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
Typically they are 5.14 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. 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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12. Adding Binary Numbers 1 +
carries
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13. 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
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14. Biggest unsigned integers
4 bit: 1111  15 =
8 bit:  255 = 28 – 1
16 bit:  65535= 216 – 1
32 bit:  = 232 – 1
Etc.
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15. Bigger Numbers You can represent larger numbers 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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16. Negative numbers
Negative x is the number that when added to x gives zero
Ignoring overflow the two eight-bit numbers above sum to zero 1
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17. Two’s Complement Step 1: exchange 1’s and 0’s
Step 2: add 1 (to the lowest bit only) 1 1 1
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18. 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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19. 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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20. 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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21. 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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22. 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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23. 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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24. 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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25. Addition Truth Table INPUT OUTPUT A B Sum A XOR B Carry A AND B 11
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26. Multiplication: Shift and add1  +
shift
shift
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27. Fractions Similar to what we’re used to with decimal numbers 3.14159 =
3 · · · · · · 10-5 =
1 · · · · · · · · 2-6
(  )
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28. 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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29. 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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30. 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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31. Checking: Go to Programs/Accessories/Calculator
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32. Put the calculator in Scientific view
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33. Enter number, put into binary mode
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34. 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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35. Checking fractional part: Enter digits found in binary mode
Note that the leading zero does not display.
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36. Convert to decimal mode, then
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37. Divide by 2 raised to the number of digits (in this case 7, including leading zero)
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38. In most cases it will not be exact
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39. 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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40. 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 place in the number.
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41. 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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42. 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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43. 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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44. 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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45. 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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46. Floating Point Rules
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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47. 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 
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48. 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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49. 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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50. Biasing
Actually the exponent is not represented as shown on the previous slide
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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51. 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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52. Biasing (Cont.) With 8 bits, the bias is 128.
We had to shift 7 times to the left, corresponding to an exponent of +7
We add that to the bias 128+7=135
That is the number we put in the exponent portion:
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53. 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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54. Small floats
Assume we use 8 bits, 4 for the mantissa and 4 for the exponent (neglecting sign). What is the smalles 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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55. One more fine point
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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56. 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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57. 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 on these special purposes is “Not a number” (NaN) which is the floating point version of overflow.
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