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Math 371 Lecture
§2.1: Congruence and Congruence Classes
We review the notion of congruence mod n from Math 290, and revisit the arithmetic of the set Zn of all congruence classes of integers modulo n.
Definition. Let a, b, n be integers with n > 0. We say a is congruent to b modulo n,
written a ≡ b (mod n), if n | (a − b). Congruence mod n is a relation on Z.
Theorem 2.1 For a positive integer n, and integers a, b, c, we have
(1) a ≡ a (mod n) (congruence mod n is reflexive), (2) if a ≡ b (mod n), then b ≡ a (mod n) (congruence mod n is symmetric), and (3) if a ≡ b (mod n) and b ≡ c (mod n), then a ≡ c (mod n) (congruence mod n is transitive). Remark. Thus congruence mod n is an equivalence relation on Z. Proof. (1) Since a − a = 0 is divisible of n, that is a − a = n0, then a ≡ a (mod n). (2) Suppose a ≡ b (mod n). Then a − b = nk for some k ∈ Z, and so b − a = n(−k). This says that b ≡ a (mod n). (3) Suppose a ≡ b (mod n) and b ≡ c (mod n). Then a − b = nk for some k ∈ Z and b − c = nl for some l ∈ Z. It follows that a − c = (a − b) + (b − c) = nk + nl = n(k + l).
This says that a ≡ c (mod n). Some of the arithmetic of congruence mod n is very similar to the corresponding arith- metic of Z.
Theorem 2.2. If a ≡ b (mod n) and c ≡ d (mod n), then
(1) a + c ≡ b + d (mod n), and (2) ac ≡ bd (mod n).
Proof. We suppose that a − b = nk for some k ∈ Z and c − d = nl for some l ∈ Z. (1) We have
(a + c) − (b + d) = (a − b) + (c − d) = nk + nl = n(k + l).
This says that a + c ≡ b + d (mod n). (2) We have
ac − bd = ac − bc + bc − bd = (a − b)c + b(c − d) = nkc + bnl = n(kc + bl).
This says that ac ≡ bd (mod n).
Definition. Let n be a positive integer. The congruence class of a ∈ Z modulo n is the
set [a] = {b ∈ Z : b ≡ a (mod n)}.
Because b ≡ a (mod n) means b − a = nk for some k ∈ Z, we have
[a] = {a + kn : k ∈ Z}.
For example, when n = 6, we have
[1] = { 1 , 1 ± 6 , 1 ± 12 ,... } = {... , − 11 , − 5 , 1 , 7 , 13 ,... }.
Theorem 2.3. a ≡ c (mod n) if and only if [a] = [c].
Proof. Suppose that a ≡ c (mod n). We show [a] = [c] by showing the inclusions [a] ⊆ [c] and [c] ⊆ [a]. For b ∈ [a] we have b ≡ a (mod n). Since a ≡ c (mod n) we have by transitivity that b ≡ c (mod n). Thus b ∈ [c]. For b ∈ [c], we have b ≡ c (mod n). Since a ≡ c (mod n), we have by symmetry that c ≡ a (mod n). Thus by transitivity we have b ≡ a (mod n), meaning b ∈ [a]. Now suppose that [a] = [c]. By reflexivity, a ≡ a (mod n), so that a ∈ [a], and hence a ∈ [c]. Then we have a ≡ c (mod n). Recall for two sets A and C we have three possibilites: either A ∩ C = ∅ (A and C are disjoint), A = C (they are identical), or A ∩ C is nonempty but A 6 = C.
Corollary 2.4. Two congruence classes modulo n are either disjoint or identical.
Proof. If [a] and [c] are disjoint, there is nothing to prove. So suppose that [a] ∩ [c] is nonempty. Then there is b ∈ [a] ∩ [c], which means that b ≡ a (mod n) and b ≡ c (mod n). By symmetry and transitivity we have a ≡ c (mod n). Hence by Theorem 2.3 we have [a] = [c].
Corollary 2.5. Congruence mod n has the following two properties.
(1) If a ∈ Z and r is the remainder when a is divided by n, then [a] = [r]. (2) There are exactly n distinct congruence classes modulo n, namely [0], [1],... , [n−1].
Remark. The set {[0], [1],... , [n − 1]} is denoted by Zn.
