Associative Law of Multiplication/Euclid's Proof

Theorem

 * If a first magnitude be the same multiple of a second that a third is of a fourth, and if equimultiples be taken of the first and third, then also ex aequali the magnitudes taken will be equimultiples respectively, the one of the second and the other of the fourth.

That is, if:
 * $na, nb$ are equimultiples of $a, b$

and if:
 * $m \cdot na, m \cdot nb$ are equimultiples of $na, nb$

then:
 * $m \cdot na$ is the same multiple of $a$ that $m \cdot nb$ is of $b$

Alternatively, this can be expressed as:
 * $m \cdot na = mn \cdot a$

Proof
Let a first magnitude $A$ be the same multiple of a second $B$ that a third $C$ is of a fourth $D$.

Let equimultiples $EF, GH$ be taken of $A, C$.

We need to show that $EF$ is the same multiple of $B$ that $GH$ is of $D$.

We have that $EF$ is the same multiple of $A$ that $GH$ is of $C$.

Therefore as many magnitudes as there are in $EF$ equal to $A$, so many also are there in $GH$ equal to $C$.

Let $EF$ be divided into the magnitudes $EK, KF$ equal to $A$, and $GH$ ito the magnitudes $GL, LH$ equal to $C$.

Then the multitude of the magnitudes $EK, KF$ will be equal to the multitude of the magnitudes $GL, LH$.


 * Euclid-V-3.png

We have that $A$ is the same multiple of $B$ that $C$ is of $D$, while $EK = A$ and $GL = C$.

So $EK$ is the same multiple of $B$ that $GL$ is of $D$.

For the same reason, $KF$ is the same multiple of $B$ that $LH$ is of $D$.

So we have that:
 * a first magnitude $EK$ is the same multiple of a second $B$ that a third $GL$ is of a fourth $D$
 * a fifth $KF$ is also of the same multiple of the second $B$ that a sixth $LH$ is of the fourth $D$.

Therefore the sum of the first and fifth, $EF$, is also the same multiple of the second $B$ that the sum of the third and sixth, $GH$ is of the fourth $D$.