Relative Sizes of Successive Ratios

Theorem

 * If there be three magnitudes, and others equal to them in multitude, which taken two and two are in the same ratio, and if ex aequali the first be greater than the third, the fourth will also be greater than the sixth; if equal, equal; and if less, less.

That is, let:
 * $a : b = d : e$
 * $b : c = e : f$

Then:
 * $a > c \implies d > f$
 * $a = c \implies d = f$
 * $a < c \implies d < f$

Proof
Let there be three magnitudes $A, B, C$, and others $D, E, F$ equal to them in multitude, which taken two and two are in the same ratio:
 * $A : B = D : E$
 * $B : C = E : F$

and let $A > C$ ex aequali.

We need to show that $D > F$.

Similarly, we need to show that $A = C \implies D = F$ and $A < C \implies D < F$.


 * Euclid-V-20.png

Since $A > C$ we have from Relative Sizes of Ratios on Unequal Magnitudes that $A : B > C : B$.

But $A : B = D : E$ and $C : B = F : E$.

Therefore from Relative Sizes of Proportional Magnitudes $D : E > F : E$.

But from Relative Sizes of Magnitudes on Unequal Ratios it follows that $D > F$.

Similarly we can show that $A = C \implies D = F$ and $A < C \implies D < F$.