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

Proof: By Euclid

(related to Proposition: Prop. 10.114: Area contained by Apotome and Binomial Straight Line Commensurable with Terms of Apotome and in same Ratio)

• For let an area, the (rectangle contained) by $AB$ and $CD$, have been contained by the apotome $AB$, and the binomial $CD$, of which let the greater term be $CE$.
• And let the terms of the binomial, $CE$ and $ED$, be commensurable with the terms of the apotome, $AF$ and $FB$ (respectively), and in the same ratio.
• And let the square root of the (rectangle contained) by $AB$ and $CD$ be $G$.
• I say that $G$ is a rational (straight line).

• For let the rational (straight line) $H$ be laid down.
• And let (some rectangle), equal to the (square) on $H$, have been applied to $CD$, producing $KL$ as breadth.
• Thus, $KL$ is an apotome, of which let the terms, $KM$ and $ML$, be commensurable with the terms of the binomial, $CE$ and $ED$ (respectively), and in the same ratio [Prop. 10.112].
• But, $CE$ and $ED$ are also commensurable with $AF$ and $FB$ (respectively), and in the same ratio.
• Thus, as $AF$ is to $FB$, so $KM$ (is) to $ML$.
• Thus, alternately, as $AF$ is to $KM$, so $BF$ (is) to $LM$ [Prop. 5.16].
• Thus, the remainder $AB$ is also to the remainder $KL$ as $AF$ (is) to $KM$ [Prop. 5.19].
• And $AF$ (is) commensurable with $KM$ [Prop. 10.12].
• $AB$ is thus also commensurable with $KL$ [Prop. 10.11].
• And as $AB$ is to $KL$, so the (rectangle contained) by $CD$ and $AB$ (is) to the (rectangle contained) by $CD$ and $KL$ [Prop. 6.1].
• Thus, the (rectangle contained) by $CD$ and $AB$ is also commensurable with the (rectangle contained) by $CD$ and $KL$ [Prop. 10.11].
• And the (rectangle contained) by $CD$ and $KL$ (is) equal to the (square) on $H$.
• Thus, the (rectangle contained) by $CD$ and $AB$ is commensurable with the (square) on on $H$.
• And the (square) on $G$ is equal to the (rectangle contained) by $CD$ and $AB$.
• The (square) on $G$ is thus commensurable with the (square) on on $H$.
• And the (square) on $H$ (is) rational.
• Thus, the (square) on $G$ is also rational.
• $G$ is thus rational.
• And it is the square root of the (rectangle contained) by $CD$ and $AB$.
• Thus, if an area is contained by an apotome, and a binomial whose terms are commensurable with, and in the same ratio as, the terms of the apotome, then the square root of the area is a rational (straight line).
  ∎

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References

Adapted from (subject to copyright, with kind permission)

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