal to $2 \cdot AC \cdot CB$.
Let $MG$ be bisected at $N$.
Let $NO$ be drawn parallel to $ML$ (or $GF$, which is the same thing).
Therefore each of the rectangles $MO$ and $NF$ equals $AC \cdot CB$.
We have that $AB$ is a first bimedial which has been divided into its medials at $C$.
Therefore, by definition, $AC$ and $CB$ are medial straight lines which are commensurable in square only such that $AC \cdot CB$ is a rational rectangle.
Thus, by definition, $AC^2$ and $CB^2$ are also medial.
From:
:
and:
:
it follows that:
:$DL$ is medial.
We have that $DL$ has been applied to the rational straight line $DE$.
Therefore from :
:$MD$ is rational and incommensurable in length with $DE$.
As $2 \cdot AC \cdot CB$ is rational, so is $MF$.
We have that $MF$ is applied to the rational straight line $ML$.
Therefore from :
:$MG$ is rational straight line and commensurable in length with $ML$, which equals $DE$.
So by :
:$DM$ is incommensurable in length with $MG$.
But $DM$ and $MG$ are both rational.
Therefore $DM$ and $MG$ are rational straight lines which are commensurable in square only.
Therefore by definition $DG$ is binomial.
It remains to be proved that $DG$ is a second binomial straight line.
From :
:$AC^2 + CB^2 > 2 \cdot AC \cdot CB$
Therefore $DL > MF$.
From :
:$DM > MG$
Since:
:$AC^2$ is commensurable with $CB^2$
it follows that:
:$DH$ is commensurable with $KL$.
So from:
:
and:
:
it follows that:
:$DK$ is commensurable in length with $KM$.
Also $DK \cdot KM = MN^2$.
Thus from :
:$DM^2$ is greater than $MG^2$ by the square on a straight line commensurable in length with $DM$.
Also:
:$MG$ is commensurable in length with $DE$.
Therefore $DG$ is a second binomial straight line.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Square on Major Straight Line applied to Rational Straight Line

== Theorem ==
== Proof ==
:
Let $AB$ be a major straight line divided at $C$.
Let $AC > CB$.
Let $DE$ be a rational straight line.
Let $DEFG$ equal to $AB^2$ be applied to $DE$ producing $DG$ as its breadth.
It is to be demonstrated that $DG$ is a fourth binomial straight line.
From :
:$AB^2 = AC^2 + CB^2 + 2 \cdot AC \cdot CB$
Let the rectangle $DH$ be applied to $DE$ such that $DH = AC^2$.
Let the rectangle $KL$ be applied to $DE$ such that $KL = BC^2$.
Then the rectangle $MF$ is equal to $2 \cdot AC \cdot CB$.
Let $MG$ be bisected at $N$.
Let $NO$ be drawn parallel to $ML$ (or $GF$, which is the same thing).
Therefore each of the rectangles $MO$ and $NF$ equals $AC \cdot CB$.
We have that $AB$ is a major which has been divided at $C$.
Therefore, by definition, $AC$ and $CB$ are straight lines which are incommensurable in square such that:
:$AC^2 + CB^2$ is rational
:$AC \cdot CB$ is a medial rectangle.
Since $AC^2 + CB^2$ is rational, it follows that $DL$ is rational.
Therefore from :
:$DM$ is rational and commensurable in length with $DE$.
We have that:
:$2 \cdot AC \cdot CB$, which equals $MF$, is medial.
We also have that $MF$ is applied to the rational straight line $ML$.
Therefore from :
:$MG$ is rational and incommensurable in length with $DE$.
Therefore from :
:$DM$ is incommensurable in length with $MG$.
That is, $DM$ and $MG$ are rational straight lines which are incommensurable in length with each other.
Therefore by definition $DG$ is binomial.
It remains to be proved that $DG$ is a fourth binomial straight line.
From :
:$AC^2 + CB^2 > 2 \cdot AC \cdot CB$
Therefore $DL > MF$.
From :
:$DM > MG$
Also $DK \cdot KM = MN^2$.
Because:
:$AC^2$ is incommensurable with $CB^2$
it follows that:
:$DH$ is incommensurable with $KL$.
So from:
:
and:
:
it follows that:
:$DK$ is incommensurable in length with $KM$.
So from :
:$DM^2$ is greater than $MG^2$ by the square on a straight line incommensurable in length with $DM$.
Also:
:$DM$ and $MG$ are rational straight lines which are commensurable in square only with each other
and:
:$DM$ is commensurable in length with the rational straight line $DE$.
Therefore $DG$ is a fourth binomial straight line.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Square on Medial Straight Line

== Theorem ==
=== Lemma ===
== Proof ==
Let $A = \rho \sqrt [4] k$ be a medial straight line.
Let $BC = \sigma$ be a rational straight line.
The square on $A$ is $\rho^2 \sqrt k$.
The breadth $CD$ of the rectangle whose area is $\rho^2 \sqrt k$ and whose side is $BC$ is:
where $m$ and $n$ are integers.
Thus $CD$ can be expressed in the form:
:$CD = \sqrt {k'} \sigma$
which is commensurable in square only with $\sigma$.
By definition, therefore, $CD$ is incommensurable in length with $BC$.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory
Square on Medial Straight Line/Lemma

== Lemma for Square on Medial Straight Line ==
Algebraically:
:$a : b = a^2 : a b$
== Proof ==
:
Let $FE$ and $EG$ be straight lines.
Let the square $DF$ be described on $FE$.
Let the rectangle $GD$ be completed.
From Areas of Triangles and Parallelograms Proportional to Base:
:$FE : EG = FD : DG$
We have that $DG$ is the rectangle contained by $DE$ and $EG$.
But as $DF$ is a square, then $DE = FE$.
Thus $DG$ is the rectangle contained by $FE$ and $EG$.
So as $FE$ is to $EG$, so is the square on $EF$ to the rectangle contained by $FE$ and $EG$.
== Sources ==
* : Book $\text{X}$. Propositions
Category:Euclidean Number Theory