Multiplication of Numbers is Left Distributive over Addition

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Theorem

In the words of Euclid:

If there be any number of magnitudes whatever which are, respectively, equimultiples of any magnitudes equal in multitude, then, whatever multiple of one of the magnitudes is of one, that multiple also will all be of all.

(The Elements: Book $\text{V}$: Proposition $1$)


That is, if $m a$, $m b$, $m c$ etc. be any equimultiples of $a$, $b$, $c$ etc., then:

$m a + m b + m c + \cdots = m \paren {a + b + c + \cdots }$


Proof

Let any number of magnitudes whatever $AB, CD$ be respectively equimultiples of any magnitudes $E, F$ equal in multitude.

Then we are to show that whatever multiple $AB$ is of $E$, then that multiple will $AB + CD$ be of $E + F$.

Euclid-V-1.png

Since $AB$ is the same multiple of $E$ that $CD$ is of $F$, as many magnitudes as there are in $AB$ equal to $E$, so many also are there in $CD$ equal to $F$.

Let $AB$ be divided into the magnitudes $AG, GB$ equal to $E$, and $CH, HD$ equal to $F$.

Then the multitude of the magnitudes $AG, GB$ will be equal to the multitude of the magnitudes $CH, HD$.

Since $AG = E$ and $CH = F$ it follows that $AG = E$ and $AG + CH = E + F$.

For the same reason, $GB = E$ and $GB + HD = E + F$.

Therefore, as many magnitudes as there are in $AB$ equal to $E$, so many also are there in $AB + CD$ equal to $E + F$.

$\blacksquare$


Also see


Historical Note

This proof is Proposition $1$ of Book $\text{V}$ of Euclid's The Elements.


Sources

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