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.
For let the two incommensurable magnitudes / A B /, / B C / be added together; I say that the whole / A C / is also incommensurable with each of the magnitudes / A B /, / B C /.
For, if / A C /, / A B / are not incommensurable, some magnitude will measure them. Let it measure them, if possible, and let it be D.
Since then D measures / A C /, / A B /,
therefore it will also measure the remainder / B C /.
But it measures / A B / also;
therefore D measures / A B /, / B C /.
Therefore / A B /, / B C / are commensurable;
but they were also, by hypothesis, incommensurable:
which is impossible.
Therefore no magnitude will measure / A C /, / A B /; therefore / A C /, / A B / are incommensurable. [G.372]
Similarly we can prove that / A C /, / B C / are also incommensurable.
Therefore / A C / is incommensurable with each of the magnitudes / A B /, / B C /.
Next, let / A C / be incommensurable with one of the magnitudes / A B /, / B C /.
First, let it be incommensurable with / A B /; I say that / A B /, / B C / 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 / A B /, / B C /,
therefore it will also measure the whole / A C /.
But it measures / A B / also;
therefore D measures / A C /, / A B /.
Therefore / A C /, / A B / are commensurable;
but they were also, by hypothesis, incommensurable:
which is impossible.
Therefore no magnitude will measure / A B /, / B C /; therefore / A B /, / B C / are incommensurable. [G.372]
If to any straight line there be applied a parallelogram deficient by a squarefigure, 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 / A B / the parallelogram < A C D E > deficient by the square figure < C B F D >; I say that < A C D E > is equal to the rectangle contained by / A C /, / C B /.
This is indeed at once manifest;
for, since < C B F D > is a square,
/ C D / is equal to / C B /;
and < A C D E > is the rectangle / A C /, / C D /, that is, the rectangle / A C /, / C B /.
