If, when the less of two unequal magnitudes is continually subtracted in turn from the greater that which is left never measures the one before it, then the two magnitudes are incommensurable.

There being two unequal magnitudes *AB* and *CD,* with *AB* being the less, when the less is continually subtracted in turn from the greater, let that which is left over never measure the one before it.

I say that the magnitudes *AB* and *CD* are incommensurable.

If they are commensurable, then some magnitude *E* measures them.

Let *AB,* measuring *FD,* leave *CF* less than itself, let *CF* measuring *BG,* leave *AG* less than itself, and let this process be repeated continually, until there is left some magnitude which is less than *E.*

Suppose this done, and let there be left *AG* less than *E.*

Then, since *E* measures *AB,* while *AB* measures *DF,* therefore *E* also measures *FD.* But it measures the whole *CD* also, therefore it also measures the remainder *CF.* But *CF* measures *BG,* therefore *E* also measures *BG.* But it measures the whole *AB* also, therefore it also measures the remainder *AG,* the greater the less, which is impossible.

Therefore no magnitude measures the magnitudes *AB* and *CD.* Therefore the magnitudes *AB* and *CD* are incommensurable.

Therefore, *if, when the less of two unequal magnitudes is continually subtracted in turn from the greater that which is left never measures the one before it, then the two magnitudes are incommensurable.*

Q.E.D.

Heath claims that Euclid uses X.1 to prove this proposition, in particular, to show that antenaresis eventually leaves some magnitude which is less than *E.* It is hard to tell what Euclid thought his justification was. Since both magnitudes are multiples of *E,* whatever justification Euclid intended back in proposition VII.2 works just as well here. Euclid did, however, put X.1 just before this proposition, perhaps for an intended logical connection. If so, there is a missing statement to the effect that *GB* is greater than half of *AB,* and so forth, so that X.1 might be invoked.

Consider the 36°-72°-72° triangle constructed ABC in proposition IV.10. This triangle was used in the following proposition IV.11 to construct regular pentagons. When its base BC is
subtracted from a side AC then the remainder CD is the base of a similar triangle BCD. Likewise, when the base CD of this new triangle is subtracted from its side BD then the remainder DE is the base of yet another smaller similar triangle CDE. And so forth.
Thus, when we begin with the two lines AB : BC = BC : CD = CD : DE = DE : EF = ...
Thus, according to this proposition, the two quantities Cutting the line |

This proposition is used in the next one.