Relative Sizes of Ratios on Unequal Magnitudes

Theorem
That is:
 * $a > b \implies a : c > b : c$
 * $a > b \implies c : a < c : b$

Proof
Let $AB, C$ be unequal magnitudes, and let $AB$ be greater.

Let $D$ be another arbitrary magnitude.

We are to show that $AB$ has to $D$ a greater ratio than $C$ has to $D$, and $D$ has to $C$ a greater ratio than it has to $AB$.

We have that $AB > C$, so let $BE = C$.

Then from, the lesser of the magnitudes $AE, EB$, if multiplied, will eventually be greater than $D$.

There are two cases.


 * First, let $AE < EB$.


 * Euclid-V-8a.png

Let $AE$ be multiplied by some number, and let $FG$ be a multiple of it which is greater than $D$.

Then whatever multiple $FG$ is of $AE$, let $GH$ be made the same multiple of $EB$ and $K$ of $C$.

Let $L := 2D, M := 3D, N := 4D \ldots$ until one of these multiples is greater than $K$.

Suppose $M \le K$ while $N > K$.

WE have that $FG$ is the same multiple of $AE$ that $GH$ is of $EB$.

So from Multiplication of Numbers is Left Distributive over Addition, $FG$ is the same multiple of $AE$ that $FHK$ is of $AB$.

But $FG$ is the same multiple of $AE$ that $K$ is of $C$.

Therefore $FH, K$ are equimultiples of $AB, C$.

Again, we have that $GH$ is the same multiple of $EB$ that $K$ is of $C$, and $EB = C$.

Therefore $GH = K$.

But $K \ge M$ and so $GH \ge M$.

Also, $FG > D$ so $FH > D + M$.

But $D + M = N$ by the construction of $M$ and $N$, and $FH > M + D$.

So $FH > N$ while $K \le N$.

Also, $FH, K$ are equimultiples of $AB, C$ while $N$ is a multiple of some arbitrary $D$.

So from Ratios of Equal Magnitudes $AB$ has a greater ratio to $D$ than $C$ has to $D$.

Next, note that with the same construction, we can show similarly that $N > K$ while $N \le FH$.

Also we have that $N$ is a multiple of $D$, while $FH, K$ are other equimultiples of $AB, C$.

Therefore $D$ has a greater ratio to $C$ than $D$ has to $AB$.


 * Second, let $AE > EB$.

From, $EB$, if multiplied, will eventually be greater than $D$.


 * Euclid-V-8b.png

Let it be so multiplied, and let $GH$ be a multiple of $EB$ and greater than $D$.

Whatever multiple $GH$ is of $EB$, let $FG$ be made the same multiple of $AE$, and $K$ of $C$.

Then we can prove similarly that $FH, K$ are equimultiples of $AB, C$.

Similarly, let $M, N$ be consecutive multiples of $D$ such that $M \le FG$ and $N > FG$.

But $GH > D$, therefore $FH > D + M$, that is, $FH > N$.

Now $K \le N$ inasmuch as $FG$ also, which is greater than $GH$, that is, than $K$, is not in excess of $N$.

Hence the result.