Relative Sizes of Successive Ratios

Theorem
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 $A < C \implies D < F$.