# Associative Law of Multiplication

## Theorem

Let $\mathbb F$ be one of the standard number sets: $\N, \Z, \Q, \R$ and $\C$.

Then:

- $\forall x, y, z \in \mathbb F: x \times \paren {y \times z} = \paren {x \times y} \times z$

That is, the operation of multiplication on the standard number sets is associative.

### Natural Number Multiplication is Associative

The operation of multiplication on the set of natural numbers $\N$ is associative:

- $\forall x, y, z \in \N: \paren {x \times y} \times z = x \times \paren {y \times z}$

### Integer Multiplication is Associative

The operation of multiplication on the set of integers $\Z$ is associative:

- $\forall x, y, z \in \Z: x \times \paren {y \times z} = \paren {x \times y} \times z$

### Rational Multiplication is Associative

The operation of multiplication on the set of rational numbers $\Q$ is associative:

- $\forall x, y, z \in \Q: x \times \paren {y \times z} = \paren {x \times y} \times z$

### Real Multiplication is Associative

The operation of multiplication on the set of real numbers $\R$ is associative:

- $\forall x, y, z \in \R: x \times \paren {y \times z} = \paren {x \times y} \times z$

### Complex Multiplication is Associative

The operation of multiplication on the set of complex numbers $\C$ is associative:

- $\forall z_1, z_2, z_3 \in \C: z_1 \paren {z_2 z_3} = \paren {z_1 z_2} z_3$

### 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:

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

## 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 $GH$ into 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 $D$.

$\blacksquare$

## Also see

## Historical Note

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

## Sources

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