Associative Law of Multiplication

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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 $GL, LH$.

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 $D$.

$\blacksquare$


Also see


Historical Note

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


Sources

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