# Greatest Common Divisor of Three Numbers

## Theorem

In the words of Euclid:

Given three numbers not prime to one another, to find their greatest common measure.

## Proof

Let $A, B, C$ be the three given (natural) numbers not prime to one another. Let the GCD $D$ of $A, B$ be found, by the Euclidean Algorithm.

Then $D$ either divides or does not divide $C$.

First, suppose $D$ divides $C$.

Then as it also divides $A$ and $B$, it is a common divisor of $A, B$ and $C$.

Suppose $D$ is not the GCD of $A, B, C$.

Then some number $E$ is a common divisor of $A, B$ and $C$ such that $E > D$.

But then $E$ is a common divisor of $A$ and $B$ and so from the porism to Euclid's Algorithm $E$ also divides $D$.

That means $E \le D$ which contradicts $E > D$.

So $D$ is the GCD of $A, B, C$ after all.

Next, suppose $D$ is not a divisor of $C$.

Since $A, B, C$ are not prime to one another, some number will divide them.

That number which divides $A, B, C$ will also divide $A, B$ and so from the porism to Euclid's Algorithm will also divide $D$.

But it also divides $C$.

So some number divides both $D$ and $C$, and so $C$ and $D$ are not prime to one another.

Let the GCD of $C$ and $D$ be $E$, by the Euclidean Algorithm.

Since $E$ divides $D$ and $D$ divides $A, B$, then $E$ divides $A, B$.

But $E$ divides $C$, and so $E$ divides $A, B, C$.

So $E$ is a common divisor of $A, B$ and $C$.

Now suppose $E$ is not the GCD of $A, B, C$.

Then some number $F$ such that $F > E$ divides $A, B, C$.

It follows that $F$ divides $A$ and $B$, and so from the porism to Euclid's Algorithm will also divide $D$, the GCD of $A, B$.

So $F$ divides $D$ and $F$ also divides $C$.

So $F$ is a common divisor of $D$ and $C$.

So by the porism to Euclid's Algorithm, $F$ divides the GCD of $C, D$, which is $E$.

But this contradicts the supposition that $F > E$.

So no such $F$ can exist, and so $E$ is the GCD of $A, B, C$.

$\blacksquare$

## Historical Note

This proof is Proposition $3$ of Book $\text{VII}$ of Euclid's The Elements.