1. Dr Damian Conway Room 132 Building 26
Real Number Representation (Lecture 25 of the Introduction to Computer Programming series)
Dr Damian Conway
Room 132
Building 26

2. Some Terminology
All digits in a number following any leading zeros are significant digits:

3. Some Terminology
The scientific notation for real numbers is: mantissa  base exponent

4. Some Terminology
The mantissa is always normalized between 1 and the base (i.e. exactly one significant figure before the point):
Normalized Unnormalized   B.139FC  B1.39FC    2-1

5. Some Terminology
The precision of a number is how many digits (or bits) we use to represent it.
For example:

6. Representing numbers
A real number n is represented by a floating-point approximation n*
The computer uses 32 bits (or more) to store each approximation.
It needs to store the mantissa, the sign of the mantissa, and the exponent (with its sign).

7. Representing numbers
So it has to allocate some of its 32 bits to each task.
The standard way to do this (specified by IEEE standard 754) is:

8. Representing numbers 23 bits for the mantissa;
1 bit for the mantissa's sign (i.e. the mantissa is signed magnitude);
The remaining 8 bits for the exponent.

9. Representing numbers 23 bits for the mantissa;
1 bit for the mantissa's sign (i.e. the mantissa is signed magnitude);
The remaining 8 bits for the exponent.

10. Representing numbers 23 bits for the mantissa;
1 bit for the mantissa's sign (i.e. the mantissa is signed magnitude);
The remaining 8 bits for the exponent.

11. Representing numbers 23 bits for the mantissa;
1 bit for the mantissa's sign (i.e. the mantissa is signed magnitude);
The remaining 8 bits for the exponent.

12. Representing the mantissa
Since the mantissa has to be in the range 1 ≤ mantissa < base, if we use base 2 the digit before the decimal has to be a 1.
So we don't have to worry about storing it!
That way we get 24 bits of precision using only 23 bits.

13. Representing the mantissa
Those 24 bits of precision are equivalent to a little over 7 decimal digits:

14. Representing the mantissa
Suppose we want to represent :
That means that we can only represent it as: (if we truncate) (if we round)

15. Representing the mantissa
Even if the computer appears to give you more than seven decimal places, only the first seven are meaningful.
For example: #include <math.h> main() { float pi = 2 * asin(1); printf("%.35f\n", pi); }

16. Representing the mantissa
On my machine this prints out:

17. Representing the mantissa
On my machine this prints out:

18. Representing the exponent
The exponent is represented as an excess-127 number.
That is:  –  –    +128

19. Representing the exponent
However, the IEEE standard restricts exponents to the range: –126 ≤ exponent ≤ +127
The exponents –127 and +128 have special meanings (basically, zero and infinity respectively)

20. Floating point overflow
Just like the integer representations in the previous lecture, floating point representations can overflow:                  
 10128

21. Floating point overflow
Just like the integer representations in the previous lecture, floating point representations can overflow:                  
 10128

22. Floating point overflow
Just like the integer representations in the previous lecture, floating point representations can overflow:                   ∞

23. Floating point underflow
But floating point numbers can also get too small:  ÷                  

24. Floating point underflow
But floating point numbers can also get too small:  ÷                  

25. Floating point underflow
But floating point numbers can also get too small:  ÷                 

26. Floating point addition
Five steps to add two floating point numbers:
Express them with the same exponent (denormalize)
Add the mantissas
Adjust the mantissa to one digit/bit before the point (renormalize)
Round or truncate to required precision.
Check for overflow/underflow

27. Floating point addition example
y =  106

28. Floating point addition example
1. Same exponents: x =  107
y =  107

29. Floating point addition example
2. Add mantissas: x =  107
y =  107
x+y =  107

30. Floating point addition example
3. Renormalize sum: x =  107
y =  107
x+y =  108

31. Floating point addition example
4. Trucate or round: x =  107
y =  107
x+y =  108

32. Floating point addition example
5. Check overflow and underflow: x =  107
y =  107
x+y =  108

33. Floating point addition example 2
y =  10-5

34. Floating point addition example 2
1. Same exponents: x =  10-5
y =  10-5

35. Floating point addition example 2
2. Add mantissas: x =  10-5
y =  10-5
x+y =  10-5

36. Floating point addition example 2
3. Renormalize sum: x =  10-5
y =  10-5
x+y =  10-8

37. Floating point addition example 2
4. Trucate or round: x =  10-5
y =  10-5
x+y =  10-8 (no change)

38. Floating point addition example 2
5. Check overflow and underflow: x =  10-5
y =  10-5
x+y =  10-8

39. Floating point addition example 2
Question: should we believe these zeroes? x =  10-5
y =  10-5
x+y =  10-8

40. Floating point multiplication
Five steps to multiply two floating point numbers:
Multiply mantissas
Add exponents
Renormalize mantissa
Round or truncate to required precision.
Check for overflow/underflow

41. Floating point multiplication example
y =  10-3

42. Floating point multiplication example
1&2. Multiply mantissas/add exponents: x =  105
y =  10-3
x  y =  102

43. Floating point multiplication example
3. Renormalize product: x =  105
y =  10-3
x  y =  103

44. Floating point multiplication example
4. Trucate or round: x =  105
y =  10-3
x  y =  103

45. Floating point multiplication example
4. Trucate or round: x =  105
y =  10-3
x  y =  103

46. Floating point multiplication example
5. Check overflow and underflow: x =  105
y =  10-3
x  y =  103

47. Limitations Float-point representations only approximate real numbers.
The normal laws of arithmetic don't always hold (even less often than for integer representations).
For example, associativity is not guaranteed:

48. Limitations
x =  103
y =  103
z =  100

49. Limitations x = 3.002  103 x+y = 2.000  100 y = -3.000  103
z =  100

50. Limitations x = 3.002  103 x+y = 2.000  100 y = -3.000  103
(x+y)+z =  100
z =  100

51. Limitations
x =  103
y =  103
z =  100

52. Limitations x = 3.002  103 y = -3.000  103 y+z = -2.993  103

53. Limitations x = 3.002  103 x+(y+z) = 0.009  103 y = -3.000  103

54. Limitations x = 3.002  103 x+(y+z) = 9.000  100 y = -3.000  103

55. Limitations x = 3.002  103 x+(y+z) = 9.000  100 y = -3.000  103

56. Limitations Consider the other laws of arithmetic:
Commutativity (additive and multiplicative)
Associativity
Distributivity
Identity (additive and multiplicative)
Spend some time working out which ones (if any!) always hold for floating- point numbers.

57. Reading (for the very keen)
Goldberg, D., What Every Computer Scientist Should Know About Floating- Point Arithmetic, ACM Computing Surveys, Vol.23, No.1, March 1991.