© 2010 Pearson Prentice Hall. All rights reserved. CHAPTER 13 Mathematical Systems

© 2010 Pearson Prentice Hall. All rights reserved Rotational Symmetry, Groups, and Clock Arithmetic

© 2010 Pearson Prentice Hall. All rights reserved. Objectives 1.Recognize rotational symmetry. 2.Determine if a mathematical system is a group. 3.Understand clock systems as groups. 4.Understand congruence in a modulo m system. 5.Perform additions in a modulo m system. 3

© 2010 Pearson Prentice Hall. All rights reserved. Rotational Symmetry A symmetry of an object is a motion that moves the object back onto itself. In symmetry, you cannot tell, at the end of the motion, that the object has been moved. If it takes m equal turns to restore an object to its original position and each of these turns is a figure that is identical to the original figure, the object has m-fold rotational symmetry. 4

© 2010 Pearson Prentice Hall. All rights reserved. Rotational Symmetry A pinwheel has fourfold rotational symmetry. 5

© 2010 Pearson Prentice Hall. All rights reserved. Groups 6

© 2010 Pearson Prentice Hall. All rights reserved. Example 1: Showing that a Mathematical System is a Group ●−1−11 −1−11−1−1 1−1−11 The table to the right is a mathematical system for the set {−1, 1} under the binary operation of multiplication. a.Show that the mathematical system is a group. b.Show that the group is commutative. 7

© 2010 Pearson Prentice Hall. All rights reserved. Example 1 continued Solution: We verify the four requirements. 1. The Closure Property. The set {−1, 1} is closed under the binary operation of multiplication because the entries in the body of the table are all elements of the set. 2. The Associative Property. Is (a b) c = a (b c) for all elements a, b, and c of the set? Let us select some values for a, b, and c. Let a = 1, b = −1 and c = −1. 8

© 2010 Pearson Prentice Hall. All rights reserved. Example 1 continued Then and The associative property holds for a = 1, b = −1 and c = −1. Next, we must check every possible choice of three elements from the set in order for the system to be associative. We leave this to the student! 9

© 2010 Pearson Prentice Hall. All rights reserved. Example 1 continued 3.The Identity Property. Look for the element that does not change anything when used in multiplication. The element 1 does not change anything: Thus, 1 is the identity element. The identity property is satisfied because 1 is contained in the set {−1, 1}. 10

© 2010 Pearson Prentice Hall. All rights reserved. Example 1 continued 4.The Inverse Property. When an element operates on its inverse, the result is the identity element. Because the identity element is 1, what must each element be multiplied by to obtain 1? Using 1 in each row, we see that, so the inverse of −1 is −1., so the inverse of 1 is 1. Because each element in {−1, 1} has an inverse within the set, the inverse property is satisfied. 11

© 2010 Pearson Prentice Hall. All rights reserved. Example 1 continued a. Since all requirements are satisfied the mathematical system is a group. b.To show that the group is commutative, we must show the commutative property is satisfied. Is a b = b a for all elements a and b of {−1, 1}? We examine Hence,, and the order of multiplication does not matter. Thus, {−1, 1} is a commutative group under multiplication. 12

© 2010 Pearson Prentice Hall. All rights reserved. Clock Arithmetic & Groups Clock addition is defined as follows: Add by moving the hour hand in a clockwise direction. –The symbol + is used to designate clock addition. Example: The figure below illustrates that = 7, = 5, and = 3. 13

© 2010 Pearson Prentice Hall. All rights reserved. Example 2: The 12-Hour Clock System as a Group Show that the 12-hour clock system is a commutative group. Solution: We must check the five requirements - closure, associative, identity, inverse, and the commutative property - for a commutative group. 1.The Closure Property. The set {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} is closed under the operation of clock addition because the entries in the body are all elements of the set. 14

© 2010 Pearson Prentice Hall. All rights reserved. Example 2 continued 15

© 2010 Pearson Prentice Hall. All rights reserved. Example: 2 continued 2.The Associative Property. Is (a + b) + c = a + (b + c) for all elements a, b, and c of the set? Select a = 4, b = 7, and c = 9. The associative property holds for a = 4, b = 7, and c = 9. Again, we must check every possible choice of three elements from the set. We leave this to the student. 16

© 2010 Pearson Prentice Hall. All rights reserved. Example 2 continued 3.The Identity Property. Look for the element that does not change anything when used in clock addition. Notice the identity is 0 because = 0, = 1, = 2, = 3, etc. Thus, 0 must be the identity element. 17

© 2010 Pearson Prentice Hall. All rights reserved. Example 2 continued 4.The Inverse Property. When an element operates on its inverse, the result is the identity element. Because the identity element is 0, we can find the inverse for every element in the set such that “element + ? = 0”. 18

© 2010 Pearson Prentice Hall. All rights reserved. Example 2 continued 5.The Commutative Property. Is a + b = b + a for all elements a and b of {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}? Let’s check a few cases: and both yield and both yield 1. Order of clock addition does not matter for any case - the operation is commutative. 19

© 2010 Pearson Prentice Hall. All rights reserved. Modular Systems We describe m-fold rotational symmetry with a modulo m system. Such a system consists of m integers, starting with 0 and ending with m − 1. Example: The 12-hour clock system is a modulo 12 system. If we had a 7-hour clock, then = 2 in mod 7 or modulo 7 system. 20

© 2010 Pearson Prentice Hall. All rights reserved. Congruence Modulo m 21

© 2010 Pearson Prentice Hall. All rights reserved. Example 3: Understanding Congruence Is ? In other words, “If 22 is divided by 6, is the remainder 4?” Since the remainder is 4, the answer is Yes. 22

© 2010 Pearson Prentice Hall. All rights reserved. Modular Addition 23

© 2010 Pearson Prentice Hall. All rights reserved. Example 4: Modular Additions Find the sum (2 + 4)(mod 7). Solution: To find the sum, we first add to get 6. Because this sum is less than 7, then 24

© 2010 Pearson Prentice Hall. All rights reserved. Example 4 continued Find the sum (9 + 10)(mod 12). Solution: To find the sum, we first add to get 19. Because this sum is greater than 12, then divide 19 by 12. The remainder, 7, is the desired sum. Thus, 25