l6 ARITHMETIC^ 



. Method of Proof. 



1. Draw a line below the uppermost number, and sup- 

 pose it cut off. 



2. Add all the rest together, and set their sum under 

 the number to be proved. 



3. Add 



Demon. Let there be any number, as 3467 ; this separated 

 into its several pans becomes 3000+ 400 -|- 60 -j- 7 ; but 3000=3 



XiocGr=3X999-i-i=5X 999+3. In like manner 400=4 X 

 $9+4, and 60=6x9+6. Therefore 34<57=3 X99.9+3+4X 

 99+4+6x9+6+7=3X999+4X99+6x9 + 3 + 4+6+7. 

 And^ = 3ii999+4X99 + 6X9 3+4+6+2: j^^ 



9 9 9 



999 + 4X99 + 6x9 is evidently divisible by 9 ; therefore 3467 

 tiivided by 9 will leave the same remainder as 3+4+6+7 di- 

 vided by 9 ; and the same vv'ill hold for any Other number what- 

 ever. Q^E. D. 



The same may be demonstrated universally thus : 

 Demon. Let iV= any number whatever, a, h^ r, &c. the digits 

 of which it is composed, and«=r as many cyphers as j, the highest 

 digit, is places from unity. Then N-=ia with «, o's -f-^ with 



K — i,o's+rwIth« — 2, o^s, &c. by the nature of notation ; 



=^X«— I, 9's+fl+^X«--2, 9's+^-f rxn— 3, 9*s + r, &c. 



=5X«— i,9's+^X«— 2,9's+cX«— 3, 9*s, &c. J^a-i^h-^-Cy 



Zee. hutaxn—i) 9's+^X« — 2, 9's+^x« — 3, 9's,&c. is plain- 

 ly divisible by 9 ; therefore N divided by 9 will leave the same 

 remainder, as ^+^+f, &c. divided by 9. Q;_E. D. 



In the very same manner, this property may be shown to belong 

 to the number three ; but the preference is usually given to the 

 number 9, on account of its being more convenient in practice. 



Now from the demonstration here given, the reason of the rule 

 itself is evident ; for the excess of nines in two or more numbers 

 being taken separately, and the excess of nines taken also out of 



the 



