Every non-zero $p$-adic number ($p$ is a positive rational prime number) can be uniquely written in , formally as a Laurent series,

 $\xi=a_{-m}p^{-m}+a_{-m+1}p^{-m+1}+\cdots+a_{0}+a_{1}p+a_{2}p^{2}+\cdots$

where  $m\in\mathbb{N}$,  $0\leqq a_{k}\leqq p-1$  for all $k$’s, and at least one of the integers $a_{k}$ is positive.  In addition, we can write:  $0=0+0p+0p^{2}+\cdots$

The field $\mathbb{Q}_{p}$ of the $p$-adic numbers is the completion of the field $\mathbb{Q}$ with respect to its $p$-adic valuation (http://planetmath.org/PAdicValuation); thus $\mathbb{Q}$ may be thought the subfield (prime subfield) of $\mathbb{Q}_{p}$.  We can call the elements of  $\mathbb{Q}_{p}\!\setminus\!\mathbb{Q}$  the proper $p$-adic numbers.

If, e.g.,  $p=2$,  we have the 2-adic or, according to G. W. Leibniz, dyadic numbers, for which every $a_{k}$ is 0 or 1.  In this case we can write the sum expression for $\xi$ in the reverse and use the ordinary positional (http://planetmath.org/Base3) (i.e., dyadic) figure system (http://planetmath.org/Base3).  Then, for example, we have the rational numbers

 $-1=...111111,$
 $1=...0001,$
 $6.5=...000110.1,$
 $\frac{1}{5}=...00110011001101.$

(You may check the first by adding 1, and the last by multiplying by  5 = …000101.) All 2-adic rational numbers have periodic binary expansion (http://planetmath.org/DecimalExpansion).  Similarly as the decimal (http://planetmath.org/DecimalExpansion) (according to Leibniz: decadic) expansions of irrational real numbers are aperiodic, the proper 2-adic numbers also have aperiodic binary expansion, for example the 2-

 $\alpha=...1000010001001011.10111.$

The 2-adic fractional numbers have some bits “1” after the dyadic point “.” (in continental Europe: comma “$,$”), the 2-adic integers have none.  The 2-adic integers form a subring of the 2-adic field $\mathbb{Q}_{2}$ such that $\mathbb{Q}_{2}$ is the quotient field of this ring.

Every such 2-adic integer $\varepsilon$ whose last bit is “1”, as  $-3/7=...11011011011$, is a unit of this ring, because the division$1\colon\!\varepsilon$  clearly gives as quotient a integer (by the way, the divisions of the binary expansions in practice go from right to left and are very comfortable!).

Those integers ending in a “0” are non-units of the ring, and they apparently form the only maximal ideal in the ring (which thus is a local ring).  This is a principal ideal $\mathfrak{p}$, the generator of which may be taken  $...00010=10$ (i.e., two).  Indeed, two is the only prime number of the ring, but it has infinitely many associates, a kind of copies, namely all expansions of the form  $...10=\varepsilon\cdot 10$.  The only non-trivial ideals in the ring of 2-adic integers are  $\mathfrak{p},\,\mathfrak{p}^{2},\,\mathfrak{p}^{3},\,\ldots$  They have only 0 as common element.

All 2-adic non-zero integers are of the form $\varepsilon\cdot 2^{n}$ where  $n=0,\,1,\,2,\,\ldots$.  The values  $n=-1,\,-2,\,-3,\,\ldots$  here would give non-integral, i.e. fractional 2-adic numbers.

If in the binary of an arbitrary 2-adic number, the last non-zero digit “1” corresponds to the power $2^{n}$, then the 2-adic valuation of the 2-adic number $\xi$ is given by

 $|\xi|_{2}=2^{-n}.$