(related to Proposition: Prop. 10.036: Binomial is Irrational)

- 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.

- 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.∎

**Fitzpatrick, Richard**: Euclid's "Elements of Geometry"