Relative Sizes of Elements in Perturbed Proportion

Theorem

As Euclid defined it:

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.

(The Elements: Book V: Proposition $21$)

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 $D > F$.

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 $A < C \implies D < F$.

$\blacksquare$

Historical Note

This is Proposition 21 of Book V of Euclid's The Elements.