Floating Point Numbers

Real Numbers: pi = 3.14159265... e = 2.71828...

Scientific Notation: has a single digit to the left of the decimal point.

A number in Scientific Notation with no leading 0s is called a Normalised Number: 1.0 × 10^-8

Not in normalised form: 0.1 × 10^-7 or 10.0 × 10^-9

Can also represent binary numbers in scientific notation: 1.0 × 2^-3

Computer arithmetic that supports such numbers is called Floating Point.

The form is 1.xxxx… × 2^yy…

Using normalised scientific notation

1. Simplifies the exchange of data that includes floating-point numbers
2. Simplifies the arithmetic algorithms to know that the numbers will always be in this form
3. Increases the accuracy of the numbers that can be stored in a word, since each unnecessary leading 0 is replaced by another significant digit to the right of the decimal point

Representation of Floating-Point numbers

-1^S × M × 2^E

Bit No	Size	Field Name
31	1 bit	Sign (S)
23-30	8 bits	Exponent (E)
0-22	23 bits	Mantissa (M)

A Single-Precision floating-point number occupies 32-bits, so there is a compromise between the size of the mantissa and the size of the exponent.

These chosen sizes provide a range of approx:

± 10^-38 ... 10³⁸

• Overflow

  The exponent is too large to be represented in the Exponent field

• Underflow

  The number is too small to be represented in the Exponent field

To reduce the chances of underflow/overflow, can use 64-bit Double-Precision arithmetic

Bit No	Size	Field Name
63	1 bit	Sign (S)
52-62	11 bits	Exponent (E)
0-51	52 bits	Mantissa (M)

providing a range of approx

± 10^-308 ... 10³⁰⁸

These formats are called ...

IEEE 754 Floating-Point Standard

Since the mantissa is always 1.xxxxxxxxx in the normalised form, no need to represent the leading 1. So, effectively:

Since zero (0.0) has no leading 1, to distinguish it from others, it is given the reserved bitpattern all 0s for the exponent so that hardware won't attach a leading 1 to it. Thus:

• Zero (0.0) = 0000...0000
• Other numbers = -1^S × (1 + Mantissa) × 2^E

If we number the mantissa bits from left to right m1, m2, m3, ...

mantissa = m1 × 2^-1 + m2 × 2^-2 + m3 × 2^-3 + ....

Negative exponents could pose a problem in comparisons.

For example (with two's complement):

	Sign	Exponent	Mantissa
1.0 × 2-1	0	11111111	0000000 00000000 00000000
1.0 × 2+1	0	00000001	0000000 00000000 00000000

With this representation, the first exponent shows a "larger" binary number, making direct comparison more difficult.

To avoid this, Biased Notation is used for exponents.

If the real exponent of a number is X then it is represented as (X + bias)

IEEE single-precision uses a bias of 127. Therefore, an exponent of

-1 is represented as -1 + 127 = 126 = 011111102

0 is represented as  0 + 127 = 127 = 011111112

+1 is represented as +1 + 127 = 128 = 100000002

+5 is represented as +5 + 127 = 132 = 100001002

So the actual exponent is found by subtracting the bias from the stored exponent. Therefore, given S, E, and M fields, an IEEE floating-point number has the value:

-1S × (1.0 + 0.M) × 2E-bias

(Remember: it is (1.0 + 0.M) because, with normalised form, only the fractional part of the mantissa needs to be stored)

Floating Point Addition

Add the following two decimal numbers in scientific notation:

8.70 × 10^-1 with 9.95 × 10¹

1. Rewrite the smaller number such that its exponent matches with the exponent of the larger number.

   8.70 × 10^-1 = 0.087 × 10¹

2. Add the mantissas

   9.95 + 0.087 = 10.037 and write the sum 10.037 × 10¹

3. Put the result in Normalised Form

   10.037 × 10¹ = 1.0037 × 10² (shift mantissa, adjust exponent)

   check for overflow/underflow of the exponent after normalisation

4. Round the result

   If the mantissa does not fit in the space reserved for it, it has to be rounded off.

   For Example: If only 4 digits are allowed for mantissa

   1.0037 × 10² ===> 1.004 × 10²

(only have a hidden bit with binary floating point numbers)

Example addition in binary

Perform 0.5 + (-0.4375)

0.5 = 0.1 × 2⁰ = 1.000 × 2^-1 (normalised)

-0.4375 = -0.0111 × 2⁰ = -1.110 × 2^-2 (normalised)

1. Rewrite the smaller number such that its exponent matches with the exponent of the larger number.

   -1.110 × 2^-2 = -0.1110 × 2^-1

2. Add the mantissas:

   1.000 × 2^-1 + -0.1110 × 2^-1 = 0.001 × 2^-1

3. Normalise the sum, checking for overflow/underflow:

   0.001 × 2^-1 = 1.000 × 2^-4

   -126 <= -4 <= 127 ===> No overflow or underflow

4. Round the sum:

   The sum fits in 4 bits so rounding is not required

   Check: 1.000 × 2^-4 = 0.0625 which is equal to 0.5 - 0.4375

Correct!

Floating Point Multiplication

Multiply the following two numbers in scientific notation by hand:

1.110 × 10¹⁰ × 9.200 × 10^-5

1. Add the exponents to find

   New Exponent = 10 + (-5) = 5

   If we add biased exponents, bias will be added twice. Therefore we need to subtract it once to compensate: (10 + 127) + (-5 + 127) = 259 259 - 127 = 132 which is (5 + 127) = biased new exponent

2. Multiply the mantissas

   1.110 × 9.200 = 10.212000

   Can only keep three digits to the right of the decimal point, so the result is

   10.212 × 10⁵

3. Normalise the result

   1.0212 × 10⁶

4. Round it

   1.021 × 10⁶

Example multiplication in binary:

1.000 × 2^-1 × -1.110 × 2^-2

1. Add the biased exponents

   (-1 + 127) + (-2 + 127) - 127 = 124 ===> (-3 + 127)

2. Multiply the mantissas

   1.000 × 1.110 ----------- 0000 1000 1000 + 1000 ----------- 1110000 ===> 1.110000

   The product is 1.110000 × 2-3

   Need to keep it to 4 bits 1.110 × 2-3

3. Normalise (already normalised)

   At this step check for overflow/underflow by making sure that

   -126 <= Exponent <= 127

   1 <= Biased Exponent <= 254

4. Round the result (no change)
5. Adjust the sign.

   Since the original signs are different, the result will be negative

   -1.110 × 2^-3

Further Reading

last updated: 2-Dec-04 Ian Harries <ih@doc.ic.ac.uk>