Relative Sizes of Proportional Magnitudes

Theorem

 * If a first magnitude have to a second the same ratio as a third to a fourth, and the third have to the fourth a greater ratio than a fifth has to a sixth, the first will also have to the second a greater ratio than the fifth to the sixth.

That is:
 * $a : b = c : d, c : d > e : f \implies a : b > e : f$

Proof
Let a first magnitude $A$ have to a second $B$ the same ratio as a third $C$ to a fourth $D$.

Let the third $C$ have to the fourth $D$ a greater ratio than a fifth $E$ has to a sixth $F$.


 * Euclid-V-13.png

We have that $C : D > E : F$.

From, there will be some equimultiples of $C, E$ and other arbitrary equimultiples of $D, F$ such that the multiple of $C$ is in excess of the multiple of $D$, while the multiple of $E$ is not in excess of the multiple of $F$.

Let these equimultiples be taken.

Let $G, H$ be equimultiples of $C, E$, and $K, L$ be other arbitrary equimultiples of $D, F$, so that $G > K$ but $H \le L$.

Whatever multiple $G$ is of $C$, let $M$ be also that multiple of $A$.

Also, whatever multiple $K$ is of $D$, let $N$ be also that multiple of $B$.

Now we have that $A : B = C : D$ and of $A, C$ equimultiples $M, G$ have been taken.

We also have that of $B, D$ other arbitrary equimultiples $N, K$ have been taken.

Therefore:
 * $M > N \implies G > K$
 * $M = N \implies G = K$
 * $M < N \implies G < K$

from.

But $G > K$ and so $M > N$.

But $H \le L$, and:
 * $M, H$ are equimultiples $A, E$
 * $N, L$ are other, arbitrary equimultiples $B, F$.

Therefore from, $A : B > E : F$.