Straight Line Commensurable with Bimedial Straight Line is Bimedial and of Same Order
Theorem
In the words of Euclid:
- A straight line commensurable in length with a bimedial straight line is itself also bimedial and the same in order.
(The Elements: Book $\text{X}$: Proposition $67$)
Proof
Let $AB$ be bimedial.
Let $CD$ be commensurable in length with $AB$.
It is to be shown that $CD$ is bimedial, and that the order of $CD$ is the same as the order of $AB$.
Let $AB$ be divided into its medials by $E$.
Let $AE$ be the greater medial.
By definition, $AE$ and $EB$ are medial straight lines which are commensurable in square only.
Using Proposition $12$ of Book $\text{VI} $: Construction of Fourth Proportional Straight Line, let it be contrived that:
- $AB : CD = AE : CF$
Therefore by Proposition $19$ of Book $\text{V} $: Proportional Magnitudes have Proportional Remainders:
- $EB : FD = AB : CD$
But $AB$ is commensurable in length with $CD$.
Therefore from Proposition $11$ of Book $\text{X} $: Commensurability of Elements of Proportional Magnitudes:
- $AE$ is commensurable in length with $CF$
and:
- $EB$ is commensurable in length with $FD$.
But by hypothesis $AE$ and $EB$ are medial.
Therefore by Proposition $23$ of Book $\text{X} $: Straight Line Commensurable with Medial Straight Line is Medial:
- $CF$ and $FD$ are medial.
From Proposition $11$ of Book $\text{V} $: Equality of Ratios is Transitive:
- $AE : CF = EB : FD$
Therefore by Proposition $16$ of Book $\text{V} $: Proportional Magnitudes are Proportional Alternately:
- $AE : EB = CF : FD$
But by hypothesis $AE$ and $EB$ are commensurable in square only.
Therefore by Proposition $11$ of Book $\text{X} $: Commensurability of Elements of Proportional Magnitudes:
- $CF$ and $FD$ are commensurable in square only.
But $CF$ and $FD$ are medial.
Therefore, by definition, $CD$ is bimedial.
It remains to be demonstrated that $CD$ is of the same order as $AB$.
We have that:
- $AE : EB = CF : FD$
Therefore:
- $AE^2 : AE \cdot EB = CF^2 : CF \cdot FD$
Therefore by Proposition $16$ of Book $\text{V} $: Proportional Magnitudes are Proportional Alternately:
- $AE^2 : CF^2 = AE \cdot EB : CF \cdot FD$
But:
- $AE^2$ is commensurable with $CF^2$.
Therefore $AE \cdot EB$ is commensurable with $CF \cdot FD$.
Suppose $AB$ is a first bimedial.
Then $AE \cdot EB$ is rational.
It follows that $CF \cdot FD$ is rational.
Thus by definition $CD$ is a first bimedial.
Suppose otherwise that $AB$ is a second bimedial.
Then $AE \cdot EB$ is medial.
It follows that $CF \cdot FD$ is medial.
Thus by definition $CD$ is a second bimedial.
Thus $CD$ is of the same order as $AB$.
$\blacksquare$
Historical Note
This proof is Proposition $67$ 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