Enter question here... In the first chapter of Disquisitiones Arithmeticae, Gaus
ID: 3081496 • Letter: E
Question
Enter question here...
In the first chapter of Disquisitiones Arithmeticae, Gauss introduced the concept of congruence. Here is Gauss' definition: we define a = b (mod n) if n divides the difference a - b; in other words, a - b = kn for some integer k. Now any positive integer n can be written in decimal form: n = ak10k +ak-1 + ak-110k-1 + ... + a110 + a0 where a le aj le 9 for all j. Prove that n = a0 + ai + ... + ak(mod 9). Consequently, 9 divides n if and only 9 divides the sum of the digits aj. n = a0 - ai + a2 - ... + (-1 )kak(mod 11). Consequently 11 divides n if 11 divides the alternating sum of the digits aj.Explanation / Answer
WE HAVE
N=A0+10A1+............+[10^(K-1)][A(K-1))+[10^K][A^K]................1
PROPOSITION
GIVEN 9 DIVIDES N ....
THAT IS
N=9P..................................................2
WHERE P IS AN INTEGER
TO PROVE
A0+A1+.........+[A(K-1)]+[A(K)]..................IS DIVISIBLE BY 9
THAT IS TO PROVE
A0+A1+.........+[A(K-1)]+[A(K)] = 9Q...................................3
PROOF
FROM 1 AND 2 WE HAVE
N=A0+10A1+............+[10^(K-1)][A(K-1)]+[10^K][A^K] = 9P
[A0+(9+1)A1+...........+[9*(111.....K-1 TIMES)+1][A(K-1)]+[9*(111......K TIMES)+1][A^K] =9P
[A0+A1+..........+A(K-1)+A(K)] + 9[A1+A2+..........A(K-1)+A(K)]=9P
[A0+A1+..........+A(K-1)+A(K)] = 9P - 9[A1+A2+..........A(K-1)+A(K)]
[A0+A1+..........+A(K-1)+A(K)] = 9 [P - {A1+A2+..........A(K-1)+A(K)}] = 9*Q..........SAY
WHERE
Q= [P - {A1+A2+..........A(K-1)+A(K)}] IS AN INTEGER
SINCE P,A1,A2,.....ARE ALL INTEGERS
HENCE EQN. 3 IS PROVED .....
==========================
CONVERSE ....
PROPOSITION
GIVEN
A0+A1+.........+[A(K-1)]+[A(K)]..................IS DIVISIBLE BY 9
THAT IS GIVEN
A0+A1+.........+[A(K-1)]+[A(K)] = 9Q...................................3
TO PROVE
9 DIVIDES N ....
THAT IS
N=9P..................................................2
WHERE P IS AN INTEGER
PROOF
FROM 1 WE HAVE
N=A0+10A1+............+[10^(K-1)][A(K-1)]+[10^K][A^K]
N=[A0+(9+1)A1+...........+[9*(111.....K-1 TIMES)+1][A(K-1)]+[9*(111......K TIMES)+1][A^K]
N= [A0+A1+..........+A(K-1)+A(K)] + 9[A1+A2+..........A(K-1)+A(K)]
FROM 3 WE GET
N = 9Q + 9[A1+A2+..........A(K-1)+A(K)] = 9 P ..............................SAY
WHERE
P = [Q + {A1+A2+..........A(K-1)+A(K)}] IS AN INTEGER
SINCE Q,A1,A2,.....ARE ALL INTEGERS
HENCE EQN.2 IS PROVED .....
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IN A SIMILAR WAY YOU CAN PROVE THE II RESULT BY WRITING
10=11-1...ETC.... , TAKING SUM OF ALTERNATE DIGITS
IF YOU ARE IN DIFFICULTY PLEASE COME BACK
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