# Associative Law of Multiplication

## Theorem

On all the number systems:

- natural numbers $\N$
- integers $\Z$
- rational numbers $\Q$
- real numbers $\R$
- complex numbers $\C$

the operation of multiplication is associative:

- $a \times \paren {b \times c} = \paren {a \times b} \times c$

### Euclid's Statement

In the words of Euclid:

*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.*

(*The Elements*: Book $\text{V}$: Proposition $3$)

That is, if:

- $n a, n b$ are equimultiples of $a, b$

and if:

- $m \cdot n a, m \cdot nb$ are equimultiples of $n a, n b$

then:

- $m \cdot n a$ is the same multiple of $a$ that $m \cdot n b$ is of $b$

This can also be expressed as:

- $m \cdot n a = m n \cdot a$

## Proof

This is demonstrated in these pages:

- Natural Number Multiplication is Associative
- Integer Multiplication is Associative
- Rational Multiplication is Associative
- Real Multiplication is Associative
- Complex Multiplication is Associative

$\blacksquare$

## Euclid's 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 $GL, LH$.

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 $D$.

$\blacksquare$

## Historical Note

This theorem is Proposition $3$ of Book $\text{V}$ of Euclid's *The Elements*.

## Sources

- 1960: Walter Ledermann:
*Complex Numbers*... (previous) ... (next): $\S 1.1$. Number Systems: $\text{V}.$ - 1965: J.A. Green:
*Sets and Groups*... (previous) ... (next): $\S 4.2$. Commutative and associative operations: Example $64$ - 1965: Seth Warner:
*Modern Algebra*... (previous) ... (next): $\S 2$: Example $2.1$ - 1998: David Nelson:
*The Penguin Dictionary of Mathematics*(2nd ed.) ... (previous) ... (next): Entry:**associative**