Commensurability is Transitive Relation
Theorem
In the words of Euclid:
- Magnitudes commensurable with the same magnitude are commensurable with one another also.
(The Elements: Book $\text{X}$: Proposition $12$)
Proof
Let $A$ and $B$ be magnitudes which are both commensurable with another magnitude $C$.
As $A$ is commensurable with $C$, from Ratio of Commensurable Magnitudes:
Let it have the ratio which $D$ has to $E$, where $D$ and $E$ are numbers.
As $C$ is commensurable with $B$, from Ratio of Commensurable Magnitudes:
Let it have the ratio which $F$ has to $G$, where $F$ and $G$ are numbers.
From Construction of Sequence of Numbers with Given Ratios, let the numbers $H$, $K$, $L$ be assigned the ratios:
- $D : E = H : K$
- $F : G = K : L$
So:
| \(\ds A : C\) | \(=\) | \(\ds D : E\) | ||||||||||||
| \(\ds D : E\) | \(=\) | \(\ds H : K\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds A : C\) | \(=\) | \(\ds H : K\) | Equality of Ratios is Transitive |
and:
| \(\ds C : B\) | \(=\) | \(\ds F : G\) | ||||||||||||
| \(\ds F : G\) | \(=\) | \(\ds K : L\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds C : B\) | \(=\) | \(\ds K : L\) | Equality of Ratios is Transitive |
Then:
| \(\ds C : B\) | \(=\) | \(\ds K : L\) | ||||||||||||
| \(\ds A : C\) | \(=\) | \(\ds H : K\) | ||||||||||||
| \(\ds \leadsto \ \ \) | \(\ds A : B\) | \(=\) | \(\ds H : L\) | Equality of Ratios Ex Aequali |
That is, $A$ has to $B$ the ratio which a number has to a number.
From Magnitudes with Rational Ratio are Commensurable:
- $A$ and $B$ are commensurable with each other.
$\blacksquare$
Historical Note
This proof is Proposition $12$ 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