Least Ratio of Numbers

Theorem

 * Given as many (natural) 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 they are the least of those which have the same ratio with them.

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

Then by 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, $E, F, G$ 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 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:
 * $E D = H M$

Therefore by :
 * $E : H = M : D$

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 $A, B, C$.