Complement arithmetic is a system of mathematical techniques used in the field of digital electronics to perform various arithmetic operations mainly, subtraction.

Here, we will cover the following most widely used types of complements in digital systems −

9's Complement
10's Complement
1's Complement
2's Complement
7's Complement
8's Complement
15's Complement
16's Complement

Let us discuss each of these complements in detail along with their application in arithmetic operations.

What is 9's Complement?

In digital electronics, the 9's complement is a type of complement used to perform subtraction of decimal numbers using a digital system. Thus, 9's complement is related to the decimal number system.

9's complement is used to perform subtraction because it simplifies the subtraction operation.
The 9's complement of a given decimal number is found by subtracting each digit of the number from 9.

The following table shows the 9's complement of each decimal digit −

Decimal Digit	9's Complement
0	9 0 = 9
1	9 1 = 8
2	9 2 = 7
3	9 3 = 6
4	9 4 = 5
5	9 5 = 4
6	9 6 = 3
7	9 7 = 2
8	9 8 = 1
9	9 9 = 0

Let us understand it with the help of examples.

Example 1

Find the 9's complement of the decimal number 7824.450.

Solution

Here is the step-by-step process for finding 9's complement of the given decimal number −

The 9's complement of 7 = 9 7 = 2
The 9's complement of 8 = 9 8 = 1
The 9's complement of 2 = 9 2 = 7
The 9's complement of 4 = 9 4 = 5
The 9's complement of 4 = 9 4 = 5
The 9's complement of 5 = 9 5 = 4
The 9's complement of 0 = 9 0 = 9

Thus, the 9's complement of the decimal number 7824.450 is 2175.549.

Example 2

Find the 9's complement of 45608.

Solution

The 9's complement of the decimal number 45608 is given below −

The 9's complement of 4 = 9 4 = 5.
The 9's complement of 5 = 9 5 = 4.
The 9's complement of 6 = 9 6 = 3.
The 9's complement of 0 = 9 0 = 9.
The 9's complement of 8 = 9 8 = 1.

Thus, the 9's complement of 45608 is 54391.

What is 10's Complement?

In digital electronics, the 10's complement is another type of complement used to perform subtraction of decimal numbers. Again, the purpose of the 10's complement is to simplify the decimal subtraction operation.

There are two methods for finding the 10's complement of a decimal number −

Method I − To find the 10's complement of a given decimal number, firstly we find the 9's complement by subtracting each digit of the number from 9. Then, we add 1 to the 9's complement to obtain the 10's complement, i.e.,

10s Complement = 9s Complement + 1

The 10's complement of each decimal digit using this method is given in the following table −

Decimal Digit	9's Complement
0	9 0 = 9 + 1 = 10 = 0 (Ignore the carry)
1	9 1 = 8 + 1 = 9
2	9 2 = 7 + 1 = 8
3	9 3 = 6 + 1 = 7
4	9 4 = 5 + 1 = 6
5	9 5 = 4 + 1 = 5
6	9 6 = 3 + 1 = 4
7	9 7 = 2 + 1 = 3
8	9 8 = 1 + 1 = 2
9	9 9 = 0 + 1 = 1

Method II − In this method, we can use the following formula to find the 10's complement of a given decimal number,

10s Complement = 10^N Number

Where, N is the number of digits in the decimal number.

Let us understand the process of finding the 10's complement through examples.

Example 1

Find the 10's complement of the decimal number 4872.

Solution

The 10's complement of 4872 can be determined as follows −

Finding the 9's complement of 4872,

9999 4872 = 5127

Adding 1 to the 9's complement to obtain the 10's complement,

5127 + 1 = 5128

So, the 10's complement of 4872 is 5128.

Example 2

Find the 10's complement of 2478.98.

Solution

The 10's complement of 2478.98 can be found as given below −

Finding the 9's complement of 2478.98,

9999.99 2478.98 = 7521.01

Adding 1 to the 9's complement to obtain the 10's complement,

7521.01 + 1 = 7521.02

Hence, the 10's complement of 7521.01 is 7521.02.

Example 3

Find the 10's complement of 58942.

Solution

The 10's complement of 58942 is given below −

10's Complement of 58942 = 10⁵ 58942

10's Complement of 58942 = 100000 58942 = 41058

Thus, the 10's complement of 58942 is 41058.

What is 1's Complement?

In digital electronics, the 1's complement is a type of complement used to simplify the subtraction of binary numbers. Also, the 1's complement is used to represent the negative of a given binary number.

We can find the 1's complement of a binary number by changing all the 0s to 1s and all the 1s to 0s in the number.

We can also find the 1's complement of a binary number by subtracting each bit of the number from 1.

However, there is a major issue associated with the 1's complement that is it has two representations for 0. Where, 00000000 represents the positive zero and its 1's complement is 11111111 that represents 0, but it is called negative zero.

Let us consider some examples to understand the process of finding the 1's complement of binary numbers.

Example 1

Find the 1's complement of 101101.

Solution

The 1's complement of 101101 can be obtained as follows −

Method I − By flipping each bit −

The 1's complement of 1 = 0
The 1's complement of 0 = 1
The 1's complement of 1 = 0
The 1's complement of 1 = 0
The 1's complement of 0 = 1
The 1's complement of 1 = 0

Method II − By subtract each bit from 1 −

111111 101101 = 010010

Hence, the 1's complement of 101101 is 010010.

Example 2

Find the 1's complement of 101101101.

Solution

The 1's complement of given binary number is,

1s Complement = 111111111 101101101 = 010010010

So, the 1's complement of 101101101 is 010010010.

What is 2's Complement?

In digital electronics, the 2's complement is a concept widely used to perform binary subtraction using a digital system.

Here are the following three methods that can be used to determine the 2's complement of a given binary number −

Method I − By finding the 1's complement and then adding 1 to the 1's complement, i.e.,

2s Complement = 1s Complement + 1

Method II − By subtracting the given binary number from 2^N, i.e.,

2s Complement = 2^N Number

Where, "N" is the number of bits in the number.

Method III − Starting from the least significant bit (LSB), copy down the bits up to and including the first 1 bit encountered and then complement the remaining bits.

Let us understand the process of finding the 2's complement of binary numbers through examples.

Example 1

Find the 2's complement of 1100111.

Solution

We can find the 2's complement of 1100111 as follows −

Method I − Using 1's complement −

1s complement of 1100111 = 0011000

Adding 1 to 1's complement to obtain the 2's complement,

0011000 + 1 = 0011001

Method II − Using 2's complement formula −

2's Complement = 2⁷ 1100111 = 128 1100111

2's Complement = 10000000 1100111 = 0011001

Method III − By copying down the bits starting from the LSB up to and including the first 1 bit −

Example 2

Find the 2's complement of 11001100.

Solution

The 2's complement of 11001100 can be obtained as follows −

Method I − Using 1's complement −

1's complement of 11001100 = 00110011

2's complement = 1's complement + 1

2's complement = 00110011 + 1

Therefore,

2s complement = 00110100

Method II − By subtracting the number from 2^N −

2's complement = 2⁸ - 11001100

2's complement = 100000000 11001100 = 00110100

Method III − By copying down bits up to first 1 bit −

What is 7's Complement?

In digital electronics, the 7's complement is a concept used to simplify the octal subtraction. The 7's complement of a given octal number can be obtained by subtracting each digit of the number from 7.

The 7's complement of each octal digit is given in the following table −

Octal Digit	7's Complement
0	7 0 = 7
1	7 1 = 6
2	7 2 = 5
3	7 3 = 4
4	7 4 = 3
5	7 5 = 2
6	7 6 = 1
7	7 7 = 0

Let us consider some examples to understand the process of finding 7's complement of an octal number.

Example 1

Find the 7's complement of the octal number 3152.

Solution

The 7's complement of 3152 can be obtained as follows −

The 7's complement of 3 = 7 3 = 4.
The 7's complement of 1 = 7 1 = 6.
The 7's complement of 5 = 7 5 = 2.
The 7's complement of 2 = 7 2 = 5.

Hence, the 7's complement of 3152 is 4625.

Example 2

Find the 7's complement of the octal number 427102.

Solution

The 7's complement of the given number is determined as given below −

777777 427102 = 350675

Thus, the 7's complement of 427102 is 350675.

What is 8's Complement?

The 8's complement is another type of complement concept used to simplify the octal subtraction. Actually, it is similar to that of 10's complement in the decimal number system.

We can find the 8's complement of a given octal number as follows −

Find the 7's complement of the given octal number by subtracting each digit of the number from 7.
Add 1 to the 7's complement.
Result will be the 8's complement of the given octal number.

Thus,

8s Complement = 7s Complement + 1

The following table shows the 8's complement of each octal digit −

Octal Digit	8's Complement
0	7 0 = 7 + 1 = 10 = 0 (Ignore the carry)
1	7 1 = 6 + 1 = 7
2	7 2 = 5 + 1 = 6
3	7 3 = 4 + 1 = 5
4	7 4 = 3 + 1 = 4
5	7 5 = 2 + 1 = 3
6	7 6 = 1 + 1 = 2
7	7 7 = 0 + 1 = 1

Let us understand the process of finding the 8's complement with the help of examples.

Example 1

Find the 8's complement of 4257.

Solution

The 8's complement of 4257 can be found as given below −

7's complement of 4257 = 7777 4257 = 3520

8's complement = 7's complement + 1

8's complement = 3520 + 1 = 3521

Thus, the 8's complement of 4257 is 3521.

Example 2

Find the 8's complement of 77201.

Solution

The 8's complement of given octal number can be determined as follows −

7's complement of 77201 = 77777 77201 = 00576

8's complement = 7's complement + 1

8's complement = 00576 + 1 = 00577

So, the 8's complement of 77201 is 00577.

What is 15's Complement?

In hexadecimal number system, the 15's complement is a complement concept used to simplify the subtraction operation of hexadecimal numbers. The 15's complement is similar to the 9's complement in decimal number system.

To find the 15's complement of a given hexadecimal number, we subtract each digit of the number from 15 (F).

The 15's complement of each hexadecimal digit is given in the following table −

Hexadecimal Digit	15's Complement
0	F 0 = F
1	F 1 = E
2	F 2 = D
3	F 3 = C
4	F 4 = B
5	F 5 = A
6	F 6 = 9
7	F 7 = 8
8	F 8 = 7
9	F 9 = 6
A	F A = 5
B	F B = 4
C	F C = 3
D	F D = 2
E	F E = 1
F	F F = 0

The following examples demonstrate the process of finding 15's complement of the hexadecimal numbers.

Example 1

Find the 15's complement of the hexadecimal number A259C.

Solution

The 15's complement of A259C can be obtained as follows −

The 15's complement of A = F A = 5.
The 15's complement of 2 = F 2 = D.
The 15's complement of 5 = F 5 = A.
The 15's complement of 9 = F 9 = 6.
The 15's complement of C = F C = 3.

So, the 15's complement of A259C is 5DA63.

Example 2

Find the 15's complement of 1BCFA.

Solution

The 15's complement of the given hexadecimal number is,

FFFFF 1BCFA = E4305

So, the 15's complement of hexadecimal number 1BCFA is E4305.

What is 16's Complement?

In hexadecimal arithmetic, we can also determine the 16's complement of a given hexadecimal number. The 16's complement is a concept used to simplify the subtraction operation of hexadecimal numbers.

We can determine the 16's complement of a given hexadecimal number as described below −

Find the 15's complement of the given hexadecimal number.
Add 1 to the 15's complement obtained. This gives the 16's complement of the hexadecimal number.

Therefore,

16's Complement = 15's Complement + 1

The 16's complement of each hexadecimal digit is given in the following table −

Hexadecimal Digit	15's Complement
0	F 0 = F + 1 = 10 = 0 (Ignore the carry)
1	F 1 = E + 1 = F
2	F 2 = D + 1 = E
3	F 3 = C + 1 = D
4	F 4 = B + 1 = C
5	F 5 = A + 1 = B
6	F 6 = 9 + 1 = A
7	F 7 = 8 + 1 = 9
8	F 8 = 7 + 1 = 8
9	F 9 = 6 + 1 = 7
A	F A = 5 + 1 = 6
B	F B = 4 + 1 = 5
C	F C = 3 + 1 = 4
D	F D = 2 + 1 = 3
E	F E = 1 + 1 = 2
F	F F = 0 + 1 = 1

Let us take some examples to understand the process of finding the 16's complement of hexadecimal numbers.

Example 1

Find the 16's complement of 1ABDF7.

Solution

The 16's complement of the given hexadecimal number can be determined as follows −

15's complement of 1ABDF7 = FFFFFF 1ABDF7 = E54208

16's complement = 15's complement + 1

16's complement = E54208 + 1 = E54209

Thus, the 16's complement of 1ABDF7 is E54209.

Example 2

Find the 16's complement of ABC.

Solution

The 16's complement of ABC is,

15's complement of ABC = FFF ABC = 543

16's complement = 15's complement + 1

16's complement = 543 + 1 = 544

Thus, the 16's complement of ABC is 544.

This is all about finding different types of complements used in digital electronics.

Now, let us see their application in performing subtraction operations.

Subtraction using 9's Complement

The 9's complement can be used to perform subtraction of decimal numbers. In this method, the difference of two decimal numbers is obtained by adding the 9's complement of the subtrahend to the minuend.

Let us understand the subtraction using 9's complement through an example.

Example 1

Subtract (517)₁₀ from (729)₁₀.

Solution

In this example, we have,

Minuend = 729

Subtrahend = 517

Finding the 9's complement of 517, we get

999 517 = 482

Now, adding 729 and 482 to obtain the difference of 729 and 517, we get,

729 + 482 = 1211

There is an end around carry, indicating the result is positive and is obtained by adding the end around carry to the LSD of intermediate result to obtain the final result, i.e.,

211 + 1 = 212

So, the difference of 729 and 517 is 212.

Example 2

Subtract (203)₁₀ from (159)₁₀ using 9's complement method.

Solution

In this example,

Minuend = 159

Subtrahend = 203

Taking 9's complement of 203, we get,

999 203 = 796

Adding 159 and 796, we get,

159 + 796 = 955

There is no end around carry. Thus, the final result is negative and obtained by taking the 9's complement of 955, i.e.,

999 955 = 44

Thus, the final result of subtraction 159 203 = 44.

Subtraction using 10's Complement

We can also perform decimal subtraction using 10's complement. The step-by-step procedure to perform decimal subtraction using 10's complement is given below −

Step 1 − Consider the decimal subtraction, X Y. Where, X is minuend and Y is subtrahend.
Step 2 − Find 10's complement of Y.
Step 3 − Add X and 10's complement of Y.
Step 4 − If there is an end-around carry, the result will be positive and final result is obtained by discarding the carry. If there is no end-around carry, it indicates that the result is negative and is obtained by taking 10's complement of the intermediate result and assign a negative sign before it.

Let us understand the decimal subtraction using 10's complement with the help of examples.

Example 1

Subtract (599)₁₀ from (875)₁₀ using 10's complement arithmetic.

Solution

In this example, we have,

Minuend = 875

Subtrahend = 599

Finding 10's complement of 599, we get,

10s complement of 599 = 9s complement + 1

Therefore,

10s complement of 599 = (999 599) + 1 = 401

Adding 875 and 401, we get,

875 + 401 = 1276

There is an end-around carry, showing that the result is positive and is obtained by discarding the carry.

Thus, the difference of 875 and 599 is 276.

Example 2

Subtract (307)₁₀ from (279)₁₀ using 10's complement arithmetic.

Solution

We have,

Minuend = 279

Subtrahend = 307

Taking 10's complement of 307, we get,

10s complement of 307 = (999 307) + 1 = 693

Adding 279 and 693, we get,

279 + 693 = 972

There is no end-around carry, indicating that the result is negative. The final result is obtained by taking the 10's complement of 972 i.e.,

10's complement of 972 = (999 - 972) + 1 = 28

Hence, the final result is 28.

Subtraction using 1's Complement

The 1's complement is used in binary subtraction operation.

The subtraction of two binary numbers, say X and Y i.e., X Y, can be performed using 1's complement as per the following steps −

Step 1 − Find the 1's complement of subtrahend (Y).
Step 2 − Add X and 1's complement of Y.
Step 3 − If there is an end-around carry, it indicates the result is positive and final result is obtained by adding the end-around carry to the LSB of the intermediate result. If there is no end-around carry, the result is negative and obtained by taking 1's complement of the intermediate result and put a negative sign in front of it.

Consider the following examples to understand the binary subtraction using 1's complement arithmetic.

Example 1

Subtract (111)₂ from (1011)₂ using 1's complement.

Solution

In this example, we have,

Minuend = 1011

Subtrahend = 0111

Finding 1's complement of subtrahend,

1s complement of 0111 = 1000

Adding 1011 and 1000, we get,

1011 + 1000 = 1 0011

There is an end-around carry, indicating that the result is positive. The final result is obtained by adding this end-around carry to LSB of the intermediate result (0011) i.e.,

0011 + 1 = 0100

Hence, the binary difference of 1011 and 111 is 100.

Example 2

Subtract (1100)₂ from (111)₂ using 1's complement arithmetic.

Solution

We are given,

Minuend = 0111

Subtrahend = 1100

Finding the 1's complement of the subtrahend,

1s complement of 1100 = 0011

Adding 0111 and 0011, we get,

0111 + 0011 = 1010

There is no end-around carry, indicating that the result is negative and is obtained by taking 1's complement of 1010 i.e.,

1s complement of 1010 = 0101

Hence, the binary difference of 111 and 1100 is -101.

Subtraction using 2's Complement

The 2's complement is also used to perform binary subtraction operation using digital systems. The step-by-step procedure to perform the subtraction of two binary numbers, say X and Y, i.e., (X Y) is given as follows −

Step 1 − Find the 2's complement of subtrahend (Y).
Step 2 − Add X and 2's complement of Y.
Step 3 − If there is an end-around carry, it indicates the result is positive and the final result is obtained by ignoring the end-around carry. If there is no end-around carry, the result is negative and obtained by taking 2's complement of the intermediate result and put a negative sign in front of it.

Let us see some examples to understand binary subtraction using 2's complement arithmetic.

Example 1

Subtract (101)₂ from (1100)₂ using 2's complement arithmetic.

Solution

In this example, we are given,

Minuend = 1100

Subtrahend = 0101

Taking 2's complement of the subtrahend, we get,

2's complement of 0101 = (1111 - 0101) + 1 = 1011

Adding 1100 and 1011, we get,

1100 + 1011 = 1 0111

There is an end-around carry, indicating that the result is positive and the final result is obtained by ignoring this end around carry.

Thus, the binary difference of 1100 and 101 is 111.

Example 2

Subtract (1010)₂ from (0110)₂ using 2's complement arithmetic.

Solution

The given numbers are,

Minuend = 0110

Subtrahend = 1010

Taking 2's complement of the subtrahend, we get,

2's complement of 1010 = (1111 - 1010) + 1 = 0110

Adding minuend and 2's complement of subtrahend, we get,

0110 + 0110 = 1100

Since, there is no end-around carry, indicating that the result is negative. The final result is obtained by taking 2's complement of intermediate result, i.e.,

2s complement of 1100 = (1111 - 1100) + 1 = 0100

Thus, the binary difference of 0110 and 1010 is -100.

Subtraction using 7's Complement

The 7's complement arithmetic can be used to perform subtraction of octal numbers. Here are the steps involved in performing octal subtraction using 7's complement.

Let we want to subtract octal number Y from X i.e., X Y, then

Step 1 − Find the 7's complement of subtrahend (Y).
Step 2 − Add X and 7's complement of Y.
Step 3 − If there is an end-around carry, it indicates the result is positive and the final result is obtained by adding the end-around carry to the intermediate result. If there is no end-around carry, the result is negative and is obtained by taking 7's complement of the intermediate result and put a negative sign in front of it.

Let us understand octal subtraction using 7's complement arithmetic.

Example 1

Subtract (540)₈ from (721)₈ using 7's complement arithmetic.

Solution

Given numbers are,

Minuend = 721

Subtrahend = 540

Taking 7's complement of the subtrahend,

7s complement of 540 = 777 540 = 237

Adding 7's complement and minuend, we get,

721 + 237 = 1 160

There is an end-around carry, indicating that the result is positive and the final result is obtained by adding this end-around carry to the intermediate result, i.e.,

160 + 1 = 161

Thus, the octal difference of 721 and 540 is 161.

Example 2

Subtract (310)₈ from (121)₈ using 7's complement method.

Solution

In this example, the given numbers are,

Minuend = 121

Subtrahend = 310

Taking 7's complement of the subtrahend, we get,

7s complement of 310 = 777 310 = 467

Adding minuend and the 7's complement of the subtrahend, i.e.,

121 + 467 = 610

Since, there is no end around carry, indicating the result is negative and is obtained by taking 7's complement of the intermediate result i.e.,

7s complement of 610 = 777 610 = 167

Hence, the octal difference of 121 and 310 is -167.

Subtraction using 8's Complement

The 8's complement is another technique that used to perform octal subtraction. The step-by-step procedure to perform octal subtraction using 8's complement is explained below −

Step 1 − If an octal subtraction is defined as X Y. Then, find the 8's complement of subtrahend (Y).
Step 2 − Add X and 8's complement of Y.
Step 3 − If there is an end-around carry, it indicates the result is positive and the final result is obtained by ignoring the end-around carry. If there is no end-around carry, the result is negative and is obtained by taking the 8's complement of the intermediate result and put a negative sign in front of it.

Let us understand the octal subtraction using 8's complement method through examples.

Example 1

Subtract (103)₈ from (712)₈ using 8's complement arithmetic.

Solution

We are given,

Minuend = 712

Subtrahend = 103

Finding the 8's complement of subtrahend, we get,

8s complement of 103 = (777 - 103) + 1 = 675

Adding the minuend and 8's complement of subtrahend, we get,

712 + 675 = 1607

There is an end-around carry. The final result is obtained by ignoring the end-around carry.

Thus, the octal difference of 712 and 103 is 607.

Example 2

Subtract (471)₈ from (206)₈ using the 8's complement method.

Solution

In this example, we have,

Minuend = 206

Subtrahend = 471

Finding the 8's complement of subtrahend,

8s complement of 471 = (777 - 471) + 1 = 307

Adding the minuend and 8's complement of subtrahend, we get,

206 + 307 = 515

Since, there is no end-around carry, hence the final result is negative and is obtained by taking the 8's complement of the intermediate result i.e.,

8's complement of 515 = (777 - 515) + 1 = 263

Hence, the octal difference of 206 and 471 is -263.

Subtraction using 15's Complement

The 15's complement is used to perform subtraction of hexadecimal numbers. If we want to subtract a hexadecimal number Y from X, then we follow the steps given below −

Step 1 − Find the 15's complement of subtrahend (Y).
Step 2 − Add X and 15's complement of Y.
Step 3 − If there is an end-around carry, it shows the result is positive and the final result is obtained by adding the end-around carry to the intermediate result. If there is no end-around carry, the result is negative and is obtained by taking 15's complement of the intermediate result and put a negative sign in front of it.

The following examples demonstrate the process of performing hexadecimal subtraction using 15's complement arithmetic.

Example 1

Subtract (1920)₁₆ from (E57A)₁₆ using 15's complement arithmetic.

Solution

The given numbers are,

Minuend = E57A

Subtrahend = 1920

Finding 15's complement of the subtrahend,

15s complement of subtrahend = FFFF 1920 = E6DF

Adding the minuend and 15's complement of subtrahend, we get,

E57A + E6DF = 1 CC59

There is an end around carry showing that the result is positive and is obtained by adding the end-around carry to the intermediate result, i.e.,

CC59 + 1 = CC5A

Hence, the hexadecimal difference of E57A and 1920 is CC5A.

Example 2

Subtract (DC25)₁₆ from (A209)₁₆ using 8's complement arithmetic.

Solution

Given numbers are,

Minuend = A209

Subtrahend = DC25

Finding 15's complement of subtrahend,

15s complement of DC25 = FFFF DC25 = 23DA

Adding minuend and 15's complement of subtrahend, we get,

A209 + 23DA = C5E3

Since, there is no end-around carry. The result is negative and is obtained by taking 15's complement of intermediate result, i.e.,

15s complement of C5E3 = FFFF C5E3 = 3A1C

Hence, the hexadecimal difference of A209 and DC25 us -3A1C.

Subtraction using 16's Complement

The 16's complement is also used to perform hexadecimal subtraction. The steps involved in hexadecimal subtraction using 16's complement is explained here −

Step 1 − If a hexadecimal subtraction is defined as X Y. Then, find the 16's complement of subtrahend (Y).
Step 2 − Add X and 16's complement of Y.
Step 3 − If there is an end-around carry, it indicates the result is positive and the final result is obtained by ignoring the end-around carry. If there is no end-around carry, the result is negative and is obtained by taking the 16's complement of the intermediate result and put a negative sign in front of it.

Let us see some examples to understand the hexadecimal subtraction using 16's complement.

Example 1

Subtract (E7C)₁₆ from (F9D)₁₆ using 16's complement arithmetic.

Solution

The given hexadecimal numbers are

Minuend = F9D

Subtrahend = E7C

Taking 16's complement of the subtrahend, we get,

16s complement of E7C = (FFF E7C) + 1 = 184

Adding minuend and 16's complement of subtrahend, we get,

F9D + 184 = 1121

There is an end-around carry, indicating the result is positive. The final result is obtained ignoring the end-around carry.

Hence, the hexadecimal difference of F9D and E7C is 121.

Example 2

Subtract (FF2)₁₆ from (AC5)₁₆ using 16's complement method.

Solution

The given numbers are,

Minuend = AC5

Subtrahend = FF2

Taking the 16's complement of the subtrahend, we get,

16s complement of FF2 = (FFF FF2) + 1 = 00E

Adding the minuend and 16's complement of subtrahend, we get,

AC5 + 00E = AD3

There is no end-around carry, indicating that the result is negative. The final result is obtained by taking the 16's complement of the intermediate result, as follows −

16s complement of AD3 = (FFF AD3) + 1 = 52D

Hence, the hexadecimal difference of AC5 and FF2 is -52D.

Conclusion

In conclusion, complement arithmetic is a method used in digital electronics for simplifying subtraction operations. In this chapter, we explained the different types of complements and their application in subtraction operations along with solved examples.

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