Proof: By Euclid

For if not, $A$ is surely either equal to or less than $B$.
In fact, $A$ is not equal to $B$.
For (then) $A$ and $B$ would each have the same ratio to $C$ [Prop. 5.7].
But they do not.
Thus, $A$ is not equal to $B$.
Neither, indeed, is $A$ less than $B$.
For (then) $A$ would have a lesser ratio to $C$ than $B$ (has) to $C$ [Prop. 5.8].
But it does not.
Thus, $A$ is not less than $B$.
And it was shown not (to be) equal either.
Thus, $A$ is greater than $B$.
So, again, let $C$ have a greater ratio to $B$ than $C$ (has) to $A$.
I say that $B$ is less than $A$.
For if not, (it is) surely either equal or greater.
In fact, $B$ is not equal to $A$.
For (then) $C$ would have the same ratio to each of $A$ and $B$ [Prop. 5.7].
But it does not.
Thus, $A$ is not equal to $B$.
Neither, indeed, is $B$ greater than $A$.
For (then) $C$ would have a lesser ratio to $B$ than (it has) to $A$ [Prop. 5.8].
But it does not.
Thus, $B$ is not greater than $A$.
And it was shown that (it is) not equal (to $A$) either.
Thus, $B$ is less than $A$.
Thus, for (magnitudes) having a ratio to the same (magnitude), that (magnitude which) has the greater ratio is (the) greater.
And that (magnitude) to which the latter (magnitude) has a greater ratio is (the) lesser.
(Which is) the very thing it was required to show.
∎

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