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Transcript
2.5 Division of Integers Recall that division is the inverse operation for multiplication. For example: 36 ÷ 12 = 3 since 3 • 12 = 36 15 ÷ 3 = 5 since 5 • 3 = 15 Now we consider division of integers. To compute !45 ÷ 5 , note that if !45 ÷ 5 = x , then 5 • x = !45 , thus x = !9 . Similarly, for 36 ÷ (!4) we have 36 ÷ (!4) = x , so !4 • x = 36 , and thus x = !9 . Thus one negative number results in a quotient which is negative. If we are dividing two negative numbers such as !28 ÷ (!7) , let !28 ÷ (!7) = x and so !7 • x = !28 , thus x = 4 . So dividing two negative numbers results in a positive number. In summary, when dividing two numbers, the sign is positive if the two numbers have the same sign, and negative if the two numbers have different signs. This is an easy rule to remember, since this rule of signs is identical to that of multiplication. An alternate notation used to represent division of integers is a a , where = a ÷ b . b b Example 1 Divide the two integers. a. b. c. d. Solution a. !48 ÷ 6 !36 ÷ (!3) 108 !9 !68 !4 Since the quotient involves one negative number, the quotient will be negative. Computing the value: !48 ÷ 6 = ! ( 48 ÷ 6 ) rule for signs = !8 b. dividing Since the quotient involves two negative numbers, the quotient will be positive. Computing the value: !36 ÷ (!3) = + ( 36 ÷ 3) rule for signs = 12 dividing 97 c. Since the quotient involves one negative number, the quotient will be negative. Computing the value: 108 108 =! rule for signs !9 9 = !12 dividing d. Since the quotient involves two negative numbers, the quotient will be positive. Computing the value: !68 68 =+ rule for signs !4 4 = 17 dividing When the number 0 is involved with division, we must be careful. Beginning with the case 0÷6= 0 6 Recall we are looking for the value of x such that 0 =x 6 For this to be true, we must have 0=6•x x=0 Thus 0 ÷ 6 = 0 = 0 . What happens if we reverse the situation? Consider 6 6÷0= Again, we are looking for the value of x such that 6 =x 0 98 6 0 For this to be true, we must have 6=0•x 6=0 Clearly this is impossible, so the desired value of x cannot exist, and we say that 6 is 0 undefined, or not defined. One more case to consider is 0÷0= 0 0 Using the same argument, we are looking for the value of x such that 0 =x 0 For this to be true, we must have 0=0•x 0=0 0 is 0 indeterminate, or not determined. We can summarize these division properties with 0 in the following table, where x ≠ 0. But this last statement is always true, so the desired value of x can be anything. We say that General 0 0÷x= x x x÷0= 0 0 0 ÷ 0= 0 Example Conclusion 0÷8 0 5÷0 undefined 0÷0 indeterminate We can now determine solutions to equations which involve division, as the following example illustrates. 99 Example 2 Determine whether or not the given integer value is a solution to the equation. a. b. c. d. Solution x = 8 ; x = 32 4 x = !9 ; x = 45 !5 36 = !4 ; y = !9 y !32 ÷ a = !4 ; a = !8 a. Substitute x = 32 into the equation and determine whether the statement is true: 32 =8 4 8=8 Since this last statement (8 = 8) is true, x = 32 is a solution to the equation x = 8. 4 b. Substitute x = 45 into the equation and determine whether the statement is true: 45 = !9 !5 !9 = !9 Since this last statement (–9 = –9) is true, x = 45 is a solution to the x equation = !9 . !5 c. Substitute y = !9 into the equation and determine whether the statement is true: 36 = !4 !9 !4 = !4 Since this last statement (–4 = –4) is true, y = !9 is a solution to the 36 = !4 . equation y 100 d. Substitute a = !8 into the equation and determine whether the statement is true: !32 ÷ (!8) = !4 4 = !4 Since this last statement (4 = –4) is false, a = !8 is not a solution to the equation !32 ÷ a = !4 . Recall that a geometric sequence is a sequence of numbers (called terms) in which the same number (called the common ratio) is multiplied by each term to obtain the next term. For example, with the sequence 1, 3, 9, 27, … from the last section, the common ratio was noted to be 3. Another way to find the common ratio is by using division. Note that: 3 9 27 = 3, = 3, =3 1 3 9 Since each successive division results in the number 3, the common ratio is 3, and thus the next term in the sequence is 27 • 3 = 81 . This provides an alternate approach to finding terms of a geometric sequence. Example 3 Find the common ratio and the next term for each geometric sequence. a. b. c. d. Solution –4, –8, –16, … 1, –3, 9, … –5, 30, –180, … –13, –156, –1872, … a. Dividing terms to find the common ratio: !8 =2 !4 !16 =2 !8 The common ratio is 2, so the next term is !16 • 2 = !32 . b. Dividing terms to find the common ratio: !3 = !3 1 9 = !3 !3 The common ratio is –3, so the next term is 9 • (!3) = !27 . 101 c. Dividing terms to find the common ratio: 30 = !6 !5 !180 = !6 30 The common ratio is –6, so the next term is !180 • (!6) = 1080 . d. Dividing terms to find the common ratio: !156 = 12 !13 !1872 = 12 !156 The common ratio is 12, so the next term is !1872 • 12 = !22, 464 . Because there is no commutative property for division, when interpreting language the order must be carefully adhered to. If we say “find the quotient of –12 and 3”, we mean to compute !12 = !4 . The next example will further illustrate the terminology used with division. 3 Example 4 Convert each written phrase to a mathematical computation, then compute the required quantity. a. b. c. d. Solution a. b. Find the quotient of –48 and 4. Divide –32 by –8. Find the quotient of –125 and –5. Divide –6 into 120. Performing the division: !48 48 =! 4 4 = !12 rule for signs Performing the division: !32 32 =+ !8 8 =4 rule for signs dividing dividing 102 c. d. Performing the division: !125 125 =+ !5 5 = 25 rule for signs dividing Performing the division: 120 120 =! rule for signs !6 6 = !20 dividing The final topic in this section is finding the average of a group of numbers. The average of a group of numbers is defined to be the sum of that group divided by the amount of numbers in the group. For example, given the numbers 2, 5, 7, 6, their sum is 2 + 5 + 7 + 6 = 20, so the average is given by average = Example 5 Find the average of each set of numbers. a. b. c. d. Solution 2 + 5 + 7 + 6 20 = =5 4 4 5, 6, 8, 9 –4, 8, 2, 6 –3, 5, –6, 2, –8 6, –3, 8, –4, 8 a. There are 4 numbers, so the average is: 5 + 6 + 8 + 9 28 average = = =7 4 4 b. There are 4 numbers, so the average is: !4 + 8 + 2 + 6 12 average = = =3 4 4 c. There are 5 numbers, so the average is: !3 + 5 + (!6) + 2 + (!8) !10 average = = = !2 5 5 103 d. There are 5 numbers, so the average is: 6 + (!3) + 8 + (!4) + 8 15 average = = =3 5 5 Terminology division of integers indeterminate undefined average Exercise Set 2.5 Divide the two integers. If a quotient is undefined or indeterminate, be sure to state so. 1. 3. 5. 7. 9. 11. 13. 15. 17. !56 ÷ 8 72 ÷ (!3) 0 ÷ (!5) !14 ÷ 0 !45 ÷ (!9) !24 !4 !450 30 600 !25 0 0 2. 4. 6. 8. 10. 12. 14. 16. 18. !48 ÷ 12 50 ÷ (!5) !7 ÷ 0 0 ÷ (!23) !78 ÷ (!2) !120 !15 !150 15 400 !25 !68 0 Determine whether or not the given integer value is a solution to the equation. 19. 21. 23. 25. x = 23 ; x = 115 5 x = !8 ; x = !32 !4 x = 12 ; x = !108 !9 !48 = !4 ; x = !12 x 20. 22. 24. 26. 104 x = 23 ; x = !115 5 x = !8 ; x = 32 !4 x = 12 ; x = 108 !9 !48 = !4 ; x = 12 x 27. 29. !102 = 6 ; y = !17 y a = !3 ; a = !12 !4 28. 30. !102 = 6 ; y = 17 y a = !3 ; a = 12 !4 Find the common ratio and the next term for each geometric sequence. 31. 33. 35. 37. 39. 41. 43. 3, 6, 12, … 2, –4, 8, … –1, 3, –9, … 5, 60, 720, … 4, –100, 2500, … –3, 48, –768, … –4, –52, –676, … 32. 34. 36. 38. 40. 42. 44. 1, 5, 25, … –2, –4, –8, … –1, 4, –16, … 8, 120, 1800, … 5, –65, 845, … –6, 72, –432, … –2, 2, –2, … Convert each written phrase to a mathematical computation, then compute the required quantity. 45. 47. 49. 51. 53. 55. Find the quotient of –65 and –5. Divide 240 by –15. Divide –13 into 273. Find the quotient of 900 and –25. Divide –1620 by 45. Divide 12 into –456. 46. 48. 50. 52. 54. 56. Find the quotient of –48 and –12. Divide 800 by –40. Divide 14 into –378. Find the quotient of –345 and 15. Divide –2925 by –45. Divide –15 into 390. 58. 60. 62. 64. 66. 68. 32, 14, 25, 21 –13, 5, –4, –8 5, 13, –23, –14, –26 4, –8, 5, –3, 2 8, –14, –13, 5, 2, –6 46, –59, –42, –38, 16, –61 Find the average of each set of numbers. 57. 59. 61. 63. 65. 67. 13, 12, 14, 25 8, –7, –5, –12 12, 8, –7, –16, –32 1, –5, –6, 2, –12 12, –13, –14, 5, –6, –8 24, –13, 38, –47, 52, –36 105 Answer each question as true or false, where x and y represent integers. If it is false, give a specific example to show that it is false. If it is true, explain why. x < 0 , then x < 0 or y < 0. 70. y x 71. If > 0 , then x > 0 or y > 0. 72. y x 73. If = 0 , then x = 0. 74. y x 75. If y = 0 and x ≠ 0, then = 0 . 76. y x 77. If x = 0 and y = 0, then is undefined. y x 78. If x = 0 and y = 0, then is indeterminate. y 69. If x < 0 , then x < 0 and y < 0. y x If > 0 , then x > 0 and y > 0. y x If = 0 , then y = 0. y x If x = 0 and y ≠ 0, then = 0 . y If Answer each question. 79. A start-up internet company loses $525,000 during their first five years. How much did they lose (on average) each year? 80. A new business loses $5,520 during their first year of operation. How much did they lose (on average) each month? 81. Dennis is a frequent trader of stocks. During one month he had gains of $3240 and $1570, and losses of $675, $1450, and $2185. What was his average gain (or loss) for each trade that month? 82. John is a new trader of stocks. During one month he had gains of $1560 and $720, and losses of $865, $1890, and $4250. What was his average gain (or loss) for each trade that month? 83. Are undefined and indeterminate the same thing? Explain your answer. 84. Is the average of a set of integers necessarily an integer? Explain your answer. 106