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# Math1003 1.17 - Truncation, Rounding, Overflow, & Conversion Error

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### Math1003 1.17 - Truncation, Rounding, Overflow, & Conversion Error

1. 1. 1.1710110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Truncation, Rounding, Overﬂow, and Conversion Error MATH1003
2. 2. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Goal To be able to explain and demonstrate the concepts of truncation, rounding, overﬂow, and conversion error. MATH1003
3. 3. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 The computer is imperfect MATH1003
4. 4. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 The computer is imperfect No matter how large a computer is, it still has a limited amount of storage. MATH1003
5. 5. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 The computer is imperfect No matter how large a computer is, it still has a limited amount of storage. Consider the result of dividing 2 by 3. 2/ MATH1003 3
6. 6. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 The computer is imperfect No matter how large a computer is, it still has a limited amount of storage. Consider the result of dividing 2 by 3. 0.666666 is a repeating number. 2/ MATH1003 3
7. 7. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 The computer is imperfect No matter how large a computer is, it still has a limited amount of storage. Consider the result of dividing 2 by 3. 0.666666 is a repeating number. Regardless of how many bits we use to store this number, it will get “cut off” at some point. No computer can accurately store this number. 2/ MATH1003 3
8. 8. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Truncation 2/ MATH1003 3
9. 9. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Truncation To truncate a number means to simply ignore the extra digits that the computer cannot store. 2/ MATH1003 3
10. 10. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Truncation To truncate a number means to simply ignore the extra digits that the computer cannot store. Truncate the following to 3 signiﬁcant digits 0.2349 2/ MATH1003 3
11. 11. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Truncation To truncate a number means to simply ignore the extra digits that the computer cannot store. Truncate the following to 3 signiﬁcant digits 0.234 2/ MATH1003 3
12. 12. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Truncation To truncate a number means to simply ignore the extra digits that the computer cannot store. Truncate the following to 5 signiﬁcant digits 0.666666666666 2/ MATH1003 3
13. 13. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Truncation To truncate a number means to simply ignore the extra digits that the computer cannot store. Truncate the following to 5 signiﬁcant digits 0.66666 2/ MATH1003 3
14. 14. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Truncation To truncate a number means to simply ignore the extra digits that the computer cannot store. Truncate the following to 8 signiﬁcant binary digits 0.1010101111000101 2/ MATH1003 3
15. 15. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Truncation To truncate a number means to simply ignore the extra digits that the computer cannot store. Truncate the following to 8 signiﬁcant binary digits 0.10101011 2/ MATH1003 3
16. 16. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Real Numbers 0 0 1 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 Represent 0.02510 in IEEE standard 1. 0.02510 = 0.00000112 0.000001100110011 1.100110011001 1.1001 x 2-6 2. normalized as 1.1001 x 2-6 3. set the sign bit 4. store -6 in the exponent section as (-6 + 127 = 121) 011110012 5. store the normalized binary form MATH1003
17. 17. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Overﬂow Error MATH1003
18. 18. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Overﬂow Error Let’s assume that we are using 8 bits and 2’s complement to store integers (1 sign bit and 7 bits for the number). MATH1003
19. 19. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Overﬂow Error Let’s assume that we are using 8 bits and 2’s complement to store integers (1 sign bit and 7 bits for the number). Calculate 12610 + 1310 MATH1003
20. 20. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Overﬂow Error Let’s assume that we are using 8 bits and 2’s complement to store integers (1 sign bit and 7 bits for the number). Calculate 12610 + 1310 12610 + 1310 MATH1003
21. 21. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Overﬂow Error Let’s assume that we are using 8 bits and 2’s complement to store integers (1 sign bit and 7 bits for the number). Calculate 12610 + 1310 12610 011111102 + 1310 +000011012 MATH1003
22. 22. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Overﬂow Error Let’s assume that we are using 8 bits and 2’s complement to store integers (1 sign bit and 7 bits for the number). Calculate 12610 + 1310 12610 011111102 + 1310 +000011012 100010112 MATH1003
23. 23. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Overﬂow Error Let’s assume that we are using 8 bits and 2’s complement to store integers (1 sign bit and 7 bits for the number). Calculate 12610 + 1310 12610 011111102 + 1310 +000011012 bit n sig 100010112 MATH1003
24. 24. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Overﬂow Error Let’s assume that we are using 8 bits and 2’s complement to store integers (1 sign bit and 7 bits for the number). Calculate 12610 + 1310 12610 011111102 this is -11710 in 2’s complement, but the answer should be + 1310 +000011012 13910. What happened? bit n sig 100010112 MATH1003
25. 25. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Overﬂow Error Let’s assume that we are using 8 bits and 2’s complement to store integers (1 sign bit and 7 bits for the number). Calculate 12610 + 1310 12610 01111110sign 7 bits can only store Remember that the ﬁrst bit is used as a 2 + 1310 up to 127. +000011012 bit. bit n sig 100010112 MATH1003
26. 26. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Overﬂow Error Let’s assume that we are using 8 bits and 2’s complement to store integers (1 sign bit and 7 bits for the number). Calculate 12610 + 1310 12610 011111102 7 bits can only store up to 127. + 1310 We have an overﬂow +000011012 problem. bit n sig 100010112 MATH1003
27. 27. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Rounding 2/ MATH1003 3
28. 28. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Rounding An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation of the number. 2/ MATH1003 3
29. 29. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Rounding An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation of the number. Round the following to 2 signiﬁcant digits 2/ MATH1003 3
30. 30. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Rounding An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation of the number. Round the following to 2 signiﬁcant digits 2.53 2/ MATH1003 3
31. 31. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Rounding An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation of the number. Round the following to 2 signiﬁcant digits 2.53 2/ MATH1003 3
32. 32. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Rounding An alternative to truncation is rounding, where the last digit is “adjusted” to give a moreSince 3 is less than 5, accurate representation of the number. will be like this truncation Round the following to 2 signiﬁcant digits 2.53 2/ MATH1003 3
33. 33. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Rounding An alternative to truncation is rounding, where the last digit is “adjusted” to give a moreSince 3 is less than 5, accurate representation of the number. will be like this truncation Round the following to 2 signiﬁcant digits 2.5 2/ MATH1003 3
34. 34. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Rounding An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation of the number. Round the following to 2 signiﬁcant digits 2.5 2/ MATH1003 3
35. 35. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Rounding An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation of the number. Round the following to 3 signiﬁcant digits 17.948 2/ MATH1003 3
36. 36. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Rounding An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation of the number. Round the following to 3 signiﬁcant digits 17.948 2/ MATH1003 3
37. 37. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Rounding An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation Since 4 is less than 5, of the number.this will be like truncation Round the following to 3 signiﬁcant digits 17.948 2/ MATH1003 3
38. 38. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Rounding An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation of the number. Round the following to 3 signiﬁcant digits 17.9 2/ MATH1003 3
39. 39. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Rounding An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation of the number. Round the following to 2 signiﬁcant digits -0.002463 2/ MATH1003 3
40. 40. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Rounding An alternative to truncation is rounding, where the last Since we are digit is “adjusted” to give a more accurate representation concerned only about the signiﬁcant of the number. digits, we will only consider these Round the following to 2 signiﬁcant digits digits -0.002463 2/ MATH1003 3
41. 41. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Rounding An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation of the number. Round the following to 2 signiﬁcant digits -0.002463 2/ MATH1003 3
42. 42. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Rounding An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate 6representation Since is greater than or equal to 5, we of the number.“round up” the 4 to its left to 5 Round the following to 2 signiﬁcant digits -0.002463 2/ MATH1003 3
43. 43. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Rounding An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation of the number. Round the following to 2 signiﬁcant digits -0.0025 2/ MATH1003 3
44. 44. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Rounding An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation of the number. Round the following to 5 signiﬁcant digits 0.173 2/ MATH1003 3
45. 45. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Rounding An alternative to truncation is rounding, that this isthe last Note where a repeating number. digit is “adjusted” to give a more accurateexpand to at least We should representation of the number. signiﬁcant digits before 6 rounding to 5 signiﬁcant Round the following to 5 signiﬁcant digits digits 0.173 2/ MATH1003 3
46. 46. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Rounding An alternative to truncation is rounding, that this isthe last Note where a repeating number. digit is “adjusted” to give a more accurateexpand to at least We should representation of the number. signiﬁcant digits before 6 rounding to 5 signiﬁcant Round the following to 5 signiﬁcant digits digits 0.173737 2/ MATH1003 3
47. 47. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Rounding An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation of the number. Round the following to 5 signiﬁcant digits 0.173737 2/ MATH1003 3
48. 48. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Rounding An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation of the number. Round the following to 5 signiﬁcant digits 0.173737 2/ MATH1003 3
49. 49. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Rounding An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation Since 7 is greater than or equal to 5, we of the number. “round up” the 3 to its left to 4 Round the following to 5 signiﬁcant digits 0.173737 2/ MATH1003 3
50. 50. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Rounding An alternative to truncation is rounding, where the last digit is “adjusted” to give a more accurate representation of the number. Round the following to 5 signiﬁcant digits 0.17374 2/ MATH1003 3
51. 51. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Conversion Error Let’s represent 0.110 with 4 bytes in IEEE standard form. 0.1 MATH1003
52. 52. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Conversion Error Let’s represent 0.110 with 4 bytes in IEEE standard form. 1. 0.110 = 0.000112 0.1 MATH1003
53. 53. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Conversion Error Let’s represent 0.110 with 4 bytes in IEEE standard form. 1. 0.110 = 0.000112 2. normalize 0.000112 = 1.10011 x 2-4 0.1 MATH1003
54. 54. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Conversion Error Let’s represent 0.110 with 4 bytes in IEEE standard form. 1. 0.110 = 0.000112 1.10011 x 2-4 2. normalize 0.000112 = 1.10011 x 2-4 0.1 MATH1003
55. 55. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Conversion Error Let’s represent 0.110 with 4 bytes in IEEE standard form. 1. 0.110 = 0.000112 1.10011 x 2-4 2. normalize 0.000112 = 1.10011 x 2-4 3. set the sign bit 0.1 MATH1003
56. 56. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Conversion Error 0 Let’s represent 0.110 with 4 bytes in IEEE standard form. 1. 0.110 = 0.000112 1.10011 x 2-4 2. normalize 0.000112 = 1.10011 x 2-4 3. set the sign bit 0.1 MATH1003
57. 57. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Conversion Error 0 Let’s represent 0.110 with 4 bytes in IEEE standard form. 1. 0.110 = 0.000112 1.10011 x 2-4 2. normalize 0.000112 = 1.10011 x 2-4 3. set the sign bit 4. store -4 in the exponent section 0.1 MATH1003
58. 58. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Conversion Error 0 Let’s represent 0.110 with 4 bytes in IEEE standard form. 1. 0.110 = 0.000112 1.10011 x 2-4 2. normalize 0.000112 = 1.10011 x 2-4 3. set the sign bit -4 + 127 = 123 4. store -4 in the exponent section 123 = 011110112 0.1 MATH1003
59. 59. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Conversion Error 0 0 1 1 1 1 0 1 1 Let’s represent 0.110 with 4 bytes in IEEE standard form. 1. 0.110 = 0.000112 1.10011 x 2-4 2. normalize 0.000112 = 1.10011 x 2-4 3. set the sign bit -4 + 127 = 123 4. store -4 in the exponent section 123 = 011110112 0.1 MATH1003
60. 60. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Conversion Error 0 0 1 1 1 1 0 1 1 Let’s represent 0.110 with 4 bytes in IEEE standard form. 1. 0.110 = 0.000112 1.10011 x 2-4 2. normalize 0.000112 = 1.10011 x 2-4 3. set the sign bit 4. store -4 in the exponent section 5. store the normalized binary form 0.1 MATH1003
61. 61. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Conversion Error 0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 Let’s represent 0.110 with 4 bytes in IEEE standard form. 1. 0.110 = 0.000112 1.10011 x 2-4 2. normalize 0.000112 = 1.10011 x 2-4 3. set the sign bit 4. store -4 in the exponent section 5. store the normalized binary form 0.1 MATH1003
62. 62. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Conversion Error 0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 0.1 MATH1003
63. 63. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Conversion Error 0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 Now, let’s convert this back to decimal 0.1 MATH1003
64. 64. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Conversion Error 0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 Now, let’s convert this back to decimal 1. since the sign bit is 0, we know this is a positive number 0.1 MATH1003
65. 65. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Conversion Error 0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 Now, let’s convert this back to decimal 1. since the sign bit is 0, we know this is a positive number 2. the exponent section is 01111011 (= 123) 0.1 MATH1003
66. 66. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Conversion Error 0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 Now, let’s convert this back to decimal 1. since the sign bit is 0, we know this is a positive number 2. the exponent section is 01111011 (= 123) 3. the decimal exponent is 123 - 127 = -4 0.1 MATH1003
67. 67. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Conversion Error 0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 Now, let’s convert this back to decimal 1. since the sign bit is 0, we know this is a positive number 2. the exponent section is 01111011 (= 123) 3. the decimal exponent is 123 - 127 = -4 4. from the number section, we have 1.10011001100110011001100 0.1 MATH1003
68. 68. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Conversion Error 0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 Now, let’s convert this back to decimal 1. since the sign bit is 0, we know this is a positive number 2. the exponent section is 01111011 (= 123) 3. the decimal exponent is 123 - 127 = -4 4. from the number section, we have 1.10011001100110011001100 5. therefore the number is 1. 10011001100110011001100 x 2-4 0.1 MATH1003
69. 69. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Conversion Error 0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 Now, let’s convert this back to decimal 1. since the sign bit is 0, we know this is a positive number 2. the exponent section is 01111011 (= 123) 3. the decimal exponent is 123 - 127 = -4 4. from the number section, we have 1.10011001100110011001100 5. therefore the number is 1. 10011001100110011001100 x 2-4 6. 1. 10011001100110011001100 = 1.59999990463 (approximately) 0.1 MATH1003
70. 70. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Conversion Error 0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 Now, let’s convert this back to decimal 1. since the sign bit is 0, we know this is a positive number 2. the exponent section is 01111011 (= 123) 3. the decimal exponent is 123 - 127 = -4 4. from the number section, we have 1.10011001100110011001100 5. therefore the number is 1. 10011001100110011001100 x 2-4 6. 1. 10011001100110011001100 = 1.59999990463 (approximately) 7. 1.59999990463 x 2-4 = 0.099999994 0.1 MATH1003
71. 71. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001 Conversion Error 0 0 1 1 1 1 0 1 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 1 1 0 0 Now, let’s convert this back to decimal 1. since the sign bit is 0, we know this is a positive number 2. the exponent section is 01111011 (= 123) This is a 3. the decimal exponent is 123 - 127 = -4 conversion error: 4. from the number section, we have 1.10011001100110011001100 0.099999994 ≠ 0.1 5. therefore the number is 1. 10011001100110011001100 x 2-4 6. 1. 10011001100110011001100 = 1.59999990463 (approximately) 7. 1.59999990463 x 2-4 = 0.099999994 0.1 MATH1003