Ratios of Fractions in Lowest Terms

Theorem
Let $a, b, c, d \in \Z_{>0}$ be positive integers.

Let $\dfrac a b$ be in canonical form.

Let $\dfrac a b = \dfrac c d$.

Then:
 * $a \mathrel \backslash c$

and:
 * $b \mathrel \backslash d$

where $\backslash$ denotes divisibility.

Proof
Let $CD, EF$ be the least (natural) numbers of those which have the same ratio with $A, B$.

We need to show that $CD$ measures $A$ the same number of times that $EF$ measures $B$.


 * Euclid-VII-20.png

Suppose $CD$ is an aliquant part of $A$.

Then from and, $EF$ is also the same aliquant part of $B$ that $CD$ is of $A$.

Therefore as many aliquant parts of $A$ as there are in $CD$, so many aliquant parts of $B$ are there also in $EF$.

Let $CD$ be divided into the aliquant parts of $A$, namely $CG, GD$ and $EF$ into the aliquant parts of $B$, namely $EH, HF$.

Thus the multitude of $CG, GD$ will be equal to the multitude of $EH, HF$.

We have that the numbers $CG, GD$ are equal to one another, and the numbers $EH, HF$ are also equal to one another,

Therefore $CG : EH = GD : HF$.

So from, as one of the antecedents is to one of the consequents, so will all the antecedents be to all the consequents.

Therefore $CG : EH = CD : EF$.

Therefore $CG, EH$ are in the same ratio with $CD, EF$ being less than they.

This is impossible, for by hypothesis $CD, EF$ are the least numbers of those which have the same ratio with them.

So $CD$ is not an aliquant part of $A$.

Therefore from, $CD$ is an aliquot part of $A$.

Also, from and, $EF$ is the same aliquot part of $B$ that $CD$ is of $A$.

Therefore $CD$ measures $A$ the same number of times that $EF$ measures $B$.