Construction of Components of Major
Theorem
In the words of Euclid:
- To find two straight lines incommensurable in square which make the sum of the squares on them rational but the rectangle contained by them medial.
(The Elements: Book $\text{X}$: Proposition $33$)
Lemma
In the words of Euclid:
- Let $ABC$ be a right-angled triangle having the angle $A$ right, and let the perpendicular $AD$ be drawn;
I say that the rectangle $CB, BD$ is equal to the square on $BA$, the rectangle $BC, CD$ equal to the square on $CA$, the rectangle $BD, DC$ equal to the square on $AD$, and, further, the rectangle $BC, AD$ equal to the rectangle $BA, AC$.
(The Elements: Book $\text{X}$: Proposition $33$ : Lemma)
Proof
Let $AB$ and $BC$ be rational straight lines which are commensurable in square only such that:
- $AB^2 = BC^2 + \rho^2$
such that $\rho$ is incommensurable in length with $AB$.
Let $BC$ be bisected at $D$.
From Construction of Parallelogram Equal to Given Figure Exceeding a Parallelogram:
Let a parallelogram be applied to $AB$ equal to the square on either of $BD$ or $DC$, and deficient by a square.
Let this parallelogram be the rectangle contained by $AE$ and $EB$.
Let the semicircle $AFB$ be drawn with $AB$ as the diameter.
Let $EF$ be drawn perpendicular to $AB$.
Join $AF$ and $FB$.
We have that $AB > BC$ such that $AB^2 = BC^2 + \rho^2$ such that $\rho$ is incommensurable in length with $AB$.
We also have that the rectangle contained by $AE$ and $EB$ equals the parallelogram on $AB$ equal to $\dfrac {BC} 4$.
Thus from Condition for Incommensurability of Roots of Quadratic Equation:
- $AE$ is incommensurable in length with $EB$.
We have that:
- $AE : BE = AB \cdot AE : AB \cdot BE$
From Lemma to Proposition $33$ of Book $\text{X} $: Construction of Components of Major:
- $AB \cdot AE = AF^2$
and:
- $AB \cdot BE = BF^2$
Therefore $AF^2$ and $BF^2$ are incommensurable.
By definition, $AF$ and $BF$ are therefore incommensurable in square.
As $AB$ is a rational straight line, it follows by definition that $AB^2$ is a rational area.
From Pythagoras's Theorem:
- $AB^2 = \left({AF + FB}\right)^2$
Thus $\left({AF + FB}\right)^2$ is also a rational area.
Therefore $AF + FB$ is rational.
From Lemma to Proposition $33$ of Book $\text{X} $: Construction of Components of Major:
- $AE \cdot EB = EF^2$
and by hypothesis:
- $AE \cdot EB = BD^2$
then:
- $FE = BD$
Therefore:
- $BC = 2 FE$
Thus $AB \cdot BC$ is commensurable with $AB \cdot EF$.
But from Medial is Irrational:
- $AB \cdot BC$ is medial.
Therefore by Porism to Proposition $23$ of Book $\text{X} $: Straight Line Commensurable with Medial Straight Line is Medial:
- $AB \cdot EF$ is also medial.
But from Lemma to Proposition $33$ of Book $\text{X} $: Construction of Components of Major:
- $AB \cdot EF = AF \cdot FB$
Therefore $AF \cdot FB$ is also medial.
But it has been proved that $AF + FB$ is rational.
Therefore we have found two rational straight lines which are incommensurable in square whose sum of squares is rational, but such that the rectangle contained by them is medial.
$\blacksquare$
Also see
Historical Note
This proof is Proposition $33$ of Book $\text{X}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 3 (2nd ed.) ... (previous) ... (next): Book $\text{X}$. Propositions