Relative Sizes of Elements in Perturbed Proportion

Theorem

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

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

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 together are in the same ratio.

Let the proportion of them be perturbed, that is:
 * $A : B = E : F$
 * $B : C = D : E$

Let $A > C$.

Then we need to show that $D > F$.


 * Euclid-V-21.png

We have that $A > C$.

So from Relative Sizes of Ratios on Unequal Magnitudes $A : B > C : B$.

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

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

But from Relative Sizes of Magnitudes on Unequal Ratios $F < D$ and so $D > F$.

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