Definition 7 now says w:x > y:z when
Of course, y:z is called the lesser ratio.
When defining greater and lesser, there are a number of properties that should be verified. These are various transitivities and the law of trichotomy, some of the same properities of greater and lesser that magnitudes have. (See the Guide for the Common Notions.)
If u:v = w:x, and w:x < y:z, then u:v < y:z.
If u:v < w:x, and w:x < y:z, then u:v < y:z.
Euclid only has the first property, which is is proposition V.13. Its proof depends only on the definitions. The second is so much like it that it isn't mentioned but it is used in the same way as the third. The third one is quite easy to prove.
Euclid missed the law of trichotomy for magnitudes in his list of Common Notions, and he missed it for ratios, too. The side of the law which says at most one of the three cases can occur is first used in proposition V.9, while the side which says at least one occurs is first used in V.10.
The side of the law which says at most one of the three cases can occur is fairly easy to prove.
From the definitions themselves it is clear that w:x > y:z contradicts w:x = y:z. The first says there are n and m such that nw > mx but ny is not greater than mz, while the second concludes from nw > mx that ny > mz yielding a contradiction based on the law of trichotomy for magnitudes. Similarly w:x < y:z contradicts w:x = y:z.
Once transitivity has been shown, it can be shown that w:x > y:z contradicts w:x < y:z, for then w:x > w:x, and that contradicts w:x = w:x. (There are also proofs that don't depend on transitivity.)
The other side of the law of trichotomy, the one that says at least one of the three cases holds, is a bit harder to prove, and it depends on treating V.Def.4 as an axiom of comparability. In fact, it is false without it. First, a proof using V.Def.4 as an axiom, then a counterexample to show that's necessary.
A proof of trichotomy. Let w:x and y:z be any two ratios. We need to show that one of the three cases holds. We'll assume the ratios aren't the same and show one of them is greater than the other. When they're not the same, then there are numbers m and n such that
We have three cases to consider, and two of them are easy. In one case, nw > mx but not ny > mz, so for that case w:x > y:z. In another case, nw < mx but not ny < mz, so for that case w:x < y:z.
Consider now the last case: nw = mx but ny does not equal mz. Then one of ny and mz is greater, say ny > mz. Now using V.Def.4 as an axiom, there is some some number k such that k(ny - mz) > z. Since k(ny - mz) = kny - kmz, therefore kny > kmz + z, that is, kny > (km + 1)z. But knw = kmx, and kmx < (km + 1)x. Therefore, (kn)w < (km + 1)x. But (kn)y > (km + 1)z. Therefore w:x < y:z. Q.E.D.
A counterexample to trichotomy. This counterexample has infinitesimals, so it doesn't satisfy the axiom of comparability, that is, V.Def.4 treated as an axiom. Let y be an infinitesimal with respect to x, that is, for any number n, ny < x. We'll show that the two ratios x:x and x:(x + y) do not satisfy the law of trichotomy. First, though, note that the second ratio does satisfy V.Def.4 as a definition since twice each of x and x + y is greater than the other.
Now, the ratios x:x and x:(x + y) are not equal since 2x = 2x but 2x < 2(x + y). Next, x:(x + y) is not greater than x:x, since nx > m(x + y) implies nx > mx. Finally, x:x is not greater than x:(x + y), for if nx > mx, then n > m, so nx is not less than mx > x, and since y is an infinitesimal with respect to x, x > my, therefore nx - mx > my, that is, nx > m(x + y).
In summary, since we have a proof of trichotomy that uses V.Def.4 as an axiom of comparability, and a counterexample of trichotomy that violates the axiom of comparability, we can conclude that any proof trichotomy requires the axiom of comparability.
Book V Introduction - Definitions V.Def.5 and V.Def.6 - Definitions V.Def.8 through V.Def.10.