Medial Area not greater than Medial Area by Rational Area
Theorem
In the words of Euclid:
- A medial area does not exceed a medial area by a rational area.
(The Elements: Book $\text{X}$: Proposition $26$)
Proof
Let $\rho \cdot \sqrt k \rho$ and $\rho \cdot \sqrt \lambda \rho$ be two medial areas which have been applied to the same rational number $\rho$.
Their difference is then:
- $\paren {\sqrt k - \sqrt \lambda} \rho^2$
Let $x = \sqrt k \rho$ and $y = \sqrt \lambda \rho$.
Let:
- $\rho \paren {x - y} = \rho z$
Suppose $\rho z$ is a rational area.
Then $z$ must be a rational straight line and:
- $(1): \quad z \frown \rho$
where $\frown$ denotes commensurability in length.
Since $\rho x$ and $\rho y$ are medial areas, it follows that $x$ and $y$ are rational and:
- $(2): \quad x \smile \rho$ and $y \smile \rho$
where $\smile$ denotes incommensurability in length.
Then from $(1)$ and $(2)$:
- $y \smile z$
Then:
\(\ds y : z\) | \(=\) | \(\ds y^2 : y z\) | ||||||||||||
\(\ds y^2\) | \(\smile\) | \(\ds y z\) |
But:
\(\ds y^2 + z^2\) | \(\frown\) | \(\ds y^2\) | ||||||||||||
\(\ds 2 y z\) | \(\frown\) | \(\ds y z\) | ||||||||||||
\(\ds \leadsto \ \ \) | \(\ds y^2 + z^2\) | \(\smile\) | \(\ds 2 y z\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds \paren {y + z}^2\) | \(\smile\) | \(\ds \paren {y^2 + z^2}\) | |||||||||||
\(\ds \leadsto \ \ \) | \(\ds x^2\) | \(\smile\) | \(\ds \paren {y^2 + z^2}\) |
But $y^2 + z^2$ is rational.
Therefore $x^2$ must be irrational.
Therefore $x$ must be irrational.
But from $(2)$ $x$ is rational.
The result follows from this contradiction.
$\blacksquare$
Historical Note
This proof is Proposition $26$ 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