Alternate Ratios of Multiples

Theorem

As Euclid defined it:

If an unit measure any (natural) number, and another (natural) number measure any other (natural) 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 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 Book VII Proposition 12: 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 is Proposition 15 of Book VII of Euclid's The Elements.