Proof: By Euclid

For let the two rational (straight lines), $AB$ and $BC$, (which are) commensurable in square only, be laid down together.
I say that the whole (straight line), $AC$, is irrational.

fig036e

For since $AB$ is incommensurable in length with $BC$ - for they are commensurable in square only - and as $AB$ (is) to $BC$, so the (rectangle contained) by $ABC$ (is) to the (square) on $BC$, the (rectangle contained) by $AB$ and $BC$ is thus incommensurable with the (square) on on $BC$ [Prop. 10.11].
But, twice the (rectangle contained) by $AB$ and $BC$ is commensurable with the (rectangle contained) by $AB$ and $BC$ [Prop. 10.6].
And (the sum of) the (squares) on $AB$ and $BC$ is commensurable with the (square) on on $BC$ - for the rational (straight lines) $AB$ and $BC$ are commensurable in square only [Prop. 10.15].
Thus, twice the (rectangle contained) by $AB$ and $BC$ is incommensurable with (the sum of) the (squares) on $AB$ and $BC$ [Prop. 10.13].
And, via composition, twice the (rectangle contained) by $AB$ and $BC$, plus (the sum of) the (squares) on $AB$ and $BC$ - that is to say, the (square) on $AC$ [Prop. 2.4] - is incommensurable with the sum of the (squares) on $AB$ and $BC$ [Prop. 10.16].
And the sum of the (squares) on $AB$ and $BC$ (is) rational.
Thus, the (square) on $AC$ [is] [irrational]bookofproofs$2083 [ [Def. 10.4] ]bookofproofs$2084.
Hence, $AC$ is also irrational [Def. 10.4] - let it be called a binomial (straight line).
(Which is) the very thing it was required to show.
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