1. Floating Point Representations
Updated: 06/03/2010
Floating Point Representations

2. Decimal Floating Point Numbers
1.5
Decimal fractions
¾ -> 0.75
1/100 -> 0.01

3. Scientific Notation How can we more compactly represent these values?
1,000,000,000
-> 1.0 x 10^9
-> 2.5 x 10^-5
What are the two parts of a scientific notation value called?
Mantissa and Exponent
Normalize mantissa so it has one digit left of the decimal point

4. Floating point numbers
Floating point numbers such as 7.519, -0.01, and 4.3x108 are represented using the IEEE 754 standard format
Floating point is represented using a mantissa and exponent
Example: 7.51x25
The mantissa is 7.51
The exponent is > Note: exponent is a power of 2
A set number of bits is assigned to represent the mantissa and exponent 1
8 bits
23 bits
mantissa
exponent
sign bit
32 bit single precision 1
11 bits
52 bits
mantissa
exponent
sign bit
64 bit double precision

5. Rounding
Not every floating point value can be represented exactly in binary using a finite number of bits
Question: What are some examples?
1/3 = …
PI = 3.141….
In these cases, must round to the nearest number that can be represented
If a number is halfway between two possible representable values, then round to the one whose least-significant digit is even

6. Examples of Rounding
Round each of these numbers to two significant digits.
> 1.3 Choose 1.3 since is nearer to 1.3 than 1.4
> 79 Choose 79 since it’s nearer than 78
> is halfway between 12 and 13
Choose 12 since its least significant digit is even
> is halfway between 13 and 14
Choose 14 since its least significant digit is even

7. Fractional binary numbers
Fractional binary numbers use the familiar decimal place-value representation, but with a base of 2 instead of 10
Example:
11.101b = 1x21 + 1x20 + 1x x x2-3
= / / /8 =
= 3.625

8. Exercises Convert this binary fraction into decimal
Answer: 5 + 0*1/2 + 1*1/4 + 1*1/8 = 5.375
Express the decimal value 6.5 as a binary fraction
Answer: 110.1
Express the decimal value as a binary fraction
Answer:

9. Normalized Mantissa for Scientific Notation
Scientific notation numbers express the mantissa with one digit to the left of the decimal point.
Given your original number
Shift the decimal point left or right until one non-zero digit is to the left of the decimal point
For each shift left increase the power of ten exponent by 1
For each shift right decrease power of ten exponent by 1
Examples:
102.5 x 104 = x 106
7589 x 105 = x 108
x 100 = 4.5 x 10-3

10. Normalized Binary Mantissa
Binary Fraction Normalized IEEE 754 Mantissa
(shift binary point left 1 place)
(no shift)
(shift binary point left 2)
(shift binary point right 3)
Question: What do all of the normalized binary mantissas have in common?

11. Fractional representation of mantissa
What do all of the normalized binary mantissas have in common?
The one bit to the left of the binary point is always a 1
So if we use 23 bits for the single-precision mantissa, we can “save” a bit by not storing this leading 1
So simply discard the lead 1 after normalizing the binary mantissa

12. Mantissa Example
What is the binary representation of the mantissa in IEEE 754 for 6.25?
Solution:
6.25 = 1x22 + 1x21 + 0x20 + 0x x2-2
= b
Shift the binary point as far to the left as possible until the bit to the left of the binary point is 1
110.01b --> b (Shift left by 2 places)
This shift gives us the assumed 1 bit in the integer part of the mantissa fractional representation...effectively gains one additional bit of representation
Mantissa encodes only bits to the right of the binary point
1001b

13. Mantissa Example Continued...
What is the binary representation of the mantissa in IEEE 754 for 6.25?
Solution...:
Keeping only the bits to the right of the binary point...
1001b
Sign extend the the 4-bits into 23 bits for single precision
Append the extra bits to the right for a binary fraction
Our imaginary binary point

14. Updating the Exponent 6.25 = 110.01b has an implied exponent of 20
Following the IEEE 754 convention of shifting the binary point to the left, in this case by 2 positions has the effect of updating the exponent
1.1001b (Following shift left of binary point by 2 positions)
For each left shift binary point = Add 1 to binary exponent
6.25 = b x 22

15. Updating the Binary Exponent
Binary Fraction Normalized Binary Exponents
1.01 x x 20
x x 21
x x 23
x x 2-3
x x 2-5

16. Representing the exponent in IEEE 754
The exponent is represented as a biased integer
For single precision add 127 to the value of the normalized base ten integer exponent
For double precision add 1023 to the value of the normalized base ten integer exponent

17. Representing the exponent in IEEE 754
The exponent is represented as a biased integer
For single precision add 127 to the value of the exponent
For double precision add 1023 to the value of the exponent
Example:
How would the values -45 and 123 be represented in the 8-bit biased format for single precision?
Answer:
= = b
= 250 = b

18. Encoding the Biased Binary Exponent
Binary Fraction Normalized Exponents Biased Exponent
1.01 x x = 127
x x = 128
x x = 130
x x = 124
x x = 122
Encode each biased exponent as an unsigned 8-bit number.
Encode each biased exponent in 8-bit two’s complement.
Suppose you had to rapidly sort by exponents, which format would be more efficient?

19. Floating Point Example #1
Recall that 6.25 = b x 22
Encode 6.25 as a 32-bit single precision binary number
Sign bit = 0
Mantissa = (encoding omits assumed lead 1)
Exponent = = 129 =
Encode using 32-bit single precision binary format
Sign bit
Exponent
Mantissa

20. Floating point example #2
What is the value of the single-precision floating-point number represented by the following 32-bit binary encoding?
Sign bit = 0 Encoded Exponent = = 128
Encoded Mantissa =
Subtract the added bias of 127 to reveal an exponent = 1
Mantissa =
Mantissa = 1.11 (Replace the assumed 1 before the binary point)
Mantissa = 1.11 = 1x20 + 1x x2-2 = 1.75
Value = 1.75 x 21 = 3.5

21. Floating Point Example #3
-6.25 = b x 22
Encode as a 32-bit single precision binary number
Sign bit = 1 (Use signed magnitude for mantissa)
Mantissa = (encoding omits assumed lead 1)
Exponent = = 129 =
Encode using 32-bit single precision binary format
Sign bit
Exponent
Mantissa

22. Exercise
Exercise 2.18 (a) on page 42 of Computer Architecture by N. Carter
What value is represented by this IEEE single precision value?

23. Exercise: Solution
What value is represented by this IEEE single precision value?
Sign bit = 1 Encoded Exponent = = 122
Encoded Mantissa =
Subtract added bias of 127 from encoded exponent
Actual exponent is -5
Mantissa = = .1
Mantissa = (Add back the assumed 1 before the binary point)
Mantissa = - 1 x x2-1 = -1.5
Value = -1.5 x 2-5 = -1.5 x (1/32) =

24. IEEE 754 Single Precision Range
Smallest positive normalized number
x 2-126
Largest normalized number
x 2127

25. Representing 1.0

26. Representing 0.0
The assumed 1 bit in the mantissa gains an extra bit of precision
But zero cannot be represented exactly since a mantissa of 0 is interpreted as 1.0
The IEEE 754 standard specifies that zero is represented using an exponent of 0 with a mantissa of 0.

27. NaN
NaN = Not a Number
Special value used to represent a value produced by an error condition such as overflow, underflow, or divide by zero
NaN is represented by all 1’s in the exponent field and a non-zero mantissa field
Any math operation using NaN results in NaN
Example: NaN = NaN

28. Infinity
IEEE 754 represents infinity using all 1’s in the exponent and a fraction field of 0.
The sign bit designates positive or negative infinity

29. Floating Point Addition (Decimal Example)
Example: x x10-1
Step 1: Shift decimal point of smaller number to the left until its updated exponent matches the exponent of the larger number
1.610x10-1  x101
Step 2: Add the mantissas (Assume only 4 significant digits)
9.999 x 101
x 101
x 101
Step 3: Re-normalize to get one non-zero digit left of decimal point
x 101  x 102
Step 4: Round the mantissa to 4 significant digits
x 102  x 102

30. Floating Point Addition Example
Use single-precision floating point to compute
0.25 (base 10) = (1/4) = 0.01 = 1.0 x 2-2
1.5 (base 10) = 1 + (1/2) = 1.1 x 20
Shift binary point of smaller number to the left so exponents match
1.0 x 2-2  0.01 x 20

31. Floating Point Addition Example (continued)
Use single-precision floating point to compute
Next, add the mantissas, both with exponent of 0
0.01 x 20
+1.10 x 20
1.11 x 20

32. Floating Point Addition Example (Continued)
Use single-precision floating point to compute
0.01 x 20
+1.10 x 20
1.11 x 20
Encode result using 32-bit single precision
Sign bit = 0
Mantissa = (23 bits)
Exponent = = 127 =
The 32-bit single precision encoding is...

33. Floating Point Addition Exercise: Solution
2.20 (b) Use single precision to compute
147.5 (base 10) = (1/2) = =
Convert to normalized mantissa format
x 20  x 27
Shifted binary point 7 places to the left
See Computer Architecture by N. Carter, page 43

34. Floating Point Addition Exercise: Solution
2.20 (b) Use single precision to compute
0.25 (base 10) = (1/4) = 0.01
Convert to normalized mantissa format
0.01 x 20  1.0 x 2-2
Shift binary point 2 places to the right

35. Floating Point Addition Exercise: Solution
2.20 (b) Use single precision to compute
x 27
+ 1.0 x 2-2
Shift binary point of smaller number to left to match exponent (7) of the larger number
1.0 x 2-2  x 27
Shift binary point 9 places to the left to go from exponent of -2 to 7

36. Floating Point Addition Exercise: Solution
2.20 (b) Use single precision to compute
Add the mantissas, both expressed with exponent 7
x 27
x 27
x 27

37. Floating Point Addition Exercise: Solution
2.20 (b) Use single precision to compute
Encode the result x 27 in single precision
Sign bit = 0 since result is positive
Mantissa = (23 bits)
Exponent = = 134 =
The 32-bit single precision encoding is...

38. Addition with Negative Values
If a value is negative, you must first convert the negative value into two’s complement
Example:
-0.111
Convert to two’s complement by...
1.000 inverting all bits
adding 1
1.001
Use the two’s complement version of the value when adding the mantissas. Discard the carry overflow bit.

39. Addition with Negative Value(s)
1.000 x 20 (1.0 in base ten)
x 2-1 (-0.5 in base ten)
Move binary point of the smaller number so exponents match
x 20 (-0.5 in base ten)
Convert mantissa of -0.5 into two’s complement then add
x 2-1 (-0.5 in base ten)
x 2-1 Two’s complement addition discards carry overflow bit
The sum is x 2-1
Normalize the exponent to get sum of x 2-1 (0.5 base ten)

40. Floating Point Addition (page 282 of H&P)
Example: Compute (base 10) using binary arithmetic.
0.5 (base 10) = 0.1 x 20
Normalize to get 1 to left of the binary point
0.5 = 0.1 x 20 = 1.0x2-1
= = - ((1/4) + (1/8) + (1/16))
Normalize to get 1 to the left of the binary point
 1.11 x 2-2

41. Floating Point Addition (page 282 of H&P)
Compute
0.5 = 0.1 x 20 = 1.0x2-1 +
= 1.11 x 2-2
Step 1: Shift binary point of smaller number to the left until its updated exponent matches the exponent of the larger number
1.11 x 2-2  x 2-1
Step 2: Add the mantissas * Convert negative value to two’s
1.000 (1.0 decimal) complement then add
( decimal) * Discard carry overflow bit
0.001 (0.125 decimal)
0.001 x 2-1

42. Floating Point Addition (page 282 of H&P)
Step 3: Normalize to get 1 to left of binary point
0.001 x 2-1  1.0 x 2-4
Exponent of -4 lies between 127 and -126 (range of single precision exponents)...therefore no overflow or underflow
Express exponent in biased notation by adding 127
Encoded exponent = = 123
Step 4: Round to 23 binary digits of mantissa precision
1.0 x 2-4 (no rounding needed)

43. Floating point multiplication
Multiply the mantissas and add the exponents
Result = (mantissa1 x mantissa2) + 2(exp1 + exp2)
Example (in decimal)
5x103 x 2x106 = 10x109
If the mantissa is >= 10 then shift the mantissa down 1 place (divide by 10) and increment the result exponent
10x109 = 1x1010

44. Floating point multiplication
Since the IEEE 754 uses biased integers to represent the exponent, the bias must be considered when adding the exponents
Add the two biased integer exponents, then subtract the bias value from the result
Example: Add biased +127 exponents of 150 and 45
Break down the exponents to see the bias values of 127
150 = ( ) 45 = ( )
Add the biased exponents: = 195
Subtract the bias of 127: – 127 = 68 result biased exponent
Check it: 68 – 127 = Actual exponent of -59 =

45. Floating Point Multiplication: Example
Exercise 2.20 (a) Use IEEE single precision to compute 32 x 16.
32 (base 10) = x 20 (binary)
Convert to normalized binary mantissa format
x 20  1.0 x 25 (Shift binary point 5 places to left)
Exponent = = 132
16 (base 10) = x 20 (binary)
x 20  1.0 x 24 (Shift binary point 4 places to left)
Exponent = = 131
See Computer Architecture by N. Carter, page 43

46. Floating Point Multiplication: Example
Exercise 2.20 (a) Use IEEE single precision to compute 32 x 16.
1.0 x 25
Exponent = = 132
1.0 x 24
Exponent = = 131
Multiply mantissas: 1.0 x 1.0
1.0
x1.0 Count number of bits right of binary point of operands
0 0 Place binary point two places from left of product
1.0 0
Add +127 biased exponents: – 127 = 136
Actual unbiased exponent = 136 – 127 = 9
Product = 1.0 x 29 = 512

47. Floating Point Multiplication: Example
Exercise 2.20 (a) Use IEEE single precision to compute 32 x 16.
1.0 x 25
Exponent = = 132
1.0 x 24
Exponent = = 131
Multiply mantissas: 1.0 x 1.0 = 1.0 (binary)
Add +127 biased exponents: – 127 = 136
Actual unbiased exponent = 136 – 127 = 9
Product = 1.0 x 29 = 512
Sign Bit = 0
Mantissa =
Exponent = 136 =
The encoded IEEE 754 single precision number is...

48. Floating Point Multiplication Exercise: Solution
2.20 (c) Compute x 8 using single-precision binary.
0.125 (base 10) = x 20 = 1.0 x 2-3 (Normalized binary mantissa)
8 (base 10) = x 20 = 1.0 x 23 (Normalized binary mantissa)
See Computer Architecture by N. Carter, page 43

49. Floating Point Multiplication Exercise: Solution
2.20 (c) Compute x 8 using single-precision binary.
1.0 x 2-3 Biased exponent = = 124
1.0 x 23 Biased exponent = = 130
Multiply mantissas: 1.0 x 1.0 = 1.0 (binary)
Add biased exponents: – 127 = 127
Actual exponent: 127 – 127 = 0
Sign Bit = 0
Mantissa =
Exponent = 127 =
is the encoded binary number

50. Floating Point Multiplication Exercise
Multiply 0.75 x 32 using IEEE 754 single-precision format
0.75 = 0.11 x 20
Normalized 1.1 x 2-1 Biased exponent = = 126
32 = x 20
Normalized 1.0 x 25 Biased exponent = = 132
Multiply the mantissas
1.1 To place the binary point...
x1.0 Count number of bits to right of binary points
0 0 of the two operands 1.1 and 1.0
Total of 2 places so place binary point
two places from the left in the product

51. Floating Point Multiplication Exercise
Multiply 0.75 x 32 using IEEE 754 single-precision format
Multiply the mantissas
1.1
x1.0
0 0
+ 1 1
1.1 0
Add the biased exponents:
– 127 = 131 (unbiased exponent is 4)

52. Floating Point Multiplication Exercise
Multiply 0.75 x 32 using IEEE 754 single-precision format
Product of the mantissas
1.10
Add the biased exponents:
– 127 = 131 (unbiased exponent is 4)
The product is already normalized
Encode the product using IEEE 32-bit format
Sign bit = 0
Exponent = 131 =
Mantissa =

53. Rounding of Floating Point Numbers
Accurate rounding requires the hardware to use a few extra bits to hold intermediate results
Use these extra bits to decide how to round when the final result is stored in the 32-bit single precision or 64-bit double precision format
The IEEE 754 standard uses up to three additional bits called the guard, round, and sticky bits to assist in accurate rounding
See pages of Computer Organization and Design

54. Rounding of Floating Point Numbers
Compute this base ten addition rounding all intermediate values to three significant digits
2.56 x 100
x 102
First shift the decimal point of the top number to align the exponents
0.02 x 102 Rounding to three digits looses information
2.36 x 102

55. Rounding of Floating Point Numbers
Compute this base ten addition using intermediate values that keep an extra two digits
2.56 x 100
x 102
First shift the decimal point of the top number to align the exponents
x 102 Intermediate values use two extra bits
x 102
x 102
Use extra two bits to round the result down to three significant digits
2.37 x 102

56. Java Applets for IEEE Floating Point
A Java applet that converts decimal numbers to IEEE single or double precision encodings can be found at...
This applet may be used to make up your own sample problems to convert between decimal and IEEE format and to check the result of other calculations in IEEE floating point format
Interactive floating point addition demo