When negative numbers are expressed in binary addition using 2’s complement the addition of binary numbers becomes easier. This operation is almost similar to that in 1’s complement system and is explained with examples given below:
A. Addition of a positive number and a negative number.
We consider the following cases.
Case I: When the positive number has a greater magnitude
In this case the carry which will be generated is discarded and the final result is the result of addition.
The following examples will illustrate this method in binary addition using 2’s complement:
In a 5bit register find the sum of the following by using 2’s complement:
(i) 1011 and 0101
Solution:
+ 1 0 1 1 ⇒ 0 1 0 1 1
Hence the sum is + 0110.
(ii) + 0111 and – 0011.
Solution:
+ 0 1 1 1 ⇒ 0 0 1 1 1Hence the sum is + 0100.
Case II: When the negative number is greater.
When the negative numbers is greater no carry will be generated in the sign bit. The result of addition will be negative and the final result is obtained by taking 2’s complement of the magnitude bits of the result.
The following examples will illustrate this method in binary addition using 2’s complement:
In a 5bit register find the sum of the following by using 2’s complement:
(i) + 0 0 1 1 and  0 1 0 1
Solution:
+ 0 0 1 1 ⇒ 0 0 0 1 1
2’s complement of 1110 is (0001 + 0001) or 0010.
Hence the required sum is  0010.
(ii) + 0 1 0 0 and  0 1 1 1
Solution:
+ 0 1 0 0 ⇒ 0 0 1 0 02’s complement of 1101 is 0011.
Hence the required sum is – 0011.
B. When the numbers are negative.
When two negative numbers are added a carry will be generated from the sign bit which will be discarded. 2’s complement of the magnitude bits of the operation will be the final sum.
The following examples will illustrate this method in binary addition using 2’s complement:
In a 5bit register find the sum of the following by using 2’s complement:
(i) – 0011 and – 0101
Solution:
 0 0 1 1 ⇒ 1 1 1 0 1 (2’s complement)
2’s complement of 1000 is (0111 + 0001) or 1000.
Hence the required sum is – 1000.
(ii) 0111 and – 0010.
Solution:
 0 1 1 1 ⇒ 1 1 0 0 1 (2’s complement)
2’s complement of 0111 is 1001.
Hence the required sum is – 1001.
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