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 $n$:
- $m a : n b = m c : n d$
In the words of Euclid:
- 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.
(The Elements: Book $\text{V}$: Proposition $4$)
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 $E : G = F : H$.
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 $3$: Associative Law of Multiplication, $K$ 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 $H$.
$\blacksquare$
Modern Proof
From the definition of a ratio, we have:
\(\ds a : b\) | \(=\) | \(\ds c : d\) | ||||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \frac a b\) | \(=\) | \(\ds \frac c d\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \frac m n \frac a b\) | \(=\) | \(\ds \frac m n \frac c d\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds \frac {m a} {n b}\) | \(=\) | \(\ds \frac {m c} {n d}\) | |||||||||||
\(\ds \leadstoandfrom \ \ \) | \(\ds m a : n b\) | \(=\) | \(\ds m c : n d\) |
$\blacksquare$
Historical Note
This proof is Proposition $4$ of Book $\text{V}$ of Euclid's The Elements.