# Alternate Ratios of Multiples

## Theorem

In the words of Euclid:

*If an unit measure any number, and another number measure any other number the same number of times, alternately also, the unit will measure the third number the same number of times that the second measures the fourth.*

(*The Elements*: Book $\text{VII}$: Proposition $15$)

## Proof

Let the unit $A$ measure any (natural) number $BC$.

Let another (natural) number $D$ measure any other (natural) number $EF$ the same number of times.

We need to show that $A$ measures $D$ the same number of times that $BC$ measures $EF$.

We have that $A$ measures $BC$ the same number of times that $D$ measures $EF$.

So as many units as there are in $BC$ there are numbers equal to $D$ in $EF$.

Let $BC$ be divided into the units in it: $BG, GH, HC$.

Let $EF$ be divided into the numbers in it equal to $D$: $EK, KL, LF$.

So the multitude of $BG, GH, HC$ will equal the multitude of $EK, KL, LF$.

We have that $BG = GH = HC$ and $EK = KL = LF$.

So $BG : EK = GH : KL = HC : LF$.

So from Proposition $12$ of Book $\text{VII} $: Ratios of Numbers is Distributive over Addition, $BG : EK = BG + GH + HC : EK + KL + LF$.

That is, $BG : EK = BC : EF$.

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

So $A : D = BC : EF$.

So $A$ measures $D$ the same number of times that $BC$ measures $EF$.

$\blacksquare$

## Historical Note

This proof is Proposition $15$ 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