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

Proof: By Euclid

(related to Proposition: Prop. 10.063: Square on Major Straight Line applied to Rational Straight Line)

• Let $AB$ be a major (straight line) having been divided at $C$, such that $AC$ is greater than $CB$, and (let) $DE$ (be) a rational (straight line).
• And let the parallelogram $DF$, equal to the (square) on $AB$, have been applied to $DE$, producing $DG$ as breadth.
• I say that $DG$ is a fourth binomial (straight line).

• Let the same construction be made as that shown previously.
• And since $AB$ is a major (straight line), having been divided at $C$, $AC$ and $CB$ are incommensurable in square, making the sum of the squares on them rational, and the (rectangle contained) by them medial [Prop. 10.39].
• Therefore, since the sum of the (squares) on $AC$ and $CB$ is rational, $DL$ is thus rational.
• Thus, $DM$ (is) also rational, and commensurable in length with $DE$ [Prop. 10.20].
• Again, since twice the (rectangle contained) by $AC$ and $CB$ - that is to say, $MF$ - is medial, and is (applied to) the rational (straight line) $ML$, $MG$ is thus also rational, and incommensurable in length with $DE$ [Prop. 10.22].
• $DM$ is thus also incommensurable in length with $MG$ [Prop. 10.13].
• $DM$ and $MG$ are thus rational (straight lines which are) commensurable in square only.
• Thus, $DG$ is a binomial (straight line) [Prop. 10.36].
• So we must show that (it is) also a fourth (binomial straight line).
• So, similarly to the previous (propositions), we can show that $DM$ is greater than $MG$, and that the (rectangle contained) by $DKM$ is equal to the (square) on $MN$.
• Therefore, since the (square) on $AC$ is incommensurable with the (square) on on $CB$, $DH$ is also incommensurable with $KL$.
• Hence, $DK$ is also incommensurable with $KM$ [Prop. 6.1], [Prop. 10.11].
• And if there are two unequal straight lines, and a parallelogram equal to the fourth part of the (square) on the lesser, falling short by a square figure, is applied to the greater, and divides it into (parts which are) incommensurable (in length) , then the square on the greater will be larger than (the square on) the lesser by the (square) on (some straight line) incommensurable in length with the greater [Prop. 10.18].
• Thus, the square on $DM$ is greater than (the square on) $MG$ by the (square) on (some straight line) incommensurable (in length) with ($DM$).
• And $DM$ and $MG$ are rational (straight lines which are) commensurable in square only.
• And $DM$ is commensurable (in length) with the (previously) laid down rational (straight line) $DE$.
• Thus, $DG$ is a fourth binomial (straight line) [Def. 10.8] .
• (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"