Kitap 10
1
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.
1
But DF is equal to C; therefore C is also greater than AK.
1
Therefore AK is less than C.
1
Therefore there is left of the magnitude AB the magnitude AK which is less than the lesser magnitude set out, namely C. Q. E. D.
1
And the theorem can be similarly proved even if the parts subtracted be halves.
PROPOSITION 2.
2
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, the magnitudes will be incommensurable.
2
For, there being two unequal magnitudes AB, CD, and 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, CD are incommensurable.
2
For, if they are commensurable, some magnitude will measure them.
2
Let a magnitude measure them, if possible, and let it be E; 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.
2
Suppose this done, and let there be left AG less than E.
2
Then, since E measures AB, while AB measures DF, therefore E will also measure FD.
2
But it measures the whole CD also; therefore it will also measure the remainder CF.
