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

Proof: By Euclid

(related to Proposition: Prop. 10.101: Square on Straight Line which produces Medial Whole with Rational Area applied to Rational Straight Line)

• Let $AB$ be that (straight line) which with a rational (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 fifth apotome.

• Let $BG$ be an attachment to $AB$.

• Thus, the straight lines $AG$ and $GB$ are incommensurable in square, making the sum of the squares on them medial, and twice the (rectangle contained) by them rational [Prop. 10.77].
• And let $CH$, equal to the (square) on $AG$, have been applied to $CD$, and $KL$, equal to the (square) on $GB$.
• The whole of $CL$ is thus equal to the (sum of the squares) on $AG$ and $GB$.
• And the sum of the (squares) on $AG$ and $GB$ together is medial.
• Thus, $CL$ is medial.
• And it has been 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 the whole of $CL$ 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].
• Therefore, let $FM$ have been cut in half at $N$.
• And let $NO$ have been drawn through $N$, parallel to either of $CD$ or $ML$.
• Thus, $FO$ and $NL$ are each equal to the (rectangle contained) by $AG$ and $GB$.
• And since twice the (rectangle contained) by $AG$ and $GB$ is rational, and [is] equal to $FL$, $FL$ is thus rational.
• And it is applied to the rational (straight line) $EF$, producing $FM$ as breadth.
• Thus, $FM$ is rational, and commensurable in length with $CD$ [Prop. 10.20].
• And since $CL$ is medial, and $FL$ rational, $CL$ is thus incommensurable with $FL$.
• And as $CL$ (is) to $FL$, so $CM$ (is) to $MF$ [Prop. 6.1].
• $CM$ is thus incommensurable in length with $MF$ [Prop. 10.11].
• And both are 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 fifth (apotome) .
• For, similarly (to the previous propositions), we can show that the (rectangle contained) by $CKM$ is equal to the (square) on $NM$ - that is to say, to the fourth part of the (square) on $FM$.
• And since the (square) on $AG$ is incommensurable with the (square) on on $GB$, and the (square) on $AG$ (is) equal to $CH$, and the (square) on $GB$ to $KL$, $CH$ (is) thus 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].
• Therefore, since $CM$ and $MF$ are two unequal straight lines, and (some area), equal to the fourth part of the (square) on $FM$, has been applied to $CM$, falling short by a square figure, and divides it into incommensurable (parts), the square on $CM$ is thus greater than (the square on) $MF$ by the (square) on (some straight line) incommensurable (in length) with ($CM$) [Prop. 10.18].
• And the attachment $FM$ is commensurable with the (previously) laid down rational (straight line) $CD$.
• Thus, $CF$ is a fifth apotome [Def. 10.15] .
• (Which is) the very thing it was required to show.
  ∎

Thank you to the contributors under CC BY-SA 4.0!

Github:

non-Github:

@Fitzpatrick

References

Adapted from (subject to copyright, with kind permission)

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