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

Proof: By Euclid

(related to Proposition: Prop. 10.033: Construction of Components of Major)

• Let the two rational (straight lines) $AB$ and $BC$, (which are) commensurable in square only, be laid out such that the square on the greater, $AB$, is larger than (the square on) the lesser, $BC$, by the (square) on (some straight line which is) incommensurable (in length) with ($AB$) [Prop. 10.30].
• And let $BC$ have been cut in half at $D$.
• And let a parallelogram equal to the (square) on either of $BD$ or $DC$, (and) falling short by a square figure, have been applied to $AB$ [Prop. 6.28], and let it be the (rectangle contained) by $AEB$.
• And let the semicircle $AFB$ have been drawn on $AB$.
• And let $EF$ have been drawn at right angles to $AB$.
• And let $AF$ and $FB$ have been joined.¹

• And since $AB$ and $BC$ are [two] unequal straight lines, and the square on $AB$ is greater than (the square on) $BC$ by the (square) on (some straight line which is) incommensurable (in length) with ($AB$).
• And a parallelogram, equal to one quarter of the (square) on $BC$ - that is to say, (equal) to the (square) on half of it - (and) falling short by a square figure, has been applied to $AB$, and makes the (rectangle contained) by $AEB$.
• $AE$ is thus incommensurable (in length) with $EB$ [Prop. 10.18].
• And as $AE$ is to $EB$, so the (rectangle contained) by $BA$ and $AE$ (is) to the (rectangle contained) by $AB$ and $BE$.
• And the (rectangle contained) by $BA$ and $AE$ (is) equal to the (square) on $AF$, and the (rectangle contained) by $AB$ and $BE$ to the (square) on $BF$ [Prop. 10.32 lem.] .
• The (square) on $AF$ is thus incommensurable with the (square) on on $FB$ [Prop. 10.11].
• Thus, $AF$ and $FB$ are incommensurable in square.
• And since $AB$ is rational, the (square) on $AB$ is also rational.
• Hence, the sum of the (squares) on $AF$ and $FB$ is also rational [Prop. 1.47].
• And, again, since the (rectangle contained) by $AE$ and $EB$ is equal to the (square) on $EF$, and the (rectangle contained) by $AE$ and $EB$ was assumed (to be) equal to the (square) on $BD$, $FE$ is thus equal to $BD$.
• Thus, $BC$ is double $FE$.
• And hence the (rectangle contained) by $AB$ and $BC$ is commensurable with the (rectangle contained) by $AB$ and $EF$ [Prop. 10.6].
• And the (rectangle contained) by $AB$ and $BC$ (is) medial [Prop. 10.21].
• Thus, the (rectangle contained) by $AB$ and $EF$ (is) also medial [Prop. 10.23 corr.] .
• And the (rectangle contained) by $AB$ and $EF$ (is) equal to the (rectangle contained) by $AF$ and $FB$ [Prop. 10.32 lem.] .
• Thus, the (rectangle contained) by $AF$ and $FB$ (is) also medial.
• And the sum of the squares on them was also shown (to be) rational.
• Thus, the two straight lines, $AF$ and $FB$, (which are) incommensurable in square, have been found, making the sum of the squares on them rational, and the (rectangle contained) by them medial.
• (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"

Footnotes