1. Binary Numbers
Material on Data Representation can be found in Chapter 2 of Computer Architecture (Nicholas Carter)
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2. Why Binary?
Maximal distinction among values  minimal corruption from noise.
Imagine taking the same physical attribute of a circuit, e.g. a voltage lying between 0 and 5 volts, to represent a number.
The overall range can be divided into any number of regions.
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3. Don’t sweat the small stuff
For decimal numbers, fluctuations must be less than 0.25 volts.
For binary numbers, fluctuations must be less than 1.25 volts.
5 volts
0 volts
Decimal
Binary
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4. Range actually split in three
High
Forbidden range
Low
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5. It doesn’t matter ….
Two of the standard voltages coming from a computer’s power supply are ideally supposed to be 5.00 volts and volts
Measurements often reveal values that are slightly off – e.g volts or volts or some such value.
So what, who cares.
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6. How to represent big integers
Use positional weighting, same as with decimal numbers
205 = 2  100
= 127 + 126 + 025 + 0 23 + 122 + 021 + 1 = = 205
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7. Converting 205 to Binary
205/2 = 102 with a remainder of 1, place the 1 in the least significant digit position
Repeat 102/2 = 51, remainder 0 1 1
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8. Iterate 51/2 = 25, remainder 1 25/2 = 12, remainder 11 1
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9. Iterate 6/2 = 3, remainder 0 3/2 = 1, remainder 1 1/2 = 0, remainder 11 1 1
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10. Recap 205 1 127 + 126 + 025 + 024 + 123 + 122 + 021 + 120
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11. Finite representation
Typically we just think computers do binary math.
But an important distinction between binary math in the abstract and what computers do is that computers are finite.
There are only so many flip-flops or logic gates in the computer.
When we declare a variable, we set aside a certain number of flip-flops (bits of memory) to hold the value of the variable. And this limits the values the variable can have.
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12. Same number, different representation
5 using 8 bits
5 using 16 bits
5 using 32 bits
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13. Adding Binary Numbers
Same as decimal; if the sum of digits in a given position exceeds the base (10 for decimal, 2 for binary) then there is a carry into the next higher position 1 3 9 + 5 7 4
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14. Adding Binary Numbers 1 +
carries 39 35 74
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15. Uh oh, overflow
What if you use a byte (8 bits) to represent an integer
A byte may not be enough to represent the sum of two such numbers. 1
170
204
118???
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16. Biggest unsigned* integers
4 bit: 1111  15 =
8 bit:  255 = 28 – 1
16 bit:  65535= 216 – 1
32 bit:  = 232 – 1
Etc.
*If one uses all of the bits available to represent only positive counting numbers, one is said to be working with unsigned integers.
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17. Bigger Numbers
High-level languages often offer a hierarchy of types that differ in the number of bits used.
You can represent larger numbers than allowed by the highest type in the hierarchy by using more words.
You just have to keep track of the overflows to know how the lower numbers (less significant words) are affecting the larger numbers (more significant words).
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18. Negative numbers
Negative x is the number that when added to x gives zero
Ignoring overflow the two eight-bit numbers above add up to zero 1
 x
 -x
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19. Two’s Complement: a two-step procedure for finding -x from x
Step 1: exchange 1’s and 0’s
Step 2: add 1 (to the lowest bit only) 1
 x 1 1
 -x
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20. Sign bit
With the two’s complement approach, all positive numbers start with a 0 in the left-most, most-significant bit and all negative numbers start with 1.
So the first bit is called the sign bit.
But note you have to work harder than just strip away the first bit.
IS NOT the 8-bit version of –1
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21. Add 1’s to the left to get the same negative number using more bits
-5 using 8 bits
-5 using 16 bits
-5 using 32 bits
When the numbers represented are whole numbers (positive or negative), they are called just integers or signed.
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22. 3-bit signed and unsigned7 1 6 5 4 3 2 3 1 2 -1 -2 -3 -4
Think of driving a brand new car in reverse. What would happen to the odometer?
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23. Biggest signed integers
4 bit: 0111  7 =
8 bit:  127 = 27 – 1
16 bit:  32767= 215 – 1
32 bit:  = 231 – 1
Etc.
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24. Most negative signed integers
4 bit: 1000  -8 = - 23
8 bit:  = - 27
16 bit:  = - 215
32 bit:  = - 231
Etc.
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25. Riddle 1 Is it 214? Or is it – 42? Or is it Ö? Or is it …?
It’s a matter of interpretation
How was it declared? 1
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26. Hexadecimal Numbers
Even moderately sized decimal numbers end up as long strings in binary.
Hexadecimal numbers (base 16) are often used because the strings are shorter and the conversion to binary is easier.
There are 16 digits: 0-9 and A-F.
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27. Decimal  Binary  Hex 0  0000  0 1  0001  1 2  0010  2
3  0011  3
4  0100  4
5  0101  5
6  0110  6
7  0111  7
8  1000  8
9  1001  9
10  1010  A
11  1011  B
12  1100  C
13  1101  D
14  1110  E
15  1111  F
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28. Binary to Hex
Break a binary string into groups of four bits (nibbles).
Convert each nibble separately. 1 E C 9
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29. Numbers from Logic
All of the numerical operations we have talked about are really just combinations of logical operations.
E.g. the adding operation is just a particular combination of logic operations
Possibilities for adding two bits
0+0=0 (with no carry)
0+1=1 (with no carry)
1+0=1 (with no carry)
1+1=0 (with a carry)
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30. Addition Truth Table INPUT OUTPUT A B Sum A XOR B Carry A AND B 11
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31. Multiplication: Shift and add1  +
shift
shift
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32. Fractions Similar to what we’re used to with decimal numbers 3.14159 =
3 · · · · · · 10-5 =
1 · · · · 2-2
+ 1 · · · 2-5
+ 1 · 2-6
(  )
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33. Converting decimal to binary II
98.61
Integer part
98 / 2 = 49 remainder 0
49 / 2 = 24 remainder 1
24 / 2 = 12 remainder 0
12 / 2 = 6 remainder 0
6 / 2 = 3 remainder 0
3 / 2 = 1 remainder 1
1 / 2 = 0 remainder 1
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34. Converting decimal to binary III
98.61
Fractional part
0.61  2 = 1.22
0.22  2 = 0.44
0.44  2 = 0.88
0.88  2 = 1.76
0.76  2 = 1.52
0.52  2 = 1.04
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35. Another Example (Whole number part)
Integer part
123 / 2 = 61 remainder 1
61 / 2 = 30 remainder 1
30 / 2 = 15 remainder 0
15 / 2 = 7 remainder 1
7 / 2 = 3 remainder 1
3 / 2 = 1 remainder 1
1 / 2 = 0 remainder 1
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36. Checking: Go to All Programs/Accessories/Calculator
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37. Put the calculator in Programmer view
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38. Enter number, put into binary mode
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39. Another Example (fractional part)
Fractional part
0.456  2 = 0.912
0.912  2 = 1.824
0.824  2 = 1.648
0.648  2 = 1.296
0.296  2 = 0.592
0.592  2 = 1.184
0.184  2 = 0.368 … …
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40. Convert to decimal mode, then
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41. Edit/Copy result. Switch to Scientific View. Edit/Paste
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42. Divide by 2 raised to the number of digits (in this case 7, including leading zero)1 2 3 4
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43. Finally hit the equal sign. In most cases it will not be exact
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44. Other way around
Multiply fraction by 2 raised to the desired number of digits in the fractional part. For example
.456  27 =
Throw away the fractional part and represent the whole number
58 111010
But note that we specified 7 digits and the result above uses only 6. Therefore we need to put in the leading 0
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45. Fixed point
If one has a set number of bits reserved for representing the whole number part and another set number of bits reserved for representing the fractional part of a number, then one is said to be using fixed point representation.
The point dividing whole number from fraction has an unchanging (fixed) place in the number.
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46. Limits of the fixed point approach
Suppose you use 4 bits for the whole number part and 4 bits for the fractional part (ignoring sign for now).
The largest number would be =
The smallest, non-zero number would be = .0625
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47. Floating point representation
Floating point representation allows one to represent a wider range of numbers using the same number of bits.
It is like scientific notation.
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48. Scientific notation
Used to represent very large and very small numbers.
Ex. Avogadro’s number
  1023 particles 
Ex. Fundamental charge e
  C
 C
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49. Scientific notation: all of these are the same number
=  100
 10 =  101
 100 =  102
 103
 104
Rule: Shift the point to the left and increment the power of ten.
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50. Small numbers
 10-1
 10-2
 10-3
 10-4
 10-5
1.234  10-6
Rule: shift point to the right and decrement the power.
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51. Floating Point Rules
We’ll use a set of rules that are close but not quite the same as the IEEE 754 standards for floating point representation.
Starting with the fixed point binary representation, shift the point and increase the power (of 2 now that we’re in binary).
Shift so that the number has no whole number part and also so that the first fractional bit (the half’s place) has a 1.
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52. Floats
SHIFT expression so it is just under 1 and keep track of the number of shifts
 27
Express the number of shifts in binary 
We’re not done yet so this exponent will change.
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53. Mantissa and Exponent and Sign
(Significand) Mantissa
Exponent
The number may be negative, so there a bit (the sign bit) reserved to indicate whether the number is positive or negative
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54. Small numbers
 2-4
The power (a.k.a. the exponent) could be negative so we have to be able to deal with that.
Floating point numbers use a procedure known as biasing to handle the negative exponent problem.
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55. Biasing
Actually the exponent is not represented as shown on the previously.
There were 8 bits used to represent the exponent on the previous slide, that means there are 256 numbers that could be represented.
Since the exponent could be negative (to represent numbers less than 1), we choose half of the range to be positive and half to be negative , i.e to 127.
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56. Biasing (Cont.)
In biasing, one does not use 2’s complement or a sign bit.
Instead one adds a bias (equal to the magnitude of the most negative number) to the exponents and represents the result of that addition.
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57. Biasing (Cont.)
With 8 bits, the bias is 128 (= 27 that is 2 raised to the number of bits used for the exponent minus one).
In our previous example, we had to shift 7 times to the left, corresponding to an exponent of +7.
We add that shift to the bias 128+7=135.
That is the number we put in the exponent portion: 135 
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58. Big floats
Assume we use 8 bits, 4 for the mantissa and 4 for the exponent (neglecting sign). What is the largest float?
Mantissa: Exponent 1111
 27
=120
(Compare this to the largest fixed-point number using the same amount of space )
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59. Small floats
Assume we use 8 bits, 4 for the mantissa and 4 for the exponent (neglecting sign). What is the smallest float?
Mantissa: Exponent 0000
0.5  2-8 =
(Compare this to the smallest fixed-point number using the same amount of space .0625)
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60. Adding Floats
Consider adding the following numbers expressed in scientific notation
 103
 10-2
The first step is to re-express the number with the smaller magnitude so that it has the same exponent as the other number.
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61. Adding Floats (Cont.) 1.212121  10-2 0.1212121  10-1
 100
 101
 102
 103
The number was shifted 5 times (3-(-2)).
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62. Adding Floats (Cont.)
When the exponents are equal the mantissas can be added.
 103
 103
=  103
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63. Rounding
In a computer there are a finite number of bits used to represent a number.
When the smaller floating-point number is shifted to make the exponents equal, some of the less significant bits are lost.
This loss of information (precision) is known as rounding.
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64. One more fine point about floating-point representation
As discussed so far, the mantissa (significand) always starts with a 1.
When storage was expensive, designers opted not to represent this bit, since it is always 1.
It had to be inserted for various operations on the number (adding, multiplying, etc.), but it did not have to be stored.
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65. Still another fine point
When we assume that the mantissa must start with a 1, we lose 0.
Zero is too important a number to lose, so we interpret the mantissa of all zeros and exponent of all zeros as zero
Even though ordinarily we would assume the mantissa started with a one that we didn’t store.
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66. Yet another fine point
In the IEEE 754 format for floats, you bias by one less (127) and reserve the exponents and for special purposes.
One of these special purposes is “Not a number” (NaN) which is the floating point version of overflow.
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67. An example
Represent as a float using 23 bits for the mantissa, 8 for the exponent and one for the sign.
Convert the whole number magnitude 9087 to binary:
That uses up 14 of the 23 bits for the mantissa, leaving 9 for the fractional part.
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68. An example (Cont.)
Multiply the fractional part by 29 and convert whole number part of that to binary, make sure in uses 9 bits (add leading 0’s if it doesn’t).
.8735  29 =
447 
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69. An example (Cont.)
 214
Mantissa
Exponent =142 
Sign bit 1 (because number was negative)
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70. Example 2
No whole number part. Begin by using all 23 mantissa bits for the fractional part.
 223 =
64201 
Only uses 16 places, means that so far number starts with 7 zeros. But float mantissas are supposed to start with 1.
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71. Example 2 (Cont.)
23+7
 230 = 
Above is mantissa
Exponent 128 – 7 = 121 
Sign bit 0 (positive number)
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72. References Computer Architecture, Nicholas Carter
Computer Systems: Organization and Architecture, John Carpinelli
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