AS PRACTISED BY THE ARABS. 77 



(b'.) Then 3, the second found figure of the Root, expounds b (Par. 20) 

 and since 2 expounds a, and there is but one figure in b, so by Lem. 6) 

 a p 4 b is expounded by 23. And since p by Par. 21) r= a p 4 b so p is 

 expounded by 23. And hence 23 is the highest approximate Root of 

 A <p c 4 B or P or 166,571,800, the two first periods of the given number. 



As is easily tried, for 23 6 is 148,035,889, which is less, and 

 24* = 191,102,976, which is greater than 166,571,800. 



(&.) Then 84,035,889 the second Subtrahend expounds 6 a 5 p 5 b 

 4 15 a 4 <p* b z -j- 20 a 3 p 3 b 3 4 15 a* p z 6 4 4 6 a p b s 4 b 5 . 



(d'.) Since by Par. 19.) A p 6 4 B = a 6 p* 4 R p 6 4 B so A p 6 4 B 



— a* p tf =r R p 6 -f B. Subtract from both sides of this equation the 

 second Subtrahend by Art. c',) and it becomes A p 6 4 B — a 6 p 6 



— 6a 5 <p 5 b — 15 a 4 <p 4 5* — 20 a 3 p 3 & 3 — 15 a* # b* — 6 a p b 5 — b 6 



— R p 6 4 B — (6 a 5 p s 6 + 15 a 4 £ 4 Zr -J- 20 a 3 <p 3 £ 3 + 15 a 3 <f b 4 4 6a p b 5 4 £*) 

 = A p 6 4 B — (a 6 p 6 4 6 a 5 p 5 6 4 15 a 4 £ 4 6* 4 20 a 3 £, 3 6 3 4 15 a* p* b* 

 4 6 a pb 5 4 6 6 ) = (A?) 6 4 B) — (a^4Z») 6 =:byPar.21)toP'— jw 6 = R'. 

 Then since R p 6 4 B is expounded by 102,571,800 and 6a s p s b 4 15a>*6 3 

 4 20 a 3 p 3 Z> 3 4 15 a- p" b* + 6 a p b 5 4 b 6 is expounded by 84,035,889 

 so (R p 6 4 B) — (6 a 5 p 5 & 4 15 a 4 #* 6 3 4 20 a 3 p 3 & 3 4 15 a- p 2 b* 

 + 6 a p b s 4 b 6 ) or P — j9 6 or R' is expounded by 102,571,800 — 

 84,035,889 = 18,535,911 the second Remainder. 



(e'.) Since C is expounded by 758,593 and consists of 6 figures, so 

 by Lemma 6,) 18,535,91 1,758,593 expounds R' p 6 4' C, and is the second 

 Resolvend. 



(f.) Since a p 4 b or jo is expounded by 23, so j9 <p is expounded by 

 230 and;/ p> or p~pf by 230* and/» 3 p" or fj\ 3 by 230", &c. hence 6 ^ 5 ^ s 



