Unqeual Numbers

Two unequal numbers (being) laid down, and the lesser being continually subtracted, in turn, from the greater, if the remainder never measures the (number) preceding it, until a unit remains, then the original numbers will be prime to one another.

For two [unequal] numbers, AB and CD, the lesser being continually subtracted, in turn, from the greater, let the remainder never measure the (number) preceding it, until a unit remains. I say that AB and CD are prime to one another—that is to say, that a unit alone measures (both) AB and CD.

For if AB and CD are not prime to one another then some number will measure them. Let (some number) measure them, and let it be E. And let CD measuring BF leave F A less than itself, and let AF measuring DG leave GC less than itself, and let GC measuring F H leave a unit, HA.

In fact, since E measures CD, and CD measures BF, E thus also measures BF .† And (E) also measures the whole of BA. Thus, (E) will also measure the remainder AF.

AF measures DG. Thus, E also measures DG.

(E) also measures the whole of DC. Thus, (E) will also measure the remainder CG. And CG measures F H.

Thus, E also measures F H. And (E) also measures the whole of F A. Thus, (E) will also measure the remaining unit AH, (despite) being a number.

The very thing is impossible. Thus, some number does not measure (both) the numbers AB and CD. Thus, AB and CD are prime to one another. (Which is) the very thing it was required to show.