Editing P-adic number (section)

=== 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.