| If a first magnitude has to a second the same ratio as a third to a fourth, then any equimultiples whatever of the first and third also have the same ratio to any equimultiples whatever of the second and fourth respectively, taken in corresponding order. | ||
| Let a first magnitude A have to a second B the same ratio as a third C to a fourth D, and let equimultiples E and F be taken of A and C, and G and H other, arbitrary, equimultiples of B and D.
I say that E is to G as F is to H. | ||
| Take equimultiples K and L of E and F, and other, arbitrary, equimultiples M and N of G and H. | ||
| Since E is the same multiple of A that F is of C, and equimultiples K and L of E and F have been taken, therefore K is the same multiple of A that L is of C. For the same reason M is the same multiple of B that N is of D. | V.3 | |
| And, since A is to B as C is to D, and equimultiples K and L have been taken of A and C, and other, arbitrary, equimultiples M and N of B and D, therefore, if K is in excess of M, then L is in excess of N; if it is equal, equal; and if less, less. | V.Def.5 | |
| And K and L are equimultiples of E and F, and M and N are other, arbitrary, equimultiples of G and H, therefore E is to G as F is to H. | V.Def.5 | |
| Therefore, if a first magnitude has to a second the same ratio as a third to a fourth, then any equimultiples whatever of the first and third also have the same ratio to any equimultiples whatever of the second and fourth respectively, taken in corresponding order. | ||
| Q.E.D. | ||
Note how Euclid uses the definition to prove that the two ratios pa:qb and pc:qd are the same. (Here, a and b are magnitudes of one kind, and c and d are magnitudes of another kind, but p and q are numbers.) We are given a:b = c:d. That means for any numbers m and n that
We have to prove that pa:qb = pc:qd for any numbers p and q. That means, we have to prove that for any m and n,
But that's just a special case of the given relation