# Multiples of Alternate Ratios of Equal Fractions

## Theorem

In the words of Euclid:

If a number be parts of a number, and another be the same parts of another, alternately also, whatever parts or part the first is of the third, the same parts or the same part will the second also be of the fourth.

## Proof

Let the (natural) number $AB$ be an aliquant part of the (natural) number $C$, and another $DE$ be the same aliquant part of another, $F$.

We need to show that whatever aliquant part or aliquot part $AB$ is of $DE$, the same aliquant part or aliquot part is $C$ of $F$.

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

So as many aliquant parts of $C$ as there are in $AB$, the same aliquant parts also is $DE$ of $F$.

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

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

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

From Proposition $9$ of Book $\text{VII}$: Alternate Ratios of Equal Fractions, whatever aliquot part $AG$ is of $DH$, the same aliquot part or the same aliquant part is $C$ of $F$ also.

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

So from Proposition $5$ of Book $\text{VII}$: Divisors obey Distributive Law and Proposition $6$ of Book $\text{VII}$: Multiples of Divisors obey Distributive Law, whatever aliquant part or aliquot part $AB$ is of $DE$, the same aliquant part also, or the same aliquot part, is $C$ of $F$.

$\blacksquare$

## Historical Note

This proof is Proposition $10$ of Book $\text{VII}$ of Euclid's The Elements.