MATH 57091 - Algebra for High School Teachers
Basic Properties of Congruence
Professor Donald L. White
Department of Mathematical Sciences
Kent State University
D.L. White (Kent State University)
1 / 10
Review
Recall:
Definition
Fix a positive integer n and let a and b be any integers.
We say a is congruent to b modulo n if n | a − b.
We write a ≡ b (mod n).
We also showed that
a ≡ b (mod n) ⇐⇒ a and b have the same remainder on division by n.
D.L. White (Kent State University)
2 / 10
Equivalence Relation
Our first result says that congruence mod n is an equivalence relation.
For an introduction to equivalence relations, review §1.1 of the text
or §1.4 of the Fundamental Concepts of Algebra notes.
Theorem
Fix a positive integer n and let a, b, and c be any integers.
The following properties hold:
i
[Reflexive] a ≡ a (mod n).
ii
[Symmetric] If a ≡ b (mod n), then b ≡ a (mod n).
iii
[Transitive] If a ≡ b (mod n) and b ≡ c (mod n), then a ≡ c (mod n).
Remark: All of these follow immediately from the fact that
two integers are congruent mod n if and only if
they have the same remainder on division by n.
We will prove the theorem using the definition also,
because it is a very good exercise in applying divisibility properties.
D.L. White (Kent State University)
3 / 10
Equivalence Relation
Proof:
i
Since n 6= 0, we know that n | 0, hence n | a − a.
Therefore, a ≡ a (mod n).
ii
Suppose a ≡ b (mod n), so that n | a − b.
Recall that if n | m for some integer m, then n | −m.
Hence n | a − b implies n | −(a − b), and −(a − b) = b − a.
Therefore, n | b − a, and so b ≡ a (mod n).
iii
Suppose a ≡ b (mod n) and b ≡ c (mod n).
By definition of congruence, we have n | a − b and n | b − c.
By the Combination Theorem, this implies n | (a − b) + (b − c).
Therefore, n | a − c, and so a ≡ c (mod n).
D.L. White (Kent State University)
4 / 10
Congruence Classes
Recall that by the Division Algorithm,
each integer a has a unique remainder r (with 0 6 r < n) on division by n.
We can “classify” all integers by their remainders on division by n
and obtain n classes of integers.
The classes are the integers with remainder r , for r = 0, 1, 2, . . . , n − 1.
All integers in a given class have the same remainder on division by n,
hence are congruent mod n.
These classes of integers are the congruence classes mod n.
Definition
Fix a positive integer n and let a be any integer.
The congruence class of a mod n is the set
[a]n = {b | b ≡ a (mod n)}.
D.L. White (Kent State University)
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Congruence Classes
NOTES:
If the modulus n is understood, we will write [a] for [a]n .
Since a ≡ a (mod n), we have that a is an element of [a]n ,
and so [a]n is non-empty.
Since b ≡ a (mod n) ⇔ b = a + qn for some integer q, we have
[a]n = {a + qn | q is an integer}.
There is a unique integer r in [a]n with 0 6 r < n;
namely, r is the remainder on division of a by n.
This integer r is called
the least non-negative residue of a mod n
or the principal representative of the class [a]n .
In the language of equivalence relations,
[a]n is the equivalence class of a
under the equivalence relation of congruence mod n.
D.L. White (Kent State University)
6 / 10
Congruence Classes
EXAMPLES:
1 If n = 2, the only possible remainders are 0 and 1.
Hence there are exactly two congruence classes.
All even integers are congruent to 0 mod 2, and so
[0]2 = {0, ±2, ±4, ±6, . . . }
is the set of even integers.
All odd integers are congruent to 1 mod 2, and so
[1]2 = {±1, ±3, ±5, ±7, . . . }
is the set of odd integers.
2 If n = 3, there are three congruence classes:
[0]3 = {0 + 3q | q is an integer} = {0, ±3, ±6, ±9, . . . }
[1]3 = {1 + 3q | q is an integer}
= {1, 4, 7, 10, 13, . . . } ∪ {−2, −5, −8, −11, −14, . . . }
[2]3 = {2 + 3q | q is an integer}
= {2, 5, 8, 11, 14, . . . } ∪ {−1, −4, −7, −10, −13, . . . }
D.L. White (Kent State University)
7 / 10
Congruence Classes
Our next result says that the congruence classes “partition” the integers.
Theorem
Fix a positive integer n.
If a and b are integers, then either [a]n = [b]n or [a]n ∩ [b]n = ∅.
This theorem follows immediately either
from the fact that the remainder r , 0 6 r < n, on division by n is unique
or from the general theory of equivalence relations.
Again, it is more instructive to prove it directly from the definition.
D.L. White (Kent State University)
8 / 10
Congruence Classes
Proof: We will show that if [a]n ∩ [b]n 6= ∅, then [a]n = [b]n .
If [a]n ∩ [b]n 6= ∅, then there is an integer x that is in both [a]n and [b]n .
Hence x ≡ a (mod n) and x ≡ b (mod n).
By symmetry, a ≡ x (mod n), and so by transitivity a ≡ b (mod n).
If y ∈ [a]n , then y ≡ a (mod n), and since a ≡ b (mod n),
transitivity implies y ≡ b (mod n), and so y ∈ [b]n . Therefore, [a]n ⊆ [b]n .
Similarly, we can show that if z ∈ [b]n , then z ∈ [a]n , and so [b]n ⊆ [a]n .
Finally, [a]n ⊆ [b]n and [b]n ⊆ [a]n together imply [a]n = [b]n .
D.L. White (Kent State University)
9 / 10
Congruence Classes
Corollary
Fix a positive integer n and let a and b be integers.
The following are equivalent:
i
a ≡ b (mod n)
ii
b ≡ a (mod n)
iii
a is in [b]n
iv
b is in [a]n
v
[a]n ∩ [b]n 6= ∅
vi
[a]n = [b]n
D.L. White (Kent State University)
10 / 10