Proof: By Euclid

For let the two straight lines, $AB$ and $BC$, incommensurable in square, (and) fulfilling the prescribed (conditions), be laid down together [Prop. 10.34].
I say that $AC$ is irrational.

fig036e

For since the sum of the (squares) on $AB$ and $BC$ is medial, and twice the (rectangle contained) by $AB$ and $BC$ (is) rational, the sum of the (squares) on $AB$ and $BC$ is thus incommensurable with twice the (rectangle contained) by $AB$ and $BC$.
Hence, the (square) on $AC$ is also incommensurable with twice the (rectangle contained) by $AB$ and $BC$ [Prop. 10.16].
And twice the (rectangle contained) by $AB$ and $BC$ (is) rational.
The (square) on $AC$ (is) thus irrational.
Thus, $AC$ (is) irrational [Def. 10.4] - let it be called the square root of a rational plus a medial (area) .
(Which is) the very thing it was required to show.
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