hcju/proofwikiqa huggingface dataset:
:450px 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 {{EuclidPropLink|book = II|prop = 7|title = Square of Difference}}: :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 {{EuclidPropLink|book = X|prop = 22|title = Square on Medial Straight Line}}: :$EM$ 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 {{EuclidPropLink|book = X|prop = 22|title = Square on Medial Straight Line}}: :$HM$ 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 {{EuclidPropLink|book = VI|prop = 1|title = Areas of Triangles and Parallelograms Proportional to Base}}: :$EG : HG = EM : HM$ Therefore from {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$EM$ 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 {{EuclidPropLink|book = X|prop = 79|title = Construction of Apotome is Unique}}, this is impossible. The result follows. {{qed}} {{Euclid Note|84|X}}
:400px Let $A$ be a rational straight line. Let $BG$ be commensurable in length with $A$. Therefore, by definition, $BG$ is also rational. Let $DE$ and $EF$ be square numbers such that $FD = DE - EF$ is not square. Then $ED : FD$ is not the ratio that a square number has to another square number. Using {{EuclidPropLink|book = X|prop = 6|qual=Porism|title = Magnitudes with Rational Ratio are Commensurable}}, let it be contrived that: :$ED : DF = BG^2 : CG^2$ Therefore by {{EuclidPropLink|book = X|prop = 6|title = Magnitudes with Rational Ratio are Commensurable}}: :$BG^2$ is commensurable with $CG^2$. But $BG^2$ is rational. Therefore $GC^2$ is rational. Therefore $GC$ is rational. We have that $ED : FD$ is not the ratio that a square number has to another square number. Therefore $BG^2 : GC^2$ is not the ratio that a square number has to another square number. Therefore from {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: :$BG$ is incommensurable in length with $GC$. Both $BG$ and $GC$ are rational. Therefore $BG$ and $GC$ rational straight lines which are commensurable in square only. Therefore $BC$ is an apotome. It remains to be shown that $BC$ is a first apotome. Let $H$ be a straight line such that $H^2 = BG^2 - GC^2$. We have that: :$ED : FD = BG^2 : GC^2$  So by {{EuclidPropLink|book = V|prop = 19|qual=Porism|title = Proportional Magnitudes have Proportional Remainders}}: :$DE : EF = GB^2 : H^2$ But $DE$ and $EF$ are both square numbers. So $DE$ has to $EF$ the ratio that a square number has to another square number. Therefore $GB^2$ has to $H^2$ the ratio that a square number has to another square number. Therefore from {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: :$BG$ is commensurable in length with $H$. We have that: :$BG^2 = GC^2 + H^2$ Therefore $BG^2$ is greater than $GC^2$ by the square on a straight line which is commensurable in length with $BG$. Also, the whole $BG$ is commensurable in length with the rational straight line $A$. Therefore, by definition, $BC$ is a first apotome. {{qed}} {{Euclid Note|85|X}}
:400px Let $A$ be a rational straight line. Let $GC$ be commensurable in length with $A$. Therefore, by definition, $GC$ is also rational. Let $DE$ and $EF$ be square numbers such that $DF = DE - EF$ is not square. Then $ED : FD$ is not the ratio that a square number has to another square number. Using {{EuclidPropLink|book = X|prop = 6|qual=Porism|title = Magnitudes with Rational Ratio are Commensurable}}, let it be contrived that: :$FD : DE = CG^2 : GB^2$ Therefore by {{EuclidPropLink|book = X|prop = 6|title = Magnitudes with Rational Ratio are Commensurable}}: :$CG^2$ is commensurable with $GB^2$. But $CG^2$ is rational. Therefore $GB^2$ is rational. Therefore $GB$ is rational. We have that $GC^2 : GB^2$ is not the ratio that a square number has to another square number. Therefore from {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: :$GC$ is incommensurable in length with $GB$. Both $GC$ and $GB$ are rational. Therefore $GC$ and $GB$ rational straight lines which are commensurable in square only. Therefore $BC$ is an apotome. It remains to be shown that $BC$ is a second apotome. Let $H$ be a straight line such that $H^2 = BG^2 - GC^2$. We have that: :$ED : DF = BG^2 : GC^2$  So by {{EuclidPropLink|book = V|prop = 19|qual=Porism|title = Proportional Magnitudes have Proportional Remainders}}: :$DE : EF = BG^2 : H^2$ But $DE$ and $EF$ are both square numbers. So $DE$ has to $EF$ the ratio that a square number has to another square number. Therefore $GB^2$ has to $H^2$ the ratio that a square number has to another square number. Therefore from {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: :$BG$ is commensurable in length with $H$. We have that: :$BG^2 = GC^2 + H^2$ Therefore $BG^2$ is greater than $GC^2$ by the square on a straight line which is commensurable in length with $BG$. Also, the annex $CG$ is commensurable in length with the rational straight line $A$. Therefore, by definition, $BC$ is a second apotome. {{qed}} {{Euclid Note|86|X}}
:500px 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 {{EuclidPropLink|book = X|prop = 6|qual=Porism|title = Magnitudes with Rational Ratio are Commensurable}}, 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 {{EuclidPropLink|book = X|prop = 6|title = Magnitudes with Rational Ratio are Commensurable}}: :$A^2$ 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 {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: :$A$ is incommensurable in length with $FG$. We have that: :$BC : CD = FG^2 : GH^2$ From {{EuclidPropLink|book = X|prop = 6|title = Magnitudes with Rational Ratio are Commensurable}}: :$FG^2$ 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 {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: :$FG$ 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 {{EuclidPropLink|book = V|prop = 22|title = Equality of Ratios Ex Aequali}}: :$E : CD = A^2 : HG^2$ 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 {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: :$A$ 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 {{EuclidPropLink|book = V|prop = 19|qual=Porism|title = Proportional Magnitudes have Proportional Remainders}}: :$BC : BD = FG^2 : K^2$ 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 {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: :$FG$ 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. {{qed}} {{Euclid Note|87|X}}
:400px Let $A$ be a rational straight line. Let $BG$ be commensurable in length with $A$. Therefore, by definition, $BG$ is also rational. Let $DF$ and $FE$ be square numbers such that $DE$ does not have to either $DF$ or $FE$ the ratio that a square number has to another square number. Using {{EuclidPropLink|book = X|prop = 6|qual=Porism|title = Magnitudes with Rational Ratio are Commensurable}}, let it be contrived that: :$DE : EF = BG^2 : GC^2$ Therefore by {{EuclidPropLink|book = X|prop = 6|title = Magnitudes with Rational Ratio are Commensurable}}: :$BG^2$ is commensurable with $GC^2$. But $BG^2$ is rational. Therefore $GC^2$ is rational. Therefore $GC$ is rational. We have that $DE : EF$ is not the ratio that a square number has to another square number. Therefore $BG^2 : GC^2$ is not the ratio that a square number has to another square number. Therefore from {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: :$BG$ is incommensurable in length with $GC$. Both $BG$ and $GC$ are rational. Therefore $BG$ and $GC$ rational straight lines which are commensurable in square only. Therefore $BC$ is an apotome. It remains to be shown that $BC$ is a fourth apotome. Let $H$ be a straight line such that $H^2 = BG^2 - GC^2$. We have that: :$DE : EF = BG^2 : GC^2$  So by {{EuclidPropLink|book = V|prop = 19|qual=Porism|title = Proportional Magnitudes have Proportional Remainders}}: :$ED : EF = GB^2 : H^2$ So $ED$ does not have to $EF$ the ratio that a square number has to another square number. Therefore neither does $GB^2$ have to $H^2$ the ratio that a square number has to another square number. Therefore from {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: :$BG$ is incommensurable in length with $H$. We have that: :$BG^2 = GC^2 + H^2$ Therefore $BG^2$ is greater than $GC^2$ by the square on a straight line which is incommensurable in length with $BG$. Also, the whole $BG$ is commensurable in length with the rational straight line $A$. Therefore, by definition, $BC$ is a fourth apotome. {{qed}} {{Euclid Note|88|X}}
:400px Let $A$ be a rational straight line. Let $CG$ be commensurable in length with $A$. Therefore, by definition, $CG$ is also rational. Let $DF$ and $FE$ be square numbers such that $DE$ does not have to either $DF$ or $FE$ the ratio that a square number has to another square number. Using {{EuclidPropLink|book = X|prop = 6|qual=Porism|title = Magnitudes with Rational Ratio are Commensurable}}, let it be contrived that: :$FE : ED = CG^2 : GB^2$ Therefore by {{EuclidPropLink|book = X|prop = 6|title = Magnitudes with Rational Ratio are Commensurable}}: :$GB^2$ is commensurable with $GC^2$. Therefore $GB^2$ is rational. Therefore $BG$ is rational. We have that $DE : EF$ is not the ratio that a square number has to another square number. Therefore $BG^2 : GC^2$ is not the ratio that a square number has to another square number. Therefore from {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: :$BG$ is incommensurable in length with $GC$. Both $BG$ and $GC$ are rational. Therefore $BG$ and $GC$ rational straight lines which are commensurable in square only. Therefore $BC$ is an apotome. It remains to be shown that $BC$ is a fifth apotome. Let $H$ be a straight line such that $H^2 = BG^2 - GC^2$. We have that: :$DE : EF = BG^2 : GC^2$  So by {{EuclidPropLink|book = V|prop = 19|qual=Porism|title = Proportional Magnitudes have Proportional Remainders}}: :$ED : EF = BG^2 : H^2$ So $ED$ does not have to $EF$ the ratio that a square number has to another square number. Therefore neither does $GB^2$ have to $H^2$ the ratio that a square number has to another square number. Therefore from {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: :$BG$ is incommensurable in length with $H$. We have that: :$BG^2 = GC^2 + H^2$ Therefore $GB^2$ is greater than $GC^2$ by the square on a straight line which is incommensurable in length with $GB$. Also, the annex $CG$ is commensurable in length with the rational straight line $A$. Therefore, by definition, $BC$ is a fifth apotome. {{qed}} {{Euclid Note|89|X}}
:500px 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. Let $CB : BD$ also not be the ratio that a square number has to a square number. Using {{EuclidPropLink|book = X|prop = 6|qual=Porism|title = Magnitudes with Rational Ratio are Commensurable}}, 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 {{EuclidPropLink|book = X|prop = 6|title = Magnitudes with Rational Ratio are Commensurable}}: :$A^2$ 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 {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: :$A$ is incommensurable in length with $FG$. We have that: :$BC : CD = FG^2 : GH^2$ From {{EuclidPropLink|book = X|prop = 6|title = Magnitudes with Rational Ratio are Commensurable}}: :$FG^2$ 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 {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: :$FG$ 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 sixth apotome. We have that: :$E : BC = A^2 : FG^2$. and: :$BC : CD = FG^2 : GH^2$ Therefore from {{EuclidPropLink|book = V|prop = 22|title = Equality of Ratios Ex Aequali}}: :$E : CD = A^2 : HG^2$ 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 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 {{EuclidPropLink|book = V|prop = 19|qual=Porism|title = Proportional Magnitudes have Proportional Remainders}}: :$BC : BD = FG^2 : K^2$ But $BC$ does not have to $BD$ the ratio that a square number has to another square number. Therefore $FG^2$ does not have to $K^2$ the ratio that a square number has to another square number. Therefore by {{EuclidPropLink|book = X|prop = 9|title = Commensurability of Squares}}: :$FG$ is incommensurable in length with $K$. So: :$FG^2$ is greater than $GH^2$ by the square on $K$. :$FG^2$ is greater than $GH^2$ by the square on a straight line which is incommensurable in length with $FG$. So neither $FG$ nor $GH$ is commensurable in length with the rational straight line $A$. Therefore, by definition, $FH$ is a sixth apotome. {{qed}} {{Euclid Note|90|X}}
:400px Let the area $AB$ be contained by the rational straight line $AC$ and the fifth apotome $AD$. It is to be proved that the "side" of $AB$ is a straight line which produces with a rational area a medial whole. Let $DG$ be the annex of the fifth apotome $AD$. Then, by definition: :$AG$ and $GD$ are rational straight lines which are commensurable in square only :the annex $GD$ is commensurable with the rational straight line $AC$ :the square on the whole $AG$ is greater than the square on the annex $GD$ by the square on a straight line which is incommensurable in length with $AG$. Let there be applied to $AG$ a parallelogram equal to the fourth part of the square on $GD$ and deficient by a square figure. From {{EuclidPropLink|book = X|prop = 18|title = Condition for Incommensurability of Roots of Quadratic Equation}}: :that parallelogram divides $AG$ into incommensurable parts. Let $DG$ be bisected at $E$. Let the rectangle contained by $AF$ and $FG$ be applied to $AG$ which is equal to the square on $EG$ and deficient by a square figure. Therefore $AF$ is incommensurable with $FG$. We have that $AG$ commensurable in length with $AC$, and both are rational. Therefore from {{EuclidPropLink|book = X|prop = 21|title = Medial is Irrational}}: :the rectangle $AK$ is medial. We have that $DG$ isincommensurable in length with $AC$, while both are  rational. Therefore from {{EuclidPropLink|book = X|prop = 21|title = Medial is Irrational}}:  :the rectangles $DK$ is medial. Let the square $LM$ be constructed equal to $AI$. Let the square $NO$ be subtracted from $LM$ having the common angle $\angle LPM$ equal to $FK$. Therefore from {{EuclidPropLink|book = VI|prop = 26|title = Parallelogram Similar and in Same Angle has Same Diameter}}: :the squares $LM$ and $NO$ are about the same diameter. Let $PR$ be the diameter of $LM$ and $NO$. It is to be shown that $LN$ is the "side" of the area $AB$. We have that $AK$ is medial and equals $LP^2 + PN^2$. Therefore $LP^2 + PN^2$ is medial. We have that the rectangle $DK$ is rational. But $DK$ equals twice the rectangle contained by $LP$ and $PN$. Therefore twice the rectangle contained by $LP$ and $PN$ is rational. We have that $AI$ is incommensurable with $FK$. Therefore $LP^2$ is incommensurable with $PN^2$. Therefore $LP$ and $PN$ are straight lines which are incommensurable in square such that $LP^2 + PN^2$ is medial and such that $2 \cdot LP \cdot PN$ is rational. Therefore by definition $LN$ is a straight line which produces with a rational area a medial whole. But $LN$ is the "side" of the area $AB$. Hence the result. {{qed}} {{Euclid Note|95|X|{{EuclidNoteConverse|prop=101|title=Square on Straight Line which produces Medial Whole with Rational Area applied to Rational Straight Line}}}}
:300px Let $AB$ be an apotome of a medial straight line. Let $CD$ be commensurable in length with $AB$. It is to be demonstrated that: :$CD$ is an apotome of a medial straight line and: :the order of $CD$ is the same as the order of $AB$. Let $BE$ be the annex of $CD$. Therefore by definition of apotome of a medial straight line: :$AE$ and $EB$ are medial straight lines which are commensurable in square only. From {{EuclidPropLink|book = VI|prop = 12|title = Construction of Fourth Proportional Straight Line}}, let it be contrived that: :$BE : DF = AB : CD$ From {{EuclidPropLink|book = V|prop = 12|title = Sum of Components of Equal Ratios}}: :$AE : CF = AB : CD$ and so from {{EuclidPropLink|book = V|prop = 11|title = Equality of Ratios is Transitive}}: :$AE : CF = BE : DF$ But $AB$ is commensurable in length with $CD$. Therefore from {{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: :$AE$ is commensurable in length with $CF$ and: :$BE$ is commensurable in length with $DF$. We have that $AE$ and $EB$ are medial straight lines which are commensurable in square only. From {{EuclidPropLink|book = X|prop = 23|title = Straight Line Commensurable with Medial Straight Line is Medial}}: : $CF$ and $FD$ are medial straight lines. Therefore from {{EuclidPropLink|book = X|prop = 13|title = Commensurable Magnitudes are Incommensurable with Same Magnitude}}: : $CF$ and $FD$ are medial straight lines which are commensurable in square only. Thus $CD$ is an apotome of a medial straight line. {{qed|lemma}} It remains to be shown that the order of $CD$ is the same as the order of $AB$. We have that: :$AE : CF = BE : DF$ So from {{EuclidPropLink|book = V|prop = 16|title = Proportional Magnitudes are Proportional Alternately}}: :$AE : EB = CF : FD$ Therefore: :$AE^2 : AE \cdot EB = CF^2 : CF \cdot FD$ But $AE^2$ is commensurable with $CF^2$. Therefore from: :{{EuclidPropLink|book = V|prop = 16|title = Proportional Magnitudes are Proportional Alternately}}: and: :{{EuclidPropLink|book = X|prop = 11|title = Commensurability of Elements of Proportional Magnitudes}}: we have that: :$AE \cdot EB$ is commensurable with $CF \cdot FD$. Therefore by {{EuclidDefLink|X|4|Rational Area}}: :if $AE \cdot EB$ is rational, then $CF \cdot FD$ is rational. From {{EuclidPropLink|book = X|prop = 23|qual=Porism|title = Straight Line Commensurable with Medial Straight Line is Medial}}: :if $AE \cdot EB$ is medial, then $CF \cdot FD$ is medial. Thuis $CD$ is an apotome of a medial straight line of the same as the order as $AB$. {{qed}} {{Euclid Note|104|X}}