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Subtraction of Numbers

Complement of a Number   Octal and Hexadecimal Number   Convert Decimal to Binary

Subtraction of numbers by using complement


Subtraction with r’s Complements

The direct method of subtraction taught in elementary schools uses the borrow concept. In this method, we borrow a 1 from a higher significant position when the minuend digit is smaller than the corresponding subtrahend digit. This seems to be easiest when people perform subtraction with paper and pencil. When subtraction is implemented by means of digital components, this method is found to be less efficient than the method that uses complements and addition as stated below.

     The subtraction of two positive numbers (M - N), both of base r, may be done as follows:

1.       Add the minuend M to r’s complement of the subtrahend N.

2.       Inspect the result obtained in step 1 for an end carry:

a.       If the end carry occurs, discard it.

b.      If an end carry does not occur, take r’s complement of the number obtained in step 1 and place a negative sign in front.

If you don't understand about r's complement Click Complement of a Number

The following example illustrate the procedure:

Ex1. Using 10’s complement subtract 72532 – 3250

Sol. M = 72532, N = 03250

       10’s complement of N = 96750

       Now add M + 10’s complement of N

       72352 + 96750 = 1 69282, where 1 is end carry, discard it as per step 1

       Now the answer is 69282

Ex2. Subtract 3250 – 72352

Sol. M = 03250

        N = 72352

        10’s complement of N = 27468

        Now add M + 10’s complement of N

        03250 + 27468 = 30718, There we don’t have any carry then we find 10’s complement of result 30718

        Which is -69282

Ex3. Subtract binary numbers 1010100 – 1000100 using 2’s complement

Sol. M = 1010100, N = 1000100

       2’s complement of N = 0111100

       Now add M + 2’s complement of N

       1010100 + 0111100 = 1 0010000, where 1 is end carry, discard as per step 1

       Now the answer is 0010000 or 10000

Ex4. Subtract binary numbers 1000100 – 1010100

Sol.  M = 1000100, N = 1010100

         2’s complement of N = 0101100

        Now add M + 2’s complement of N

        1000100 + 0101100 = 1110000, here we don’t have any carry then we find 2’s complement of result 1110000

        Which is -10000

 

Subtraction with (r  -1)’s complement

The procedure for subtraction with the (r - 1)’s complement is exactly the same as the one used with the r’s complement except for one variation, called ‘end-around carry,’ as shown below. The subtraction of M-N, both positive numbers in base r, may be calculated in the following manner:

1.       Add the minuend M to (r - 1)’s complement of the subtrahend N.

2.       Inspect the result obtained in step 1 for an end carry.

a.       In an end carry occurs, add 1 to the least significant digit (end-around carry).

b.      If an end carry does not occur, take the (r - 1)’s complement of the number obtained in step 1 and place negative sign in front.

The following examples illustrate the procedure.

Note we are using same examples which we have discussed in r’s complement

Ex1. M = 72532, N = 03250

Sol. 9’s complement of N = 96749

        Now add M + 9’s complement of N

        72532 + 96749 = 1 69281, where 1 is end around carry, let’s add this 1 in 69281

        The answer is 69281 + 1 = 69282

Ex2. M = 03250, N = 72532

Sol. 9’s complement of N = 27467

        Now add M + 9’s complement of N

        03250 + 27467 = 30717, here we don’t have any carry, then we find 9’s complement of 30717

        9’s complement of 30717 is -69282

Ex3. M = 1010100, N = 1000100

Sol. 1’s complement of N = 0111011

        Now add M + 1’s complement of N

        1000100 + 0111011 = 1 0001111, where 1 is end-around carry, let’s add this 1 in 0001111

        The answer is 0001111 + 1 = 0010000

Ex4. M = 1000100, N = 1010100

Sol. 1’s complement of N = 0101011

        Now add M + 1’s complement of N

        1000100 + 0101011 = 1101111, here we don’t have any carry, then we find 1’s complement of 1101111

        1’s complement of 1101111 is -10000


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