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Is it possible to convert irrational p-adic numbers to a standard number? Rationals and negative rationals are relatively straightforward, but is there a way to know that for instance $\ldots 2100121201_{p=3} = 10.0111010220\ldots_{3}$?

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If by "standard" you mean "real", it is not possible : $p$-adic numbers and real ones are really different. By the way, what do you really mean by the "decimal" $10.0111010220..._{3}$ ?

$\mathbf{R}$ i s a completion of $\mathbf{Q}$ for a precise metric, the usual one. $\mathbf{Q}_p$ is the completion for an other metric, the $p$-adic one. Those metrics give deeply different topologies, hence different behavior. Example : reals can have two decimal developments, $p$-adic numbers not ; $p$-adic have ultrametricity property, reals not ; etc. It is for that it is interesting to consider both real numbers and $p$-adic numbers. They are all the different ways to "do analysis" on $\mathbf{Q}$.

Miller's notes give a swift introduction and good references to $p$-adics. A very lovely introductory and intuitive text is the one of Alain Robert, if you read french -- otherwise, explore his bibliography.

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  • $\begingroup$ 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. $\endgroup$ – law-of-fives Nov 14 '14 at 14:55
  • $\begingroup$ @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? $\endgroup$ – Felipe Voloch Nov 14 '14 at 15:03
  • $\begingroup$ @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 $\mathbf{R}$. $\endgroup$ – Desiderius Severus Nov 14 '14 at 15:05
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P-adic numbers and real numbers are different things. They are both extensions of the rationals, but the adjoined elements are different. A given irrational p-adic number might be algebraic, for example, but there is no canonical choice of real root of its characteristic polynomial for it to correspond to. And if a p-adic number is transcendental, it simply has no real equivalent at all.

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