Relative Sizes of Magnitudes on Unequal Ratios

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Theorem

In the words of Euclid:

Of magnitudes which have a ratio to the same, that which has a greater ratio is greater, and that to which the same has a greater ratio is less.

(The Elements: Book $\text{V}$: Proposition $10$)


That is:

$a : c > b : c \implies a > b$
$c : b > c : a \implies b < a$


Proof

Let $A$ have to $C$ a greater ratio than $B$ has to $C$.

Euclid-V-10.png

Suppose $A = B$.

Then from Ratios of Equal Magnitudes $A : C = B : C$.

But by hypothesis $A : C > B : C$, so $A \ne B$.

Suppose $A < B$.

Then from Relative Sizes of Ratios on Unequal Magnitudes it would follow that $A : C < B : C$.

But by hypothesis $A : C > B : C$.

Therefore it must be that $A > B$.

$\Box$


Let $C$ have to $B$ a greater ratio than $C$ has to $A$.

Suppose $B = A$.

Then from Ratios of Equal Magnitudes $C : B = C : A$.

But by hypothesis $C : B > C : A$, so $B \ne A$.

Suppose $B > A$.

Then from Relative Sizes of Ratios on Unequal Magnitudes it would follow that $C : B < C : A$.

But by hypothesis $C : B > C : A$.

Therefore it must be that $B < A$.

$\blacksquare$


Historical Note

This proof is Proposition $10$ of Book $\text{V}$ of Euclid's The Elements.
It is the converse of Proposition $8$ of Book $\text{5} $: Relative Sizes of Ratios on Unequal Magnitudes.


Sources