$\begin{aligned}
&a = \langle x 1 0^r \rangle \\
&-a = \lnot a + 1 = x01^r + 1 = \overline{x}10^r \\
&a \& (-a) = a \& (\lnot a + 1) = (x 10^r) \& (\overline{x}10^r) = 0^{|\alpha|}10^r = 2^r
\end{aligned}$

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.
$\begin{aligned}
&-1 = \dots 1 1 1 1 \\
&-2 = -1 + -1 = \dots 1 1 1 1 + \dots 1 1 1 1 = \dots 1 1 1 0 \\
&-4 = -2 + -2 = \dots 1 1 1 0 + \dots 1 1 1 0 = \dots 1 1 0 0 \\
&-8 = -2 + -2 = \dots 1 1 0 0 + \dots 1 1 0 0 = \dots 1 0 0 0 \\
\end{aligned}$

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:
$-3 = -4 + 1 = \dots 1 1 0 0 + \dots 0 0 0 1 = \dots 1 1 0 1$

Which once again agrees with the 2's complement definition.
```
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 ```
0 - 1
=unsigned= b11111111
=signed= -1
```

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