1. Floating Point Numbers
Binary 5
Floating Point Numbers

2. Floating Point Numbers
Floating point numbers are real numbers.
In Java, this just means any numbers that aren’t integers (whole numbers)
For example…
x 109
In Java we use either double or float

3. Floating Point Numbers
A floating point number is represented in three parts:
M x BE
M  mantissa, represented as a number that is >=0 and <1
B  base (2 is binary, 10 is decimal)
E  exponent (the power of the base as a positive or negative)

4. Floating Point Numbers
For example, let’s consider what happens when we write in the format M x BE
Scientific notation uses one significant digit before the decimal point, multiplied by the base to the appropriate power
eg: x 10-2
Normalised notation has no significant digits before the decimal point, multiplied by the base to the appropriate power
eg: x 10-1

5. Floating Point Numbers
Floating point numbers are stored in normalised form as this is the most efficient way.
Examples of normalisation (decimal):
x 10-4  x 10-6
 x 104
164 x 10-2  x 101
794 x 10-5  x 10-2

6. Floating Point Numbers
Examples of normalisation (binary):
1101 x 2100  x 21000
x  x 2-011
Note: This examples have used unsigned binary, with a minus symbol where the exponent is negative

7. To “normalise” a number…
Move the decimal point to put a 0 to its left, and a non-zero digit to the right.
Count how many places the decimal point moved.
If the decimal point moved…
left, increase the exponent
right, decrease the exponent
… by the same amount.

8. Floating Point Numbers
Practice normalising the following decimal numbers:
45.1 x =
0.063 x =
x =
x =
x =

9. Floating Point Numbers
Practice normalising the following decimal numbers:
45.1 x = x 103(moved 2x to the left, so add 2)
0.063 x = 0.63 x 102(moved 1x to the right, so minus 1)
x = x 10-5
x = 0.76 x 102
x = x 10-8

10. Floating Point Numbers
Practice normalising the following binary numbers (for now, just put in a minus for negatives):
e.g x 2010 = x 2110
101.1 x =
x =
x =
x = =
Note: If no exponent is provided, then you start at 0

11. Floating Point Numbers
Practice normalising the following binary numbers
101.1 x = x 2101
x = x 2-110
x = x
x = 0.1 x 2-111
= x 2111
Note: If no exponent is provided, then you start at 0

12. Floating Point Numbers
To store a floating number we need to allocate space for the mantissa, exponent and sign
There needs to be a balance between:
Precision
Range
Total memory required for storage
Type
Size
Largest Value
Smallest Value
Precision
float
32 bits
 3.4  1038
 1.4  10-45
6-7 sig figs
double
64 bits
 1.8  10308
 4.9 
14-15 sig figs

13. Floating Point Numbers
Lets look at some examples of floating point representation using a 16-bit word
Example 1: 16-bit word with 5-bit 2’s complement exponent and normalised mantissa
0.1 x 201 To convert that to decimal == == x 2 = 1
Sign Bit of mantissa (1 bit)
Exponent in 2’s complement (5 bits)
Normalised mantissa (10 bits)
00001

14. Floating Point Numbers
Lets look at more examples:
x 20
To convert that to decimal == == x 1 =
Sign Bit of mantissa (1 bit)
Exponent in 2’s complement (5 bits)
Normalised mantissa (10 bits) 1
00000

15. Floating Point Numbers
Lets look at more examples:
As a normalised binary number, this is... x
As a regular decimal… mantissa = = exponent = = -2 decimal number = x 2-2 =
Sign Bit of mantissa (1 bit)
Exponent in 2’s complement (5 bits)
Normalised mantissa (10 bits) 1
11110

16. Floating Point Numbers
Lets look at more examples:
As a normalised binary number, this is… x
As a regular decimal… mantissa = = exponent = -13
decimal number = x 2-13
= …
Sign Bit of mantissa (1 bit)
Exponent in 2’s complement (5 bits)
Normalised mantissa (10 bits) 1
10011

17. Floating Point Numbers
Lets look at more examples:
As a normalised binary number, this is…
?x 2?
As a regular decimal… mantissa = exponent =
decimal number = =
Sign Bit of mantissa (1 bit)
Exponent in 2’s complement (5 bits)
Normalised mantissa (10 bits)
01111

18. Floating Point Numbers
Lets look at more examples:
As a normalised binary number, this is… x
As a regular decimal… mantissa = exponent = 15
decimal number = x 215
= …
Sign Bit of mantissa (1 bit)
Exponent in 2’s complement (5 bits)
Normalised mantissa (10 bits)
01111

19. Problems with floating point
Floating point numbers are often an inexact representation
Many decimals ie. 0.1, 0.2 & 0.4 are recurring binary fractions and cannot be represented exactly
When carrying out calculations, the size of the mantissa changes
Two examples using a 4-digit mantissa:
x x x x x x 103 = 0.1 x x 105 (Should be so it loses precision)

20. Helpful Hint
Always convert a decimal to binary BEFORE attempting to normalise

21. Floating Point Numbers
Represent the following binary numbers using a 16-bit representation, with…
1 bit for the exponent sign
1 bit for the mantissa sign
6 bits for the exponent
8 bits for the mantissa
0.011 x 2100
111000
10.11 x 2-111

22. Precision
When using an 8-bit two’s complement representation, which of the following decimals can be represented precisely? (that is, without any rounding off)
128.5
0.1