# Greatest Common Measure of Three Commensurable Magnitudes

## Theorem

In the words of Euclid:

*Given three commensurable magnitudes, to find their greatest common measure.*

(*The Elements*: Book $\text{X}$: Proposition $4$)

### Porism

In the words of Euclid:

*From this it is manifest that, if a magnitude measure three magnitudes, it will also measure their greatest common measure.*

Similarly too, with more magnitudes, the greatest common measure can be found, and the porism can be extended.

(*The Elements*: Book $\text{X}$: Proposition $4$ : Porism)

## Proof

Let $A, B, C$ be three commensurable magnitudes.

From Greatest Common Measure of Commensurable Magnitudes, let the greatest common measure of $A$ and $B$ be $D$.

Either $D$ measures $C$ or it does not.

Let $D$ measure $C$.

Since:

- $D$ measures $C$

and:

- $D$ measures $A$ and $B$

then:

- $D$ is a common measure for $A, B$ and $C$.

As no magnitude can measure a smaller magnitude, it follows that $D$ is the greatest common measure of $A, B$ and $C$.

Let $D$ not measure $C$.

As $A$, $B$ and $C$ are commensurable, then some magnitude $E$ will measure them by definition.

$E$ will measure both $A$ and $B$.

From the porism to Greatest Common Measure of Commensurable Magnitudes, $E$ also measures their greatest common measure $D$.

But $E$ also measures $C$.

Therefore $E$ will measure $C$ and $D$.

Therefore $C$ and $D$ are commensurable.

So from Greatest Common Measure of Commensurable Magnitudes, let $E$ be the greatest common measure of $C$ and $D$.

Since:

- $E$ measures $D$

and:

- $D$ measures $A$ and $B$

then:

- $E$ measures $A$ and $B$.

But $E$ measures $C$ also.

So $E$ is a common measure for $A, B$ and $C$.

Suppose there were some magnitude $F$ which is greater than $E$ which is also a common measure for $A, B$ and $C$.

Since $F$ measures $A, B$ and $C$, it measures $A$ and $B$.

From the porism to Greatest Common Measure of Commensurable Magnitudes, $F$ also measures the greatest common measure of $A$ and $B$.

That is, $F$ measures $C$

But $F$ measures $D$.

From the porism to Greatest Common Measure of Commensurable Magnitudes, $F$ measures the greatest common measure of $D$ and $C$.

That is, $F$ measures $E$.

But $F$ is greater than $E$ and so cannot measure $E$.

From this contradiction it follows that no magnitude greater than $E$ can measure $A, B$ and $C$.

The result follows: $E$ is the greatest common measure of $A, B$ and $C$, unless $D$ measures $C$.

If $D$ measures $C$ then $D$ is the greatest common measure of $A, B$ and $C$.

$\blacksquare$

## Historical Note

This proof is Proposition $4$ 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