A (straight line) commensurable in length with a binomial (straight line) is itself also binomial, and the same in order. * Let $AB$ be a binomial (straight line), and let $CD$ be commensurable in length with $AB$. * I say that $CD$ is a binomial (straight line), and (is) the same in order^{1} as $AB$.
(not yet contributed)
Proofs: 1
Euclid's expression "(not) being the same in order" means that the resulting irrational number is "(not) of the same kind" as that irrational number, with which it is commensurable. ↩