SEVENTH EDITION and EXPANDED SEVENTH EDITION Slide 10-1
Chapter 10 Mathematical Systems
10.1 Groups
Definitions A mathematical system consists of a set of elements and at least one binary operation. A binary operation is an operation, or rule, that can be performed on two and only two elements of a set. Slide 10-4
Properties For elements a, b, and c Commutative property Associate property Addition a + b = b + a (a + b) + c = a + (b + c) Multiplication ab = ba (ab)c = a(bc) Slide 10-5
Closure If a binary operation is performed on any two elements of a set and the result is an element of the set, then that set is closed (or has closure) under the given binary operation. Slide 10-6
Identity Element An identity element is an element in a set such that when a binary operation is performed on it and any given element in the set, the result is the given element. Additive identity element is 0. Multiplicative identity element is 1. Slide 10-7
Inverses When a binary operation is performed on two elements in a set and the result is the identity element for the binary operation, each element is said to be the inverse of the other. Slide 10-8
Properties of a Group Any mathematical system that meets the following four requirements is called a group. The set of elements is closed under the given operation. An identity element exists for the set under the given operation. Every element in the set has an inverse under the given operation. The set of elements is associative under the given operation. Slide 10-9
Commutative Group A group that satisfies the commutative property is called a commutative group or (abelian group). Slide 10-10
Properties of a Commutative Group A mathematical system is a commutative group if all five conditions hold. The set of elements is closed under the given operation. An identity element exists for the set under the given operation. Every element in the set has an inverse under the given operation. The set of elements is associative under the given operation. The set of elements is commutative under the given conditions. Slide 10-11
10.2 Finite Mathematical Systems
Definition A finite mathematical system is one whose set contains a finite number of elements. Example: Determine whether the clock arithmetic system under the operation of addition is a commutative group. Slide 10-13
Definition continued Closure: The set of elements in clock arithmetic is closed under the operation of addition. Identity: There is an additive identity element, namely 12. Inverse elements: Each element in the set has an inverse. Associative property: The system is associative under the operation of addition. Commutative property: The commutative property of addition is true for clock arithmetic. The system satisfies the five properties required for a mathematical system. Thus, clock arithmetic under the operation of addition is a commutative or abelian group. Slide 10-14
10.3 Modular Arithmetic
Definition A modulo m system consists of m elements, 0 through m 1, and a binary operation. a is congruent to b modulo m, written a b(mod m), if a and b have the same remainder when divided by m. Slide 10-16
Example Determine which number from 0 to 7, the following numbers are congruent to in modulo 8. a) 66 b) 72 c) 109 Slide 10-17
Solution: a) 66 To determine the value 66 is congruent to in mod 8, divide 66 by 8 and find the remainder. Thus, 66 2 (mod 8) 8 866 64 2 remainder 66 8 = 6, remainder 2 Slide 10-18
Solutions continued b) 72 72? (mod 8) c) 109 109? (mod 8) 72 8 = 9, remainder 0 109 8 = 13, remainder 5 72 0 (mod 8) 109 5 (mod 8) Slide 10-19
Example Evaluate each in mod 6. a) 4 + 2 b) 3 2 c) 2(4) a) 4 + 2? (mod 6) 6? (mod 6) 6 6 = 1, remainder 0. Therefore, 4 + 2 0 (mod 6) Slide 10-20
Example continued b) 3 2 3 2? (mod 6) 1? (mod 6) 1 1 (mod 6) c) 4(2) 4(2)? (mod 6) 8? (mod 6) 8 6 = 1, remainder 2. Therefore, 4(2) 2 (mod 6) Slide 10-21
Find all replacements for the question mark that make the statements true. a) 4? 3(mod 5) One method is to replace the? mark with the numbers. 4 0 0(mod 5) 4 1 4(mod 5) 4 2 3(mod 5) 4 3 2(mod 5) 4 4 1(mod 5) Therefore,? = 2 since 4 2 3(mod 5). b) 3? 0(mod 5) 3 0 0(mod 5) 3 1 3(mod 5) 3 2 0(mod 5) 3 3 3(mod 5) 3 4 0(mod 5) 3 5 3(mod 5) Therefore, the 0, 2 and 4 result in true statements. Slide 10-22