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.
∎
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References
Adapted from (subject to copyright, with kind permission)
 Fitzpatrick, Richard: Euclid's "Elements of Geometry"