Introduction

Here is the key Introduction of Computer Arithmetic:

Computer arithmetic specifically refers to the design and implementation of arithmetic operations within the central processing unit (CPU) and associated hardware components.

• It involves the representation of numerical data, execution of arithmetic operations, and the handling of related issues such as precision, overflow, and speed optimization.

• An arithmetic processor is the part of a processor unit that executes arithmetic operations.

The four basic arithmetic operations are:

Note:

• The arithmetic processor is very simple if only a binary fixed-point add instruction is included.

• The arithmetic processor would be more complicated if it includes all four arithmetic operations for binary and decimal data in fixed-point and floating-point representation.

Points to be Remembered:

Negative Number Representation:

• Sign Magnitude:

Sign magnitude is a very simple representation of negative numbers. In sign magnitude the first bit is dedicated to represent the sign and hence it is called sign bit.

→ Sign bit ‘1’ represents negative sign.

→ Sign bit ‘0’ represents positive sign.

In sign magnitude representation of a n – bit number, the first bit will represent sign and rest n-1 bits represent magnitude of number.

For example:

• +25 = 011001

Where 11001 = 25

And 0 for ‘+’

• -25 = 111001

Where 11001 = 25

And 1 for ‘-‘.

Range of number represented by sign magnitude method = -(2^n-1-1) to +(2^n-1-1) (for n bit number)

But there is one problem in sign magnitude and that is we have two representations of 0

+0 = 000000

– 0 = 100000

• 2’s complement method

To represent a negative number in this form, first we need to take the 1’s complement of the number represented in simple positive binary form and then add 1 to it.

For example:

(-8)10 = (1000)2

1’s complement of 1000 = 0111

Adding 1 to it, 0111 + 1 = 1000

So, (-8)10 = (1000)2

Please don’t get confused with (8)10 =1000 and (-8)10=1000 as with 4 bits, we can’t represent a positive number more than 7. So, 1000 is representing -8 only.

Range of number represented by 2’s complement = (-2^n-1 to 2^n-1 – 1)

  Floating point representation of numbers:

• 32-bit representation floating point numbers IEEE standard

Normalization:

• Floating point numbers are usually normalized
• Exponent is adjusted so that leading bit (MSB) of mantissa is 1
• Since it is always 1 there is no need to store it
• Scientific notation where numbers are normalized to give a single digit before the decimal point like in decimal system e.g. 3.123 x 10³

For example, we represent 3.625 in 32 bit format. Changing 3 in binary=11 Changing .625 in binary

.625 X 2       1
.25 X 2         0
.5 X 2           1

Writing in binary exponent form:

3.625=11.101 X 2⁰

On normalizing:

11.101 X 2⁰=1.1101 X 2¹

On biasing exponent = 127 + 1 = 128

(128)10=(10000000) 2

For getting significand Digits after decimal = 1101 Expanding to 23 bit = 11010000000000000000000 Setting sign bit As it is a positive number, sign bit = 0 Finally we arrange according to representation 

Sign bit      exponent      significand

0            10000000      11010000000000000000000

64-bit representation floating point numbers IEEE standard

Again we follow the same procedure upto normalization. After that, we add 1023 to bias the exponent. For example, we represent -3.625 in 64 bit format. Changing 3 in binary = 11 Changing .625 in binary

.625 X 2     1

.25 X 2       0

.5 X 2         1

 Writing in binary exponent form 3.625 = 11.101 X 2⁰ On normalizing 11.101 X 2⁰ = 1.1101 X 2¹ On biasing exponent 1023 + 1 = 1024 (1024)10 = (10000000000)2 So 11 bit exponent = 10000000000 52 bit significand = 110100000000 …………. making total 52 bits Setting sign bit = 1 (number is negative) So, final representation 1 10000000000 110100000000 …………. making total 52 bits by adding further 0’s   Converting floating point into decimal Let’s convert a FP number into decimal 1 01111100 11000000000000000000000 The decimal value of an IEEE number is given by the formula: (1 -2s) * (1 + f) * 2^{( e – bias )} where

• s, f and e fields are taken as decimal here.
• (1 -2s) is 1 or -1, depending upon sign bit 0 and 1
• add an implicit 1 to the significand (fraction field f), as in formula

Again, the bias is either 127 or 1023, for single or double precision respectively.   First convert each individual field to decimal. 

• The sign bit s is 1
• The e field contains 01111100 = (124)10
• The mantissa is 0.11000 … = (0.75)10

Putting these values in formula (1 – 2) * (1 + 0.75) * 2^{124 – 127} = ( – 1.75 * 2^-3 ) = – 0.21875