Editing P-adic number (section)

== ''p''-adic expansion of rational numbers ==

The [[decimal expansion]] of a positive [[rational number]] <math>r</math> is its representation as a [[series (mathematics)|series]]
:<math>r = \sum_{i=k}^\infty a_i 10^{-i},</math>
where <math>k</math> is an integer and each <math>a_i</math> is also an [[integer]] such that <math>0\le a_i <10.</math> This expansion can be computed by [[long division]] of the numerator by the denominator, which is itself based on the following theorem: If <math>r=\tfrac n d</math> is a rational number such that <math>10^k\le r <10^{k+1},</math> there is an integer <math>a</math> such that <math>0< a <10,</math> and <math>r = a\,10^k +r',</math> with <math>r'<10^k.</math> The decimal expansion is obtained by repeatedly applying this result to the remainder <math>r'</math> which in the iteration assumes the role of the original rational number <math>r</math>.

The {{mvar|p}}-''adic expansion'' of a rational number is defined similarly, but with a different division step. More precisely, given a fixed [[prime number]] <math>p</math>, every nonzero rational number <math>r</math> can be uniquely written as <math>r=p^k\tfrac n d,</math> where <math>k</math> is a (possibly negative) integer, <math>n</math> and <math>d</math> are [[coprime integer]]s both coprime with <math>p</math>, and <math>d</math> is positive. The integer <math>k</math> is the '''{{mvar|p}}-adic valuation''' of <math>r</math>, denoted <math>v_p(r),</math> and <math>p^{-k}</math> is its '''{{mvar|p}}-adic absolute value''', denoted <math>|r|_p</math> (the absolute value is small when the valuation is large). The division step consists of writing
: {{anchor|division_step}}<math>r = a\,p^k + r'</math>
where <math>a</math> is an integer such that <math>0\le a <p,</math> and <math>r'</math> is either zero, or a rational number such that <math>|r'|_p < p^{-k}</math> (that is, <math>v_p(r')>k</math>).

The <math>p</math>-''adic expansion'' of <math>r</math> is the [[formal power series]] 
: <math>r = \sum_{i=k}^\infty a_i p^i</math>
obtained by repeating indefinitely the [[#division_step|above]] division step on successive remainders. In a {{mvar|p}}-adic expansion, all <math>a_i</math> are integers such that <math>0\le a_i <p.</math>

If <math>r=p^k \tfrac n 1</math> with <math>n > 0</math>, the process stops eventually with a zero remainder; in this case, the series is completed by trailing terms with a zero coefficient, and is the representation of <math>r</math> in [[base-N|base-{{mvar|p}}]].

The existence and the computation of the {{mvar|p}}-adic expansion of a rational number results from [[Bézout's identity]] in the following way. If, as above, <math>r=p^k \tfrac n d,</math> and <math>d</math> and <math>p</math> are coprime, there exist integers <math>t</math> and <math>u</math> such that <math>t d+u p=1.</math> So
: <math>r=p^k \tfrac n d(t d+u p)=p^k n t + p^{k+1}\frac{u n}d.</math>
Then, the [[Euclidean division]] of <math>n t</math> by <math>p</math> gives
: <math>n t=q p+a,</math>
with <math>0\le a <p.</math>
This gives the division step as 
: <math>\begin{array}{lcl}
r & = & p^k(q p+a) + p^{k+1}\frac {u n}d \\
& = & a p^k +p^{k+1}\,\frac{q d+u n} d, \\
\end{array}</math>
so that in the iteration
: <math>r' = p^{k+1}\,\frac{q d+u n} d</math>
is the new rational number.

The uniqueness of the division step and of the whole {{mvar|p}}-adic expansion is easy: if <math>p^k a_1 + p^{k+1}s_1=p^k a_2 + p^{k+1}s_2,</math> one has <math>a_1-a_2=p(s_2-s_1).</math> This means <math>p</math> divides <math>a_1-a_2.</math> Since <math>0\le a_1 <p</math> and <math>0\le a_2 <p,</math> the following must be true: <math>0\le a_1</math> and <math>a_2<p.</math> Thus, one gets <math>-p < a_1-a_2 < p,</math> and since <math>p</math> divides <math>a_1-a_2</math> it must be that <math>a_1=a_2.</math>

The {{mvar|p}}-adic expansion of a rational number is a series that converges to the rational number, if one applies the definition of a [[convergent series]] with the {{mvar|p}}-adic absolute value.
In the standard {{mvar|p}}-adic notation, the digits are written in the same order as in a [[Positional notation#Base of the numeral system|standard base-{{mvar|p}} system]], namely with the powers of the base increasing to the left. This means that the production of the digits is reversed and the limit happens on the left hand side.

The {{mvar|p}}-adic expansion of a rational number is eventually [[periodic function|periodic]]. [[Converse (logic)|Conversely]], a series <math display=inline>\sum_{i=k}^\infty a_i p^i,</math> with <math>0\le a_i <p</math> converges (for the {{mvar|p}}-adic absolute value) to a rational number [[if and only if]] it is eventually periodic; in this case, the series is the {{mvar|p}}-adic expansion of that rational number. The [[mathematical proof|proof]] is similar to that of the similar result for [[repeating decimal]]s.

=== Example ===
Let us compute the 5-adic expansion of <math>\tfrac 13.</math> Bézout's identity for 5 and the denominator 3 is <math>2\cdot 3 + (-1)\cdot 5 =1</math> (for larger examples, this can be computed with the [[extended Euclidean algorithm]]). Thus
: <math>\frac 13= 2+5(\frac {-1}3).</math>
For the next step, one has to expand <math>-1/3</math> (the factor 5 has to be viewed as a "[[arithmetic shift|shift]]" of the {{mvar|p}}-adic valuation, similar to the basis of any number expansion, and thus it should not be itself expanded).  To expand <math>-1/3</math>, we start from the same Bézout's identity and multiply it by <math>-1</math>, giving
: <math>-\frac 13=-2+\frac 53.</math>
The "integer part" <math>-2</math> is not in the right interval. So, one has to use [[Euclidean division]] by <math>5</math> for getting <math>-2= 3-1\cdot 5,</math> giving
: <math>-\frac 13=3-5+\frac 53 = 3-\frac {10}3 = 3 +5 (\frac{-2}3),</math>
and the expansion in the first step becomes
: <math>\frac 13= 2+5\cdot (3 + 5 \cdot (\frac{-2}3))= 2+3\cdot 5 + \frac {-2}3\cdot 5^2.</math>

Similarly, one has 
: <math>-\frac 23=1-\frac 53,</math>
and
: <math>\frac 13=2+3\cdot 5 + 1\cdot 5^2 +\frac {-1}3\cdot 5^3.</math>

As the "remainder" <math>-\tfrac 13</math> has already been found, the process can be continued easily, giving coefficients <math>3</math> for [[parity (mathematics)|odd]] powers of five, and <math>1</math> for [[parity (mathematics)|even]] powers.
Or in the standard 5-adic notation
: <math>\frac 13= \ldots 1313132_5 </math>
with the [[ellipsis]] <math> \ldots </math> on the left hand side.

=== Positional notation ===
It is possible to use a [[positional notation]] similar to that which is used to represent numbers in [[radix|base]] {{mvar|p}}.

Let <math display = inline>\sum_{i=k}^\infty a_i p^i</math> be a normalized {{mvar|p}}-adic series, i.e. each <math>a_i</math> is an integer in the interval <math>[0,p-1].</math> One can suppose that <math>k\le 0</math> by setting <math>a_i=0</math> for <math>0\le i <k</math> (if <math>k>0</math>), and adding the resulting zero terms to the series.

If <math>k\ge 0,</math> the positional notation consists of writing the <math>a_i</math> consecutively, ordered by decreasing values of {{mvar|i}}, often with {{mvar|p}} appearing on the right as an index:
: <math>\ldots a_n \ldots a_1{a_0}_p</math>
So, the computation of the [[#Example|example above]] shows that 
: <math>\frac 13= \ldots 1313132_5,</math>
and 
: <math>\frac {25}3= \ldots 131313200_5.</math>
When <math>k<0,</math> a separating dot is added before the digits with negative index, and, if the index {{mvar|p}} is present, it appears just after the separating dot. For example,
: <math>\frac 1{15}= \ldots 3131313._52,</math>
and
: <math>\frac 1{75}= \ldots 1313131._532.</math>

If a {{mvar|p}}-adic representation is finite on the left (that is, <math>a_i=0</math> for large values of {{mvar|i}}), then it has the value of a nonnegative rational number of the form <math>n p^v,</math> with <math>n,v</math> integers. These rational numbers are exactly the nonnegative rational numbers that have a finite representation in [[radix|base]] {{mvar|p}}. For these rational numbers, the two representations are the same.