Medial Area not greater than Medial Area by Rational Area

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In the words of Euclid:

A medial area does not exceed a medial area by a rational area.

(The Elements: Book $\text{X}$: Proposition $26$)


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 $y = \sqrt \lambda \rho$.


$\rho \paren {x - y} = \rho z$

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 $(2)$:

$y \smile z$


\(\ds y : z\) \(=\) \(\ds y^2 : y z\)
\(\ds y^2\) \(\smile\) \(\ds y z\)


\(\ds y^2 + z^2\) \(\frown\) \(\ds y^2\)
\(\ds 2 y z\) \(\frown\) \(\ds y z\)
\(\ds \leadsto \ \ \) \(\ds y^2 + z^2\) \(\smile\) \(\ds 2 y z\)
\(\ds \leadsto \ \ \) \(\ds \paren {y + z}^2\) \(\smile\) \(\ds \paren {y^2 + z^2}\)
\(\ds \leadsto \ \ \) \(\ds x^2\) \(\smile\) \(\ds \paren {y^2 + z^2}\)

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.


Historical Note

This proof is Proposition $26$ of Book $\text{X}$ of Euclid's The Elements.