Transcript Binary Real Numbers

Binary Real Numbers
Introduction


Computers must be able to represent real
numbers (numbers w/ fractions)
Two different ways:



Fixed-point
Floating-point
NOTE: Everything in binary uses powers of
two
Decimal Review

Digits to the right of the decimal point correspond to
negative powers of 10
102
101
100
.
10-1
10-2
100
10
1
.
0.1
0.01
Binary Fractions
2-1
2-2
2-3
2-4
2-5
2-6
0.5
0.25
0.125
0.0625
0.03125
0.0015625
1/2
1/4
1/8
1/16
1/32
1/64
Fixed Point Notation
1.
2.
Multiply each 1 by the corresponding power
of 2
Add up the resulting powers of 2
Example:
11.012 = 2 + 1 + ¼ = 3.2510
00111.0102 = 4 + 2 + 1 + ¼ = 7.2510
Floating-Point Notation


Floating-point notation is essentially the
computer’s way of storing a number that has
been normalized
3 different parts of any number:




Mantissa: normalized number
Exponent: power to which the base is raised
Sign: of both mantissa and exponent
Decimal Example:
12.5 = 0.125 x 102  normalized!
Normalization Steps
1.
2.
3.
Beginning with a fixed point number
Normalize the number such that the radix
point (decimal point) is all the way to the left
(produces the mantissa)
Multiply the resulting number by the base
raised to an exponent
Floating-Point Example
What is 12.5 in floating-point representation?
1.
Convert 12.5 to binary fixed point
12.510 = 1100.12
2.
3.
Normalize the number by moving the radix point,
producing the mantissa
1100.12 = 0.11001 * 24
Fill in the bits for each of the three parts of any
real number:
1.
2.
3.
4.
Sign (2 bits)
Mantissa (# bits varies)
Exponent (# bits varies)
NOTE: 2’s complement may be applied to the
mantissa or the exponent if either are negative
Placing the Bits

Assume you have the following:




1 bit for the mantissa sign
8 bits for the mantissa
1 bit for the exponent sign
6 bits for the exponent
SM M M M M M M M M S E E E E E E E

Example:
0.11001 * 24  0 00011001 0 000100
Another Example
Convert -12.510 to binary:
1.
Convert 12.5 to fixed point  01100.1
2.
Normalize  0.11001 * 24
3.
Convert exponent base to binary: 4  0100
4.
2s complement the mantissa by flipping bits
and adding 1:
011001  100111
5.
Final number  1 00111 0100
Upper & Lower Bounds

Assume you have the following:







1 bit for the mantissa sign
8 bits for the mantissa
1 bit for the exponent sign
6 bits for the exponent
What is the upper bound for the floating-point
number?
What is the lower bound for the floating-point
number?
What happens if we convert a floating-point
number to an integer?
Integers vs. Floating-point

integers:



smaller range than floating-point
all numbers within the range are 100% accurate
floating-point


large range of numbers
not all numbers within the range can be
represented accurately

Example: 2.9999999999999 repeating
Possible Errors

truncation error


overflow error


round off errors using floating-point numbers
because not all real numbers can be represented
accurately
attempting to represent a number that is greater
than the upper bound for the given number of bits
underflow error

attempting to represent a number that is less than
the lower bound for the given number of bits