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.