hcju/proofwikiqa huggingface dataset:
Let $\rho \cdot \sqrt k \rho$ and $\rho \cdot \sqrt \lambda \rho$ be two medial areas which have been applied to the same rational number $\rho$. Their difference is then: :$\paren {\sqrt k - \sqrt \lambda} \rho^2$ Let $x = \sqrt k \rho$ and $y = \sqrt \lambda \rho$. Let: :$\rho \paren {x - y} = \rho z$ Suppose $\rho z$ is a rational area. Then $z$ must be a rational straight line and: :$(1): \quad z \frown \rho$ where $\frown$ denotes commensurability in length. Since $\rho x$ and $\rho y$ are medial areas, it follows that $x$ and $y$ are rational and: :$(2): \quad x \smile \rho$ and $y \smile \rho$ where $\smile$ denotes incommensurability in length. Then from $(1)$ and $(2)$: :$y \smile z$ Then: {{begin-eqn}} {{eqn | l = y : z       | r = y^2 : y z       | c =  }} {{eqn | l = y^2       | o = \smile       | r = y z       | c =  }} {{end-eqn}} But: {{begin-eqn}} {{eqn | l = y^2 + z^2       | o = \frown       | r = y^2       | c =  }} {{eqn | l = 2 y z       | o = \frown       | r = y z       | c =  }} {{eqn | ll= \leadsto       | l = y^2 + z^2       | o = \smile       | r = 2 y z       | c =  }} {{eqn | ll= \leadsto       | l = \paren {y + z}^2       | o = \smile       | r = \paren {y^2 + z^2}       | c =  }} {{eqn | ll= \leadsto       | l = x^2       | o = \smile       | r = \paren {y^2 + z^2}       | c =  }} {{end-eqn}} But $y^2 + z^2$ is rational. Therefore $x^2$ must be irrational. Therefore $x$ must be irrational. But from $(2)$ $x$ is rational. The result follows from this contradiction. {{qed}} {{Euclid Note|26|X}}
Let $\rho$ and $\rho \sqrt k$ be rational straight lines which are commensurable in square only. Their mean proportional is $\rho \sqrt [4] k$ which is medial. Let $x$ be such that: :$\rho : \rho \sqrt k = \rho \sqrt [4] k : x$ which gives: :$x = \rho k^{3/4}$ We have that: :$\rho \frown \!\! - \rho \sqrt k$ where $\frown \!\! -$ denotes commensurability in square only. Thus: :$\rho \sqrt [4] k \frown \!\! - \rho k^{3/4}$ From Straight Line Commensurable with Medial Straight Line is Medial it follows that $\rho k^{3/4}$ is also medial. {{qed}}
Let $\rho$, $\rho \sqrt k$ and $\rho \sqrt \lambda$ be rational straight lines which are commensurable in square only. Take the mean proportional $\rho \sqrt [4] k$ of $\rho$ and $\rho \sqrt k$, which is medial. Let $x$ be such that: :$\rho \sqrt k : \rho \sqrt \lambda = \rho \sqrt [4] k : x$ which gives: :$x = \dfrac {\rho \sqrt \lambda} {\sqrt [4] k}$ We have that: :$\rho \sqrt k \frown \!\! - \rho \sqrt \lambda$ where $\frown \!\! -$ denotes commensurability in square only. Thus: :$\rho \sqrt [4] k \frown \!\! - x$ From Straight Line Commensurable with Medial Straight Line is Medial it follows that $\rho k^{3/4}$ is also medial. Then: {{begin-eqn}} {{eqn | l = \rho \sqrt \lambda : x       | r = \rho \sqrt k : \rho \sqrt [4] k       | c =  }} {{eqn | r = \rho \sqrt [4] k : \rho       | c =  }} {{eqn | ll= \implies       | l = x \cdot \rho \sqrt [4] k       | r = \rho^2 \sqrt \lambda       | c = which is a medial area }} {{end-eqn}} {{qed}}
Let $\rho$ be a rational straight line. Let $m$ and $n$ be natural numbers such that $m^2 - n^2$ is not square. Let $x$ be a straight line such that: :$(1): \quad m^2 : \paren {m^2 - n^2} = \rho^2 : x^2$ Thus: :$x^2 = \dfrac {m^2 - n^2} {m^2} \rho^2$ and so: :$x = \rho \sqrt {1 - k^2}$ where $k = \dfrac n m$. From $(1)$: :$x^2 \frown \rho^2$ where $\frown$ denotes commensurability in length. Thus $x$ is a rational straight line, but: :$x \smile \rho$ where $\smile$ denotes incommensurability in length. From $(1)$: :$ m^2 : n^2 = \rho^2 : \rho^2 - x^2$ so that: :$\sqrt {\rho^2 - x^2} \frown \rho$ and in fact: :$\sqrt {\rho^2 - x^2} = k \rho$ where $k$ is a rational number. {{qed}} {{Euclid Note|29|X}}
Let $\rho$ be a rational straight line. Let $m$ and $n$ be natural numbers such that $m^2 + n^2$ is not square. Let $x$ be such that: :$m^2 + n^2 : m^2 = \rho^2 + x^2$ and so: :$x^2 = \dfrac {m^2} {m^2 + n^2} \rho^2$ or: :$x = \dfrac {\rho} {\sqrt {1 + k^2} }$ for some rational $k$. Then $\rho$ and $\dfrac {\rho} {\sqrt {1 + k^2} }$ fit the condition. {{qed}} {{Euclid Note|30|X}}
Let $A$ and $B$ be rational straight lines which are commensurable in square only. Let $A^2 > B^2$ such that: :$A^2 = B^2 + \rho^2$ where $\rho$ is a straight line which is commensurable in length with $A$. This can be done using Construction of Rational Straight Lines Commensurable in Square Only whose Square Differences Commensurable with Greater. Let $C$ be a straight line such that $C^2 = A B$. From Medial is Irrational, $A B$ is medial. Therefore $C^2$ is medial. So by definition $C$ is medial. Let $C D = B^2$. By hypothesis $B^2$ is rational. Therefore $C D$ is rational. We have that: :$A : B = A^2 : A B$ But: :$C^2 = A B$ and: :$C D = B^2$ and so: :$A : B = C^2 : C D$ But as: :$C^2 : C D = C : D$ it follows that: :$A : B = C : D$ But $A$ is commensurable in square only with $B$. Therefore by Commensurability of Elements of Proportional Magnitudes: :$C$ is commensurable in square only with $D$. We have that $C$ is medial. Therefore by Straight Line Commensurable with Medial Straight Line is Medial, $D$ is medial. We have that: :$A : B = C : D$ and: : $A^2 = B^2 + \rho^2$ where $\rho$ is a straight line which is commensurable in length with $A$. Therefore Commensurability of Squares on Proportional Straight Lines: :$C^2 = D^2 + \sigma^2$ where $\sigma$ is a straight line which is commensurable in length with $C$. Therefore two medial straight lines $C$ and $D$ have been found which are commensurable in square only and which contain a rational rectangle. Further, $C^2$ is greater than $D^2$ by the square on a straight line commensurable in length with $C$. Similarly it can be proved that the square on $C$ exceeds the square on $D$ by the square on a straight line incommensurable in length with $C$, when the square on $A$ is greater than the square on $B$ by the square on a straight line incommensurable in length with $A$. {{qed}} {{Euclid Note|31|X}}
Let $A$, $B$ and $C$ be rational straight lines which are commensurable in square only. Let $A^2 > C^2$ such that: :$A^2 = C^2 + \rho^2$ where $\rho$ is a straight line which is commensurable in length with $A$. This can be done using Construction of Rational Straight Lines Commensurable in Square Only whose Square Differences Commensurable with Greater. Let $D$ be a straight line such that $D^2 = A B$. From Medial is Irrational, $A B$ is medial. Therefore $D^2$ is medial. So by definition $D$ is medial. Let $D E = B C$. Since: :$A B : B C = A : C$ and: :$D^2 = A B$ and: :$D E = B C$ then: :$A : C = D^2 : D E$ So as: :$D^2 : D E = D : E$ it follows that: :$A : C = D : E$ But $A$ is commensurable in square only with $C$. Therefore by Commensurability of Elements of Proportional Magnitudes: :$D$ is commensurable in square only with $E$. We have that $D$ is medial. Therefore by Straight Line Commensurable with Medial Straight Line is Medial, $E$ is medial. We have that: :$A : C = D : E$ and: :$A^2 = C^2 + \rho^2$ where $\rho$ is a straight line which is commensurable in length with $A$. Therefore Commensurability of Squares on Proportional Straight Lines: :$D^2 = E^2 + \sigma^2$ where $\sigma$ is a straight line which is commensurable in length with $D$. We have that: : $B C = D E$ while from Medial is Irrational: :$B C$ is medial. Therefore $D E$ is also medial. Therefore two medial straight lines $D$ and $E$ have been found which are commensurable in square only and which contain a medial rectangle. Further, $D^2$ is greater than $E^2$ by the square on a straight line commensurable in length with $D$. Similarly it can be proved that the square on $D$ exceeds the square on $E$ by the square on a straight line incommensurable in length with $D$, when the square on $A$ is greater than the square on $C$ by the square on a straight line incommensurable in length with $A$. {{qed}} {{Euclid Note|32|X}}
:500px From {{EuclidPropLink|book=X|prop=30|title=Construction of Rational Straight Lines Commensurable in Square Only whose Square Differences Incommensurable with Greater}}: Let $AB$ and $BC$ be rational straight lines which are commensurable in square only such that: : $AB^2 = BC^2 + \rho^2$ such that $\rho$ is incommensurable in length with $AB$. Let $BC$ be bisected at $D$. From Construction of Parallelogram Equal to Given Figure Exceeding a Parallelogram: Let a parallelogram be applied to $AB$ equal to the square on either of $BD$ or $DC$, and deficient by a square. Let this parallelogram be the rectangle contained by $AE$ and $EB$. Let the semicircle $AFB$ be drawn with $AB$ as the diameter. Let $EF$ be drawn perpendicular to $AB$. Join $AF$ and $FB$. We have that $AB > BC$ such that $AB^2 = BC^2 + \rho^2$ such that $\rho$ is incommensurable in length with $AB$. We also have that the rectangle contained by $AE$ and $EB$ equals the parallelogram on $AB$ equal to $\dfrac {BC} 4$. Thus from Condition for Incommensurability of Roots of Quadratic Equation: :$AE$ is incommensurable in length with $EB$. We have that: :$AE : BE = AB \cdot AE : AB \cdot BE$ From {{EuclidPropLink|book=X|prop=33|qual=Lemma|title=Construction of Components of Major}}: :$AB \cdot AE = AF^2$ and: :$AB \cdot BE = BF^2$ Therefore $AF^2$ and $BF^2$ are incommensurable. By definition, $AF$ and $BF$ are therefore incommensurable in square. As $AB$ is a rational straight line, it follows by definition that $AB^2$ is a rational area. From Pythagoras's Theorem: :$AB^2 = \left({AF + FB}\right)^2$ Thus $\left({AF + FB}\right)^2$ is also a rational area. Therefore $AF + FB$ is rational. From {{EuclidPropLink|book=X|prop=33|qual=Lemma|title=Construction of Components of Major}}: :$AE \cdot EB = EF^2$ and by hypothesis: :$AE \cdot EB = BD^2$ then: :$FE = BD$ Therefore: : $BC = 2 FE$ Thus $AB \cdot BC$ is commensurable with $AB \cdot EF$. But from Medial is Irrational: :$AB \cdot BC$ is medial. Therefore by {{EuclidPropLink|book=X|prop=23|qual=Porism|title=Straight Line Commensurable with Medial Straight Line is Medial}}: :$AB \cdot EF$ is also medial. But from {{EuclidPropLink|book=X|prop=33|qual=Lemma|title=Construction of Components of Major}}: :$AB \cdot EF = AF \cdot FB$ Therefore $AF \cdot FB$ is also medial. But it has been proved that $AF + FB$ is rational. Therefore we have found two rational straight lines which are incommensurable in square whose sum of squares is rational, but such that the rectangle contained by them is medial. {{qed}}
:500px From {{EuclidPropLink|book=X|prop=31|title=Construction of Medial Straight Lines Commensurable in Square Only containing Rational Rectangle whose Square Differences Commensurable with Greater}}: Let $AB$ and $BC$ be medial straight lines which are commensurable in square only such that: : $AB^2 = BC^2 + \rho^2$ such that $\rho$ is incommensurable in length with $AB$. Let the semicircle $ADB$ be drawn with $AB$ as the diameter. Let $BC$ be bisected at $E$. From Construction of Parallelogram Equal to Given Figure Exceeding a Parallelogram: Let a parallelogram be applied to $AB$ equal to the square on either of $BE$ or $EC$, and deficient by a square. Let this parallelogram be the rectangle contained by $AF$ and $FB$. From {{EuclidPropLink|book=X|prop=18|title=Condition for Incommensurability of Roots of Quadratic Equation}}: :$AF$ is incommensurable in length with $FB$. Let $FD$ be drawn perpendicular to $AB$. Join $AD$ and $DB$. We have that $AF$ is incommensurable in length with $FB$. So from {{EuclidPropLink|book=X|prop=11|title=Commensurability of Elements of Proportional Magnitudes}}: :$BA \cdot AF$ is incommensurable with $AB \cdot BF$. From {{EuclidPropLink|book=X|prop=33|qual=Lemma|title=Construction of Components of Major}}: :$BA \cdot AF = AD^2$ and: :$AB \cdot BF = DB^2$ Therefore $AD^2$ and $DB^2$ are incommensurable. As $AB$ is medial, it follows by definition that $AB^2$ is a medial area. From Pythagoras's Theorem: :$AB^2 = \left({AD + DB}\right)^2$ Thus $\left({AD + DB}\right)^2$ is also a medial area. Therefore $AF + FB$ is medial. As $BC = 2 DF$: :$AB \cdot BC = 2 AB \cdot FD$ But $AB \cdot BC$ is a rational area. Therefore from {{EuclidPropLink|book=X|prop=6|title=Magnitudes with Rational Ratio are Commensurable}}: :$AB \cdot FD$ is a rational area. But from {{EuclidPropLink|book=X|prop=33|qual=Lemma|title=Construction of Components of Major}}: :$AB \cdot FD = AD \cdot DB$ Thus $AD \cdot DB$ is a rational area. Therefore we have found two straight lines which are incommensurable in square whose sum of squares is medial, but such that the rectangle contained by them is rational. {{qed}}
:500px From {{EuclidPropLink|book=X|prop=32|title=Construction of Medial Straight Lines Commensurable in Square Only containing Medial Rectangle whose Square Differences Commensurable with Greater}}: Let $AB$ and $BC$ be medial straight lines which are commensurable in square only such that: : $(1): \quad$ $AB$ and $BC$ contain a medial rectangle : $(2): \quad AB^2 = BC^2 + \rho^2$ such that $\rho$ is incommensurable in length with $AB$. Let the semicircle $ADB$ be drawn with $AB$ as the diameter. Let $BC$ be bisected at $E$. From Construction of Parallelogram Equal to Given Figure Exceeding a Parallelogram: Let a parallelogram be applied to $AB$ equal to the square on either of $BE$ or $EC$, and deficient by a square. Let this parallelogram be the rectangle contained by $AF$ and $FB$. Let $FD$ be drawn perpendicular to $AB$. Join $AD$ and $DB$. From {{EuclidPropLink|book=X|prop=18|title=Condition for Incommensurability of Roots of Quadratic Equation}}: :$AF$ is incommensurable in length with $FB$. So from {{EuclidPropLink|book=X|prop=11|title=Commensurability of Elements of Proportional Magnitudes}}: :$BA \cdot AF$ is incommensurable with $AB \cdot BF$. From {{EuclidPropLink|book=X|prop=33|qual=Lemma|title=Construction of Components of Major}}: :$BA \cdot AF = AD^2$ and: :$AB \cdot BF = DB^2$ Therefore $AD^2$ and $DB^2$ are incommensurable. ... this far As $AB$ is medial, it follows by definition that $AB^2$ is a medial area. From Pythagoras's Theorem: :$AB^2 = \left({AD + DB}\right)^2$ Thus $\left({AD + DB}\right)^2$ is also a medial area. Therefore $AD + DB$ is medial. Since $AF \cdot FB = BE^2$ and also $AF \cdot FB = DF^2$: :$BE^2 = DF^2$ As $BC = 2 DF$: :$AB \cdot BC = 2 AB \cdot FD$ But $AB \cdot BC$ is a medial area. Therefore from {{EuclidPropLink|book=X|prop=6|title=Magnitudes with Rational Ratio are Commensurable}}: :$AB \cdot FD$ is a medial area. But from {{EuclidPropLink|book=X|prop=33|qual=Lemma|title=Construction of Components of Major}}: :$AB \cdot FD = AD \cdot DB$ Thus $AD \cdot DB$ is a medial area. Since: :$AB$ is incommensurable in length with $BC$ and :$CB$ is commensurable in length with $BE$ from {{EuclidPropLink|book=X|prop=13|title=Commensurable Magnitudes are Incommensurable with Same Magnitude}}: :$AB$ is incommensurable in length with $BE$. By {{EuclidPropLink|book=X|prop=11|title=Commensurability of Elements of Proportional Magnitudes}}: :$AB^2 = AB \cdot BE$ From Pythagoras's Theorem: :$AD^2 + DB^2 = AB^2$ But from {{EuclidPropLink|book=X|prop=33|qual=Lemma|title=Construction of Components of Major}}: :$AB \cdot FD = AD \cdot DB$ and: :$AD \cdot DB = AD \cdot BE$ Therefore $AD^2 + DB^2$ is incommensurable with $AD \cdot DB$. Therefore we have found two straight lines $AD$ and $DB$ which are incommensurable in square whose sum of squares is medial, but such that the rectangle contained by them is medial. {{qed}} {{Euclid Note|35|X}}
Let $AB$ and $BC$ be two rational straight lines which are commensurable in square only. By definition, $AB$ and $BC$ are incommensurable in length. We have: :$AB : BC = AB \cdot BC : BC^2$ Therefore from {{EuclidPropLink|book=X|prop=11|title=Commensurability of Elements of Proportional Magnitudes}}: : $AB \cdot BC$ is incommensurable with $BC^2$. But by {{EuclidPropLink|book=X|prop=6|title=Magnitudes with Rational Ratio are Commensurable}}: :$2 AB \cdot BC$ is commensurable with $AB \cdot BC$. We have that $AB$ and $BC$ are commensurable in square. So from {{EuclidPropLink|book=X|prop=15|title=Commensurability of Sum of Commensurable Magnitudes}}: :$AB^2 + BC^2$ is commensurable with $BC^2$. So from {{EuclidPropLink|book=X|prop=13|title=Commensurable Magnitudes are Incommensurable with Same Magnitude}}: :$2 AB \cdot BC$ is incommensurable with $AB^2 + BC^2$. Thus from {{EuclidPropLink|book=II|prop=4|title=Square of Sum}}: :$AC^2 = \left({AB + BC}\right)^2 = 2 AB \cdot BC + AB^2 + BC^2$ is incommensurable with $AB^2 + BC^2$. But $AB^2 + BC^2$ is rational. Therefore $AC^2$ is irrational. Thus by {{EuclidDefLink|X|4|Rational Area}}: :$AC$ is irrational. Such a straight line is called '''binomial'''. {{qed}} {{Euclid Note|36|X}}
Let $AB$ and $BC$ be medial straight lines which are commensurable in square only. Let $AB$ and $BC$ contain a rational rectangle. By definition, $AB$ and $BC$ are incommensurable in length. We have: :$AB : BC = AB \cdot BC : BC^2$ :$AB : BC = AB^2 : AB \cdot BC$ Therefore from {{EuclidPropLink|book=X|prop=11|title=Commensurability of Elements of Proportional Magnitudes}}: : $AB \cdot BC$ is incommensurable with $AB^2$ and : $AB \cdot BC$ is incommensurable with $BC^2$. But by {{EuclidPropLink|book=X|prop=6|title=Magnitudes with Rational Ratio are Commensurable}}: :$2 AB \cdot BC$ is commensurable with $AB \cdot BC$. We have that $AB$ and $BC$ are commensurable in square. So from {{EuclidPropLink|book=X|prop=15|title=Commensurability of Sum of Commensurable Magnitudes}}: :$AB^2 + BC^2$ is commensurable with $BC^2$. So from {{EuclidPropLink|book=X|prop=13|title=Commensurable Magnitudes are Incommensurable with Same Magnitude}}: :$2 AB \cdot BC$ is incommensurable with $AB^2 + BC^2$. Thus from {{EuclidPropLink|book=X|prop=16|title=Incommensurability of Sum of Incommensurable Magnitudes}}: :$2 AB \cdot BC + AB^2 + BC^2$ is incommensurable with $AB \cdot BC$. We have that $AB$ and $BC$ contain a rational rectangle. Thus $AB \cdot BC$ is rational. From {{EuclidPropLink|book=II|prop=4|title=Square of Sum}}: :$AC^2 = \left({AB + BC}\right)^2 = 2 AB \cdot BC + AB^2 + BC^2$ Thus from {{EuclidDefLink|X|4|Rational Area}}: :$AC$ is irrational. Such a straight line is called '''first bimedial'''. {{qed}} {{Euclid Note|37|X}}
:400px Let $AB$ and $BC$ be medial straight lines which are commensurable in square only. Let $AB$ and $BC$ contain a medial rectangle. Let $DE$ be a rational straight line. Using {{EuclidPropLink|book = I|prop = 45|title = Construction of Parallelogram in Given Angle equal to Given Polygon}}: :Let $DF$ be a parallelogram set out on $DE$ equal to the square on $AC$. Let its breadth be $DG$. From {{EuclidPropLink|book = II|prop = 4|title = Square of Sum}}: :$AC^2 = AB^2 + BC^2 + 2 \cdot AB \cdot BC$ Let $EH$ be a rectangle applied to $DE$ whose area equals $AB^2 + BC^2$. The rectangle $HF$ which remains from $DF$ having had $EH$ removed is therefore equal to $2 \cdot AB \cdot BC$. Since each of $AB$ and $BC$ are medial, both of $AB^2$ and $BC^2$ are also medial. By hypothesis, $2 \cdot AB \cdot BC$ is also medial. As: :$EH = AB^2 + BC^2$ and: :$FH = 2 \cdot AB \cdot BC$ each of the rectangles $EH$ and $FH$ is medial. Each of $EH$ and $FH$ is applied to the rational straight line $DE$. From {{EuclidPropLink|book = X|prop = 22|title = Square on Medial Straight Line}}: :Each of $DH$ and $HG$ is rational and incommensurable in length with $DE$. We have that $AB$ is incommensurable in length with $BC$. We also have that: :$AB : BC = AB^2 : AB \cdot BC$ So by {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$AB^2$ is incommensurable with $AB \cdot BC$. But from {{EuclidPropLink|book = X|prop = 15|title = Commensurability of Sum of Commensurable Magnitudes}}: :$AB^2 + BC^2$ is commensurable with $AB^2$ and from {{EuclidPropLink|book = X|prop = 15|title = Magnitudes with Rational Ratio are Commensurable}}: :$2 \cdot AB \cdot BC$ is commensurable with $AB \cdot BC$. Therefore from {{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}}: :$AB^2 + BC^2$ is incommensurable with $2 \cdot AB \cdot BC$. But: :$EH = AB^2 + BC^2$ and: :$HF = 2 \cdot AB \cdot BC$ Therefore $EH$ is incommensurable with $HF$. So from: :{{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}} and: :{{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}} it follows that $DH$ is incommensurable in length with $HG$. Therefore $DH$ and $HG$ are rational straight lines which are commensurable in square only. By {{EuclidPropLink|book = X|prop = 11|title = Binomial is Irrational}}: :$DG$ is irrational. But $DE$ is rational. From {{EuclidPropLink|book = X|prop = 20|title = Quotient of Rationally Expressible Numbers is Rational}}: :The rectangle contained by an irrational and a rational straight line is irrational. Therefore $DF$ is irrational. From {{EuclidDefLink|X|4|Rational Area}}: :The side of a square equal to an irrational area is irrational. But $AC$ is the side of a square equal to $DF$. Therefore $AC$ is irrational. Such a straight line is called '''second bimedial'''. {{qed}} {{Euclid Note|38|X}}
From {{EuclidPropLink|book = X|prop = 33|title = Construction of Components of Major}}: :Let $AB$ and $BC$ be straight lines which are incommensurable in square which make the sum of the squares on them rational, but the rectangle contained by them medial. We have that the rectangle contained by $AB$ and $BC$ is medial. From: : {{EuclidPropLink|book = X|prop = 6|title = Magnitudes with Rational Ratio are Commensurable}} and: : {{EuclidPropLink|book = X|prop = 23|qual = Porism|title = Straight Line Commensurable with Medial Straight Line is Medial}}: it follows that: :$2 \cdot AB \cdot BC$ is medial. But $AB^2 + BC^2$ is rational. Therefore $2 \cdot AB \cdot BC$ is incommensurable with $AB^2 + BC^2$. So by {{EuclidPropLink|book = X|prop = 16|title = Incommensurability of Sum of Incommensurable Magnitudes}}: :$AC^2 = \left({AB + BC}\right)^2 = AB^2 + BC^2 + 2 \cdot AB \cdot BC$ is also incommensurable with $AB^2 + BC^2$. Therefore $AC^2$ is irrational. Hence $AC$ is irrational. Such a straight line is called '''major'''. {{qed}} {{Euclid Note|39|X}}
From {{EuclidPropLink|book = X|prop = 34|title = Construction of Components of Side of Rational plus Medial Area}}: :Let $AB$ and $BC$ be straight lines which are incommensurable in square which make the sum of the squares on them medial but the rectangle contained by them rational. We have that: :$AB^2 + BC^2$ is medial and from {{EuclidPropLink|book = X|prop = 6|title = Magnitudes with Rational Ratio are Commensurable}}: :$2 \cdot AB \cdot BC$ is rational. So: :$AB^2 + BC^2$ is incommensurable with $2 \cdot AB \cdot BC$. So from {{EuclidPropLink|book = X|prop = 16|title = Incommensurability of Sum of Incommensurable Magnitudes}}: :$AC^2 = \left({AB + BC}\right)^2 = AB^2 + BC^2 + 2 \cdot AB \cdot BC$ is also incommensurable with $2 \cdot AB \cdot BC$. But $2 \cdot AB \cdot BC$ is rational. Therefore $AC^2$ is irrational. Hence $AC$ is irrational. Such a straight line is called the '''side of a rational plus a medial area'''. {{qed}} {{Euclid Note|40|X}}