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From: janw@inmet.UUCP
Newsgroups: net.math
Subject: Re: Need proof for density problem
Message-ID: <5700010@inmet.UUCP>
Date: Sun, 29-Sep-85 18:18:00 EDT
Article-I.D.: inmet.5700010
Posted: Sun Sep 29 18:18:00 1985
Date-Received: Thu, 3-Oct-85 04:39:06 EDT
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Nf-From: inmet!janw    Sep 29 18:18:00 1985


[ Written 12:17 am  Sep 23, 1985 by southard@unc in inmet:net.math]

> Is the set of numbers of the form 2^m * 3^n (that's 2 to the m power times
> 3 to the n power) where m and n are integers, dense in the positive
> rational numbers?

Your problem was the last thing I read before the hurricane hit.
So it gave me something to think about during the blackout.
The answer to the problem is *yes*, proof follows:

Denote the set of numbers of the form 2^m * 3^n with Z.

(1) Obviously, multiplying or dividing two elements of Z
produces an element of Z.

In particular, 1/x, where x is a number of our type, also belongs to it.

(2) Lemma :
 1 is a condensation point of Z. In other words, there is a
sequence of elements of Z, all different from 1, tending to 1.
This lemma is proved below, in (4); for now assume it proved.

Then it follows, as a corollary, that there exists a sequence of
elements of Z, all *less than 1*, tending to 1.
To prove the corollary, take any sequence of the kind whose
existence is established by Lemma, and replace
each member  z  of it that is >1 , with 1/z.

(3) Using this corollary, we can prove the main proposition.
Namely, that for every alpha & beta such that 0 < alpha < beta,
there exists a z belonging to Z such that alpha < z < beta.
By Lemma and its corollary in (2), there exists an A such that 
alpha/beta < A < 1 and A belongs to Z.
Obviously, there exists an element B of Z such that beta < B.
Now consider the sequence b[i] = B * A^i (i = 0, 1, ...).
Each member of this sequence belongs to Z. For sufficiently
large i, b[i] < beta (since A < 1). Consider the *last*
element b[t] such that b[t] >= beta. Then, since A  > alpha / beta,
alpha < b[t+1] < beta, 
Q. E. D.

(4) Now we have to prove the Lemma stated in (2).
Consider two sequences a[i], b[i] (i = 0, 1, ...)
defined as follows:

a[0] = 2/3 ; b[0] = 4/3 ;
for every i, a[i+1] = a[i] * b[i] iff a[i] * b[i] < 1;
	     else a[i+1] = a[i];
similarly,
for every i, b[i+1] = a[i] * b[i] iff a[i] * b[i] > 1;
	     else b[i+1] = b[i];

Note : The case a[i] * b[i] = 1 will never arise since all
the numerators in a[i], b[i] are powers of 2, while all the
denominators are powers of 3. For the same reason 
no a[i] or b[i] equals 1.

Intervals ( a[i] , b[i] )
form a sequence of shrinking intervals straddling 1 :
a[i] < 1 < b[i] for all i.

a[i] monotonically increase, while b[i] monotonically decrease.
It follows that there are limits A <= 1 and B >= 1, to which
a[i] and b[i], respectively, tend.

Let us prove that at least one of these limits equals 1
(actually, both do).
Since all the numbers a[i], b[i] belong to Z, this will 
prove the lemma.

Assume the contrary : A<1 1, or A * B < 1.
  
  Assume the former: A*B > 1.
   But then, for sufficiently large i, say, for i >= t >0,
   a[i] * b[i] > 1. Therefore, for all i >=t,
   b[i+1] = a[i] * b[i]  < A * b[i].
   Thus, for i > t, b[i] < b[t] * A^(i-t), 
   and, since, by hypothesis, A < 1,
   b[i] tends to 0, which is absurd.
   
  For the case A*B < 1, the proof is symmetrical to this:
   Assume A*B < 1.
   But then, for sufficiently large i, say, for i >= t >0,
   a[i] * b[i] < 1. Therefore, for all i >=t,
   a[i+1] = a[i] * b[i]  >  a[i] * B.
   Thus, for i > t, a[i] > a[t] * B^(i-t), 
   and, since, by hypothesis, B > 1,
   a[i] tends to infinity, which is absurd.
   
 The Lemma is proved.

	Jan Wasilewsky, at Intermetrics, Inc.
	733 Concord Ave, Cambridge, MA