Commensurability of Sum of Commensurable Magnitudes
Theorem
In the words of Euclid:
- If two commensurable magnitudes be added together, the whole will also be commensurable with each of them; and, if the whole be commensurable with one of them, the original magnitudes will also be commensurable.
(The Elements: Book $\text{X}$: Proposition $15$)
Proof
Let $AB$ and $BC$ be commensurable magnitudes which are added together to make $AC$.
Since $AB$ and $BC$ are commensurable, then some magnitude $D$ will measure them both.
Since $D$ measures both $AB$ and $BC$, $D$ also measures $AC$.
That is, $D$ measures $AB$, $BC$ and $AC$.
Therefore from Book $\text{X}$ Definition $1$: Commensurable, $AC$ is commensurable with each of the magnitudes $AB$ and $BC$.
$\Box$
Let $AC$ be commensurable with $AB$.
Since $AC$ and $AB$ are commensurable, then some magnitude $D$ will measure them both.
Since $D$ measures both $AC$ and $AB$, $D$ also measures the remainder $BC$.
That is, $D$ measures $AB$, $BC$ and $AC$.
Therefore from Book $\text{X}$ Definition $1$: Commensurable, $AB$ and $BC$ are commensurable.
$\blacksquare$
Historical Note
This proof is Proposition $15$ 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