Proposition: Prop. 10.053: Construction of Sixth Binomial Straight Line

(Proposition 53 from Book 10 of Euclid's “Elements”)

To find a sixth binomial (straight line). * Let the two numbers $AC$ and $CB$ be laid down such that $AB$ does not have to each of them the ratio which (some) square number (has) to (some) square number. * And let $D$ also be another number, which is not square, and does not have to each of $BA$ and $AC$ the ratio which (some) square number (has) to (some) square number either [Prop. 10.28 lem. I] . * And let some rational straight line $E$ be laid down. * And let it have been contrived that as $D$ (is) to $AB$, so the (square) on $E$ (is) to the (square) on $FG$ [Prop. 10.6 corr.] . * Thus, the (square) on $E$ (is) commensurable with the (square) on on $FG$ [Prop. 10.6]. * And $E$ is rational. * Thus, $FG$ (is) also rational. * And since $D$ does not have to $AB$ the ratio which (some) square number (has) to (some) square number, the (square) on $E$ thus does not have to the (square) on $FG$ the ratio which (some) square number (has) to (some) square number either. * Thus, $E$ (is) incommensurable in length with $FG$ [Prop. 10.9]. * So, again, let it have be contrived that as $BA$ (is) to $AC$, so the (square) on $FG$ (is) to the (square) on $GH$ [Prop. 10.6 corr.] . * The (square) on $FG$ (is) thus commensurable with the (square) on on $HG$ [Prop. 10.6]. * The (square) on $HG$ (is) thus rational. * Thus, $HG$ (is) rational. * And since $BA$ does not have to $AC$ the ratio which (some) square number (has) to (some) square number, the (square) on $FG$ does not have to the (square) on $GH$ the ratio which (some) square number (has) to (some) square number either. * Thus, $FG$ is incommensurable in length with $GH$ [Prop. 10.9]. * Thus, $FG$ and $GH$ are rational (straight lines which are) commensurable in square only. * Thus, $FH$ is a binomial (straight line) [Prop. 10.36]. * So, we must show that (it is) also a sixth (binomial straight line).

fig053e

Modern Formulation

If the rational straight line has unit length then the length of a sixth binomial straight line is \[\sqrt{\alpha}+\sqrt{\beta},\]

where $\alpha,\beta$ denote positive rational numbers.

Notes

This, and the sixth apotome, whose length according to [Prop. 10.90] is \[\sqrt{\alpha}-\sqrt{\beta},\] are the roots of the quadratic function \[x^2- 2\,\sqrt{\alpha}\,x+(\alpha-\beta)=0,\] where $\alpha,\beta$ denote positive rational numbers.

Proofs: 1