To find a fourth binomial (straight line). * Let the two numbers $AC$ and $CB$ be laid down such that $AB$ does not have to $BC$, or to $AC$ either, the ratio which (some) square number (has) to (some) square number [Prop. 10.28 lem. I] . * And let the rational (straight line) $D$ be laid down. * And let $EF$ be commensurable in length with $D$. * Thus, $EF$ is also a rational (straight line). * 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.] . * Thus, the (square) on $EF$ is commensurable with the (square) on on $FG$ [Prop. 10.6]. * Thus, $FG$ is also a rational (straight line). * And since $BA$ does not have to $AC$ the ratio which (some) square number (has) to (some) square number, 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]. * Thus, $EF$ and $FG$ are rational (straight lines which are) commensurable in square only. * Hence, $EG$ is a binomial (straight line) [Prop. 10.36]. * So, I say that (it is) also a fourth (binomial straight line).
If the rational straight line has unit length then the length of a fourth binomial straight line is \[\alpha+\frac \alpha{\sqrt{1+\beta}},\]
where \(\alpha,\beta\) denote positive rational numbers.
This, and the fourth apotome, whose length according to [Prop. 10.88] is \[\alpha-\frac \alpha{\sqrt{1+\beta}}\] are the roots of the quadratic function \[x^2- 2\,\alpha\,x+\frac{\alpha^2\,\beta}{1+\beta}=0,\]
where \(\alpha,\beta\) denote positive rational numbers.
Proofs: 1
Propositions: 1
