To find a fifth apotome. * Let the rational (straight line) $A$ be laid down, and let $CG$ be commensurable in length with $A$. * Thus, $CG$ [is] a rational (straight line). * And let the two numbers $DF$ and $FE$ be laid down such that $DE$ again does not have to each of $DF$ and $FE$ the ratio which (some) square number (has) to (some) square number. * And let it have been contrived that as $FE$ (is) to $ED$, so the (square) on $CG$ (is) to the (square) on $GB$. * Thus, the (square) on $GB$ (is) also rational [Prop. 10.6]. * Thus, $BG$ is also rational. * And since as $DE$ is to $EF$, so the (square) on $BG$ (is) to the (square) on $GC$. * And $DE$ does not have to $EF$ the ratio which (some) square number (has) to (some) square number. * The (square) on $BG$ thus does not have to the (square) on $GC$ the ratio which (some) square number (has) to (some) square number either. * Thus, $BG$ is incommensurable in length with $GC$ [Prop. 10.9]. * And they are both rational (straight lines). * $BG$ and $GC$ are thus rational (straight lines which are) commensurable in square only. * Thus, $BC$ is an apotome [Prop. 10.73]. * So, I say that (it is) also a fifth (apotome) .
This proposition proves that the fifth apotome has length \[\alpha\,(\sqrt{1+\beta}-1),\]
where \(\alpha,\beta\) denote positive rational numbers.
See also [Prop. 10.52].
Proofs: 1
Propositions: 1