Number theory and Group theory

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+7_a211² +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 7a_i º 1 modulo 11 is a_i = 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 + ...

Here we provide the code for the p-adic inverse. The value of p and the lenght of the expansion can be modified.

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.