Proof: By Euclid
(related to Proposition: Prop. 10.075: Second Apotome of Medial is Irrational)
- For let the rational (straight line) $DI$ be laid down.
- And let $DE$, equal to the (sum of the squares) on $AB$ and $BC$, have been applied to $DI$, producing $DG$ as breadth.
- And let $DH$, equal to twice the (rectangle contained) by $AB$ and $BC$, have been applied to $DI$, producing $DF$ as breadth.
- The remainder $FE$ is thus equal to the (square) on $AC$ [Prop. 2.7].
- And since the (squares) on $AB$ and $BC$ are medial and commensurable (with one another), $DE$ (is) thus also medial [Prop. 10.15], [Prop. 10.23 corr.] .
- And it is applied to the rational (straight line) $DI$, producing $DG$ as breadth.
- Thus, $DG$ is rational, and incommensurable in length with $DI$ [Prop. 10.22].
- Again, since the (rectangle contained) by $AB$ and $BC$ is medial, twice the (rectangle contained) by $AB$ and $BC$ is thus also medial [Prop. 10.23 corr.] .
- And it is equal to $DH$.
- Thus, $DH$ is also medial.
- And it has been applied to the rational (straight line) $DI$, producing $DF$ as breadth.
- $DF$ is thus rational, and incommensurable in length with $DI$ [Prop. 10.22].
- And since $AB$ and $BC$ are commensurable in square only, $AB$ is thus incommensurable in length with $BC$.
- Thus, the square on $AB$ (is) also incommensurable with the (rectangle contained) by $AB$ and $BC$ [Prop. 10.21 lem.] , [Prop. 10.11].
- But, the (sum of the squares) on $AB$ and $BC$ is commensurable with the (square) on on $AB$ [Prop. 10.15], and twice the (rectangle contained) by $AB$ and $BC$ is commensurable with the (rectangle contained) by $AB$ and $BC$ [Prop. 10.6].
- Thus, twice the (rectangle contained) by $AB$ and $BC$ is incommensurable with the (sum of the squares) on $AB$ and $BC$ [Prop. 10.13].
- And $DE$ is equal to the (sum of the squares) on $AB$ and $BC$, and $DH$ to twice the (rectangle contained) by $AB$ and $BC$.
- Thus, $DE$ [is] [incommensurable]bookofproofs$1095 with $DH$.
- And as $DE$ (is) to $DH$, so $GD$ (is) to $DF$ [Prop. 6.1].
- Thus, $GD$ is incommensurable with $DF$ [Prop. 10.11].
- And they are both rational (straight lines).
- Thus, $GD$ and $DF$ are rational (straight lines which are) commensurable in square only.
- Thus, $FG$ is an apotome [Prop. 10.73].
- And $DI$ (is) rational.
- And the (area) contained by a rational and an irrational (straight line) is irrational [Prop. 10.20], and its square root is irrational.
- And $AC$ is the square root of $FE$.
- Thus, $AC$ is an irrational (straight line) [Def. 10.4] .
- And let it be called the second apotome of a medial (straight line).
- (Which is) the very thing it was required to show.
∎
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References
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"
