# Divisors obey Distributive Law

## Theorem

In the words of Euclid:

If a number be a part of a number, and another be the same part of another, the sum will also be the same part of the sum that the one is of the one.

In modern algebraic language:

$\displaystyle a = \frac 1 n b, c = \frac 1 n d \implies a + c = \frac 1 n \left({b + d}\right)$

## Euclid's Proof

Let the (natural) number $A$ be an aliquot part of the (natural) number $BC$.

Let the (natural) number $D$ be the same aliquot part of another (natural) number $EF$ that $A$ is of $BC$.

We need to show that $A + D$ is the same aliquot part of $BC + EF$.

We have that whatever aliquot part $A$ is of $BC$, then $D$ is also the same aliquot part of $EF$.

Therefore as many numbers as there are in $BC$ equal to $A$, so many numbers are there in $EF$ equal to $D$.

Let $BC$ be divided into the numbers equal to $A$, that is, $BG, GC$.

Let $EF$ be divided into the numbers equal to $D$, that is, $EH, HF$.

Then the multitude of $BG, GC$ will be equal to the multitude of $EH, HF$.

Since $BG = A$ and $EH = D$, then $BG + EH = A + D$.

For the same reason, $GC + HF = A + D$.

Therefore, as many numbers as there are in $BC$ equal to $A$, so many are there also in $BC + EF$ equal to $A + D$.

Therefore, whatever multiple $BC$ is of $A$, the same multiple also is $BC + EF$ of $A + D$.

Therefore, whatever aliquot part $A$ is of $BC$, the same aliquot part also is $A + D$ of $BC + EF$.

$\blacksquare$

## Modern Proof

A direct application of the Distributive Property:

$\dfrac 1 n b + \dfrac 1 n d = \dfrac 1 n \paren {b + d}$

$\blacksquare$

## Historical Note

This theorem is Proposition $5$ of Book $\text{VII}$ of Euclid's The Elements.