COMPLIMENTS:
For a number system with base r (with r > 1), there are two types of compliments:
 r’s compliment
 (r1) ‘s compliment
r’s Compliment: If we have a positive number N with integer part of n digits then we define r’s compliment as
10’s compliment of number (5432)_{ 10} is 10^{4} 5432 = 10000 – 5432 = (4568)_{ 10}
[value of n is 4]
10’s compliment of number (0.5432)_{ 10} is 10^{0} 0.5432 = 1 – 0.5432 = (0.4568)_{ 10}
[value of n is 0 as there is no integer part in this number]
10’s compliment of number (54.32)_{ 10} is 10^{2} 54.32 = 100 – 54.32 = (45.68)_{ 10} [value of n is 2]
2’s compliment of number (110.01)_{ 2} is (2^{3})_{2}– 110.01_{2}= 1000 – 110.01 = 001.11 [value of n is 3]
Eg. Find 2’s compliment of (110.01)_{ 2} (value from above example)
Ans: Following the rule we get the answer as 000.1110
(r1) ‘s compliment: If we have a positive number N with base r and integer part with digits n and fractional part of m digits, we define (r1) ‘s compliment as r^{n }– r^{m} N
9’s compliment of number (5432)_{ 10} is 10^{4} 10^{0}  5432 = 10000 – 1  5432 = (4567)_{ 10}
[Value of n is 4 and of m is 0]
9’s compliment of number (0.5432)_{ 10} is 10^{0} 10^{4}  0.5432 = 1 – 0.0001  0.5432 = (0.4567)_{ 10}
[Value of n is 0 as there is no integer part in this number and of m is 4]
9’s compliment of number (54.32)_{ 10} is 10^{2 } 10^{2}  54.32 = 100 – 0.01  54.32 = (45.67)_{ 10} [Value of n is 2 and of m is 2]
1’s compliment of number (110.01)_{ 2} is (2^{3})_{10}– (2^{2})_{10 } 110.01_{2}= 1000 – 0.01  110.01 = 001.10 [Value of n is 3 and of m is 2]
Applying the rule to above example we get the answer as 001.10, the sameJ
Relation between the r’s Compliment and (r1) ‘s compliment: We can get r’s Compliment from (r1) ‘s compliment by just adding r^{m} and viceversa by subtracting r^{m}
Compliment of compliment: If we take r’s compliment of r’s compliment we get the original number as
Let the original number be N
So r’s compliment is r^{n }– N and compliment of compliment is r^{n }– (r^{n }– N) = N which is the original number
and the same is for (r1) ‘s compliment.
Let the original number be N
So r’s compliment is r^{n }– r^{m}  N and compliment of compliment is r^{n }– r^{m} – (r^{n }– r^{m}  N ) = N which is the original number
1’s compliment and 2’s compliment: Following table would show the representation of decimal 7 to +8 in 1’s and 2’s compliment. In the signed number , we take the 4^{th} bit as signed bit and make this bit one(‘1’) when ever we have negative number & make it zero(‘0’) when we have a positive number and rest of the 3 bits are used to represent magnitude.
Comparison of 1’s compliment and 2’s compliment:
 1’s compliment is easier to implement than 2’s compliment
 While for the subtraction, 2’s compliment is better because for 2’s compliment we need only 1 arithmetic operation while 2 arithmetic operations are needed for 1’s compliment.
 2’s compliment has only one representation for zero while 1’s compliment has two representation for zero as shown above
