To find a first binomial (straight line). * Let two numbers $AC$ and $CB$ be laid down such that their sum $AB$ has to $BC$ the ratio which (some) square number (has) to (some) square number, and does not have to $CA$ the ratio which (some) square number (has) to (some) square number [Prop. 10.28 lem. I] . * And let some rational (straight line) $D$ be laid down. * And let $EF$ be commensurable in length with $D$. * $EF$ is thus also rational [Def. 10.3] . * And let it have been contrived that as the number $BA$ (is) to $AC$, so the (square) on $EF$ (is) to the (square) on $FG$ [Prop. 10.6 corr.] . * And $AB$ has to $AC$ the ratio which (some) number (has) to (some) number. * Thus, the (square) on $EF$ also has to the (square) on $FG$ the ratio which (some) number (has) to (some) number. * Hence, the (square) on $EF$ is commensurable with the (square) on on $FG$ [Prop. 10.6]. * And $EF$ is rational. * Thus, $FG$ (is) also rational. * And since $BA$ does not have to $AC$ the ratio which (some) square number (has) to (some) square number, thus the (square) on $EF$ does not have to the (square) on $FG$ the ratio which (some) square number (has) to (some) square number either. * Thus, $EF$ is incommensurable in length with $FG$ [Prop. 10.9]. * $EF$ and $FG$ are thus rational (straight lines which are) commensurable in square only. * Thus, $EG$ is a binomial (straight line) [Prop. 10.36]. * I say that (it is) also a first (binomial straight line).
If the rational straight line has unit length then the length of a first binomial straight line is \[\alpha+\alpha\sqrt{1-\beta^{\,2}},\]
where \(\alpha,\beta\) denote positive rational numbers.
This, and the first apotome, whose length according to [Prop. 10.85] is \[\alpha-\alpha\,\sqrt{1-\beta^{\,2}},\] are the roots of the quadratic function \[x^2- 2\,\alpha\,x+\alpha^2\,\beta^{\,2}=0, \]
where \(\alpha,\beta\) denote positive rational numbers.
Proofs: 1
Propositions: 1