To find a first apotome. * Let the rational (straight line) $A$ be laid down. * And let $BG$ be commensurable in length with $A$. * $BG$ is thus also a rational (straight line). * And let two square numbers $DE$ and $EF$ be laid down, and let their difference $FD$ be not square [Prop. 10.28 lem. I] . * Thus, $ED$ does not have to $DF$ the ratio which (some) square number (has) to (some) square number. * And let it have been contrived that as $ED$ (is) to $DF$, so the square on $BG$ (is) to the square on $GC$ [Prop. 10.6 corr.] . * Thus, the (square) on $BG$ is commensurable with the (square) on on $GC$ [Prop. 10.6]. * And the (square) on $BG$ (is) rational. * Thus, the (square) on $GC$ (is) also rational. * Thus, $GC$ is also rational. * And since $ED$ does not have to $DF$ 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). * Thus, $BG$ and $GC$ are 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 first (apotome).
This proposition proves that the first apotome has length \[\alpha-\alpha\,\sqrt{1-\beta^{\,2}},\]
where \(\alpha,\beta\) denote positive rational numbers.
See also [Prop. 10.48].
Proofs: 1
Propositions: 1
