**(r – 1) ’s compliment: (M – N) _{r}**

This is similar to r’s compliment. There is a difference while dealing with the final carry we get

**If we have both M & N positives, then**

The Procedure for doing subtraction using (r – 1) ’s compliment is as follow:

- Take (r – 1) ’s compliment of subtrahend N
- Add it to minuend M
- If we get a carry, add 1 to the result otherwise take (r – 1)’s compliment of the result and place a –ve sign in front of it.

**If we have negative M & positive N, then **i.e. – m – n where m & n are magnitudes of M&N

The Procedure for doing subtraction using (r – 1) ’s compliment is as follow:

- Take (r – 1) ’s compliment of subtrahend N
- Add it to minuend M

If we get a carry, add 1 to the result and also take (r – 1’s compliment and place a –ve sign in front of it otherwise if there is no carry then do nothing

**Eg. 1010100 _{2} – 1000100_{2}**

1’s compliment of 1000100 is 0111011 and then adds to 1010100

**Eg. 1000100 _{2} – 1010100_{2}**

1’s compliment of 1010100 is 0101011 and it is added to 1000100

As we don’t have carry so we take 1’s compliment of the result and put the –ve sign in front

So answer is -0010000

**We don’t have to worry about whether M is larger or N is larger, carry which we get takes care of this thing. So one has to just follow the procedure and we’ll get the result.**