Binary 5Floating Point Numbers

Floating Point Numbers Floating point numbers are real

numbers. In Java, this just means any numbers

that aren’t integers (whole numbers) For example…

• 2.86 -0.5 4.000• -0.03 3.1415926 7.2 x 109

In Java we use either double or float

Floating Point Numbers A floating point number is

represented in 3 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)

Floating Point Numbers M x BE

eg: 0.06779

Scientific notation uses one significant digit before the decimal point, multiplied by the base to the appropriate power

eg: 6.779 x 10-2

Normalised notation has no significant digits before the decimal point, multiplied by the base to the appropriate power

eg: 0.6779 x 10-1

Floating Point Numbers Floating point numbers are stored in

normalised form as this is the most efficient way.

Examples of normalisation (decimal):• 0.00156 x 10-4 0.156 x 10-6

• 1732.1 0.17321 x 104

• 164 x 10-2 0.164 x 101

• 794 x 10-5 0.794 x 10-2

Floating Point Numbers Examples of normalisation (binary):

• 1101 x 2100 0.1101 x 21000

• 1101.001 x 2-111 0.1101001 x 2-011

• Note: This examples have used unsigned binary, with a minus symbol where the exponent is negative

Floating Point Numbers Normalising a number:

• Move the decimal point to the left or right to ensure that there is a 0 to the left of the decimal point and a number 1-9 on the right of the decimal point

• Count how many places the decimal point moved

• If the decimal point moved to the left (decreasing the size of the mantissa) you increase the exponent (add to the powers)

• If the decimal point moved to the right (increasing the size of the mantissa) you decrease the exponent (subtract the powers)

Floating Point Numbers Practice normalising the following

decimal numbers: Left + / Right -

• 45.1 x 101 = • 0.063 x 103 =• -178.135 x 10-8 = • 0.0076 x 104 = • 1200.21 x 10-12 =

Floating Point Numbers Practice normalising the following

decimal numbers: Left + / Right -

• 45.1 x 101 = 0.451 x 103(moved 2x to the left, so add 2)

• 0.063 x 103 = 0.45163 x 102(moved 1x to the

right, so minus 1)

• -178.135 x 10-8 = -0.178135 x 10-5 • 0.0076 x 104 = 0.76 x 102 • 1200.21 x 10-12 = 0.120021 x 10-8

Floating Point Numbers Practice normalising the following

binary numbers (just put in a minus for

negatives): Left + / Right -

• 101.1 x 210 = • -0.01101 x 2-101 =• 1000.01 x 21100 = • 0.0001 x 2-100 = • -1101001 =

Note: If no exponent is provided, then you start at 0

Floating Point Numbers Practice normalising the following

binary numbers (just put in a minus for

negatives): Left + / Right -

• 101.1 x 210 = 0.1011 x 2101 • -0.01101 x 2-101 = -0.1101x 2-110

• 1000.01 x 21100 = 0.100001 x 210000

• 0.0001 x 2-100 = 0.1 x 2-111

• -1101001 = -0.1101001 x 2111

Note: If no exponent is provided, then you start at 0

Floating Point Numbers Floating Point Numbers require the

computer to store the mantissa, exponent and sign

There needs to be a balance between:• Precision• Range• Total memory required for storageType 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 10-324 14-15 sig figs

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 decimal0.1 == 0.5201 == 20.5 x 2 = 1

Sign Bit of mantissa (1 bit)

Exponent in 2’s complement (5 bits)

Normalised mantissa (10 bits)

0 00001 10 0000 0000

Floating Point Numbers Lets look at more examples:

-0.1101 x 20 To convert that to decimal

-0.1101 == -0.812520 == 1-0.8125 x 1 = -0.8125

Sign Bit of mantissa (1 bit)

Exponent in 2’s complement (5 bits)

Normalised mantissa (10 bits)

1 00000 11 0100 0000

Floating Point Numbers Lets look at more examples:

-0.101 x 211110 To convert that to decimal

-0.101 == -0.62511110 == -22-2 == 0.25-0.625 x 0.25 = -0.15625

Sign Bit of mantissa (1 bit)

Exponent in 2’s complement (5 bits)

Normalised mantissa (10 bits)

1 11110 10 1000 0000

Floating Point Numbers Problems with floating point numbers

• 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

• Also when carrying out calculations, the size of the mantissa changes

• Two examples using a 4 digit mantissa: 0.6152 x 103 0.6152 x 105

- 0.6151 x 103 x 0.6151 x 105

----------------------------- -----------------------------0.0001 x 103 = 0.1 x 100 0.3875 x 105 (Should be 0.387471..so it

loses precision)

Floating Point Numbers Complete info booklet questions:

Page 13

Helpful Hint:Always convert a decimal to binary BEFORE attempting to normalise