Timeline for Converting p-adic to decimal
Current License: CC BY-SA 3.0
5 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Nov 14, 2014 at 15:05 | comment | added | Desiderius Severus | @anonymous Yes, but you cannot discriminate between differents roots of the same polynomial (here maybe by a "sign" argument, but no more when le degree is 3 or higher). Another example : $2X^2+X+2$ has factorization in $\mathbf{Z}_2$, but not in $\mathbf{Q}$ or in $\mathbf{R}$. | |
| Nov 14, 2014 at 15:03 | comment | added | Felipe Voloch | @anonymous Your equation has two roots in each of $\mathbb{R}$ and $\mathbb{Q}_3$ and there are two ways of matching them up and no canonical choice. Why should you match the positive real root with the root that is $1$ modulo $3$ and not with the other one? | |
| Nov 14, 2014 at 14:56 | vote | accept | law-of-fives | ||
| Nov 14, 2014 at 14:55 | comment | added | law-of-fives | Thank you for this answer, Dydo. I knew that p-adic numbers weren't just another kind of representation, but I thought since you can use p-adic number to solve the same problems—for instance the two numbers I gave are both roots of $p(x)=x^2-10$ in 3-adics and base 3 respectively—then these are "the same" numbers in some sense, and so might be convertible. | |
| Nov 14, 2014 at 14:46 | history | answered | Desiderius Severus | CC BY-SA 3.0 |