Why we need to add 1 while doing 2's complement

Question

The 2's complement of a number which is represented by N bits is 2^N-number.
For example: if number is 7 (0111) and i'm representing it using 4 bits then, 2's complement of it would be (2^N-number) i.e. (2^4 -7)=9(1001)

7==> 0111
1's compliment of 7==> 1000
1000
+  1
-------------
1001 =====> (9)

While calculating 2's complement of a number, we do following steps: 1. do one's complement of the number 2. Add one to the result of step 1.

I understand that we need to do one's complement of the number because we are doing a negation operation. But why do we add the 1?

This might be a silly question but I'm having a hard time understanding the logic. To explain with above example (for number 7), we do one's complement and get -7 and then add +1, so -7+1=-6, but still we are getting the correct answer i.e. +9

Answer 1

Your error is in "we do one's compliment and get -7". To see why this is wrong, take the one's complement of 7 and add 7 to it. If it's -7, you should get zero because -7 + 7 = 0. You won't.

The one's complement of 7 was 1000. Add 7 to that, and you get 1111. Definitely not zero. You need to add one more to it to get zero!

The negative of a number is the number you need to add to it to get zero.

If you add 1 to ...11111, you get zero. Thus -1 is represented as all 1 bits.

If you add a number, say x, to its 1's complement ~x, you get all 1 bits.

Thus:
~x + x = -1

Add 1 to both sides:
~x + x + 1 = 0

Subtract x from both sides:
~x + 1 = -x

Answer 2

The +1 is added so that the carry over in the technique is taken care of.

Take the 7 and -7 example.

If you represent 7 as 00000111

In order to find -7:

Invert all bits and add one

11111000 -> 11111001

Now you can add following standard math rules:

  00000111
+ 11111001
 -----------
  00000000

For the computer this operation is relatively easy, as it involves basically comparing bit by bit and carrying one.

If instead you represented -7 as 10000111, this won't make sense:

   00000111
+  10000111
  -----------
   10001110 (-14)

To add them, you will involve more complex rules like analyzing the first bit, and transforming the values.

A more detailed explanation can be found here.

Answer 3

Short answer: If you don't add 1 then you have two different representations of the number 0.

Longer answer: In one's complement

• the values from 0000 to 0111 represent the numbers from 0 to 7

• the values from 1111 to 1000 represent the numbers from 0 to -7

  • since their inverses are 0000 and 0111.

There is the problem, now you have 2 different ways of writing the same number, both 0000 and 1111 represent 0.

If you add 1 to these inverses they become 0001 and 1000 and represent the numbers from -1 to -8 therefore you avoid duplicates.

Answer 4

I'm going to directly answer what the title is asking (sorry the details aren't as general to everyone as understanding where flipping bits + adding one comes from).

First let motivate two's complement by recalling the fact that we can carry out standard (elementary school) arithmetic with them (i.e. adding the digits and doing the carrying over etc). Easy of computation is what motivates this representation (I assume it means we only 1 piece of hardware to do addition rather than 2 if we implemented subtraction differently than addition, and we do and subtract differently in elementary school addition btw).

Now recall the meaning of each of the digit's in two's complements and some binary numbers in this form as an example (slides borrowed from MIT's 6.004 course):

Now notice that arithmetic works as normal here and the sign is included inside the binary number in two's complement itself. In particular notice that:

1111....1111 + 0000....1 = 000....000 i.e. -1 + 1 = 0

Using this fact let's try to derive what the two complement representation for -A should be. So the problem to solve is:

Q: Given the two's complement representation for A what is the two's complement's representation for -A?

To do this let's do some algebra using values we know:

A + (-A) = 0 = 1 + (-1) = 11...1 + 00000...1 = 000...0 now let's make -A the subject expressed in terms of numbers expressed in two's complement: -A = 1 + (-1 - A) = 000.....1 + (111....1 - A) where A is in two's complements. So what we need to compute is the subtraction of -1 and A in two's complement format. For that we notice how numbers are represented as a linear combination of it's bases (i.e. 2^i):

1*-2^N-1 + 1 * 2^N-1 + ... 1 = -1

a_N * -2^N-1 + a_N-1 * 2^N-1 + ... + a_0 = A

--------------------------------------------- (subtract them)

a_N-1 * -2^N-1 + a_N-1 -1 * 2^N-1 + ... + a_0 -1 = A

which essentially means we subtract each digit for it's corresponding value. This ends up simply flipping bits which results in the following:

-A = 1 + (-1 - A) = 1 + ~ A

where ~ is bit flip. This is why you need to bit flip and add 1.

---

Remark:

I think a comment that was helpful to me is that complement is similar to inverse but instead of giving 0 it gives 2^N (by definition) e.g. with 3 bits for the number A we want A+~A=2^N so 010 + 110 = 1000 = 8 which is 2^3. At least that clarifies what the word "complement" is suppose to mean here as it't not just the inverting of the meaning of 0 and 1.

---

If you've forgotten what two's complement is perhaps this will be helpful: What is “2's Complement”?

---

Cornell's answer that I hope to read at some point: https://www.cs.cornell.edu/~tomf/notes/cps104/twoscomp.html#whyworks