If two incommensurable magnitudes be added together, the whole will also be incommensurable with each of them; and, if the whole be incommensurable with one of them, the original magnitudes will also be incommensurable.
若将两个不可公度的量相加，则整体与每个量均不可公度；且若整体与其中一个量不可公度，则原两量亦不可公度。
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For let the two incommensurable magnitudes AB, BC be added together; I say that the whole AC is also incommensurable with each of the magnitudes AB, BC. For, if CA, AB are not incommensurable, some magnitude will measure them. Let it measure them, if possible, and let it be D. Since then D measures CA, AB, therefore it will also measure the remainder BC. But it measures AB also; therefore D measures AB, BC. Therefore AB, BC are commensurable; but they were also, by hypothesis, incommensurable: which is impossible.
设两不可公度量AB、BC相加得AC。假设CA与AB可公度，则存在某量D同时量尽CA和AB，从而D也量尽余量BC。但D量尽AB，故D量尽AB和BC，即AB与BC可公度，与假设矛盾。因此CA与AB不可公度。同理可证AC与CB亦不可公度。
Therefore no magnitude will measure CA, AB; therefore CA, AB are incommensurable. [X. Def. 1] Similarly we can prove that AC, CB are also incommensurable. Therefore AC is incommensurable with each of the magnitudes AB, BC. Next, let AC be incommensurable with one of the magnitudes AB, BC. First, let it be incommensurable with AB; I say that AB, BC are also incommensurable. For, if they are commensurable, some magnitude will measure them.
故AC与AB、BC均不可公度。
Let it measure them, and let it be D. Since then D measures AB, BC. therefore it will also measure the whole AC. But it measures AB also; therefore D measures CA, AB. Therefore CA, AB are commensurable; but they were also, by hypothesis, incommensurable: which is impossible. Therefore no magnitude will measure AB, BC; therefore AB, BC are incommensurable.
反之，设AC与AB不可公度。假设AB与BC可公度，则存在某量D量尽AB和BC，从而D也量尽整体AC。但D量尽AB，故D量尽CA和AB，即CA与AB可公度，与假设矛盾。因此AB与BC不可公度。
[X. Def. 1] Therefore etc. LEMMA. If to any straight line there be applied a parallelogram deficient by a square figure, the applied parallelogram is equal to the rectangle contained by the segments of the straight line resulting from the application. For let there be applied to the straight line AB the parallelogram AD deficient by the square figure DB; I say that AD is equal to the rectangle contained by AC, CB. This is indeed at once manifest; for, since DB is a square, DC is equal to CB; and AD is the rectangle AC, CD, that is, the rectangle AC, CB.
综上，命题得证。