Proof: By Euclid
(related to Proposition: Prop. 10.010: Construction of Incommensurable Lines)
- Let $A$ be the given straight line.
-
So it is required to find two straight lines incommensurable with $A$, the one (incommensurable) in length only, the other also (incommensurable) in square.
-
For let two numbers, $B$ and $C$, not having to one another the ratio which (some) square number (has) to (some) square number - that is to say, not (being) similar plane (numbers) - have been taken.
- And let it be contrived that as $B$ (is) to $C$, so the square on $A$ (is) to the square on $D$.
- For we learned (how to do this) [Prop. 10.6 corr.] .
- Thus, the (square) on $A$ (is) commensurable with the (square) on on $D$ [Prop. 10.6].
- And since $B$ does not have to $C$ the ratio which (some) square number (has) to (some) square number, the (square) on $A$ thus does not have to the (square) on $D$ the ratio which (some) square number (has) to (some) square number either.
- Thus, $A$ is incommensurable in length with $D$ [Prop. 10.9].
- Let the (straight line) $E$ (which is) in mean proportion3 to $A$ and $D$ have been taken [Prop. 6.13].
- Thus, as $A$ is to $D$, so the square on $A$ (is) to the (square) on $E$ [Def. 5.9] .
- And $A$ is incommensurable in length with $D$.
- Thus, the square on $A$ is also incommensurble with the square on $E$ [Prop. 10.11].
- Thus, $A$ is incommensurable in square with $E$.
- Thus, two straight lines, $D$ and $E$, (which are) incommensurable with the given straight line $A$, have been found, the one, $D$, (incommensurable) in length only, the other, $E$, (incommensurable) in square, and, clearly, also in length.
- (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)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"
