# Least Ratio of Numbers

## Theorem

In the words of Euclid:

Given as many numbers as we please, to find the least of those which have the same ratio with them.

## Proof

Let $A, B, C$ be the given numbers, as many as we please.

$A, B, C$ are either coprime or not.

Let $A, B, C$ be coprime.

Then by Proposition $21$ of Book $\text{VII}$: Coprime Numbers form Fraction in Lowest Terms they are the least of those which have the same ratio with them.

Let $A, B, C$ not be coprime.

Then by Proposition $3$ of Book $\text{VII}$: Greatest Common Divisor of Three Numbers their GCD $D$ can be found.

As many times as $D$ measures each of $A, B, C$, let those be $E, F, G$ respectively.

That is:

$D E = A, D F = B, D G = C$

By Proposition $20$ of Book $\text{VII}$: Ratios of Fractions in Lowest Terms, $E, F, G$ are in the same ratio with $A, B, C$.

Suppose $E, F, G$ are not the least of those which have the same ratio with $A, B, C$.

Let those numbers be $H, K, L$.

Then $H$ measures $A$ the same number of times that $K, L$ measure $B, C$.

Let this number of times be $M$.

We have that $H$ measures $A$ according to the units of $M$.

It follows from Proposition $16$ of Book $\text{VII}$: Natural Number Multiplication is Commutative that $M$ also measures $A$ according to the units of $H$.

For the same reason, $M$ also measures $B$ and $C$ according to the units of $K$ and $L$.

Therefore $M$ measures $A$.

That is:

$A = H M$

For the same reason:

$A = E D$

That is:

$E D = H M$
$E : H = M : D$

But $E > H$ and so $M > D$.

Also, $M$ measures $A$, $B$ and $C$.

But by hypothesis $D$ is the greatest common measure of $A, B, C$.

Therefore there cannot be any numbers less than $E, F, G$ which are in the same ratio with $A, B, C$.

Therefore $E, F, G$ are the least of those which are in the same ratio with $A, B, C$.

$\blacksquare$

## Historical Note

This proof is Proposition $33$ of Book $\text{VII}$ of Euclid's The Elements.