Multiples of Terms in Equal Ratios

Theorem
Let $a, b, c, d$ be quantities.

Let $a : b = c : d$ where $a : b$ denotes the ratio between $a$ and $b$.

Then for any numbers $m$ and $n$:
 * $m a : n b = m c : n d$


 * If a first magnitude have to a second the same ratio as a third to a fourth, any equimultiples whatever of the first and third will also have the same ratio to any equimultiples whatever of the second and fourth respectively, taken in corresponding order.

Euclid's Proof
Let a first magnitude $A$ have to a second magnitude $B$ the same ratio as a third $C$ to a fourth $D$.

Let equimultiples $E, F$ be taken of $A, C$, and let different equimultiples $G, H$ be taken of $B, D$.

We need to show that $E : G = F : H$.


 * Euclid-V-4.png

Let equimultiples $K, L$ be taken of $E, F$ and other arbitrary equimultiples $M, N$ be taken of $G, H$.

We have that $E$ is the same multiple of $A$ that $L$ is of $C$.

So from Proposition 3: Multiplication of Numbers is Associative, $K$ is the same multiple of $A$ that $L$ is of $C$.

For the same reason, $M$ is the same multiple of $B$ that $N$ is of $D$.

We have that:
 * $A$ is to $B$ as $C$ is to $D$
 * of $A, C$ equimultiples $K, L$ have been taken
 * of $B, D$ other equimultiples $M, N$ have been taken.

So from the definition of equality of ratios:
 * if $K$ is in excess of $M$, $L$ is also in excess of $N$
 * if $K$ is equal to $M$, $L$ is equal to $N$
 * if $K$ is less than $M$, $L$ is less than $N$

But $K, L$ are equimultiples of $E, F$.

Therefore as $E$ is to $G$, so is $F$ to $H$.

Modern Proof
From the definition of a ratio, we have: