From: "Pierre Abbat" <phma@pop.trellis.net>
Subject: Re: p-Adic Numbers
Date: 7 Nov 97 19:54:13 GMT
Newsgroups: sci.math

> The sequence n!/(n! + 1) converges to 1 in Q and to 0 in *every* Q_p.

And here's a sequence that converges to 0 in Q, 1 in dyadics, and 0 in all
other p-adics:
a[n]=n!/2^m, where 2^m divides n! but 2^(m+1) does not
b[n]=2^(n^2)
sequence is a[n]/(a[n]+b[n]).

phma
==============================================================================
From: astephan@students.uiuc.edu (adam louis stephanides)
Newsgroups: sci.math
Subject: Re: p-Adic Numbers
Date: 9 Nov 1997 22:02:22 GMT

"Pierre Abbat" <phma@pop.trellis.net> writes:

>> The sequence n!/(n! + 1) converges to 1 in Q and to 0 in *every* Q_p.

>And here's a sequence that converges to 0 in Q, 1 in dyadics, and 0 in all
>other p-adics:
>a[n]=n!/2^m, where 2^m divides n! but 2^(m+1) does not
>b[n]=2^(n^2)
>sequence is a[n]/(a[n]+b[n]).

In fact, what I stated on an earlier post can be generalized.  Denote the
i-th prime as p_i.  Then for any real number r, and any sequence of "adics"
a_i, where a_i is a p_i-adic number, there is a sequence of rationals
that converges to r in the reals and to a_i in the p_i-adics for all i.

Proof: I'll do the proof for the case where the a_i are p_i-adic integers;
the general case isn't much more difficult, but introduces an extra
complication.  For any n, denote by f(n) the number (p_1*p_2*...*p_n)^n.
The Chinese Remainder Theorem guarantees that there exists a rational q_n,
with denominator (2^n)*f(n) + 1, whose distance in the reals from r is
less than 1/2^n and which is p_i-adically "congruent" to a_i mod p_i^n,
for 1 =< i =< n.  The sequence q_n then converges to r in the reals and
to a_i in the p_i-adics for all i.

--Adam 
==============================================================================

From: Kurt Foster <kfoster@rmi.net>
Newsgroups: sci.math
Subject: Re: Incompatibility between different p-adics?
Date: 24 Nov 1998 16:25:57 GMT

In <365A5F84.2D2D7812@math.ucla.edu>, Mike Oliver said:
. For a prime p, let Z_p be the ring of p-adic integers (i.e. they have a
. base-p expansion extending infinitely many places to the left), and let
. Q_p be its quotient field (whose elements have a natural base-p
. expansion with infinitely many places to the left of the radix point,
. and finitely many to the *right*).

. How different are these structures for different values of p?
.
  Depends how much of the "structure" you're worried about.  The ordinary
integers are a dense additive subgroup of Z_p for any given prime p. 
Similarly for the ordinary nonzero rationals in the multiplicative group
of the nonzero elements of Q_p.  For each p, the p-adic integers/rationals
are a completion of the ordinary integers/rationals in the p-adic
topology.
  However, the topologies for different values of p are inequivalent, and
this is a well known basic result that should be covered in most any
introductory treatment of valuations or p-adic fields and rings.  The
usual statement is to the effect that, given distinct primes p and q,
there is an x in Q such that |x|_p < 1 and |x|_q > 1.  The powers of such
an x give a sequence converging to 0 in one topology, but to infinity in
the other.