Question

1. Obtain the 1’s complement, 2’s complement and sign magnitude system representation in 7 bits for the following decimal numbers:

a) 15₁₀

b) -21₁₀

c) 35₁₀

d) -2710

2. Use 1’s and 2’s complement system to perform the following calculations and mention if
there will be overflow or not:
a) 1100 – 0101
b) 1010 + 0100
c) 01100 + 00111

Answer 1

a)
15
Since this is a positive number. we can directly convert this into binary
Step 1. Divide 15 successively by 2 until the quotient is 0
15/2 = 7, remainder is 1
7/2 = 3, remainder is 1
3/2 = 1, remainder is 1
1/2 = 0, remainder is 1
Read remainders from the bottom to top as 1111
1's complement: 0001111
2's complement: 0001111
sign-magnitude: 0001111

b)
-21
This is negative. so, follow these steps to convert this into a 2's complement binary
Step 1. Divide 21 successively by 2 until the quotient is 0
21/2 = 10, remainder is 1
10/2 = 5, remainder is 0
5/2 = 2, remainder is 1
2/2 = 1, remainder is 0
1/2 = 0, remainder is 1
Read remainders from the bottom to top as 10101
So, 21 in normal binary is 0010101

1's complement:
-------------------
flip all the bits
   0010101 is flipped to 1101010
1's complement: 1101010

2's complement:
-------------------
Add 1 to above result
1101010 + 1 = 1101011
2's complement: 1101011

sign-magnitude:
-----------------
set 1 as left most bit
so, 0010101 is 1010101
sign-magnitude: 1010101

c)
35
Since this is a positive number. we can directly convert this into binary
Step 1. Divide 35 successively by 2 until the quotient is 0
35/2 = 17, remainder is 1
17/2 = 8, remainder is 1
8/2 = 4, remainder is 0
4/2 = 2, remainder is 0
2/2 = 1, remainder is 0
1/2 = 0, remainder is 1
Read remainders from the bottom to top as 100011
1's complement: 0100011
2's complement: 0100011
sign-magnitude: 0100011

d)
This is negative. so, follow these steps to convert this into a 2's complement binary
Step 1:
Step 1. Divide 27 successively by 2 until the quotient is 0
27/2 = 13, remainder is 1
13/2 = 6, remainder is 1
6/2 = 3, remainder is 0
3/2 = 1, remainder is 1
1/2 = 0, remainder is 1
Read remainders from the bottom to top as 11011
So, 27 in normal binary is 0011011
1's complement:
-------------------
flip all the bits
   0011011 is flipped to 1100100
1's complement: 1100100

2's complement:
-------------------
Add 1 to above result
1100100 + 1 = 1100101
2's complement: 1100101

sign-magnitude:
-----------------
set 1 as left most bit
so, 0011011 is 1011011
sign-magnitude: 1011011