Multiples of Divisors Obey Distributive Law

Theorem

As Euclid defined it:

If a (natural) number be parts of a (natural) number, and another be the same parts of another, the sum will also be the same parts of the sum that the one is of the one.

(The Elements: Book VII: Proposition $6$)

In modern algebraic language:

$\displaystyle a = \frac m n b, c = \frac m n d \implies a + c = \frac m n \left({b + d}\right)$

Euclid's Proof

Let the (natural) number $AB$ be parts of the (natural) number $C$.

Let the (natural) number $DE$ be the same parts of another (natural) number $F$ that $AB$ is of $C$.

We need to show that $AB + DE$ is the same parts of $C + F$.

We have that whatever parts $AB$ is of $C$, $DE$ is also the same parts as $F$.

It follows that as many parts of $C$ as there are in $AB$, so many parts of $F$ are there also in $DE$.

Let $AB$ be divided into the parts of $C$, namely $AG, GB$, and $DE$ into the parts of $F$, namely $DH, HE$.

Thus the multitude of $AG, GB$ will be equal to the multitude of $DH, HE$.

We have that whatever part $AG$ is of $C$, the same part is $DH$ of $F$ also.

Therefore, from Book VII Proposition 5: Divisors Obey Distributive Law, whatever part $AG$ is of $C$, the same part also is $AG + DH$ of $C + F$ also.

For the same reason, whatever part $GB$ is of $C$, the same part also is $GB + HE$ of $C + F$.

Therefore whatever parts $AB$ is of $C$, the same parts also is $AB + DE$ of the $C + F$

$\blacksquare$

Modern Proof

A direct application of the Distributive Property:

$\displaystyle \frac m n b + \frac m n d = \frac m n \left({b + d}\right)$

$\blacksquare$

Historical Note

This is Proposition 6 of Book VII of Euclid's The Elements.