For, since AB is greater than C , let BE be made equal to C ; then the less of the magnitudes AE , EB , if multiplied, will sometime be greater than D . [[V. Def. 4](elem.5.def.4)]

[

Case I.]
First, let AE be less than EB ; let AE be multiplied, and let FG be a multiple of it which is greater than D ; then, whatever multiple FG is of AE , let GH be made the same multiple of EB and K of C ; and let L be taken double of D , M triple of it, and successive multiples increasing by one, until what is taken is a multiple of D and the first that is greater than K . Let it be taken, and let it be N which is quadruple of D and the first multiple of it that is greather than K .

Then, since K is less than N first, therefore K is not less than M .

And, since FG is the same multiple of AE that GH is of EB , therefore FG is the same multiple of AE that FH is of AB . [[V. 1](elem.5.1)]

But FG is the same multiple of AE that K is of C ; therefore FH is the same multiple of AB that K is of C ; therefore FH , K are equimultiples of AB , C .

Again, since GH is the same multiple of EB that K is of C , and EB is equal to C , therefore GH is equal to K .

But K is not less than M ; therefore neither is GH less than M .

And FG is greater than D ; therefore the whole FH is greater than D , M together.

But D , M together are equal to N , inasmuch as M is triple of D , and M , D together are quadruple of D , while N is also quadruple of D ; whence M , D together are equal to N .

But FH is greater than M , D ; therefore FH is in excess of N , while K is not in excess of N .

And FH , K are equimultiples of AB , C , while N is another, chance, multiple of D ; therefore AB has to D a greater ratio than C has to D . [[V. Def. 7](elem.5.def.7)]

I say next, that D also has to C a greater ratio than D has to AB .

For, with the same construction, we can prove similarly that N is in excess of K , while N is not in excess of FH .

And N is a multiple of D , while FH , K are other, chance, equimultiples of AB , C ; therefore D has to C a greater ratio than D has to AB . [[V. Def. 7](elem.5.def.7)]

[

Case 2.]
Again, let AE be greater than EB .

Then the less, EB , if multiplied, will sometime be greater than D . [[V. Def. 4](elem.5.def.4)]

Let it be multiplied, and let GH be a multiple of EB and greater than D ; and, whatever multiple GH is of EB , let FG be made the same multiple of AE , and K

of C .

Then we can prove similarly that FH , K are equimultiples of AB , C ; and, similarly, let N be taken a multiple of D but the first that is greater than FG , so that FG is again not less than M .

But GH is greater than D ; therefore the whole FH is in excess of D , M , that is, of N .

Now K is not in excess of N , inasmuch as FG also, which is greater than GH , that is, than K , is not in excess of N .

And in the same manner, by following the above argument, we complete the demonstration.

Therefore etc. Q. E. D.