hcju/proofwikiqa huggingface dataset:
Let $a \in \N$ be an even perfect number. We can extract the highest power of $2$ out of $a$ that we can, and write $a$ in the form: :$a = m 2^{n-1}$ where $n \ge 2$ and $m$ is odd. Since $a$ is perfect and therefore $\map \sigma a = 2 a$: {{begin-eqn}}      {{eqn| l = m 2^n      | r = 2 a      | c = }} {{eqn| r = \map \sigma a      | c = }} {{eqn| r = \map \sigma {m 2^{n - 1} }      | c = }} {{eqn| r = \map \sigma m \map \sigma {2^{n - 1} }      | c = Sigma Function is Multiplicative }} {{eqn| r = \map \sigma m {2^n - 1}      | c = Sigma Function of Power of Prime }} {{end-eqn}} So: :$\map \sigma m = \dfrac {m 2^n} {2^n - 1}$ But $\map \sigma m$ is an integer and so $2^n - 1$ divides $m 2^n$. From Consecutive Integers are Coprime, $2^n$ and $2^n - 1$ are coprime. So from Euclid's Lemma $2^n - 1$ divides $m$. Thus $\dfrac m {2^n - 1}$ divides $m$. Since $2^n - 1 \ge 3$ it follows that: :$\dfrac m {2^n - 1} < m$ Now we can express $\map \sigma m$ as: :$\map \sigma m = \dfrac {m 2^n} {2^n - 1} = m + \dfrac m {2^n - 1}$ This means that the sum of all the divisors of $m$ is equal to $m$ itself plus one other divisor of $m$. Hence $m$ must have exactly two divisors, so it must be prime by definition. This means that the other divisor of $m$, apart from $m$ itself, must be $1$. That is: :$\dfrac m {2^n - 1} = 1$ Hence the result. {{qed}}
Let $A, B, C$ be three commensurable magnitudes. From Greatest Common Measure of Commensurable Magnitudes, let the greatest common measure of $A$ and $B$ be $D$. Either $D$ measures $C$ or it does not. Let $D$ measure $C$. Since: :$D$ measures $C$ and: :$D$ measures $A$ and $B$ then: :$D$ is a common measure for $A, B$ and $C$. As no magnitude can measure a smaller magnitude, it follows that $D$ is the greatest common measure of $A, B$ and $C$. Let $D$ not measure $C$. As $A$, $B$ and $C$ are commensurable, then some magnitude $E$ will measure them by definition. $E$ will measure both $A$ and $B$. From the porism to Greatest Common Measure of Commensurable Magnitudes, $E$ also measures their greatest common measure $D$. But $E$ also measures $C$. Therefore $E$ will measure $C$ and $D$. Therefore $C$ and $D$ are commensurable. So from Greatest Common Measure of Commensurable Magnitudes, let $E$ be the greatest common measure of $C$ and $D$. Since: :$E$ measures $D$ and: :$D$ measures $A$ and $B$ then: :$E$ measures $A$ and $B$. But $E$ measures $C$ also. So $E$ is a common measure for $A, B$ and $C$. Suppose there were some magnitude $F$ which is greater than $E$ which is also a common measure for $A, B$ and $C$. Since $F$ measures $A, B$ and $C$, it measures $A$ and $B$. From the porism to Greatest Common Measure of Commensurable Magnitudes, $F$ also measures the greatest common measure of $A$ and $B$. That is, $F$ measures $C$ But $F$ measures $D$. From the porism to Greatest Common Measure of Commensurable Magnitudes, $F$ measures the greatest common measure of $D$ and $C$. That is, $F$ measures $E$. But $F$ is greater than $E$ and so cannot measure $E$. From this contradiction it follows that no magnitude greater than $E$ can measure $A, B$ and $C$. The result follows: $E$ is the greatest common measure of $A, B$ and $C$, unless $D$ measures $C$. If $D$ measures $C$ then $D$ is the greatest common measure of $A, B$ and $C$. {{qed}} {{Euclid Note|4|X}}
Let $A$ and $B$ be commensurable magnitudes. By definition of commensurable, some magnitude will measure them both. Let $C$ be a common measure of $A$ and $B$. Let $D$ be the number of times $C$ is the measure of $A$. Let $E$ be the number of times $C$ is the measure of $B$. Since: :$C$ measures $A$ according to the number of units of $D$ and :the unit measures $D$ according to the number of units of $D$ it follows that the unit measures $D$ the same number of times $C$ measures $A$. From Ratios of Fractions in Lowest Terms: :$\dfrac C A = \dfrac 1 D$ and: :$\dfrac A C = \dfrac D 1 = D$ Similarly, since: :$C$ measures $B$ according to the number of units of $E$ and :the unit measures $E$ according to the number of units of $E$ it follows that the unit measures $E$ the same number of times $C$ measures $B$. From Ratios of Fractions in Lowest Terms: :$\dfrac C B = \dfrac 1 E$ So by Equality of Ratios Ex Aequali: :$\dfrac A B = \dfrac D E$ Hence the result. {{qed}} {{Euclid Note|5|X}}
Let $A$ and $B$ be magnitudes which have to one another the ratio which the number $D$ has to the number $E$. Let $A$ be divided into as many equal parts as there are units in $D$. Let $C$ be equal to one of those parts. Let $F$ be made up of as many magnitudes equal to $C$ as there are units in $E$. Since: : there are in $A$ as many magnitudes equal to $C$ as there are units in $D$ then: :whatever part the unit is of $D$, the same part is $C$ of $A$. From Ratios of Fractions in Lowest Terms: :$\dfrac C A = \dfrac 1 D$ and so from the porism to Ratios of Equal Magnitudes: :$\dfrac A C = \dfrac D 1 = D$ Since: : there are in $F$ as many magnitudes equal to $C$ as there are units in $E$ then from Ratios of Fractions in Lowest Terms: :$\dfrac C F = \dfrac 1 E$ From Equality of Ratios Ex Aequali: :$\dfrac A F = \dfrac D E$ But: :$\dfrac D E = \dfrac A B$ and so from Equality of Ratios is Transitive: :$\dfrac A B = \dfrac A F$ Therefore $A$ has the same ratio to each of the magnitudes $B$ and $F$. Therefore from Magnitudes with Same Ratios are Equal: :$B = F$ But $C$ measures $F$. Therefore $C$ also measures $B$. Further, $C$ also measures $A$. Therefore $C$ measures both $A$ and $B$. Therefore $A$ is commensurable with $B$. {{qed}} {{Euclid Note|6|X}}
Let $A$ and $B$ be incommensurable magnitudes. Suppose $A$ and $B$ have the ratio which a number has to a number Then by {{EuclidPropLink|book=X|prop=6|title=Magnitudes with Rational Ratio are Commensurable}}, $A$ and $B$ are commensurable. From this contradiction follows the result. {{qed}} {{Euclid Note|7|X}}
Let $A$ and $B$ be magnitudes which do not have to one another the ratio which a number has to a number. Suppose $A$ and $B$ are commensurable. Then from Ratio of Commensurable Magnitudes it follows that $A$ and $B$ have to one another the ratio which a number has to a number. From this contradiction follows the result. {{qed}} {{Euclid Note|8|X}}
Let $A$ and $B$ be commensurable in length. Then $A$ has to $B$ the ratio which a number has to a number. Let $\dfrac A B = \dfrac C D$. From Similar Polygons are composed of Similar Triangles: Porism: :$\dfrac {A^2} {B^2} = \paren {\dfrac C D}^2 = \dfrac {C^2} {D^2}$ {{qed|lemma}} Let: :$\dfrac {A^2} {B^2} = \dfrac {C^2} {D^2}$ Then it follows that: :$\dfrac A B = \dfrac C D$ and from Magnitudes with Rational Ratio are Commensurable it follows that $A$ and $B$ are commensurable in length. {{qed|lemma}} Let $A$ and $B$ be incommensurable in length. Suppose: :$\dfrac {A^2} {B^2} = \dfrac {C^2} {D^2}$ where $C$ and $D$ are integers. Then $A$ and $B$ would be commensurable in length, which they are not. So the square on $A$ and the square on $B$ do not have to one another the ratio which a square number has to a square number. {{qed|lemma}} Let the square on $A$ not have to the square on $B$ not have the ratio which a square number has to a square number. Suppose $A$ was commensurable in length with $B$. Then the square on $A$ would have to the square on $B$ the ratio which a square number has to a square number, which it does not. So $A$ is not commensurable in length with $B$. {{qed}} {{Euclid Note|9|X}}
Let $A$ be the assigned straight line. Let $B$ and $C$ be two numbers which do not have to one another the ratio which a square number has to a square number. That is, from the lemma, that $B$ and $C$ are not similar plane numbers. From Magnitudes with Rational Ratio are Commensurable: Porism, let $D$ be constructed such that: :$\dfrac {A^2} {D^2} = \dfrac B C$ Therefore from Magnitudes with Rational Ratio are Commensurable, the square on $A$ is commensurable with the square on $D$. Since: : $B$ does not have to one $C$ the ratio which a square number has to a square number then: : $A^2$ does not have to one $D^2$ the ratio which a square number has to a square number. Therefore from Commensurability of Squares $A$ is incommensurable in length with $D$. Let a mean proportional $E$ be taken between $A$ and $D$. From {{EuclidDefLink|V|9|Duplicate Ratio}}: :$\dfrac A D = \dfrac {A^2} {E^2}$ But $A$ is incommensurable in length with $D$. Therefore $A^2$ is incommensurable with $E^2$. Therefore $A$ is incommensurable in square with $E$. Therefore two straight lines $D$ and $E$ have been found, such that: :$D$ is incommensurable in length with $A$ :$E$ is incommensurable in square with $A$. {{qed}} {{Euclid Note|10|X|It was suggested by {{AuthorRef|Johan Ludvig Heiberg|Heiberg}}, and is generally believed, that this theorem, along with the lemma that accompanies it, is spurious.}}
Let $A$, $B$, $C$ and $D$ be four magnitudes in proportion: :$A : B = C : D$ Let $A$ be commensurable with $B$. Then from Ratio of Commensurable Magnitudes: : $A$ has to $B$ the ratio which a number has to a number. As $A : B = C : D$ it follows that: : $C$ has to $D$ the ratio which a number has to a number. From Magnitudes with Rational Ratio are Commensurable it follows that: :$C$ is commensurable with $D$. {{qed|lemma}} Let $A$ be incommensurable with $B$. Then from Incommensurable Magnitudes have Irrational Ratio: : $A$ does not have to $B$ the ratio which a number has to a number. As $A : B = C : D$ it follows that: : $C$ does not have to $D$ the ratio which a number has to a number. From Magnitudes with Irrational Ratio are Incommensurable it follows that: :$C$ is incommensurable with $D$. {{qed}} {{Euclid Note|11|X}}
Let $A$ and $B$ be magnitudes which are both commensurable with another magnitude $C$. As $A$ is commensurable with $C$, from Ratio of Commensurable Magnitudes: : $A$ has to $C$ the ratio which a number has to a number. Let it have the ratio which $D$ has to $E$, where $D$ and $E$ are numbers. As $C$ is commensurable with $B$, from Ratio of Commensurable Magnitudes: : $C$ has to $B$ the ratio which a number has to a number. Let it have the ratio which $F$ has to $G$, where $F$ and $G$ are numbers. From Construction of Sequence of Numbers with Given Ratios, let the numbers $H$, $K$, $L$ be assigned the ratios: :$D : E = H : K$ :$F : G = K : L$ So: {{begin-eqn}} {{eqn | l = A : C       | r = D : E       | c =  }} {{eqn | l = D : E       | r = H : K       | c =  }} {{eqn | ll= \implies       | l = A : C       | r = H : K       | c = Equality of Ratios is Transitive }} {{end-eqn}} and: {{begin-eqn}} {{eqn | l = C : B       | r = F : G       | c =  }} {{eqn | l = F : G       | r = K : L        | c =  }} {{eqn | ll= \implies       | l = C : B       | r = K : L       | c = Equality of Ratios is Transitive }} {{end-eqn}} Then: {{begin-eqn}} {{eqn | l = C : B       | r = K : L       | c =  }} {{eqn | l = A : C       | r = H : K        | c =  }} {{eqn | ll= \implies       | l = A : B       | r = H : L       | c = Equality of Ratios Ex Aequali }} {{end-eqn}} That is, $A$ has to $B$ the ratio which a number has to a number. From Magnitudes with Rational Ratio are Commensurable: :$A$ and $B$ are commensurable with each other. {{qed}} {{Euclid Note|12|X}}
Let $A$ and $B$ be magnitudes which are commensurable with each other. Let $A$ be incommensurable with any other magnitude $C$. Suppose $B$ is commensurable with $C$. But $A$ is commensurable with $B$. So from Commensurability is Transitive Relation it follows that $A$ is commensurable with $C$. From that contradiction it follows that $B$ cannot be commensurable with $C$. Hence the result. {{qed}} {{Euclid Note|13|X}}
Let $A$, $B$, $C$ and $D$ be four straight lines in proportion, such that: :$A : B = C : D$ Using the lemma let the straight lines $E$ and $F$ be found such that: : $A^2 = B^2 + E^2$ : $C^2 = D^2 + F^2$ As $A : B = C : D$ it follows from Similar Figures on Proportional Straight Lines that: :$A^2 : B^2 = C^2 : D^2$ But: : $E^2 + B^2 = A^2$ : $F^2 + D^2 = C^2$ Therefore: :$E^2 + B^2 : B^2 = F^2 + D^2 : D^2$ and so by Magnitudes Proportional Compounded are Proportional Separated: :$E^2 : B^2 = F^2 : D^2$ By Similar Figures on Proportional Straight Lines: :$B : E = D : F$ But: :$A : B = C : D$ Therefore from Equality of Ratios Ex Aequali: :$A : E = C : F$ From Commensurability of Elements of Proportional Magnitudes it follows that: :if $A$ is commensurable with $E$ then $C$ is also commensurable with $F$ and: :if $A$ is incommensurable with $E$ then $C$ is also incommensurable with $F$. {{qed}} {{Euclid Note|14|X}}
Let $AB$ and $BC$ be commensurable magnitudes which are added together to make $AC$. Since $AB$ and $BC$ are commensurable, then some magnitude $D$ will measure them both. Since $D$ measures both $AB$ and $BC$, $D$ also measures $AC$. That is, $D$ measures $AB$, $BC$ and $AC$. Therefore from {{EuclidDefLink|X|1|Commensurable}}, $AC$ is commensurable with each of the magnitudes $AB$ and $BC$. {{qed|lemma}} Let $AC$ be commensurable with $AB$. Since $AC$ and $AB$ are commensurable, then some magnitude $D$ will measure them both. Since $D$ measures both $AC$ and $AB$, $D$ also measures the remainder $BC$. That is, $D$ measures $AB$, $BC$ and $AC$. Therefore from {{EuclidDefLink|X|1|Commensurable}}, $AB$ and $BC$ are commensurable. {{qed}} {{Euclid Note|15|X}}
Let $AB$ and $BC$ be incommensurable magnitudes which are added together to make $AC$. Suppose $AC$ and $AB$ are not incommensurable. Then some magnitude $D$ will measure them both. Since $D$ measures both $AC$ and $AB$, $D$ also measures the remainder $BC$. That is, $D$ measures $AB$ and $BC$. Therefore from {{EuclidDefLink|X|1|Commensurable}}, $AB$ and $BC$ are commensurable. But by hypothesis $AB$ and $BC$ are incommensurable. Therefore no magnitude $D$ will measure both $AC$ and $AB$. Therefore $AC$ and $AB$ are incommensurable. In the same way it is shown that $AC$ and $BC$  are also incommensurable. {{qed|lemma}} Now let $AC$ be incommensurable with one of either $AB$ and $BC$. {{WLOG}}, let $AC$ be incommensurable $AB$. Suppose $AB$ and $BC$ are not incommensurable. Then some magnitude $D$ will measure them both. Since $D$ measures both $AB$ and $BC$, $D$ also measures the whole $AC$. But $D$ measures $AB$. Therefore $D$ measures both $AC$ and $AB$. Therefore $AC$ and $AB$ are commensurable. But by hypothesis $AB$ and $AC$ are incommensurable. Therefore no magnitude will measure both $AB$ and $BC$. Therefore $AB$ and $BC$ are incommensurable. {{qed}} {{Euclid Note|16|X}}
:300px Let $AB$ be a straight line. Let the parallelogram $AD$ be applied to $AB$ which is deficient by the square $DB$. As $DB$ is a square: : $DC = BC$ Thus $AD$ is the rectangle contained by $AC$ and $CB$. {{qed}} {{Euclid Note|17|X}}
We have that: {{begin-eqn}} {{eqn | l = x \left({a - x}\right) + \left({\frac a 2 - x}\right)^2       | r = a x - x^2 + \frac {a^2} 4 - 2 \frac a 2 x + x^2       | c =  }} {{eqn | r = \frac {a^2} 4       | c = simplifying }} {{eqn | ll= \implies       | l = 4 x \left({a - x}\right) + 4 \left({\frac a 2 - x}\right)^2       | r = a^2       | c =  }} {{eqn | ll= \implies       | l = b^2 + \left({a - 2 x}\right)^2       | r = a^2       | c = from $(1)$ }} {{eqn | ll= \implies       | l = a^2 - b^2       | r = \left({a - 2 x}\right)^2       | c =  }} {{eqn | ll= \implies       | l = \sqrt {a^2 - b^2}       | r = a - 2 x       | c =  }} {{end-eqn}} Let $a \frown b$ denote that $a$ is commensurable with $b$. === Necessary Condition === Let $\left({a - x}\right) \frown x$. From Commensurability of Sum of Commensurable Magnitudes: :$a \frown x$ From Magnitudes with Rational Ratio are Commensurable: :$x \frown 2 x$ So: {{begin-eqn}} {{eqn | l = a       | o = \frown       | r = 2 x       | c = Commensurability is Transitive Relation }} {{eqn | o = \frown       | r = \left({a - 2 x}\right)       | c = Commensurability of Sum of Commensurable Magnitudes }} {{eqn | o = \frown       | r = \sqrt {a^2 - b^2}       | c =  }} {{end-eqn}} {{qed|lemma}} === Sufficient Condition === Let $a \frown \sqrt {a^2 - b^2}$. Then: {{begin-eqn}} {{eqn | l = a       | o = \frown       | r = \sqrt {a^2 - b^2}       | c =  }} {{eqn | o = \frown       | r = \left({a - 2 x}\right)       | c =  }} {{eqn | o = \frown       | r = 2 x       | c = Commensurability of Sum of Commensurable Magnitudes }} {{end-eqn}} From Magnitudes with Rational Ratio are Commensurable: :$x \frown 2 x$ From Commensurability is Transitive Relation: :$a \frown x$ From Commensurability of Sum of Commensurable Magnitudes: : $\left({a - x}\right) \frown x$ {{qed}} {{Euclid Note|17|X}}
We have that: {{begin-eqn}} {{eqn | l = x \paren {a - x} + \paren {\frac a 2 - x}^2       | r = a x - x^2 + \frac {a^2} 4 - 2 \frac a 2 x + x^2       | c =  }} {{eqn | r = \frac {a^2} 4       | c = simplifying }} {{eqn | ll= \leadsto       | l = 4 x \paren {a - x} + 4 \paren {\frac a 2 - x}^2       | r = a^2       | c =  }} {{eqn | ll= \leadsto       | l = b^2 + \paren {a - 2 x}^2       | r = a^2       | c = from $(1)$ }} {{eqn | ll= \leadsto       | l = a^2 - b^2       | r = \paren {a - 2 x}^2       | c =  }} {{eqn | ll= \leadsto       | l = \sqrt {a^2 - b^2}       | r = a - 2 x       | c =  }} {{end-eqn}} Let: :$a \smile b$ denote that $a$ is incommensurable with $b$ :$a \frown b$ denote that $a$ is commensurable with $b$. === Necessary Condition === Let $\paren {a - x} \smile x$. From Incommensurability of Sum of Incommensurable Magnitudes: :$a \smile x$ From Magnitudes with Rational Ratio are Commensurable: :$x \frown 2 x$ So: {{begin-eqn}} {{eqn | l = a       | o = \smile       | r = 2 x       | c = Commensurable Magnitudes are Incommensurable with Same Magnitude }} {{eqn | o = \smile       | r = \paren {a - 2 x}       | c = Incommensurability of Sum of Incommensurable Magnitudes }} {{eqn | o = \smile       | r = \sqrt {a^2 - b^2}       | c =  }} {{end-eqn}} {{qed|lemma}} === Sufficient Condition === Let $a \smile \sqrt {a^2 - b^2}$. Then: {{begin-eqn}} {{eqn | l = a       | o = \smile       | r = \sqrt {a^2 - b^2}       | c =  }} {{eqn | o = \smile       | r = \paren {a - 2 x}       | c =  }} {{eqn | o = \smile       | r = 2 x       | c = Incommensurability of Sum of Incommensurable Magnitudes }} {{end-eqn}} From Magnitudes with Rational Ratio are Commensurable: :$x \frown 2 x$ From Commensurable Magnitudes are Incommensurable with Same Magnitude: :$a \smile x$ From Incommensurability of Sum of Incommensurable Magnitudes: :$\paren {a - x} \smile x$ {{qed}} {{Euclid Note|18|X}}
This lemma does little more than restate (in a considerably more wordy form) {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}. {{qed}} {{Euclid Note|19|X|It was suggested by {{AuthorRef|Johan Ludvig Heiberg|Heiberg}} that this lemma is a later interpolation, and not part of the original work by {{AuthorRef|Euclid}}. In his own edition of this work, he relegated it to the appendix.<br/><br/>{{AuthorRef|Sir Thomas L. Heath|Heath}} similarly separates it out from the canon in his case by enclosing it in brackets $[\,]$.}}
:200px Let the rational area $AC$ be applied to the rational straight line $AB$ such that $BC$ is its breadth. Let the square $AD$ be described on $AB$. From {{EuclidDefLink|X|4|Rational Area}}, $AD$ is rational. But $AC$ is also rational. Therefore $DA$ is commensurable in length with $AC$. From Areas of Triangles and Parallelograms Proportional to Base: :$BD : BC = DA : AC$ Therefore from Commensurability of Elements of Proportional Magnitudes $DB$ is also commensurable in length with $BC$. As $DB = BA$ it follows that $AB$ is also commensurable in length with $BC$. But $AB$ is rational. Therefore $BC$ is rational and commensurable in length with $AB$. {{qed}} {{Euclid Note|20|X}}
By definition, a medial is a mean proportional between two rational line segments which are commensurable in square only. Let $\rho$ and $\rho \sqrt k$ be two rational line segments which are commensurable in square only. Then: :$\rho: \rho \sqrt k = \rho^2 : \rho^2 \sqrt k$ From Commensurability of Elements of Proportional Magnitudes, $\rho^2 \sqrt k$ is incommensurable in length with $\rho^2$. From {{EuclidDefLink|X|4|Rational Area}}, $\rho^2 \sqrt k$ is an irrational area. Thus $\sqrt {\rho^2 \sqrt k} = \rho \sqrt [4] k$ is by definition an irrational straight line. By definition of mean proportional, $\rho \sqrt [4] k$ is the medial between $\rho$ and $\rho \sqrt k$ {{qed}} {{Euclid Note|21|X}}
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: {{begin-eqn}} {{eqn | l = CD       | r = \frac {\rho^2 \sqrt k} \sigma       | c =  }} {{eqn | r = \dfrac {\rho^2 \sqrt k} {\sigma^2} \sigma       | c =  }} {{eqn | r = \sqrt k \dfrac m n \sigma       | c =  }} {{end-eqn}} 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 $BC$. {{qed}} {{Euclid Note|22|X}}
:300px Let $A$ be a medial straight line. Let $B$ be a straight line which is commensurable in length with $A$. Let $CD$ be a rational straight line. Let a rectangle $CE$ be set up on $CD$ equal to the square on $A$. Let the breadth of $CE$ be $DE$. Then from Square on Medial Straight Line: : $ED$ is a rational straight line which is commensurable in length with CD. Let a rectangle $CF$ be set up on $CD$ equal to the square on $B$. Let the breadth of $CF$ be $DF$. As $A$ is commensurable in length with $B$, the square on $A$ is also commensurable with the square on $B$. But as $CE = A^2$ and $CF = B^2$, $CE$ is commensurable with $CF$. From Areas of Triangles and Parallelograms Proportional to Base: :$EC : CF = ED : DF$ From Commensurability of Elements of Proportional Magnitudes: :$ED$ is commensurable with $DF$. But $ED$ is a rational straight line which is commensurable in length with CD. By {{EuclidDefLink|X|3|Rational Line Segment}}, $DF$ is a rational straight line. By Commensurable Magnitudes are Incommensurable with Same Magnitude: : $DF$ is incommensurable in length with $DC$. Therefore $CD$ and $DF$ are rational and commensurable in square only. But from Medial is Irrational, the side of the square equal to therectangle contained by $CD$ and $DF$ is medial. Therefore $B$ is medial. {{qed}} {{Euclid Note|23|X}}
Let $x$ and $\lambda x$ be two medials such that: :$x \frown \lambda x$ where $\frown$ denotes that $x$ and $\lambda x$ are commensurable in length. By Areas of Triangles and Parallelograms Proportional to Base: :$x^2 : x \cdot \lambda x = x : \lambda x$ From Commensurability of Elements of Proportional Magnitudes:  :$x^2 \frown x \cdot \lambda x$ From Medial is Irrational we have that $x^2$ is medial. Therefore by Straight Line Commensurable with Medial Straight Line is Medial: Porism: :$x \cdot \lambda x$ is medial. {{qed}} {{Euclid Note|24|X}}
Let $A$ and $B$ be two medial straight lines. Let $A$ and $B$ be commensurable in square only. Then their lengths are of the form: :$l \left({A}\right) = \rho \sqrt [4] k$ :$l \left({B}\right) = \rho \sqrt [4] k \sqrt \lambda$ where:  : $\rho$ is a rational number : $k$ and $\lambda$ are rational numbers whose square root is irrational. The rectangle contained by $A$ and $B$ has area $R$ given by: :$R = \rho^2 \sqrt k \sqrt \lambda$ In general, $R$ is medial. However, let $\sqrt \lambda = k' \sqrt k$ where $k$ is rational. Then: :$R = \rho^2 \sqrt k k' \sqrt k = \rho^2 k k'$ which is rational. {{qed}} {{Euclid Note|25|X}}