Proof: By Euclid
(related to Proposition: Prop. 10.058: Root of Area contained by Rational Straight Line and Fifth Binomial)
- For let the area $AC$ be contained by the rational (straight line) $AB$ and the fifth binomial (straight line) $AD$, which has been divided into its (component) terms at $E$, such that $AE$ is the greater term.
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So I say that the square root of area $AC$ is the irrational (straight line which is) called the square root of a rational plus a medial (area) .
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For let the same construction be made as that shown previously.
- So, (it is) clear that $MO$ is the square root of area $AC$.
- So, we must show that $MO$ is the square root of a rational plus a medial (area) .
- For since $AG$ is incommensurable (in length) with $GE$ [Prop. 10.18], $AH$ is thus also incommensurable with $HE$ - that is to say, the (square) on $MN$ with the (square) on $NO$ [Prop. 6.1], [Prop. 10.11].
- Thus, $MN$ and $NO$ are incommensurable in square.
- And since $AD$ is a fifth binomial (straight line), and $ED$ [is] its lesser segment, $ED$ (is) thus commensurable in length with $AB$ [Def. 10.9] .
- But, $AE$ is incommensurable (in length) with $ED$.
- Thus, $AB$ is also incommensurable in length with $AE$ [$BA$ and $AE$ are [rational][bookofproofs$2083] ([straight lines][bookofproofs$645] which are) [commensurable in square][bookofproofs$2082] only] [[Prop. 10.13]]bookofproofs$2107.
- Thus, $AK$ - that is to say, the sum of the (squares) on $MN$ and $NO$ - is medial [Prop. 10.21].
- And since $DE$ is commensurable in length with $AB$ - that is to say, with $EK$ - but, $DE$ is commensurable (in length) with $EF$, $EF$ is thus also commensurable (in length) with $EK$ [Prop. 10.12].
- And $EK$ (is) rational.
- Thus, $EL$ - that is to say, $MR$ - that is to say, the (rectangle contained) by $MNO$ - (is) also rational [Prop. 10.19].
- $MN$ and $NO$ are thus (straight lines which are) incommensurable in square, making the sum of the squares on them medial, and the (rectangle contained) by them rational.
- Thus, $MO$ is the square root of a rational plus a medial (area) [Prop. 10.40].
- And (it is) the square root of area $AC$.
- (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"
