1. 1.17
10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
Truncation, Rounding,
Overflow, and
Conversion Error
MATH1003

2. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
Goal
To be able to explain and demonstrate the concepts of
truncation, rounding, overflow, and conversion error.
MATH1003

3. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
The computer is imperfect
MATH1003

4. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
The computer is imperfect
No matter how large a computer is, it still has a limited
amount of storage.
MATH1003

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. 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. 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. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
Truncation
2/
MATH1003
3

9. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
Truncation
To truncate a number means to simply ignore the extra
digits that the computer cannot store.
2/
MATH1003
3

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. 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. 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. 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. 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. 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. 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. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
Overflow Error
MATH1003

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. 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. 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. 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. 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. 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. 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. 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. 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. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
Rounding
2/
MATH1003
3

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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 10110100101011010100101010111010101111011011101111011101110111101110111011110111111010110100101011110110110101111011010100111111011010100110101001
Conversion Error
Let’s represent 0.110 with 4 bytes in IEEE standard form.
0.1
MATH1003

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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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. 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