◀ ▲ ▶Branches / Geometry / Elements-euclid / Book-10-incommensurable-magnitudes / Proof: By Euclid

Proof: By Euclid

(related to Proposition: Prop. 10.056: Root of Area contained by Rational Straight Line and Third Binomial)

• For let the area $ABCD$ be contained by the rational (straight line) $AB$ and by the third binomial (straight line) $AD$, which has been divided into its (component) terms at $E$, of which $AE$ is the greater.

• I say that the square root of area $AC$ is the irrational (straight line which is) called second bimedial.

• For let the same construction be made as previously.

• And since $AD$ is a third binomial (straight line), $AE$ and $ED$ are thus rational (straight lines which are) commensurable in square only, and the square on $AE$ is greater than (the square on) $ED$ by the (square) on (some straight line) commensurable (in length) with ($AE$), and neither of $AE$ and $ED$ [is] [commensurable in length]bookofproofs$1095 with $AB$ [Def. 10.7] .
• So, similarly to that which has been previously demonstrated, we can show that $MO$ is the square root of area $AC$, and $MN$ and $NO$ are medial (straight lines which are) commensurable in square only.
• Hence, $MO$ is bimedial.
• So, we must show that (it is) also second (bimedial).
• And, since $DE$ is incommensurable in length with $AB$ - that is to say, with $EK$ - and $DE$ (is) commensurable (in length) with $EF$, $EF$ is thus incommensurable in length with $EK$ [Prop. 10.13].
• And they are (both) rational (straight lines).
• Thus, $FE$ and $EK$ are rational (straight lines which are) commensurable in square only.
• $EL$ - that is to say, $MR$ - [is] thus medial [Prop. 10.21].
• And it is contained by $MNO$.
• Thus, the (rectangle contained) by $MNO$ is medial.
• Thus, $MO$ is a second bimedial (straight line) [Prop. 10.38].
• (Which is) the very thing it was required to show.
  ∎

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References

Adapted from (subject to copyright, with kind permission)

1. Fitzpatrick, Richard: Euclid's "Elements of Geometry"