# Alternate Ratios of Equal Fractions

## Theorem

In the words of Euclid:

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

(*The Elements*: Book $\text{VII}$: Proposition $9$)

## Proof

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

We need to show that whatever aliquot part or aliquant part $A$ is of $D$, the same aliquot part or aliquant part is $BC$ of $EF$.

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

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

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

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

Thus the multitude of $BG, GC$ equals the multitude of $EH, HF$.

We have that $BG = GC$ and $EH = HF$.

Therefore whatever aliquot part or aliquant part $BG$ is of $EH$, the same aliquot part or the same aliquant part is $GC$ of $HF$ 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, we have that whatever aliquot part or aliquant part $BG$ is of $EH$, the same aliquot part also, or the same aliquant part, is the sum $BC$ of the sum $EF$.

But $BG = A$ and $EH = D$.

So whatever aliquot part or aliquant part $A$ is of $D$, the same aliquot part or aliquant part is $BC$ of $EF$ also.

$\blacksquare$

## Historical Note

This proof is Proposition $9$ of Book $\text{VII}$ of Euclid's *The Elements*.

## Sources

- 1926: Sir Thomas L. Heath:
*Euclid: The Thirteen Books of The Elements: Volume 2*(2nd ed.) ... (previous) ... (next): Book $\text{VII}$. Propositions