Interesting Properties of Numbers. 115 



If N=16, and P=13, the quotients will be 1, 3, 11 ; the re- 

 mainders will be 3, 9j 1. 



If N=70=3 . 5 . 7, and P = 32=2^ the quotients will be 2, 3,. 

 8, 52, 16 ; the remainders will be 6, 4, 24, 16, 0. In this case the 

 process terminates. 



If N=13, and P=.ll, the quotients will be j ^}' ^^ g' 3' ^ 



The remainders will be \ ^' ^' §' f ' ^^. 



I 9, 7^, 3, 6, 1 



Now exchanging the values of N and P, that is, taking N=ll, 

 and P=13, we get the quotients > in' 1' 7' fi' «' *;' *^^® remain- 



'^^'^^ > 2, 9, 8, 10, 6, 1 

 If N=509, and P=19, we find r^ _ j^=18=P - 1, therefore 



~2" 

 the number of terms in the periods will be P— 1. And since N 

 and P are both primes, the one of the form 4?z + l, and the other 

 of the form 4h + 3, it follows that if N=19, and P = 509, the 

 number of terms in the periods will be P — 1 = 508. 



When N = 10, our process resolves itself into the usual rule for 



converting the vulgar fraction - into its equivalent decimal. 



If P=7, N being supposed 10, we find the quotients to be 

 < 0' k' «-, the remainders are ) 4' k' -i- Hence 4=0.142857 re- 

 peated in endless succession. Now it is obvious that the same 

 succession of figures must represent in decimals the value of any 

 vulgar fraction whose denominator is 7 and numerator less than 

 7 ; it is also evident that the period will commence with that quo- 

 tient which follows the remainder which is equal to the numerator 

 of the fraction ; thus f = 0.285714; | = 0. 428571 ; 1 = 0.571428; 

 4=0.714285; f = 0.857142. 



If P=17, the quotients will be < q' / ^' ,' J ,.' ^' ^, and the 



remainders will be \ ^' g' ^' ^^ ^^ g' ^^ ^ Therefore, 



_i_ = 0.0588235294117647 ; /^ = 0.1 17647058823 5294 ; 



_3_= 0.1764705882352941 ; -j-V =0.2352941176470588 ; 

 thus we could with the same period of figures represent in deci- 

 mals, the fractions y\, y\, y\, &c. 



