Let AB and C be two unequal magnitudes of which AB is the greater.
I say that, if from AB there is subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process is repeated continually, then there will be left some magnitude which is less than the magnitude C.
Some multiple DE of C is greater than AB.
Divide DE into the parts DF, FG, and GE equal to C. From AB subtract BH greater than its half, and from AH subtract HK greater than its half, and repeat this process continually until the divisions in AB are equal in multitude with the divisions in DE.
Let, then, AK, KH, and HB be divisions equal in multitude with DF, FG, and GE.
Now, since DE is greater than AB, and from DE there has been subtracted EG less than its half, and, from AB, BH greater than its half, therefore the remainder GD is greater than the remainder HA.
And, since GD is greater than HA, and there has been subtracted from GD the half GF, and from HA, HK greater than its half, therefore the remainder DF is greater than the remainder AK.
But DF equals C, therefore C is also greater than AK. Therefore AK is less than C.
Therefore there is left of the magnitude AB the magnitude AK which is less than the lesser magnitude set out, namely C.
Therefore, two unequal magnitudes being set out, if from the greater there is subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process is repeated continually, then there will be left some magnitude less than the lesser magnitude set out.
And the theorem can similarly be proven even if the parts subtracted are halves.