Download was the congruence

Congruences
One of the important notational devices used by
Gauss in his Disquisitiones Arithmeticae (1801) was
the congruence: where a, b, m are integers and m
is nonzero, he writes a ≡ b (modm), read as “a is
congruent to b modulo m”, to mean that
a ≡ b (modm) ⇔ m|(a – b) ⇔
€ same remainder when divided by m
a,b have the
Here, m is called the modulus. Congruences are
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prototypical examples of equivalence relations:
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Proposition Congruence mod m is an equivalence
relation (it is reflexive, symmetric, transitive). //
At least as important is the fact that congruence
mod m is compatible with arithmetic.
Proposition If a ≡ b (modm) and c ≡ d (mod m),
then
(1) a + c ≡ b + d (mod m);
(2) ac ≡ bd (mod m);
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(3) a k ≡ b k (mod m) for any positive integer k. //
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Proposition
(1) Reduction: If a ≡ b (modm) and n|m, then
a ≡ b (modn).
m ).
(2) Cancellation: ac ≡ bc (mod m) ⇒ a ≡ b (mod (c,m
)
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Because congruence mod m is an equivalence
relation, Z is partitioned into equivalence classes
under this relation, called more appropriately
congruence classes mod m. (Thus, every integer
belongs to exactly one congruence class mod m and
no two congruence classes have any numbers in
common.) There are exactly m congruence classes
mod m and they are determined by the m possible
remainders (or in Gauss’ terminology, residues)
r = 0, 1, … , m – 1 on division by m. These m
numbers constitute the standard residue system
(SRS) mod m, e.g. {0, 1, 2, 3, 4, 5, 6} is a SRS mod 7.
Replacing anyone of the residues by any number to
which it is congruent yields another complete
residue system (CRS), e.g., {7, 50, 30, 3, –3, 5, –1}
is a CRS mod 7.
A least absolute residue system mod m is a CRS
whose members have the smallest absolute values
possible, e.g., {–3, –2, –1, 0, 1, 2, 3} is a least
absolute residue system mod 7.
Here is an interesting application of congruence
arithmetic:
Theorem There are infinitely many primes of the
form p ≡ −1(mod 4).
Proof We argue in a manner similar to Euclid’s
proof that there are infinitely many primes.
€ Suppose there are only finitely many primes of the
desired form. List them: q1 = 3,q 2 = 7,… ,qn . Now
consider the number N = 4 q2 qn + 3, which is
certainly larger than all of the q’s and divisible by
none of them. If the prime factorization of N is
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m
N = ∏ p kek ,
k=1
then since N must be odd, all the p’s are odd. So it
must be that each pk ≡ 1(mod 4). But then on the
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one hand,
m
€N = ∏
k=1
pkek
m
≡ ∏ 1e k ≡ 1(mod 4),
k=1
while on the other, N = 4 q2 qn + 3 ≡ −1(mod 4).
Thus, 1 ≡ −1 (mod 4). This contradiction completes
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the proof. //
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