Consider the equation $x \& (-x)$ which enables us to find the largest power of 2 that divides $x$. One can prove this relatively easily from the definitions:

That is, if we state that $a = \langle x 1 0^r \rangle$ for some arbitrary $r$, we then find that $a \& (-a) = 2^r = \langle 1 0^r \rangle$, which is precisely what we need to subtract from $a$ to remove the rightmost/trailing $1$. However, I don't find this insightful. So I'm going to spend some time dwelling on $2$-adics, to find a more intuitive way to think about this.

§ 2-adics and negative numbers

In the 2-adic system, we have that:

Of course, these agree with the 2's complement representation, because the 2's complement representation simply truncates the 2-adic representation. At any rate, the point of interest is that if we now want to know how to write $-3$, we start with the "lower" number $-4$ and then add $1$ to it, giving us:

Which once again agrees with the 2's complement definition.

§ $x \& (-x)$ for powers of 2:

If we now think strictly about powers of 2, we know that, for example, $8 = \langle \dots 0 0 1 0 0 \rangle$ while $-8 = \langle \dots 1 1 1 0 0 \rangle$. Hence, $x \& (-x) = \langle 0 0 1 0 0 \rangle$. This will hold for any power of 2, so our claim that $x \& (-x)$ gives us the location of the LSB will work for any power of 2.

§ Alternative explanation for 2's complement

Start with the fact that we choose a single representation for zero:

0 ~= b0000000

Now, when we subtract 1, ask "are we in signed world or unsigned world"? If in signed world, we want the answer to be -1. If in unsigned world we want the answer to be 255.

0 - 1= b0000000 - b00000001= b11111111=unsigned= 255

If we wanted to interpret the answer as signed , then we are free to do so. This automatically tell us that

0 - 1=unsigned= b11111111=signed= -1

So, the advantage is that our operations don't care about whether the number is signed/unsigned.