Relative Sizes of Proportional Magnitudes
Theorem
In the words of Euclid:
- If a first magnitude have to a second the same ratio as a third to a fourth, and the third have to the fourth a greater ratio than a fifth has to a sixth, the first will also have to the second a greater ratio than the fifth to the sixth.
(The Elements: Book $\text{V}$: Proposition $13$)
That is:
- $a : b = c : d, c : d > e : f \implies a : b > e : f$
Proof
Let a first magnitude $A$ have to a second $B$ the same ratio as a third $C$ to a fourth $D$.
Let the third $C$ have to the fourth $D$ a greater ratio than a fifth $E$ has to a sixth $F$.
We have that $C : D > E : F$.
From Book $\text{V}$ Definition $7$: Greater Ratio, there will be some equimultiples of $C, E$ and other arbitrary equimultiples of $D, F$ such that the multiple of $C$ is in excess of the multiple of $D$, while the multiple of $E$ is not in excess of the multiple of $F$.
Let these equimultiples be taken.
Let $G, H$ be equimultiples of $C, E$, and $K, L$ be other arbitrary equimultiples of $D, F$, so that $G > K$ but $H \le L$.
Whatever multiple $G$ is of $C$, let $M$ be also that multiple of $A$.
Also, whatever multiple $K$ is of $D$, let $N$ be also that multiple of $B$.
Now we have that $A : B = C : D$ and of $A, C$ equimultiples $M, G$ have been taken.
We also have that of $B, D$ other arbitrary equimultiples $N, K$ have been taken.
Therefore:
- $M > N \implies G > K$
- $M = N \implies G = K$
- $M < N \implies G < K$
from Book $\text{V}$ Definition $5$: Equality of Ratios.
But $G > K$ and so $M > N$.
But $H \le L$, and:
- $M, H$ are equimultiples $A, E$
- $N, L$ are other, arbitrary equimultiples $B, F$.
Therefore from Book $\text{V}$ Definition $7$: Greater Ratio, $A : B > E : F$.
$\blacksquare$
Historical Note
This proof is Proposition $13$ of Book $\text{V}$ of Euclid's The Elements.
Sources
- 1926: Sir Thomas L. Heath: Euclid: The Thirteen Books of The Elements: Volume 2 (2nd ed.) ... (previous) ... (next): Book $\text{V}$. Propositions