Kitap 10
114
Let its terms be KM, ML commensurable with the terms CE, ED of the binomial straight line and in the same ratio. [X. 112]
114
But CE, ED are also commensurable with AF, FB and in the same ratio; therefore, as AF is to FB, so is KM to ML.
114
Therefore, alternately, as AF is to KM, so is BF to LM; therefore also the remainder AB is to the remainder KL as AF is to KM. [V. 19]
114
But AF is commensurable with KM; [X. 12] therefore AB is also commensurable with KL. [X. 11]
114
And, as AB is to KL, so is the rectangle CD, AB to the rectangle CD, KL; [VI. 1] therefore the rectangle CD, AB is also commensurable with the rectangle CD, KL. [X. 11]
114
But the rectangle CD, KL is equal to the square on H; therefore the rectangle CD, AB is commensurable with the square on H.
114
But the square on G is equal to the rectangle CD, AB; therefore the square on G is commensurable with the square on H.
114
But the square on H is rational; therefore the square on G is also rational; therefore G is rational.
114
And it is the side of the rectangle CD, AB.
114
Therefore etc.
114
Porism. And it is made manifest to us by this also that it is possible for a rational area to be contained by irrational straight lines. Q. E. D.
PROPOSITION 115.
115
From a medial straight line there arise irrational straight lines infinite in number, and none of them is the same as any of the preceding.
115
Let A be a medial straight line; I say that from A there arise irrational straight lines infinite in number, and none of them is the same as any of the preceding.
115
Let a rational straight line B be set out, and let the square on C be equal to the rectangle B, A; therefore C is irrational; [X. Def. 4] for that which is contained by an irrational and a rational straight line is irrational. [deduction from X. 20]
115
And it is not the same with any of the preceding; for the square on none of the preceding, if applied to a rational straight line produces as breadth a medial straight line.
