Multiplication of Numbers Distributes over Addition/Euclid's Proof 2

Theorem

As Euclid defined it:

If a first magnitude be the same multiple of a second that a third is of a fourth, and a fifth also be the same multiple of the second that a sixth is of the fourth, the sum of the first and fifth will also be the same multiple of the second that the sum of the third and sixth is of the fourth.

(The Elements: Book V: Proposition $2$)

That is:

$ma + na + pa + \cdots = \left({m + n + p + \cdots }\right) a$

Proof

Let a first magnitude, $AB$, be the same multiple of a second, $C$, that a third, $DE$, is of a fourth, $F$.

Let a fifth, $BG$, be the same multiple of $C$ that a sixth, $EH$, is of $F$.

We need to show that $AG = AB + BG$ is the same multiple of $C$ that $DH = DE + EH$ is of $F$.

We have that $AB$ is the same multiple of $C$ that $DE$ is of $F$.

It follows that as many magnitudes as there are in $AB$ equal to $C$, so many also are there in $DE$ equal to $F$.

For the same reason, as many as there are in $BG$ equal to $C$, so many also are there in $EH$ equal to $F$.

So as many as there are in the whole $AG$ equal to $C$, so many also are there in the whole $DH$ equal to $F$.

Therefore the sum of the first and fifth, $AG$, is the same multiple of the second, $C$, that the sum of the third and sixth, $DH$ is of the fourth, $F$.

$\blacksquare$

Also see

• Real Multiplication Distributes over Addition

Historical Note

This is Proposition 2 of Book V of Euclid's The Elements.