The term “Binary operation” represents the basic operations of mathematics that are performed on two operands. Basic arithmetic operations like Addition, Subtraction, Multiplication, and Division play an important role in mathematics. This page has been designed to answer questions about binary addition, including:

- What is Binary Addition? Procedure to add two binary numbers.
- Examples to Solve Binary Numbers
- Addition of Positive and Negative Binary Numbers
- Addition of Two Negative Binary Numbers

## What is Binary Addition?

The binary addition operation works similarly to the base 10 decimal system, except that it is a base 2 system. The binary system consists of only two digits, 1 and 0. Most of the functionalities of the computer system use the binary number system. The binary code uses the digits 1’s and 0’s to make certain processes turn off or on. The process of the addition operation is very familiar to the decimal system by adjusting to the base 2. Before attempting the binary addition process, we should have complete knowledge of how the place works in the binary number system. Because most of the modern digital computers and electronic circuits perform the binary operation by representing each bit as a voltage signal. The bit 0 represents the “OFF” state and the bit 1 represents the “ON” state.

Binary addition is much easier than the decimal addition when you remember the following tricks:

- 0 + 0 = 0
- 0 + 1 = 1
- 1 + 0 = 1
- 1 + 1 =10

### How do you add binary numbers?

Now, look at the example of the binary addition:101 + 101

Procedure:

101

(+) 101

Step 1: First consider the 1’s column, and add the one’s column,( 1+1 ) and it gives the result 10 as per the condition of binary addition.

Step 2: Now, leave the 0 in the one’s column and carry the value 1 to the 10’s column.

1

101

(+) 101

————–

0

Step 3: Now add 10’s place, 1+( 0 + 0 ) = 1. So, nothing carries to the 100’s place and leave the value 1 in the 10’s place

1

101

(+) 101

————-

10

Step 4: Now add the 100’s place ( 1 + 1 ) = 10. Leave the value 0 in the 100’s place and carries 1 to the 1000’s place.

1

101

(+) 101

————-

1010

So, the resultant of the addition operation is 1010.

When you cross-check the binary value with the decimal value, the resultant value should be the same.

The binary value 101 is equal to the decimal value 5

So, 5 + 5 = 10

The decimal number 10 is equal to the binary number 1010.

Consider other examples of binary additions are as follows:

Example 1: 10001 + 11101

Solution:

**1**

1 0 0 0 1

(+) 1 1 1 0 1

———————–

1 0 1 1 1 0

Example 2: 10111 + 110001

Solution:

** 1 1 1**

1 0 1 1 1

(+) 1 1 0 0 0 1

———————–

1 0 0 1 0 0 0

## Binary Addition Using 1’s Complement

The number “ 0 “ represents the positive sign

The number “ 1 “ represents the negative sign

### Addition of Positive and Negative Number

**Case 1: When a positive number has a greater magnitude**

Take the 1’s complement of the negative number and the carry is added to the resultant sum at the 1’s place. When you add the carry with the resultant, you will get the sum value.

Example:

+ 1111 and -1101

+ 1 1 1 1 = 0 1 1 1 1

– 1 1 0 1 = 1 0 0 1 0 (taking 1’s complement)

——————-

0 0 0 0 1

1

———————

0 0 0 1 0

Therefore, the solution is + 0010.

**Case 2: When a negative number has a greater magnitude**

Take the 1’s complement of the negative number and there will be no end around carry in this case. Finally, the sum is obtained by taking the 1’s complement of the resultant.

Example:

+ 1111 and -1101

– 1 1 1 1 = 1 0 0 0 0 (taking 1’s complement)

+1 1 0 1 = 0 1 1 0 1

—————-

1 1 1 0

——————

1 0 0 1 0 (taking 1’s complement)

### Addition of Two Negative Numbers

Take the 1’s complement of both the negative numbers and then add. The end around carry will appear and it will generate a number 1 in the sign bit. The sum value can be obtained by taking the 1’s complement of the resultant.

Example:

- 1010 and – 0011

- 1 0 1 0 = 1 0 1 0 1 (taking 1’s complement)
- 0 0 1 1 = 1 1 1 0 0 (taking 1’s complement)

————————–

1 0 0 0 1

1

—————————–

1 0 0 1 0

—————————-

1 1 1 0 1 (taking 1’s complement)

Therefore the solution is – 1101

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