AS PRACTISED BY THE ARABS. 65 



Let M be a surd integer to index n, and to its approximate root, so 

 that m n Z. and (to-J-1)" > M. Let M— m" = r and M = m n -f r. Then 



is M always > to -f 



n 



That is to -j- is an 



(to -f 1) n ~m n 



(m-fl) n — to°. 

 approximate w th Root of M, greater than m the integral Root. 



For as before, let n be expounded by 6. Then obviously (to -f- I) 6 — 

 m 6 r= 6 m s -}- 15 to 4 -j- 20 to 3 -J- 15 to* -j- 6 to -J- 1. Now since ¥ JM > to let 

 6 |M = to -f- a?. Then since 6 |m Z_ to -}- 1, so a? Z_ 1 and is a proper frac- 

 tion. Then M = (m+x) 6 — m 6 -\- 6m 5 x -f 15 to 4 a;* -f 20m 3 x 3 -f 15to*# 4 

 -f- 6too; s -f a;* andM — m 6 — 6m 5 x -f 15m 4 a; 3 ' -j- 20TO 3 a; 3 -f- 15TO*a; 4 -f- 6to# s 



r 6 m s a; -j- 15 to 4 x* -J- 20 to 3 a? 5 



-J- a; 6 zr r. Hence 



(to + 1) 6 — to 6 ,6 to 5 -j- 15 to 4 -J- 20 TO 3 



-f 15 to* a; 4 -\-Qmx s -\-x 6 r 



Thena? > - 



-f 15 m* -j- 6 to -f 1 (m-f-1) 6 — to 6 



If not, then x = or Z_ — First let a; =■ 



(to-j-1) 6 — m 6 . (m-J-1) 6 — to*. 



6 to 5 a? -)- 15m 4 x* -f-20m 3 a? 3 -j- 15to*# 4 -f6ma? 5 -|- a; 5 



Then a; — < and hence 



6to 5 -j- 15m 4 -J-20to 3 -f 15to* -f6»» -J- 1 



6 to s a; -f- 15 to 4 x -{- 20 to 3 a: -f- 15 to* a? -f- 6 to a? -j- a- ■=. 6 to 5 a: -J- 15 to 4 x* 



-J- 20 to 3 x 7, -|- 15 to* x 4 +6 to a: 5 -J- x 6 , an equation which is evidently 



absurd, unless x =: 1. But a? is also a proper fraction, which is absurd. 



Secondly, a fortiori x not Z_ for then also 6m s x -f- 15 to 4 x 



(TO + l) 5 — TO* 



+ 20 m 3 a- -j- 15 to* a 1 -f 6 to a? -f a- Z_ to s x + 15 to 4 a'* -f- 20 to 3 .r 3 

 + 15 to* a- 1 -}- to a s -f a; 5 , which requires x to be > 1. also absurd. 



