# Proof the integer x ba where a is a multiplicative

This preview shows pages 11–13. Sign up to view the full content.

Proof. The integer x = ba 0 , where a 0 is a multiplicative inverse of a modulo n , is clearly a solution. For any integer x 0 , we have ax 0 b (mod n ) if and only if ax 0 ax (mod n ), which by Theorem 2.3 holds if and only if x x 0 (mod n ). 2 In particular, this theorem implies that multiplicative inverses are uniquely determined modulo n . More generally, we have: Theorem 2.5 Let n be a positive integer and let a, b Z . Let d = gcd( a, n ) . If d | b , then the congruence ax b (mod n ) has a solution x , and any integer x 0 is also a solution if and only if x x 0 (mod n/d ) . If d - b , then the congruence ax b (mod n ) has no solution x . Proof. Let n, a, b, d be as defined above. For the first statement, suppose that d | b . In this case, by Theorem 2.3, we have ax b (mod n ) if and only if ( a/d ) x ( b/d ) (mod n/d ), and so the statement follows immediately from Theorem 2.4. For the second statement, assume that ax b (mod n ) for some integer x . Then since d | n , we have ax b (mod d ). However, ax 0 (mod d ), since d | a , and hence b 0 (mod d ), i.e., d | b . 2 Next, we consider systems of congruences with respect to moduli that that are relatively prime in pairs. The result we state here is known as the Chinese Remainder Theorem, and is extremely useful in a number of contexts. 6

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
Theorem 2.6 (Chinese Remainder Theorem) Let k > 0 , and let a 1 , . . . , a k Z , and let n 1 , . . . , n k be positive integers such that gcd( n i , n j ) = 1 for all 1 i < j k . Then there ex- ists an integer x such that x a i (mod n i ) ( i = 1 , . . . , k ) . Moreover, any other integer x 0 is also a solution of these congruences if and only if x x 0 (mod n ) , where n := Q k i =1 n i . Proof. Let n := Q k i =1 n i , as in the statement of the theorem. Let us also define n 0 i := n/n i ( i = 1 , . . . , k ) . It is clear that gcd( n i , n 0 i ) = 1 for 1 i k , and so let m i be a multiplicative inverse of n 0 i modulo n i for 1 i k , and define z i := n 0 i m i ( i = 1 , . . . , k ) . By construction, one sees that for 1 i k , we have z i 1 (mod n i ) and z i 0 (mod n j ) for 1 j k with j 6 = i . That is to say, for 1 i, j k , z i δ ij (mod n j ), where δ ij := 1 for i = j and δ ij := 0 for i 6 = j . Now define x := k X i =1 z i a i . One then sees that for 1 j k , x k X i =1 z i a i k X i =1 δ ij a i a j (mod n j ) . Therefore, this x solves the given system of congruences. Moreover, if x 0 x (mod n ), then since n i | n for 1 i k , we see that x 0 x a i (mod n i ) for 1 i k , and so x 0 also solves the system of congruences. Finally, if x 0 solves the system of congruences, then x 0 x (mod n i ) for 1 i k . That is, n i | ( x 0 - x ) for 1 i k . Since gcd( n i , n j ) = 1 for i 6 = j , this implies that n | ( x 0 - x ), i.e., x 0 x (mod n ). 2 2.3 Residue Classes It is easy to see that for a fixed value of n , the relation · ≡ · (mod n ) is an equivalence relation on the set Z ; that is, for all a, b, c Z , we have a a (mod n ) , a b (mod n ) implies b a (mod n ) , and a b (mod n ) and b c (mod n ) implies a c (mod n ) .
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

### What students are saying

• As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

Kiran Temple University Fox School of Business ‘17, Course Hero Intern

• I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

Dana University of Pennsylvania ‘17, Course Hero Intern

• The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

Jill Tulane University ‘16, Course Hero Intern