Subtraction Using Complements

 

Before starting Subtraction Using Complements, let us know what complement is first.

Complements:

Complements are of two types:

1. r’s Complement;

For the given positive number N in base r with an integer part of n-digits, the r’s complement of N is defined as rn-N.

2. (r-1)’s Complement;

For the given positive number N in base r with an integer part of n-digits and fractional parts of the digits, the (r-1)’s complement is defined as rn-r-m-N.

Subtraction Using Complements can be done by two ways:

 Subtraction Using r’s Complement:

The subtraction of two positive numbers (M-N) both of base r can be done as follows;

1.  Add the minuend M to the r’s complement of subtractend N.
2.  Inspect the result obtained in step 1 for end carry.

• If end carry occurs, discard it.
• If end carry doesn’t occur, take r’s complement if the number obtained in step 1 and place a negative sign in front.

A. For decimal numbers:

Example: Subtract (72532-3250) using 10’s complement.

M= 72532

N= 03250

10’s complement of N= 105-3250 = 96750

Now, adding 10’s complement of N to M,

72532+96750=169282

Here, end carry occurs as 1. So, 10’s complement= 69282

B. For binary numbers:

Example: Subtract (1000100-1010100) using 2’s complement.

M= 1000100

N= 11010100

2’s complement of N=10110

Now, adding 2’s complement of N to M

10110+1000100= 110000

Here, end carry occurs as 1, so 2’s complement = -10000

 

Subtraction Using r-1’s Complement:

 

The subtraction of two positive numbers (M-N) both of base r-1 can be done as follows;

1.  Add the minuend M to the r-1’s complement of subtractend N.
2.  Inspect the result obtained in step 1 for end carry.

• If end carry occurs, add 1 to the least significant bit, i.e. end-around carry.
• If end carry doesn’t occur, take r-1’s complement of the number obtained in step 1 and place a negative sign in front.

A. For decimal number:

Example: Subtract (453.35-321.17) using 9’s complement.

9’s complement of 321.17= 678.82

Now, Adding 9’s complement of M to N,

453.35+678.82= 1132.17

Here, end carry occur, so 9’s complement = 132.18

B. For binary numbers:

Example: Subtract (1000100-1010100) using 1’s complement.

M= 1000100

N= 11010100

1’s complement of N=101011

Now, adding 1’s complement of N to M

101011+1000100= 1101111

Here, end carry doesn’t occur as 1, so 1’s complement of 1101111= -1000 which is the required answer.