◀ ▲ ▶Branches / Geometry / Elements-euclid / Book-10-incommensurable-magnitudes / Proof: By Euclid

Proof: By Euclid

(related to Proposition: Prop. 10.113: Square on Rational Straight Line applied to Apotome)

• Let $A$ be a rational (straight line), and $BD$ an apotome.
• And let the (rectangle contained) by $BD$ and $KH$ be equal to the (square) on $A$, such that the square on the rational (straight line) $A$, applied to the apotome $BD$, produces $KH$ as breadth.
• I say that $KH$ is a binomial whose terms are commensurable with the terms of $BD$, and in the same ratio, and, moreover, that $KH$ has the same order as $BD$.

• For let $DC$ be an attachment to $BD$.
• Thus, $BC$ and $CD$ are rational (straight lines which are) commensurable in square only [Prop. 10.73].
• And let the (rectangle contained) by $BC$ and $G$ also be equal to the (square) on $A$.
• And the (square) on $A$ (is) rational.
• The (rectangle contained) by $BC$ and $G$ (is) thus also rational.
• And it has been applied to the rational (straight line) $BC$.
• Thus, $G$ is rational, and commensurable in length with $BC$ [Prop. 10.20].
• Therefore, since the (rectangle contained) by $BC$ and $G$ is equal to the (rectangle contained) by $BD$ and $KH$, thus, proportionally, as $CB$ is to $BD$, so $KH$ (is) to $G$ [Prop. 6.16].
• And $BC$ (is) greater than $BD$.
• Thus, $KH$ (is) also greater than $G$ [Prop. 5.16], [Prop. 5.14].
• Let $KE$ be made equal to $G$.
• $KE$ is thus commensurable in length with $BC$.
• And since as $CB$ is to $BD$, so $HK$ (is) to $KE$, thus, via convertion, as $BC$ (is) to $CD$, so $KH$ (is) to $HE$ [Prop. 5.19 corr.] 2.
• Let it have been contrived that as $KH$ (is) to $HE$, so $HF$ (is) to $FE$.
• And thus the remainder $KF$ is to $FH$, as $KH$ (is) to $HE$ - that is to say, [as] $BC$ (is) to $CD$ [Prop. 5.19].
• And $BC$ and $CD$ [are] [commensurable in square]bookofproofs$2082 only.
• $KF$ and $FH$ are thus also commensurable in square only [Prop. 10.11].
• And since as $KH$ is to $HE$, (so) $KF$ (is) to $FH$, but as $KH$ (is) to $HE$, (so) $HF$ (is) to $FE$, thus, also as $KF$ (is) to $FH$, (so) $HF$ (is) to $FE$ [Prop. 5.11].
• And hence as the first (is) to the third, so the (square) on the first (is) to the (square) on the second [Def. 5.9] .
• And thus as $KF$ (is) to $FE$, so the (square) on $KF$ (is) to the (square) on $FH$.
• And the (square) on $KF$ is commensurable with the (square) on on $FH$.
• For $KF$ and $FH$ are commensurable in square.
• Thus, $KF$ is also commensurable in length with $FE$ [Prop. 10.11].
• Hence, $KF$ [is] also commensurable in length with $KE$ [Prop. 10.15].
• And $KE$ is rational, and commensurable in length with $BC$.
• Thus, $KF$ (is) also rational, and commensurable in length with $BC$ [Prop. 10.12].
• And since as $BC$ is to $CD$, (so) $KF$ (is) to $FH$, alternately, as $BC$ (is) to $KF$, so $DC$ (is) to $FH$ [Prop. 5.16].
• And $BC$ (is) commensurable (in length) with $KF$.
• Thus, $FH$ (is) also commensurable in length with $CD$ [Prop. 10.11].
• And $BC$ and $CD$ are rational (straight lines which are) commensurable in square only.
• $KF$ and $FH$ are thus also rational (straight lines which are) commensurable in square only [Def. 10.3] , [Prop. 10.13].
• Thus, $KH$ is a binomial [Prop. 10.36].
• Therefore, if the square on $BC$ is greater than (the square on) $CD$ by the (square) on (some straight line) commensurable (in length) with ($BC$), then the square on $KF$ will also be greater than (the square on) $FH$ by the (square) on (some straight line) commensurable (in length) with ($KF$) [Prop. 10.14].
• And if $BC$ is commensurable in length with a (previously) laid down rational (straight line), (so) also (is) $KF$ [Prop. 10.12].
• And if $CD$ is commensurable in length with a (previously) laid down rational (straight line), (so) also (is) $FH$ [Prop. 10.12].
• And if neither of $BC$ or $CD$ (are commensurable), neither also (are) either of $KF$ or $FH$ [Prop. 10.13].
• And if the square on $BC$ is greater than (the square on) $CD$ by the (square) on (some straight line) incommensurable (in length) with ($BC$) then the square on $KF$ will also be greater than (the square on) $FH$ by the (square) on (some straight line) incommensurable (in length) with ($KF$) [Prop. 10.14].
• And if $BC$ is commensurable in length with a (previously) laid down rational (straight line), (so) also (is) $KF$ [Prop. 10.12].
• And if $CD$ is commensurable, (so) also (is) $FH$ [Prop. 10.12].
• And if neither of $BC$ or $CD$ (are commensurable), neither also (are) either of $KF$ or $FH$ [Prop. 10.13].
• $KH$ is thus a binomial whose terms, $KF$ and $FH$, [are] [commensurable (in length) ]bookofproofs$1095 with the terms, $BC$ and $CD$, of the apotome, and in the same ratio.
• Moreover, $KH$ will have the same order as $BC$ [Def. 10.5] , [Def. 10.6] , [Def. 10.7] , [Def. 10.8] , [Def. 10.9] , [Def. 10.10] .
• (Which is) the very thing it was required to show.
  ∎

Thank you to the contributors under CC BY-SA 4.0!

Github:

non-Github:

@Fitzpatrick

References

Adapted from (subject to copyright, with kind permission)

1. Fitzpatrick, Richard: Euclid's "Elements of Geometry"