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Chapter 4 Numeration and Mathematical Systems © 2008 Pearson Addison-Wesley. All rights reserved.

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Presentation on theme: "Chapter 4 Numeration and Mathematical Systems © 2008 Pearson Addison-Wesley. All rights reserved."— Presentation transcript:

1 Chapter 4 Numeration and Mathematical Systems © 2008 Pearson Addison-Wesley. All rights reserved

2 © 2008 Pearson Addison-Wesley. All rights reserved 4-4-2 Chapter 4: Numeration and Mathematical Systems 4.1 Historical Numeration Systems 4.2 Arithmetic in the Hindu-Arabic System 4.3 Conversion Between Number Bases 4.4 Clock Arithmetic and Modular Systems 4.5 Properties of Mathematical Systems 4.6Groups

3 © 2008 Pearson Addison-Wesley. All rights reserved 4-4-3 Chapter 1 Section 4-4 Clock Arithmetic and Modular Systems

4 © 2008 Pearson Addison-Wesley. All rights reserved 4-4-4 Clock Arithmetic and Modular Systems Finite Systems and Clock Arithmetic Modular Systems

5 © 2008 Pearson Addison-Wesley. All rights reserved 4-4-5 Finite Systems Because the whole numbers are infinite, numeration systems based on them are infinite mathematical systems. Finite mathematical systems are based on finite sets.

6 © 2008 Pearson Addison-Wesley. All rights reserved 4-4-6 12-Hour Clock System The 12-hour clock system is based on an ordinary clock face, except that 12 is replaced by 0 so that the finite set of the system is {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}.

7 © 2008 Pearson Addison-Wesley. All rights reserved 4-4-7 Clock Arithmetic As an operation for this clock system, addition is defined as follows: add by moving the hour hand in the clockwise direction. 0 1 2 3 4 5 6 7 8 9 10 11 5 + 3 = 8

8 © 2008 Pearson Addison-Wesley. All rights reserved 4-4-8 Example: Finding Clock Sums by Hand Rotation 0 1 2 3 4 5 6 7 8 9 10 11 Find the sum: 8 + 7 in 12-hour clock arithmetic Solution Start at 8 and move the hand clockwise through 7 more hours. Answer: 3 Plus 7 hours

9 © 2008 Pearson Addison-Wesley. All rights reserved 4-4-9 12-Hour Clock Addition Table

10 © 2008 Pearson Addison-Wesley. All rights reserved 4-4-10 12-Hour Clock Addition Properties Closure The set is closed under addition. Commutative For elements a and b, a + b = b + a. Associative For elements a, b, and c, a + (b + c) = (a + b) + c. Identity The number 0 is the identity element. Inverse Every element has an additive inverse.

11 © 2008 Pearson Addison-Wesley. All rights reserved 4-4-11 Inverses for 12-Hour Clock Addition Clock value a 01234567891011 Additive Inverse -a 01110987654321

12 © 2008 Pearson Addison-Wesley. All rights reserved 4-4-12 Subtraction on a Clock If a and b are elements in clock arithmetic, then the difference, a – b, is defined as a – b = a + (–b)

13 © 2008 Pearson Addison-Wesley. All rights reserved 4-4-13 Example: Finding Clock Differences Find the difference 5 – 9. Solution 5 – 9 = 5 + (–9)Definition of subtraction = 5 + 3 Additive inverse of 9 = 8

14 © 2008 Pearson Addison-Wesley. All rights reserved 4-4-14 Example: Finding Clock Products Find the product Solution = 8

15 © 2008 Pearson Addison-Wesley. All rights reserved 4-4-15 Modular Systems In this area the ideas of clock arithmetic are expanded to modular systems in general.

16 © 2008 Pearson Addison-Wesley. All rights reserved 4-4-16 Congruent Modulo m The integers a and b are congruent modulo m (where m is a natural number greater than 1 called the modulus) if and only if the difference a – b is divisible by m. Symbolically, this congruence is written

17 © 2008 Pearson Addison-Wesley. All rights reserved 4-4-17 Example: Truth of Modular Equations Decide whether each statement is true or false. Solution a) True. 12 – 4 = 8 is divisible by 2. b) False. 35 – 4 = 31 is not divisible by 7. c) True. 11 – 44 = –33 is divisible by 3.

18 © 2008 Pearson Addison-Wesley. All rights reserved 4-4-18 Criterion for Congruence if and only if the same remainder is obtained when a and b are divided by m.

19 © 2008 Pearson Addison-Wesley. All rights reserved 4-4-19 Example: Solving Modular Equations Solve the modular equation below for the whole number solutions. Solution Because dividing 5 by 8 has remainder 5, the equation is true only when 2 + x divided by 8 has remainder 5. After trying the values 0, 1, 2, 3, 4, 5, 6, 7 we find that only 3 works. Other solutions can be found by repeatedly adding 8: {3, 11, 19, 27, …}.


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