Construction of Medial Straight Lines Commensurable in Square Only containing Rational Rectangle whose Square Differences Commensurable with Greater
Theorem
In the words of Euclid:
- To find two medial straight lines commensurable in square only, containing a rational rectangle, and such that the square on the greater is greater than the square on the less by the square on a straight line commensurable in length with the greater.
(The Elements: Book $\text{X}$: Proposition $31$)
Proof
Let $A$ and $B$ be rational straight lines which are commensurable in square only.
Let $A^2 > B^2$ such that:
- $A^2 = B^2 + \rho^2$
where $\rho$ is a straight line which is commensurable in length with $A$.
This can be done using Construction of Rational Straight Lines Commensurable in Square Only whose Square Differences Commensurable with Greater.
Let $C$ be a straight line such that $C^2 = A B$.
From Medial is Irrational, $A B$ is medial.
Therefore $C^2$ is medial.
So by definition $C$ is medial.
Let $C D = B^2$.
By hypothesis $B^2$ is rational.
Therefore $C D$ is rational.
We have that:
- $A : B = A^2 : A B$
But:
- $C^2 = A B$
and:
- $C D = B^2$
and so:
- $A : B = C^2 : C D$
But as:
- $C^2 : C D = C : D$
it follows that:
- $A : B = C : D$
But $A$ is commensurable in square only with $B$.
Therefore by Commensurability of Elements of Proportional Magnitudes:
- $C$ is commensurable in square only with $D$.
We have that $C$ is medial.
Therefore by Straight Line Commensurable with Medial Straight Line is Medial, $D$ is medial.
We have that:
- $A : B = C : D$
and:
- $A^2 = B^2 + \rho^2$
where $\rho$ is a straight line which is commensurable in length with $A$.
Therefore Commensurability of Squares on Proportional Straight Lines:
- $C^2 = D^2 + \sigma^2$
where $\sigma$ is a straight line which is commensurable in length with $C$.
Therefore two medial straight lines $C$ and $D$ have been found which are commensurable in square only and which contain a rational rectangle.
Further, $C^2$ is greater than $D^2$ by the square on a straight line commensurable in length with $C$.
Similarly it can be proved that the square on $C$ exceeds the square on $D$ by the square on a straight line incommensurable in length with $C$, when the square on $A$ is greater than the square on $B$ by the square on a straight line incommensurable in length with $A$.
$\blacksquare$
Historical Note
This proof is Proposition $31$ 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