Why can we write almost any nonzero 2-adic integer in the same form? The title was difficult to formulate (I am open to suggestions). Two things bug me about this:


"is almost any" is imprecise, which goes against Volume 1's statement on definiteness.
When he writes, "in other words", this means Prof. Knuth explained it to me in two ways and I still do not understand it.


What is going on here in Plain English? 




  If $x$ is almost any nonzero 2-adic integer, we can write its bits in
  the form  $$x = (\alpha01^a10^b)_2 $$
  
  in other words, $x$ consists of some arbitrary (but infinite) binary
  string of $\alpha$, followed by a $0$, which is followed by $a+1$
  ones, and followed by $b$ zeros, for some $a \geq 0$ and $b \geq 0$.
  (The exceptions occur when $x = -2^b$; then $a = \infty $)
  
  Consequently 
  
  $$ \bar{x} = (\bar\alpha10^a01^b)_2 $$
  
  $$ x-1 = (\alpha01^a01^b)_2 $$
  
  $$ -x = (\bar\alpha10^a10^b)_2 $$
  
  and we can see that $\bar{x} +1 = -x = \overline{x-1}$, ...
  
  With two operations we can therefore compute relatives of $x$ in
  several useful ways:
  
  
  $x \mathbin{\&amp;} (x-1) = (\alpha01^a00^b)_2$
  $x\mathbin{\&amp;} -x = (0^\infty00^a10^b)_2$
  $x \mathbin{|} -x = (1^\infty11^a10^b)_2$
  $x \oplus -x = (1^\infty11^a00^b)_2$
  $x \mathbin{|} (x-1) = (\alpha 01^a11^b)_2$
  $x \oplus (x-1) = (0^\infty00^a11^b)_2$
  $\bar{x} \mathbin{\&amp;} (x-1) = (0^\infty00^a01^b)_2$
  ...
  $((x|(x-1))+1) \mathbin{\&amp;} x = (\alpha 00^a00^b)_2$
  



Knuth, Donald. The Art of Computer Programming. Volume 4, Fascicle 1. p. 8.
The pattern $x = (\alpha01^a10^b)_2$ is very easy to match. For example, the number $1=0^\infty1$ matches the pattern with $\alpha=0^\infty, a=b=0$. Try matching some other numbers yourself. The only thing this pattern asks for, is for the string "$01$" to appear somewhere in the sequence of bits. The only numbers to fail this condition are $0=0^\infty$ and $-2^b=1^\infty0^b$.