AS PRACTISED BY THE ARABS. 57 



(12.) But if no such number can be found then is s 6 -\-t a surd to the 



i 



6 



6th Power. If not, let s 6 -\-t be Rational, and let | s 6 +t = v. Then either 

 v is Z_ -=• or >6-. 



First let v = s then v 6 = s 6 and Z_ s 6 + £. But by supposition 



«?= j s 6 +t and also »* =$*-[-£, which is absurd. 



Second. Let v /_ s, then by similar reasoning v 6 is Z. 5 s 

 and consequently Z. s 6 -\- t and also == s 6 -\- t, which is also ab- 

 surd. 



Third Let v > s and let v zz s + w. Then v 6 = (s + w) 6 = s 6 -\- t 

 =s 6 . + 6 s s ir -j- 15 s* w 2 + 20 s 3 w 3 + 15 s z «# 4 -f- 6siv 5 + w 6 . Hence w 

 answers the conditions of par. 11, which yet by supposition no number 

 can answer, which is also absurd. v 



(13.) Let then s 6 -{- t be integers and thus irrational, and let u be the 

 greatest possible integer such that 6s 5 m + 15 s 4 u* + 20 s 3 u 3 + 15 s* u 4 

 r|- 6 su s -\- u 6 Z_ t, then is s -+- u, the greatest integral approximate 6th 

 Root of s 6 + t. That is (5 + u) 6 Z_ s 6 -f t and (5 + u + 1) 6 > s 6 + £. For 

 if not let v be an integer > s -{- it and such that v 6 Z_ s* + 2. Then since 

 v > 5 -\- it and 5 -f- u > 5 so also v > s. Let 1? — 5 -f- w; and then as before 

 v 6 — (s + w) s = s 6 -f- 6 s 5 it; + 15 s 4 iv* + 20 s 3 iv 3 -j- 15 5 s iv* -f 6 5 w 5 

 -|- w 5 and Z- s 6 -J- t. Subtract s 6 from both sides, there remains 6 s s w 

 -f 15 s 4 w* -|- 20 s 3 iv 3 + 15 s s w* -f 6 s w $ + «;« Z. *. But since w > 

 s -f- wand?; — s -\~ w so i# > u, and fulfils the conditions of par. 11, 

 consequently u both is, and is not the greatest number that fulfils these 

 conditions, which is absurd. 



