Proposition: Prop. 10.059: Root of Area contained by Rational Straight Line and Sixth Binomial

(Proposition 59 from Book 10 of Euclid's “Elements”)

If an area is contained by a rational (straight line) and a sixth binomial (straight line) then the square root of the area is the irrational (straight line which is) called the square root of (the sum of) two medial (areas). * For let the area $ABCD$ be contained by the rational (straight line) $AB$ and the sixth binomial (straight line) $AD$, which has been divided into its (component) terms at $E$, such that $AE$ is the greater term. * So, I say that the square root of $AC$ is the square root of (the sum of) two medial (areas).

Modern Formulation

If the rational straight line has unit length then this proposition states that the square root of a sixth binomial straight line is the square root of the sum of two medial area: i.e., a sixth binomial straight line has a length $\sqrt{\alpha}+\sqrt{\beta}$ whose square root can be written $\alpha^{1/4}\left(\sqrt{\frac 12+\frac{\delta}{2\sqrt{1+\delta^2}}}+\sqrt{\frac 12-\frac{\delta}{2\sqrt{1+\delta^2}}}\right),$ where $\delta^2=\frac{\alpha-\beta}\beta.$ This is the length of the square root of the sum of two medial area (see [Prop. 10.41]).

Proofs: 1

Proofs: 1 2
Propositions: 3

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