Kitap 10
65
Similarly again we can prove that the rectangle DK, KM is equal to the square on MN, and that DK is incommensurable in length with KM; and, for the same reason, the square on DM is greater than the square on MG by the square on a straight line incommensurable in length with DM.
65
And neither of the straight lines DM, MG is commensurable in length with the rational straight line DE set out.
65
Therefore DG is a sixth binomial straight line. Q. E. D.
PROPOSITION 66.
66
A straight line commensurable in length with a binomial straight line is itself also binomial and the same in order.
66
Let AB be binomial, and let CD be commensurable in length with AB; I say that CD is binomial and the same in order with AB.
66
For, since AB is binomial, let it be divided into its terms at E, and let AE be the greater term; therefore AE, EB are rational straight lines commensurable in square only. [X. 36]
66
Let it be contrived that, as AB is to CD, so is AE to CF; [VI. 12] therefore also the remainder EB is to the remainder FD as AB is to CD. [V. 19]
66
But AB is commensurable in length with CD; therefore AE is also commensurable with CF, and EB with FD. [X. 11]
66
And AE, EB are rational; therefore CF, FD are also rational.
66
And, as AE is to CF, so is EB to FD. [V. 11]
66
Therefore, alternately, as AE is to EB, so is CF to FD. [V. 16]
66
But AE, EB are commensurable in square only; therefore CF, FD are also commensurable in square only. [X. 11]
66
And they are rational; therefore CD is binomial. [X. 36]
66
I say next that it is the same in order with AB.
66
For the square on AE is greater than the square on EB either by the square on a straight line commensurable with AE or by the square on a straight line incommensurable with it.
66
If then the square on AE is greater than the square on EB by the square on a straight line commensurable with AE, the square on CF will also be greater than the square on FD by the square on a straight line commensurable with CF. [X. 14]
