Section 9.3 Modular Arithmetic

What You Will Learn Upon completion of this section, you will be able to: Solve problems involving modulo m systems. Determine whether a mathematical system defined by a modulo m system is a commutative group.

Modular and Clock Arithmetic There is one difference in notation between clock 12 arithmetic and modulo 12 arithmetic. In the modulo 12 system, the symbol 12 is replaced with the symbol 0. The set of elements {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} with the operation of addition is called a modulo 12 system or mod 12 system.

Modulo m System A modulo m system consists of m elements, { 0, 1, 2, 3, . . . , m – 1 } and a binary operation.

Modulo 7 Addition

Congruence a is congruent to b modulo m, written a ≡ b (mod m), if a and b have the same remainder when divided by m.

Modulo Classes In any modulo system, we can develop a set of modulo classes by placing all numbers with the same remainder in the appropriate modulo class. The solution to a problem in modular arithmetic, if it exists, will always be a number from 0 through m – 1, where m is the modulus of the system.

Modulo 7 Classes Every number is congruent to a number from 0 to 6 in modulo 7. For example, 24 ≡ 3 (mod 7) because 24 is in the same modulo class as 3.

Example 1: Congruence Modulo 7 Determine which number, from 0 to 6, the following numbers are congruent to in modulo 7. a) 60 b) 84 c) 412

Example 1: Congruence Modulo 7 Solution a) 60 ≡ ? (mod 7) Thus 60 ≡ 4 (mod 7).

Example 1: Congruence Module 7 Solution b) 84 ≡ ? (mod 7) 84 ÷ 7 = 12 remainder 0 Thus 84 ≡ 0 (mod 7). c) 412 ≡ ? (mod 7) 412 ÷ 7 = 58 remainder 6 Thus 412 ≡ 6 (mod 7).

Example 3: Using Modulo Classes in Subtraction Determine the positive number replacement (less than the modulus) for the question mark that makes the statement true. a) 3 – 5 ≡ ?(mod 7) b) ? – 4 ≡ 3(mod 5) c) 5 – ? ≡ 7(mod 8)

Example 3: Using Modulo Classes in Subtraction Solution a) In mod 7, adding 7, or a multiple of 7, to a number results in a sum that is in the same modulo class. We want to replace 3 with an equivalent mod 7 number that is greater than 5. We will add 7.

Example 3: Using Modulo Classes in Subtraction Solution a) 3 – 5 ≡ ? (mod 7) (3 + 7) – 5 ≡ ? (mod 7) 10 – 5 ≡ ? (mod 7) 5 ≡ ? (mod 7) 5 ≡ 5 (mod 7) Therefore, ? = 5 and 3 – 5 ≡ 5 (mod 7)

Example 3: Using Modulo Classes in Subtraction Solution b) We want to replace ? with a number less than 5. We need to determine what number, less than 5, the number 7 is congruent to in mod 5. If we subtract the modulus, 5, from 7, we obtain 2. Thus, 2 and 7 are in the same modular class. So, ? = 2.

Example 3: Using Modulo Classes in Subtraction Solution b) ? – 4 ≡ 3 (mod 5) 7 – 4 ≡ 3 (mod 5) 2 – 4 ≡ 3 (mod 5)

Example 3: Using Modulo Classes in Subtraction Solution c) In mod 8, adding 8, or a multiple of 8, to a number results in a sum that is in the same modulo class. Thus, we can add 8 to 5 so that the statement becomes

Example 3: Using Modulo Classes in Subtraction Solution c) (8 + 5) – ? ≡ 7 (mod 8) 13 – ? ≡ 7 (mod 8) We see that 13 – 6 = 7. Therefore, ? = 6 and 5 – 6 ≡ 7 (mod 8)

Example 4: Using Modulo Classes in Multiplication Determine the positive number replacement (less than the modulus) for the question mark that makes the statement true. a) 2 • ? ≡ 3(mod 5) b) 3 • ? ≡ 0(mod 6) c) 3 • ? ≡ 2(mod 6)

Example 4: Using Modulo Classes in Multiplication Solution a) One method of determining the solution is to replace the question mark with the numbers 0, 1, 2, 3, and 4 and then determine the equivalent modulo class of the product. We use the numbers 0–4 because we are working in modulo 5.

Example 4: Using Modulo Classes in Subtraction Solution a) 2 • ? ≡ 3 (mod 5) 2 • 0 ≡ 0 (mod 5) 2 • 1 ≡ 2 (mod 5) 2 • 2 ≡ 4 (mod 5) 2 • 3 ≡ 1 (mod 5) 2 • 4 ≡ 3 (mod 5) Therefore, ? = 4.

Example 4: Using Modulo Classes in Subtraction Solution b) 3 • ? ≡ 0 (mod 6) 3 • 0 ≡ 0 (mod 6) 3 • 1 ≡ 3 (mod 6) 3 • 2 ≡ 0 (mod 6) 3 • 3 ≡ 3 (mod 6) 3 • 4 ≡ 0 (mod 6) 3 • 5 ≡ 3 (mod 6) The answers are 0, 2, and 4.

Example 4: Using Modulo Classes in Subtraction Solution c) Examining the products in part (b) shows there are no values that satisfy the statement 3 • ? ≡ 2 (mod 6). The answer is “no solution.”

Modular Arithmetic and Groups Modular arithmetic systems under the operation of addition are commutative groups.