562 CHAPTER 9 Mathematical Systems Note in Table 9.12 that every number in the same modulo class differs by a multiple of the modulus, in this case a multiple of 7. Adding (or subtracting) a multiple of the modulus to (or from) a given number does not change the modulo class or congruence of the given number. For example, 3, 3 1(7), 3 2(7), + + n 3 3(7), , 3 (7) + … + are all in the same modulo class, namely, 3. We use this fact in the solution to Example 3. c) Divide 165 by 7 and observe the remainder. ÷ = 165 7 23 remainder 4 Thus, 165 4 (mod7). ≡ 7 Now try Exercise 7 RECREATIONAL MATH Cryptography n a b c d e f g h i j k o m l q p r s t u v w y z x 0 1514 16 17 18 19 21 22 23 24 25 26 20 1 2 3 4 5 6 7 8 9 10 11 13 12 Cryptography is the science of communicating in secret codes. Modern cryptography involves the disciplines of mathematics, computer science, and engineering. Cryptography is used whenever we use an ATM, log on to a computer account, or purchase items on the Internet. Modular arithmetic has long been used in cryptography (see the Did You Know ? box on page 565). One example of a coding circle is shown above. To use it, the person sending and receiving the message must know the code key to decipher the code. The code key for the following message given by the 11 numbers below is the letter j . Can you decipher this code? ( Hint : Subtract the code key from the code numbers.) The answer is upside down at the bottom of this box. 231131810 1921016424 Exercise 79 asks you to decipher another message. Math is fun. Example 2 Congruence in Modulo 9 Evaluate each of the following in modulo 9. a) 7 8 + b) 8 5 − c) 7 5 ⋅ Solution Since we are working in modulo 9, the answer in each part will be a number from 0 through 8. a) The sum, 7 8, + is 15. Since 15 is not a number from 0 through 8, we must determine the number in modulo 9 to which 15 is equivalent. 15 9 ÷ = 1, remainder 6. Therefore, 7 8 6 (mod 9). + ≡ b) The difference, 8 5, − is 3. Since 3 is a number from 0 through 8, we have 8 5 3 (mod 9). − ≡ c) The product, 7 5, ⋅ is 35. Since 35 is not a number from 0 through 8, we must determine the number in modulo 9 to which 35 is equivalent. Because 35 9 3, ÷ = remainder 8, we write 7 5 8 (mod 9). ⋅ ≡ 7 Now try Exercise 23 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) − ≡ Solution In each part, we wish to replace the question mark with a positive number less than the modulus. Therefore, in part (a) we wish to replace the ? with a positive number less than 7. In part (b), we wish to replace the ? with a positive number less than 5. In part (c), we wish to replace the ? with a positive number less than 8. 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. Thus, if we add 7, 14, 21, … to 3, the result will be a number in the same modulo class. We want to replace 3 with an equivalent mod 7 number that is greater than 5. Adding 7 to 3 yields a sum of 10, which is greater than 5. 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). − ≡ b) We wish to replace the ? with a positive number less than 5. We know that 7 4 3 − ≡ (mod 5) because 3 3 ≡ (mod 5). Therefore, we need to determine what positive number, less than 5, the number 7 is congruent to in mod 5. If Learning Catalytics Keyword: Angel-SOM-9.3 (See Preface for additional details.)
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