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

Proof: By Euclid

(related to Proposition: Prop. 10.083: Construction of that which produces Medial Whole with Rational Area is Unique)

• Let $AB$ be a (straight line) which with a rational (area) makes a medial whole, and let $BC$ be (so) attached to $AB$.
• Thus, $AC$ and $CB$ are (straight lines which are) incommensurable in square, fulfilling the (other) proscribed (conditions) [Prop. 10.77].
• I say that another (straight line) fulfilling the same (conditions) cannot be attached to $AB$.

• For, if possible, let $BD$ be (so) attached (to $AB$).
• Thus, $AD$ and $DB$ are also straight lines (which are) incommensurable in square, fulfilling the (other) prescribed (conditions) [Prop. 10.77].
• Therefore, analogously to the (propositions) before this, since by whatever (area) the (sum of the squares) on $AD$ and $DB$ exceeds the (sum of the squares) on $AC$ and $CB$, twice the (rectangle contained) by $AD$ and $DB$ also exceeds twice the (rectangle contained) by $AC$ and $CB$ by this (same area).
• And twice the (rectangle contained) by $AD$ and $DB$ exceeds twice the (rectangle contained) by $AC$ and $CB$ by a rational (area).
• For they are (both) rational (areas).
• Thus, the (sum of the squares) on $AD$ and $DB$ also exceeds the (sum of the squares) on $AC$ and $CB$ by a rational (area).
• The very thing is impossible.
• For both are medial (areas) [Prop. 10.26].
• Thus, another straight line cannot be attached to $AB$, which is incommensurable in square with the whole, and fulfills the (other) aforementioned (conditions) with the whole.
• Thus, only one (such straight line) can be (so) attached.
• (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"