Proof: By Euclid
(related to Proposition: Prop. 10.015: Commensurability of Sum of Commensurable Magnitudes)
- For let the two commensurable magnitudes AB and BC be laid down together.
- I say that the whole AC is also commensurable with each of AB and BC.

- For since AB and BC are commensurable, some magnitude will measure them.
- Let it (so) measure (them), and let it be D.
- Therefore, since D measures (both) AB and BC, it will also measure the whole AC.
- And it also measures AB and BC.
- Thus, D measures AB, BC, and AC.
- Thus, AC is commensurable with each of AB and BC [Def. 10.1] .
- And so let AC be commensurable with AB.
- I say that AB and BC are also commensurable.
- For since AC and AB are commensurable, some magnitude will measure them.
- Let it (so) measure (them), and let it be D.
- Therefore, since D measures (both) CA and AB, it will thus also measure the remainder BC.
- And it also measures AB.
- Thus, D will measure (both) AB and BC.
- Thus, AB and BC are commensurable [Def. 10.1] .
- Thus, if two magnitudes, and so on ....
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References
Adapted from (subject to copyright, with kind permission)
- Fitzpatrick, Richard: Euclid's "Elements of Geometry"