AS PRACTISED BY THE ARABS. 157 



(2 a z -\- 1) r — r % z 1 



(2 a z -J- l) c is positive so the fraction ■ — ■which expresses the 



(2az + l) 2 

 deficiency will also be negative. Now a negative deficiency is an excess. 

 That is by taking z too large, the assumed Root will be greater than the 



l 

 truth instead of less. In this case the positive fraction is greater 



(2az + \)r-r°-z°- 

 than o, and all the negative values of , though they maybe 



(2 a z + 1)2 



numbers of a greater denomination, yet as they are all less than o, so they 



l 

 are also all less than positive , which hence is still truly a maximum. 



4z 2 



(2 a z + 1) r — r 2 z 2 



66). Now since this expression when negative is the 



(2 a z + I) 2 



r z 



amount of the negative deficiency produced by assuming a -f ■ as 



2 a z + l 

 the true Square Root of a * -J- r, so if this expression have its signs chang- 



r 2 z' 2 — (2 a z + 1) r 



ed it will become — — and will be the positive excess pro- 



(2az + l) 2 



duced by the same assumption. 



For in this case since by supposition a -{- 



r z 



2 az + 1 



> a~ -f- V, SO 



instead of (a 3 -f f) — (a -) j as in Par- 64) for a deficiency we 



\ 2ac + 1 / 



/ rz \ 2 r 2 z n - - - (2 a z + 1 ) r 



have for an excess, (a -\ — («* -f- r) zz 



2 a s + 1 / (2 a z + l) 2 



as above. 



07.) Now this expression increases in value both by the increase of 

 r and of z. 



