To find a fifth binomial straight line. * Let the two numbers $AC$ and $CB$ be laid down such that $AB$ does not have to either of them the ratio which (some) square number (has) to (some) square number [Prop. 10.28 lem. II] . * And let some rational straight line $D$ be laid down. * And let $EF$ be commensurable [in length] with $D$. * Thus, $EF$ (is) a rational (straight line). * And let it have been contrived that as $CA$ (is) to $AB$, so the (square) on $EF$ (is) to the (square) on $FG$ [Prop. 10.6 corr.] . * And $CA$ does not have to $AB$ 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, $EF$ and $FG$ are rational (straight lines which are) commensurable in square only [Prop. 10.9]. * Thus, $EG$ is a binomial (straight line) [Prop. 10.36]. * So, I say that (it is) also a fifth (binomial straight line).
If the rational straight line has unit length then the length of a fifth binomial straight line is \[\alpha\,(\sqrt{1+\beta}+1),\]
where \(\alpha,\beta\) denote positive rational numbers.
This, and the fifth apotome, whose length according to [Prop. 10.89] is \[\alpha\,(\sqrt{1+\beta}-1),\] are the roots of the quadratic function \[x^2- 2\,\alpha\,\sqrt{1+\beta}\,x+\alpha^2\,\beta=0,\]
where \(\alpha,\beta\) denote positive rational numbers.
Proofs: 1
Propositions: 1
