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Congruence of Integers

Congruence of Integers. Definition:. Let n (>1) be a positive integer. For x , y  Z , we say that x is congruent to y modulo n if and only if n | x y , denoted by x  y (mod n ). Note:. x  y (mod n )  n | x y  x  y = nq for some q  Z

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Congruence of Integers

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  1. Congruence of Integers

  2. Definition: Let n (>1) be a positive integer. For x, yZ, we say that x is congruent to ymodulo n if and only if n | x y, denoted by x  y (mod n).

  3. Note: x  y (mod n)  n |x y  x y = nq for some qZ  x = y + nq  x and y yield the same remainder when each divided by n.

  4. Theorem1. • The relation of congruence modulo n is an equivalence relation on the set of integers. • Pf: Need to check (i) reflexive, (ii) symmetric and (iii) transitive.

  5. Note: The equivalence classes for congruence modulo n form a partition of Z, i.e. they separate Z into mutually disjoint subsets. These subsets are called congruence classes (or residue classes).

  6. Ex1. • When n = 4, we have 4 congruence classes as follow: • [0]={ … , 8, 4, 0, 4, 8, …..} [1]={ …., 7, 3, 1, 5, 9,……} [2]={ …., 6, 2, 2, 6, 10,……} [3]={ …., 5, 1, 3, 7, 11,……}

  7. Theorem 2. • If a  b (mod n) and xZ, then a + x  b + x (mod n) and ax  bx (mod n). • Pf:

  8. Theorem 3. • Suppose a  b (mod n), c  d (mod n). Then a+c  b+d (mod n) and acbd (mod n). • Pf:

  9. Theorem 4. • If ax  ay(mod n) and (a, n)=1, then x  y (mod n). • Pf:

  10. Use the Euclidean Algorithm to find a solution of ax  b (mod n) when (a, n) = 1.

  11. Ex2. • Consider the congruence equation 20x 14 (mod 63). • Since (20, 63) = 1,we can find two integers m, n such that 1 = 20m+63n. • Appling the Euclidean Algorithm, we have 63=203+3, 20=36+2, 3=21+1. Thus 1 =

  12. Ex2.Solve 20x 14 (mod 63)

  13. Ex2’.Use another way to find the solution of 20x  14 (mod 63).

  14. Theorem 5. If a and n are relatively prime, then the congruence equation ax  b (mod n) has an integer solution x. Moreover, any two solutions in Z are congruent modulo n.

  15. Thm5.If (a, n)=1, then the congruence ax  b (mod n) has an integer solution x. Pf:

  16. Thm5.Moreover, any two solutions ofax  b (mod n) in Z are congruent modulo n. Pf:

  17. Ex3. • Consider the equation 24x  45 (mod 12). • Suppose that there is a solution, x0, of this congruence equation. Then 24x0–45 is a multiple of 12. It is impossible, because 24x0–45 is always an odd integer which cannot be a multiple of 12. Thus the equation 24x  45 (mod 12) has no solution. • This Example tells us that the condition in Theorem5 “(a, n) = 1” is a very important condition for getting a solution of ax  b (mod n).

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