Kitap 10
50
And since, as BA is to AC, so is the square on FG to the square on GH, therefore the square on FG is greater than the square on GH.
50
Let then the squares on GH, K be equal to the square on FG; therefore, convertendo, as AB is to BC, so is the square on FG to the square on K. [V. 19, Por.]
50
But AB has to BC the ratio which a square number has to a square number; therefore the square on FG also has to the square on K the ratio which a square number has to a square number; therefore FG is commensurable in length with K. [X. 9]
50
Therefore the square on FG is greater than the square on GH by the square on a straight line commensurable with FG.
50
And FG, GH are rational straight lines commensurable in square only, and neither of them is commensurable in length with E.
50
Therefore FH is a third binomial straight line. Q. E. D.
PROPOSITION 51.
51
To find the fourth binomial straight line.
51
Let two numbers AC, CB be set out such that AB neither has to BC, nor yet to AC, the ratio which a square number has to a square number.
51
Let a rational straight line D be set out, and let EF be commensurable in length with D; therefore EF is also rational.
51
Let it be contrived that, as the number BA is to AC, so is the square on EF to the square on FG; [X. 6, Por.] therefore the square on EF is commensurable with the square on FG; [X. 6] therefore FG is also rational.
51
Now, since BA has not to AC the ratio which a square number has to a square number, neither has the square on EF to the square on FG the ratio which a square number has to a square number; therefore EF is incommensurable in length with FG. [X. 9]
51
Therefore EF, FG are rational straight lines commensurable in square only; so that EG is binomial.
51
I say next that it is also a fourth binomial straight line.
51
For since, as BA is to AC, so is the square on EF to the square on FG, therefore the square on EF is greater than the square on FG.
