Two unequal magnitudes being set out, if from the greater 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 lesser magnitude set out.
Let / A B /, C be two unequal magnitudes of which / A B / is the greater : I say that, if from / A B / 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.
Let it be multiplied, and let / D E / be a multiple ofC, and greater than / A B /; let / D E / be divided into the parts / D F /, / F G /, / G E /equal to C, from / A B / let there be subtracted / B H / greater than its half, and, from / A H /, / H K / greater than its half, and let this process be repeated continually until the divisions in / A B / are equal in multitude with the divisions in / D E /.
Let, then, / A K /, / K H /, / H B / be divisions which are equal in multitude with / D F /, / F G /, / G E /.
Now, since / D E / is greater than / A B /, and from / D E / there has been subtracted / E G / less than its half, and, from / A B /, / B H / greater than its half,
therefore the remainder / G D / is greater than the remainder / H A /.
And, since / G D / is greater than / H A /, and there has been subtracted, from / G D /, the half / G F /, and, from / H A /, / H K / greater than its half,
therefore the remainder / D F / is greater than the remainder / A K /.
But / D F / is equal to C;
therefore C is also greater than / A K /.
Therefore / A K / is less than C.
Therefore there is left of the magnitude / A B / the magnitude / A K / which is less than the lesser magnitude set out, namely C. Q. E. D.
And the theorem can be similarly proved even if the parts subtracted be halves.
