Introduction

* Binary numbers are represented with a

separate sign bit along with the magnitude.

* For example, in an 8-bit binary number, the

MSB is the sign bit and the remaining 7

bits correspond to magnitude.

Magnitude

* The magnitude part contains true binary

equivalent of the number for positive

numbers, while 2’s complement form of

the number for the negative numbers

Example

+ 13 , 0 , - 46 are represented as follows

SignSign Magnitude

Magnitude

+3 00000 1101000 1101

0 00000 0000000 0000

-46 11010 1110010 1110

Explanation

* It is important to note that the number zero is assigned

with the sign bit ‘0’ .

* Therefore, the range of numbers that can be

represented using 8-bit binary number is -128 to

+127.

* In general, the range of numbers that can be

represented by n-bit number is (-2n-1) to (+2n-1-1)

Addition in the 2’s complement system

Cases of Addition

1. When both the numbers are positive

2. When augend is a positive and addend

is a negative number

3. When augend is a negative and addend is

a positive number

4. When both the numbers are negative

Case 1Two positive numbers

Consider the addition of +29 and +19

+29 10111000 (augend)

+19 11001000 (addend)

00001100 (Sum=48)

Sign bit

Explanation

* The sign bits of both augend and addend are

zero and the sign bit of the sum=0.

* It indicating that when the sum is positive they

have the same number of bits.

Case 2Positive augend Number and Negative addend Number

Consider the addition of +39 and -22

-22 +22 [ 000 10110 ] Convert to -22 [ 111 01010]

Complement

+39 11100100 (augend)

-22 01010111 (addend)

100010001 (Sum=17)

Sign bitCarry

The carry is omitted. Then result is 0 0 0 1 0 0 0 1

Explanation

* The sign bit of addend is 1.

* A carry is generated in the last position of addition.

* This carry is always omitted.

* So the final Sum is 0 0 01 0 0 0 1

Case 3Positive addend Number and Negative augend Number

Consider the addition of -47and +29

-47 +47 [ 0 01 01110] Convert to -47 [ 110 10001]

Complement

-47 10001011 (augend)

+29 10111000 (addend)

01110111 (Sum=-18)

Sign bit

Explanation

* The result has a sign bit of 1, indicating a

negative number.

* It is in the 2’s complement form.

* The last seven bits 1101110 actually represent

the 2’s complement of the sum.

Explanation Cont.,

* The true magnitude of the sum can be found by taking the

2’s complement of 1101110, the result is 10010 (+18).

* Thus 11101110 represents -18

Case 4Two Negative Numbers

Consider the addition of -32 and -44

-32 00000111 (augend)

-44 00101011 (addend)

00101101 (Sum=-76)

Sign bit

1

Carry

The carry is discarded. Then result is 1 0 1 1 0 1 0 0

Explanation

* The true magnitude of the sum is the complement of

0110100 , i.e., 1 0001100 (-76).

* Thus, the 2’s complement addition works in every

case.

* This assumes that the decimal sum is within -128 to

+127 range. Otherwise we get an overflow.

Subtraction in the 2’s complement system

Introduction

* As in the case of addition, subtraction can

also be carried out in four possible cases.

* Subtraction by the 2’s complement system

involves addition.

Case 1Both the Numbers are positive

Consider the subtraction of +19 and +28

+19 -19 [0001 0010] Convert to +19[ 1110 1101]

Complement

Add the +28 and -19 as

+28 00111000

+19 10110111

100100001 (Sum=9)

Carry

Case 2Positive number and smaller Negative Number

Consider the subtraction of +39 and -21

-21 +21[1110 1011] Convert to -21[0001 0101]

Complement

Add the +39 and +21 as

+39 11100100

+21 10101000

00111100 (Sum=60)

Case 3Positive Number and Larger Negative Number

Consider the subtraction of +19 and -43

-43 +43[1101 0100] Convert to -43[0010 1011]

Complement

Add the +19 and +43 as

+19 11001000

+43 11010100

11111100 (Sum=62)

Case 4Both the Numbers are Negative

Consider the subtraction of +33 and -57

-57 +57[0011 1000] Convert to -57[1100 0111]

Complement

+33 -33[1101 1111] Convert to +33[0010 0001]

Complement

Add the +33 and -57 as

-57 11100011

+33 10000100

00010111 (Result=-24)

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