All negative numbers have the most significant bit set and 0 is the number with no bits set, ergo 0 must be positive since the most significant bit is not set.

Now arithmatically, this is untrue. We'll usually treat 0 as neither positive nor negative (or in certain cases both negative and positive) but bitwise, In terms of twos-complement implementation, Zero is positive. We know that since it exists in the unsigned version of the types as well.

Hopefully you'll see that some day.

While "twos complement" turns the MSB unsigned value to a negative instead of a positive. For example, 4-bit twos complement: 1000 represents -8 (in unsigned 4-bit, this supposed to be +8), 0100 represents 4, 0010 represents 2, 0001 represents 1. Some more numbers: 7 (0111), -7 (1001), 1 (0001), -1 (1111).

Intuitively, "ones complement" MSB represents a multiplication by (-1). While "twos complement" MSB adds (-N), with N = 2^(bit length - 1), in case of 4-bit twos complement it's (-2^3) or (-8). Both representation leave non-MSB bits work exactly like unsigned integer.

I find the best way to understand why 2s complement is so desirable is to write down the entire number line for e.g. 3-bit integers.

Using 1s complement, the negative numbers are backwards. 2s complement fixes this, so that arithmetic works and you can do addition and subtraction without any extra steps.

(Remember that negative numbers are less than positive numbers, so the correct way to count them is:

-8 -7 -6 -5 -4 -3 -2 -1 0 +1 +2 +3 +4 +5 +6 +7

Where -1 is the largest possible negative number)

  > I do not understand your argument about branches, how would it hinder the jump instructions?