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

Proof: By Euclid

(related to Proposition: Prop. 10.098: Square on First Apotome of Medial Straight Line applied to Rational Straight Line)

• Let $AB$ be a first apotome of a medial (straight line), 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 second apotome.

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

• Thus, $AG$ and $GB$ are medial (straight lines which are) commensurable in square only, containing a rational (area) [Prop. 10.74].
• And let $CH$, equal to the (square) on $AG$, have been applied to $CD$, producing $CK$ as breadth, and $KL$, equal to the (square) on $GB$, producing $KM$ as breadth.
• Thus, the whole of $CL$ is equal to the (sum of the squares) on $AG$ and $GB$.
• Thus, $CL$ (is) also a medial (area) [Prop. 10.15], [Prop. 10.23 corr.] .
• And it is applied to the rational (straight line) $CD$, producing $CM$ as breadth.
• $CM$ is thus rational, and incommensurable in length with $CD$ [Prop. 10.22].
• And since $CL$ is equal to the (sum of the squares) on $AG$ and $GB$, of which the (square) on $AB$ is equal to $CE$, the remainder, twice the (rectangle contained) by $AG$ and $GB$, is thus equal to $FL$ [Prop. 2.7].
• And twice the (rectangle contained) by $AG$ and $GB$ [is] [rational]bookofproofs$2083.
• Thus, $FL$ (is) rational.
• And it is applied to the rational (straight line) $FE$, producing $FM$ as breadth.
• $FM$ is thus also rational, and commensurable in length with $CD$ [Prop. 10.20].
• Therefore, since the (sum of the squares) on $AG$ and $GB$ - that is to say, $CL$ - is medial, and twice the (rectangle contained) by $AG$ and $GB$ - that is to say, $FL$ - (is) rational, $CL$ is thus incommensurable with $FL$.
• And as $CL$ (is) to $FL$, so $CM$ is to $FM$ [Prop. 6.1].
• Thus, $CM$ (is) incommensurable in length with $FM$ [Prop. 10.11].
• And they are both rational (straight lines).
• 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 second (apotome) .
• For let $FM$ have been cut in half at $N$.
• And let $NO$ have been drawn through (point) $N$, parallel to $CD$.
• Thus, $FO$ and $NL$ are each equal to the (rectangle contained) by $AG$ and $GB$.
• 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 the (square) on $AG$ is equal to $CH$, and the (rectangle contained) by $AG$ and $GB$ to $NL$, and the (square) on $BG$ to $KL$, $NL$ is thus also in mean proportion3 to $CH$ and $KL$.
• Thus, as $CH$ is to $NL$, so $NL$ (is) to $KL$ [Prop. 5.11].
• But, as $CH$ (is) to $NL$, so $CK$ is to $NM$, and as $NL$ (is) to $KL$, so $NM$ is to $MK$ [Prop. 6.1].
• Thus, as $CK$ (is) to $NM$, so $NM$ is to $KM$ [Prop. 5.11].
• The (rectangle contained) by $CK$ and $KM$ is thus equal to the (square) on $NM$ [Prop. 6.17] - that is to say, to the fourth part of the (square) on $FM$ [and since the (square) on $AG$ is commensurable with the (square) on on $BG$, $CH$ is also commensurable with $KL$ - that is to say, $CK$ with $KM$].
• Therefore, since $CM$ and $MF$ are two unequal straight lines, and the (rectangle contained) by $CK$ and $KM$, equal to the fourth part of the (square) on $MF$, has been applied to the greater $CM$, falling short by a square figure, and divides it into commensurable (parts), the square on $CM$ is thus greater than (the square on) $MF$ by the (square) on (some straight line) commensurable in length with ($CM$) [Prop. 10.17].
• The attachment $FM$ is also commensurable in length with the (previously) laid down rational (straight line) $CD$.
• $CF$ is thus a second apotome [Def. 10.16] .
• Thus, the (square) on a first apotome of a medial (straight line), applied to a rational (straight line), produces a second apotome as breadth.
• (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"