Construction of Medial Straight Lines Commensurable in Square Only containing Medial 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 medial 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 with the greater.
(The Elements: Book $\text{X}$: Proposition $32$)
Proof
Let $A$, $B$ and $C$ be rational straight lines which are commensurable in square only.
Let $A^2 > C^2$ such that:
- $A^2 = C^2 + \rho^2$
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 $D$ be a straight line such that $D^2 = A B$.
From Medial is Irrational, $A B$ is medial.
Therefore $D^2$ is medial.
So by definition $D$ is medial.
Let $D E = B C$.
Since:
- $A B : B C = A : C$
and:
- $D^2 = A B$
and:
- $D E = B C$
then:
- $A : C = D^2 : D E$
So as:
- $D^2 : D E = D : E$
it follows that:
- $A : C = D : E$
But $A$ is commensurable in square only with $C$.
Therefore by Commensurability of Elements of Proportional Magnitudes:
- $D$ is commensurable in square only with $E$.
We have that $D$ is medial.
Therefore by Straight Line Commensurable with Medial Straight Line is Medial, $E$ is medial.
We have that:
- $A : C = D : E$
and:
- $A^2 = C^2 + \rho^2$
where $\rho$ is a straight line which is commensurable in length with $A$.
Therefore Commensurability of Squares on Proportional Straight Lines:
- $D^2 = E^2 + \sigma^2$
where $\sigma$ is a straight line which is commensurable in length with $D$.
We have that:
- $B C = D E$
while from Medial is Irrational:
- $B C$ is medial.
Therefore $D E$ is also medial.
Therefore two medial straight lines $D$ and $E$ have been found which are commensurable in square only and which contain a medial rectangle.
Further, $D^2$ is greater than $E^2$ by the square on a straight line commensurable in length with $D$.
Similarly it can be proved that the square on $D$ exceeds the square on $E$ by the square on a straight line incommensurable in length with $D$, when the square on $A$ is greater than the square on $C$ by the square on a straight line incommensurable in length with $A$.
$\blacksquare$
Historical Note
This proof is Proposition $32$ 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