Proof: By Euclid

For let the rectangle $AC$ be contained by the rational straight lines $AB$ and $BC$ (which are) commensurable in square only.
I say that $AC$ is irrational, and its square root is irrational - let it be called medial.

fig020e

For let the square $AD$ have been described on $AB$.
$AD$ is thus rational [Def. 10.4] .
And since $AB$ is incommensurable in length with $BC$.
For they were assumed to be commensurable in square only.
And $AB$ (is) equal to $BD$.
$DB$ is thus also incommensurable in length with $BC$.
And as $DB$ is to $BC$, so $AD$ (is) to $AC$ [Prop. 6.1].
Thus, $DA$ [is] [incommensurable]bookofproofs$1095 with $AC$ [Prop. 10.11].
And $DA$ (is) rational.
Thus, $AC$ is irrational [Def. 10.4] .
Hence, its square root [that is to say, the square root of the square equal to it] is also irrational [Def. 10.4] - let it be called medial.
(Which is) the very thing it was required to show.
∎

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