LaTeX4Web 1.4 OUTPUT We proceed to explain how the inverse of a p-adic number can be found. This is a number of the form 1/n. Take the example of 1/7 using p=11. This can be rewritten as
7x º 1.
Then we substitute x in the p-adic expansion
7(a0+a111+a2112 +a3113 +...) = 1.
7a0+7a111+7a2112 +7a3113 + ...
7a0 º 1 mod 11.
We see by inspection that a0 equals 8. Then
7a0 = 7· 8 = 56 = 1 + 5·11.
If we put this back to the p-adic expansion we obtain
(1+5 · 11)+7a111 + 7a2112 + ... = 1.
As we saw before, 5 is carried to the next parenthesis.
1 + (5+7a1)11 + ... = 1.
Now we need to solve
5 + 7a1 º 0 mod 11.
This is equivalent to
7a1 º -5 º 6.
Since we know that the solution of 7ai º 1 modulo 11 is ai = 8 we can substitute
a1 = 8· 6 = 48 º 4 mod 11.
If we keep using this method we obtain that x = 8 + 4·11 + 9·112 + ...
LaTeX4Web 1.4 OUTPUT Here we provide the code for the p-adic inverse. The value of p and the lenght of the expansion can be modified.
 inverse.txt Size : 1.111 Kb Type : txt
LaTeX4Web 1.4 OUTPUT Here you can see an example of the output of the code. The example also uses the fraction 1/7, and you can observe that modulo 11 the expansion is the same one that we obtained theoretically.