AS PRACTISED BY THE ARABS. 123 



*p. Then since by Art. y) 240 z= 15 a 4 , with its unit put under the 

 5th place of 507,321 so by Lem. 7,) their sum in this situation = 537,321 

 -f- 240 X 10 4 = 537,321 + 2,400,000 — 2,937,321 - (20 a 3 p 3 b + 15 a" p 2 b" 

 + 6 a p b 3 + Z> 4 ) -f- 15 a 4 X p 4 — 15 a* p 4 + 20 a 3 <p 3 b + 15 a 2 p 2 b"~ + 6 a p b 3 

 + b*, and is the upper number in the Rank of the Biquadrate. 



a. Then 2,937,321 X 3 = 8,811,963 = (15 a 4 p 4 + 20 a 3 p 3 b + 15 a 2 p 2 b 2 

 + 6 a p Z> 3 -(- V) X & = 15 a 4 p 4 6 -f 20 a 3 p 3 6 2 + 15 a 2 p 2 6 3 + 6 a p 6 4 

 + b s , and is the product written in the Rank of the Biquadrate. 



\. Then since by Art. u). 192 s 6 a 5 , with its units put under the 6th 

 place of 8,811,963, so by Lem. 7), their sum in this situation = 8,811,963 

 + 192 x 10 3 — 8,811,963+ 19,200,000— 28,011,963 =(15 a 4 p 4 b + 20 a 3 p 3 b z 

 + 15 a 2 p 2 6 3 + 6 a p 5 4 + b 5 ) + 6 a 5 X p 3 = 6 a 3 p 3 + 15 a 4 p 4 6 + 20 a 3 p 3 6 2 

 + 15 a 2 p 2 £ 3 + 6 a p 6 4 + ¥ 



c_>. Then 28,011,963 X 3 = 84,035,889 = (6 a 3 p 3 + 15 a 4 p 4 & 



+ 20 a 3 p 3 6 2 + 15 a 2 p 3 & 3 + 6 a p & 4 + Z> 5 ) X b — a 5 p s b + 15 a 4 p 4 £r 



+ 20 a 3 p 3 b 3 + 15 a 3 p 3 b* + 6 a p b s + & c , and is less than 102,571,800, 

 or R p 6 + B by Art.j. 



<-. » Then since 3 is the greatest number which answers this condi- 

 tion, so 3 is the second figure of the Root, and agrees with the second 

 figure of the Root found by the European method in Par. 34, Art. b f ). 



cj And 84,035,889 expounds the second Subtrahend, which agrees 

 with the second Subtrahend found by the European method in Par. 34, 

 Art. c'. 



d> And since by Art. f) 102,571,800 = R <f + B, so 102,571,800 — 

 84,035,889 - 18,535,911 = R p 6 + B — (6 a s p s b + 15 a 4 p 4 b* + 20 a 3 



i 1 



