二进制减法-多余的数字？

Question

Just messing around with Binary Subtraction using two's complement but I would like something explained.

To perform 11 - 9 I first convert the 9 to a negative number by changing the 0's to 1's and 1's to 0's and then add 1. Then I add the negative number so it becomes 11 +(-9)

Binary

 00001011
+11110111
---------
100000010

As I understand it, I just disregard the MSB so it becomes:

00000010

Why is it the case that I can drop the MSB?

Answer 1

Because you effectively added it first.

The invert/add one combo is equivalent to subtracting from a constant "one larger than allowed":

100000000
 00001001
---------  -
 11110111

So you have changed 11 - 9 to 11 + (256 - 9) , which you can write as 11 - 9 + 256 to make it even clearer that you added that "extra bit" yourself in order to do the math, so now you have to drop it to get the result of the original expression.