For, since AB is incommensurable in length with BC — for they are commensurable in square only — and, as AB is to BC, so is the rectangle AB, BC to the square on BC,
therefore the rectangle AB, BC is incommensurable with the square on BC. [X. 11 ]
But twice the rectangle AB, BC is commensurable with the rectangle AB, BC [X. 6 ],
and the squares on AB, BC are commensurable with the square on BC — for AB, BC are rational straight lines commensurable in square only — [X. 15 ]
therefore twice the rectangle AB, BC is incommensurable with the squares on AB, BC. [X. 13 ]
And, componendo, twice the rectangle AB, BC together with the squares on AB, BC, that is, the square on AC [II. 4 ], is incommensurable with the sum of the squares on AB, BC. [X. 16 ]
But the sum of the squares on AB, BC is rational;
therefore the square on AC is irrational, so that AC is also irrational. [X. Def. 4 ]
And let it be called binomial.