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Proof: By Euclid

(related to Proposition: Prop. 10.022: Square on Medial Straight Line)

• Let $A$ be a medial (straight line), and $CB$ a rational (straight line), and let the rectangular area $BD$, equal to the (square) on $A$, have been applied to $BC$, producing $CD$ as breadth.
• I say that $CD$ is rational, and incommensurable in length with $CB$.

• For since $A$ is medial, the square on it is equal to a (rectangular) area contained by rational (straight lines which are) commensurable in square only [Prop. 10.21].
• Let the square on ($A$) be equal to $GF$.
• And the square on ($A$) is also equal to $BD$.
• Thus, $BD$ is equal to $GF$.
• And ($BD$) is also equiangular with ($GF$).
• And for equal and equiangular parallelograms, the sides about the equal angles are reciprocally proportional [Prop. 6.14].
• Thus, proportionally, as $BC$ is to $EG$, so $EF$ (is) to $CD$.
• And, also, as the (square) on $BC$ is to the (square) on $EG$, so the (square) on $EF$ (is) to the (square) on $CD$ [Prop. 6.22].
• And the (square) on $CB$ is commensurable with the (square) on on $EG$.
• For they are each rational.
• Thus, the (square) on $EF$ is also commensurable with the (square) on on $CD$ [Prop. 10.11].
• And the (square) on $EF$ is rational.
• Thus, the (square) on $CD$ is also rational [Def. 10.4] .
• Thus, $CD$ is rational.
• And since $EF$ is incommensurable in length with $EG$.
• For they are commensurable in square only.
• And as $EF$ (is) to $EG$, so the (square) on $EF$ (is) to the (rectangle contained) by $FE$ and $EG$ [see previous lemma].
• The (square) on $EF$ [is] thus incommensurable with the (rectangle contained) by $FE$ and $EG$ [Prop. 10.11].
• But, the (square) on $CD$ is commensurable with the (square) on on $EF$.
• For they are rational in square.
• And the (rectangle contained) by $DC$ and $CB$ is commensurable with the (rectangle contained) by $FE$ and $EG$.
• For they are (both) equal to the (square) on $A$.
• Thus, the (square) on $CD$ is also incommensurable with the (rectangle contained) by $DC$ and $CB$ [Prop. 10.13].
• And as the (square) on $CD$ (is) to the (rectangle contained) by $DC$ and $CB$, so $DC$ is to $CB$ [see previous lemma].
• Thus, $DC$ is incommensurable in length with $CB$ [Prop. 10.11].
• Thus, $CD$ is rational, and incommensurable in length with $CB$.
• (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"