Why does -INT_MIN = INT_MIN in a signed, two's complement representation?

Question

I still haven\'t found a reason why the lowest signed negative number doesn\'t have an equivalent signed positive number? I mean in a 3 digit binary number for simplicity 10

Answer 1

The alternative to two's complement has such a property, it's known as one's complement.
In one's complement form, the lowest possible value also has a valid positive form.

---

One's complement works by simply inverting all the bits in the number itself.
For example, we know that 0110 == 6 and in one's complement 1001 == -6. Using one's complement, we have just as many positive numbers as we do negative numbers.

But what about the bit representation 1111? Just by looking at it, we can tell that it's the "negative" form of zero (0000 = 0; 1111 = -0), but such a number doesn't make any sense and is somewhat wasteful.

Instead, we use two's complement, which is similar to one's complement, but after inverting the bits, we add one. So if we know that 0110 = 6, then the one's complement is 1001 and the two's complement is 1001 + 1 == 1010. Using two's complement, we don't have a "negative zero" because it causes an overflow.

A different way of looking at it is "if the highest bit is set, then the number is negative". That means that the positive range is [0 .. 2^(bits - 1)] and the negative range is everything else. There's the same number of positive numbers as there are negative numbers, but because (in this format) zero is considered positive, the negative range gets shifted one to [-1 .. (neg) 2^(bits - 1)].

---

Lets assume we're dealing with a 3-bit signed number in two's complement. That'd give us the following table:

BITS  VALUE
000       0
001       1
010       2
011       3

100      -4
101      -3
110      -2
111      -1

You can see that there's the same quantity of positive numbers as negative numbers, it's just that the negative numbers don't start from 0 like the positive set does.