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

Proof: By Euclid

(related to Proposition: Prop. 10.068: Straight Line Commensurable with Major Straight Line is Major)

• Let $AB$ be a major (straight line), and let $CD$ be commensurable (in length) with $AB$.
• I say that $CD$ is a major (straight line).

• Let $AB$ have been divided (into its component terms) at $E$.
• $AE$ and $EB$ are thus incommensurable in square, making the sum of the squares on them rational, and the (rectangle contained) by them medial [Prop. 10.39].
• And let (the) same (things) have been contrived as in the previous (propositions).
• And since as $AB$ is to $CD$, so $AE$ (is) to $CF$ and $EB$ to $FD$, thus also as $AE$ (is) to $CF$, so $EB$ (is) to $FD$ [Prop. 5.11].
• And $AB$ (is) commensurable (in length) with $CD$.
• Thus, $AE$ and $EB$ (are) also commensurable (in length) with $CF$ and $FD$, respectively [Prop. 10.11].
• And since as $AE$ is to $CF$, so $EB$ (is) to $FD$, also, alternately, as $AE$ (is) to $EB$, so $CF$ (is) to $FD$ [Prop. 5.16], and thus, via composition, as $AB$ is to $BE$, so $CD$ (is) to $DF$ [Prop. 5.18].
• And thus as the (square) on $AB$ (is) to the (square) on $BE$, so the (square) on $CD$ (is) to the (square) on $DF$ [Prop. 6.20].
• So, similarly, we can also show that as the (square) on $AB$ (is) to the (square) on $AE$, so the (square) on $CD$ (is) to the (square) on $CF$.
• And thus as the (square) on $AB$ (is) to (the sum of) the (squares) on $AE$ and $EB$, so the (square) on $CD$ (is) to (the sum of) the (squares) on $CF$ and $FD$.
• And thus, alternately, as the (square) on $AB$ is to the (square) on $CD$, so (the sum of) the (squares) on $AE$ and $EB$ (is) to (the sum of) the (squares) on $CF$ and $FD$ [Prop. 5.16].
• And the (square) on $AB$ (is) commensurable with the (square) on on $CD$.
• Thus, (the sum of) the (squares) on $AE$ and $EB$ (is) also commensurable with (the sum of) the (squares) on $CF$ and $FD$ [Prop. 10.11].
• And the (squares) on $AE$ and $EB$ (added) together are rational.
• The (squares) on $CF$ and $FD$ (added) together (are) thus also rational.
• So, similarly, twice the (rectangle contained) by $AE$ and $EB$ is also commensurable with twice the (rectangle contained) by $CF$ and $FD$.
• And twice the (rectangle contained) by $AE$ and $EB$ is medial.
• Therefore, twice the (rectangle contained) by $CF$ and $FD$ (is) also medial [Prop. 10.23 corr.] .
• $CF$ and $FD$ are thus (straight lines which are) incommensurable in square [Prop. 10.13], simultaneously making the sum of the squares on them rational, and twice the (rectangle contained) by them medial.
• The whole, $CD$, is thus that irrational (straight line) called major [Prop. 10.39].
• Thus, a (straight line) commensurable (in length) with a major (straight line) is major.
• (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"