Self Doubt

Question

Q1.   Find 1's and 2's complement of decimal number of 15

Q2.   Find 1's and 2's complement of decimal number of -15

Answer with Explanation will be welcomed. :)

Comments

If it's a decimal no. i.e. base 10 then you can find it's 9's and 10's complement.

If it's a binary no. i.e. base 2 then you can find it's 1's and 2's complement.

For a number represented in base 'r', we can find (r-1)'s and r's complement of that no.

Here are you asking to find out 1's complement of a number having base 10 (decimal) ?

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yes, the question here is

15 base 10 -> find 1's and 2's complement

(-15) base 10 -> find 1's and 2's complement

for example 2's complement of 3 base 10 is: (011) after converting 1's to 0's and vice versa) -> (100)-> 100+1-> 101

My main doubt over here is how to find 2's complement of negative decimal no. like (-3), (-15) etc

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We use 2's complement to subtract a no. from another no. Let suppose we want to find (a-b). What we do is find 2's complement of b and then add that with a. From this you can understand that 2's complement of a number gives us the negation of that no. i.e. 2's complement of b is (-b). That is why we do a-b=a+(2's complement of b).
So basically whenever we have to find 2's complement of a number we should know that it is asking us to find the negative representation of that no.
Here you are asking to find 2's complement of (-3). So we can do negative of (-3) is (-(-3)) i.e. +3.
Similarly for (-15) it is +15.

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From what I learnt in my computer science class, two's complement, one's complement and signed magnitude are the same only in a positive integer value. This will differ with negative integer values, where for one's complement you will have to flip the bits and for two's complement keeping the flipped bits and adding a 1 bit to getting the the negative bit to it's absolute integer.

I'm still learning, computer science student, hope it helped.

Answer 1

Let us take binary values of those decimal number first so 15 require 4 bits in the binary representation but 4 bit are not sufficient for 2's complement representation because in 4 bit we can only represent from -7 to +8 so we will take 5 bits to represent

 

now 15 in binary = 01111

+15 in 2's complement = +15 in binary

-15 in 2's complement = invert bits of +15 and add 1 so

+15 = 01111 inverting it will be 10000 adding 1 it will be 10001

thus +15 = 01111 and -15 = 10001 in 2's complement

Answer 2

let take n bit binary number.... and the ranges are

signed magnitude :           -(2^n-1-1) to + (2^n-1-1)

1's complement form :     -(2^n-1-1) to + (2^n-1-1)

2's complement form :     -(2^n-1) to + (2^n-1-1)

conclusion :-

for positive numbers, no.of digits required to represent are same in three formats. and one graterthan the normal unsigned representation.

for negative numbers, the number of digits some times (not always) vary wrt to 2's complement form and remaing (1's complement,signed magnitude)  forms. 

positive no. are same in every form i.e., 1's complement form, 2's complement form and sign magnitude form representation is same as normal unsigned binary number  

therefore

unsigned 15 = 1111

signed magnitude 15 =01111

1's complement 15 = 01111

2's complement 15 = 01111

coming to negative numbers....

signed magnitude :- just change the signed bit to 1 for positive signed magntitude representation

1's complement :- just complement the positive 1's complement representation

2's complement :- just 2'complement the positive 2's complement representation

therefore

signed magnitude -15 = 11111

1's complement form of -15 = complement(01111)=10000

2's complement form of -15 =  2's complement(01111)=10001

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For justifying my statement " for negative numbers, the number of digits some times (not always) vary wrt to 2's complement form and remaing (1's complement,signed magnitude)  forms "

-16 in three formats are

formula for signed magnitude format is ⌈ log₂^(no+1)  ⌉+1

sign magnitude format requires 6 bits  and representation is 110000

formula for signed magnitude format is ⌈ log₂^(no+1)  ⌉+1

 1's complement format requires 6 bits and representation is 101111

but formula for signed magnitude format is ⌈ log₂^(no)  ⌉+1

2's complement format requires 5 bits only  and representation is 10000

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NOTE :-  Don't confuse with 1's complement of a number and 1's complement representation of number

let take no.of digits is 5

1's complement of decimal 15 =  1's complement ( 01111 ) = 10000

1's complement form representation of decimal 15 = 01111