p-adic approximation

alekk

here an interesting pb:
let $p$ be a prime number and $v_p(n)=max \{k \geq 0 s.t. p^k | n \}$ for $|n|>0$ .
For a rational $r=\frac{a}{b}$ , $v_p(r)=v_p(a) - v_p(b)$ .

1/prove that $d_p(r_1, r_2)=p^{-v_p(r_1-r_2)}$ is a distance (with $d(a,a)=0$ )
2/let $p1,p2,..,p_n$ be n distinct prime numbers. We will define a distance on $Q^{n+1}$ .
For $x=(x_0,..,x_n)$ and $y=(y_0,..,y_n)$ in $Q^{n+1}$ ,
define their distance by: $d(x,y)=|x_0-y_0| + \sum_{1}^{n}d_{p_i}(x_i, y_i)$

The diagonal function from $Q$ in $Q^{n+1}$ is: $f(r)=(r,r,r..r)$ .
Prove that $f(Q)$ is dense in $Q^{n+1}$ .

**Posted:** Wed Mar 09, 2005 2:45 am

_________________
They misunderestimated me!

grobber

To give a quick sketch, if we are given a rational $\frac uv$ , all one need to do in order to find $\frac mn$ close to $\frac uv$ $p$ -adically, is fix $m,n$ modulo some power of $p$ (maybe a different power for each of $m,n$ ). We do this for all the primes $p_1,\ldots,p_n$ , and we find that the rational we are looking for, which we want close to a list of $n+1$ rationals in the $n+1$ metrics respectively, has the form $\frac{aM+b}{cN+d}$ , for fixed $M,N,b,d$ (i.e. its denominator and its numerator are fixed modulo two integers). Of course, rationals of that form are dense in $\mathbb Q$ with the usual metric, so we can find such a number which is close to a given rational in the usual metric as well.

To illustrate this for a prime $p$ , assume we're given $\frac uv$ in lowest terms, and we're looking for $\frac mn$ in lowest terms, close to it $p$ -adically. If $p\not|v$ , we also take $n\not|p$ (take $n\equiv 1\pmod p$ , for instance), and then all we need to do is take $m\equiv unv^{-1}\pmod {p^t}$ for some large $t$ . On the other hand, if $p|v$ , then take $n$ s.t. its $p$ -adic norm is equal to that of $v$ (i.e. the same maximal power of $p$ divides both), simplify $\frac{mv-un}{nv}$ with this power of $p$ , and we are left with $mv'-un'$ at the numerator, with $p\not|v',n'$ , so, again, we can make $mv'-un'$ divisible by powers of $p$ as large as we want by fixing $m\pmod{p^t}$ for large $t$ .

**Posted:** Wed Mar 09, 2005 8:06 am