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

Proof: By Euclid

(related to Proposition: Prop. 10.102: Square on Straight Line which produces Medial Whole with Medial Area applied to Rational Straight Line)

• Let $AB$ be that (straight line) which with a medial (area) makes a medial whole, and $CD$ a rational (straight line).
• And let $CE$, equal to the (square) on $AB$, have been applied to $CD$, producing $CF$ as breadth.

• I say that $CF$ is a sixth apotome.

• For let $BG$ be an attachment to $AB$.

• Thus, $AG$ and $GB$ are incommensurable in square, making the sum of the squares on them medial, and twice the (rectangle contained) by $AG$ and $GB$ medial, and the (sum of the squares) on $AG$ and $GB$ incommensurable with twice the (rectangle contained) by $AG$ and $GB$ [Prop. 10.78].
• Therefore, let $CH$, equal to the (square) on $AG$, have been applied to $CD$, producing $CK$ as breadth, and $KL$, equal to the (square) on $BG$.
• Thus, the whole of $CL$ is equal to the (sum of the squares) on $AG$ and $GB$.
• $CL$ [is] thus also medial.
• And it is applied to the rational (straight line) $CD$, producing $CM$ as breadth.
• Thus, $CM$ is rational, and incommensurable in length with $CD$ [Prop. 10.22].
• Therefore, since $CL$ is equal to the (sum of the squares) on $AG$ and $GB$, of which $CE$ (is) equal to the (square) on $AB$, the remainder $FL$ is thus equal to twice the (rectangle contained) by $AG$ and $GB$ [Prop. 2.7].
• And twice the (rectangle contained) by $AG$ and $GB$ (is) medial.
• Thus, $FL$ is also medial.
• And it is applied to the rational (straight line) $FE$, producing $FM$ as breadth.
• $FM$ is thus rational, and incommensurable in length with $CD$ [Prop. 10.22].
• And since the (sum of the squares) on $AG$ and $GB$ is incommensurable with twice the (rectangle contained) by $AG$ and $GB$, and $CL$ equal to the (sum of the squares) on $AG$ and $GB$, and $FL$ equal to twice the (rectangle contained) by $AG$ and $GB$, $CL$ [is] thus incommensurable with $FL$.
• And as $CL$ (is) to $FL$, so $CM$ is to $MF$ [Prop. 6.1].
• Thus, $CM$ is incommensurable in length with $MF$ [Prop. 10.11].
• And they are both rational.
• Thus, $CM$ and $MF$ are rational (straight lines which are) commensurable in square only.
• $CF$ is thus an apotome [Prop. 10.73].
• So, I say that (it is) also a sixth (apotome) .
• For since $FL$ is equal to twice the (rectangle contained) by $AG$ and $GB$, let $FM$ have been cut in half at $N$, and let $NO$ have been drawn through $N$, parallel to $CD$.
• Thus, $FO$ and $NL$ are each equal to the (rectangle contained) by $AG$ and $GB$.
• And since $AG$ and $GB$ are incommensurable in square, the (square) on $AG$ is thus incommensurable with the (square) on on $GB$.
• But, $CH$ is equal to the (square) on $AG$, and $KL$ is equal to the (square) on $GB$.
• Thus, $CH$ is incommensurable with $KL$.
• And as $CH$ (is) to $KL$, so $CK$ is to $KM$ [Prop. 6.1].
• Thus, $CK$ is incommensurable (in length) with $KM$ [Prop. 10.11].
• And since the (rectangle contained) by $AG$ and $GB$ is in mean proportion3 to the (squares) on $AG$ and $GB$ [Prop. 10.21 lem.] , and $CH$ is equal to the (square) on $AG$, and $KL$ equal to the (square) on $GB$, and $NL$ equal to the (rectangle contained) by $AG$ and $GB$, $NL$ is thus also in mean proportion3 to $CH$ and $KL$.
• Thus, as $CH$ is to $NL$, so $NL$ (is) to $KL$.
• And for the same (reasons as the preceding propositions), the square on $CM$ is greater than (the square on) $MF$ by the (square) on (some straight line) incommensurable (in length) with ($CM$) [Prop. 10.18].
• And neither of them is commensurable with the (previously) laid down rational (straight line) $CD$.
• Thus, $CF$ is a sixth apotome [Def. 10.16] .
• (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"