Why does negation flip all high bits of a number expect for the least significant bit?

Question

In his blog post on 512 bit division, Remco Bloemen makes the following assertion:

... negation flips all high bits of a number up to but not including the least significant one

Why is that?

Answer 1

This is because the EVM uses the two's complement system to store signed values.

By definition, negating a number in two's complement involves two steps:

1. Flip all bits.
2. Add 1 to the flipped value.

The 1 is added at the end, so the least significant bit remains unchanged (it goes back to its original value, in a roundabout way).