Associative Law of Multiplication/Euclid's Proof

Theorem
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 $D$.