Definiton. Let a, b, and m be integers. a is congruent to b mod m if ; that is, if
Notation: means that a is congruent to b mod m. m is called the modulus of the congruence; I will almost always work with positive moduli.
Note that if and only if . Thus, modular arithmetic gives you another way of dealing with divisibility relations.
For example:
because .
because .
Proposition. Congruence mod m is an equivalence relation:
(a) ( Reflexivity) for all a.
(b) ( Symmetry) If , then .
(c) ( Transitivity) If and , then .
Proof. Since , it follows that .
Suppose . Then , so . Hence, .
Suppose and . Then there are integers j and k such that
Add the two equations:
This implies that .
Example. (a) List the elements of the equivalence classes relative to congruence mod 3.
(b) Using 0, 1, and 2 to represent these equivalence classes, construct addition and multiplication tables mod 3.
(a) The equivalence classes are the 3 congruence classes:
Each integer belongs to exactly one of these classes. Two integers in a given class are congruent mod 3.
(If you know some group theory, you probably recognize this as constructing from .)
(b)
For example, , because as integers, and the congruence class of 3 is represented by 0. Likewise, as integers, and the congruence class of 4 is represented by 1.
I could have chosen different representatives for the classes --- say 3, -4, and 4. A choice of representatives, one from each class, is called a complete system of residues mod 3. But working mod 3 it's natural to use the numbers 0, 1, and 2 as representatives --- and in general, if I'm working mod n, the obvious choice of representatives is the set . This set is called the standard residue system mod n, and it is the set of representatives I'll usually use. Thus, the standard residue system mod 6 is .
Theorem. Suppose and . Then:
(a) and .
(b) .
Proof. I'll prove the first congruence as an example. Suppose and . Then and for some , so
This implies that .
Example. Solve the congruence
There are no "fractions" mod 11. I want to divide by 3, and to do this I need to multiply by the multiplicative inverse of 3. So I need a number k such that . A systematic way of finding such a number is to use the Extended Euclidean algorithm. In this case, I just use trial and error. Obviously, and won't work, so I'll start at :
Thus, I need to multiply the equation by 4:
Definition. x and y are multiplicative inverses mod n if .
Notation: or . Do not use fractions.
Example. (a) Find .
(b) Prove that 6 does not have a multiplicative inverse mod 8.
(a) , so .
(b) Suppose . Then
This contradiction shows that 6 does not have a multiplicative inverse mod 8.
Example. Reduce to a number in .
Example. Reduce to a number in .
, so
Example. Show that if p is prime, then
By the Binomial Theorem,
A typical coefficient is divisible by p for . So going mod p, the only terms that remain are and .
For example
The result is not true if the modulus is not prime. For example,
Copyright 2020 by Bruce Ikenaga