I say that the whole 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 is the rectangle AB, BC to the square on BC, therefore the rectangle AB, BC is incommensurable with the square on BC. [[X. 11](elem.10.11)
]

But twice the rectangle AB, BC is commensurable with the rectangle AB, BC [[X. 6](elem.10.6)
], 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](elem.10.15)
] therefore twice the rectangle AB, BC is incommensurable with the squares on AB, BC. [[X. 13](elem.10.13)
]

And, componendo, twice the rectangle AB, BC together with the squares on AB, BC, that is, the square on AC [[II. 4](elem.2.4)
], is incommensurable with the sum of the squares on AB, BC. [[X. 16](elem.10.16)
]

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](elem.10.def.4)
]

And let it be called binomial.