第10卷命题 1 · 穷竭法基本引理

Let AB, C be two unequal magnitudes of which AB is the greater: I say that, if from AB there be subtracted a magnitude greater than its half, and from that which is left a magnitude greater than its half, and if this process be repeated continually, there will be left some magnitude which will be less than the magnitude C. For C if multiplied will sometime be greater than AB. [cf. v. Def. 4] Let it be multiplied, and let DE be a multiple of C, and greater than. AB; let DE be divided into the parts DF, FG, GE equal to C, from AB let there be subtracted BH greater than its half, and, from AH, HK greater than its half, and let this process be repeated continually until the divisions in AB are equal in multitude with the divisions in DE.

设AB和C为两个不相等的量，AB较大。取C的倍数DE大于AB，且DE被分成等于C的若干段DF、FG、GE。

Let, then, AK, KH, HB be divisions which are equal in multitude with DF, FG, 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.

从AB中减去大于其一半的BH，再从AH中减去大于其一半的HK，如此继续，直到AB上的分段数与DE上的分段数相等。设AK、KH、HB为这些分段。

But DF is equal to 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.

由于DE大于AB，且从DE中减去小于其一半的EG，从AB中减去大于其一半的BH，所以余量GD大于余量HA。

Q. E. D.

又因为GD大于HA，且从GD中减去一半GF，从HA中减去大于其一半的HK，所以余量DF大于余量AK。但DF等于C，因此C大于AK，即AK小于C。