Euclid, Elements
Thomas L. Heath, Sir Thomas Little Heath, Ed.

PROPOSITION 16.

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.

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.

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. [X. Def. 1]

Therefore etc.

LEMMA.

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.

Therefore etc.